Quarter Wit Quarter Wisdom: How to Read GMAT Questions Carefully

Quarter Wit, Quarter WisdomWe all know that we need to be very careful while reading GMAT questions – that every word is important. Even small oversights can completely change an answer for you. This is what happens with many test takers who try to tackle this official question. Even though the question looks very simple, the way it is worded causes test-takers to often ignore one word, which changes the solution entirely. Let’s look at this question now:

Alice’s take-home pay last year was the same each month, and she saved the same fraction of her take-home pay each month. The total amount of money that she had saved at the end of the year was 3 times the amount of that portion of her monthly take-home pay that she did NOT save. If all the money that she saved last year was from her take-home pay, what fraction of her take-home pay did she save each month?

(A) 1/2
(B) 1/3
(C) 1/4
(D) 1/5
(E) 1/6

Let’s consider the question stem sentence by sentence:

“Alice’s take-home pay last year was the same each month, and she saved the same fraction of her take-home pay each month.”

Say Alice’s take-home pay last year was $100 each month. She saves a fraction of this every month – let the amount saved be x.

“The total amount of money that she had saved at the end of the year was 3 times the amount of that portion of her monthly take-home pay that she did NOT save.”

What would be “the total amount of money that she had saved at the end of the year”? Since Alice saves x every month, she would have saved 12x by the end of the year.

What would be “the amount of that portion of her monthly take-home pay that she did NOT save”? Note that this is going to be (100 – x). Many test takers end up using (100 – x)*12, however this equation is not correct. The key word here is “monthly” – we are looking for how much Alice does not save each month, not how much she does not save during the whole year.

The total amount of money that Alice saved at the end of the year is 3 times the amount of that portion of her MONTHLY take-home pay that she did not save. Now we know we are looking for:

12x = 3*(100 – x)
x = 20

“If all the money that she saved last year was from her take-home pay, what fraction of her take-home pay did she save each month?”

From our equation, we have determined that Alice saved $20 out of every $100 she earned every month, so she saved 20/100 = 1/5 of her take-home pay.

Therefore, the answer is D.

Often, test-takers make the mistake of writing the equation as:

12x = 3*(100 – x)*12
x = 300/4

However, this will give them the fraction (300/4)/100 = 3/4, and that’s when they will wonder what went wrong.

Be extra careful when reading GMAT questions so that precious minutes are not wasted on such avoidable errors.

Getting ready to take the GMAT? We have free online GMAT seminars running all the time. And, be sure to follow us on FacebookYouTubeGoogle+, and Twitter!

Karishma, a Computer Engineer with a keen interest in alternative Mathematical approaches, has mentored students in the continents of Asia, Europe and North America. She teaches the GMAT for Veritas Prep and regularly participates in content development projects such as this blog!

Quarter Wit, Quarter Wisdom: Solving the Fuel-Up Puzzles

Quarter Wit, Quarter WisdomWe hope you are enjoying the puzzles we have been putting up in the last few weeks. Though all of them may not be directly convertible to GMAT questions, they are great mathematical brain teasers!

(Before we tackle today’s puzzle, first take a look at our posts on how to solve pouring water puzzles, weighing puzzles, and hourglass puzzles.)

Another variety of puzzle involves distributing fuel among vehicles to reach a destination. Let’s look at this type of question today:

A military car carrying an important letter must cross a desert. There is no petrol station in the desert, and the car’s fuel tank is just enough to take it halfway across. There are other cars with the same fuel capacity that can transfer their petrol to one another. There are no canisters to carry extra fuel or rope to tow the cars.

How can the letter be delivered?

Here, we are given that a single car can only reach the midpoint of the desert on its own tank of gas. Since there are no canisters, the car cannot carry extra fuel, so it will need to be fueled up by other cars traveling along with it.

Let’s fill up 4 cars and get them to start crossing the desert together. By the time they cover a quarter of the desert, half of their fuel tanks will be empty. Hence, we will have 4 cars with half tanks, and the status of their fuel tanks will be:

(0.5, 0.5, 0.5, 0.5)

If we transfer the fuel from two of the cars into two other cars, we will have:

(1, 1, 0, 0)

The two cars with fuel in their tanks will continue to cross the desert and cover another quarter of it. Now both of the cars will have half tanks again, and they will have reached the middle of the desert:

(0.5, 0.5, 0, 0)

Now one car will transfer all of its fuel to the other car, allowing that car to have one full tank:

(1, 0, 0, 0)

That car can then carry the letter through the remaining half of the desert.

For this problem, we didn’t really care about the stalled cars in the middle of the desert since we are not required to bring them back. The only important thing is to get the letter completely across the desert. Now, how do we handle a puzzle that asks us to get all of the vehicles back, too? Let’s look at an example question with those constraints:

A distant planet “X” has only one airport located at the planet’s North Pole. There are only 3 airplanes and lots of fuel at the airport. Each airplane has just enough fuel capacity to get to the South Pole (which is diametrically opposite the North Pole). The airplanes can land anywhere on the planet and transfer their fuel to one another.

The mission is for at least one airplane to fly completely around the globe and stay above the South Pole; in the end, all of the airplanes must return to the airport at the North Pole.

For this problem, we are given that a plane with a full tank of fuel can only reach the South Pole, i.e. cover half the distance it needs to travel for the mission. We need it to take a full trip around the planet – from the North Pole, to the South pole, and back again to North Pole. Obviously, we will need more than one plane to fuel the plane which will fly above the South pole.

Let’s divide the distance from pole to pole into thirds (from the North Pole to the South Pole we have three thirds, and from the South Pole to the North Pole we have another three thirds).

Step #1: 2 airplanes will fly to the first third. A third of their fuel will be used, so the status of their fuel tanks will be:

(2/3, 2/3)

One airplane will then fuel up the other plane and go back to the airport. Now the status of their tanks is:

(3/3, 1/3)

Step #2: 2 airplanes will again fly from the airport to the first third – one airplane will fuel up the other plane and go back to the airport. So the status of these two airplanes is this:

(3/3, 1/3)

Step #3: Now there are two airplanes at the first third mark with their tanks full. They will now fly to the second third point, giving us:

(2/3, 2/3)

One of the airplanes will fuel up the second one (until its tank is full) and go back to the first third, where it will meet the third airplane (which has just come back from the airport to support it with fuel) so that they both can return to the airport.

In the meantime, the airplane at the second third, with a full tank of fuel, will fly as far as it can – over the South Pole and towards the North pole, to the last third before the airport.

Step #4: One of the two airplanes from the airport can now go to the first third (on the opposite side of the North pole as before), and share its 1/3 fuel so that both airplanes safely land back at the airport.

And that is how we can have one plane travel completely around the planet and still have all airplanes arrive back safely!

Getting ready to take the GMAT? We have free online GMAT seminars running all the time. And, be sure to follow us on FacebookYouTubeGoogle+, and Twitter!

Karishma, a Computer Engineer with a keen interest in alternative Mathematical approaches, has mentored students in the continents of Asia, Europe and North America. She teaches the GMAT for Veritas Prep and regularly participates in content development projects such as this blog!

Quarter Wit Quarter Wisdom: Solving the Weighing Puzzle (Part 2)

Quarter Wit, Quarter WisdomA couple of weeks back, we discussed how to handle puzzles involving a two pan balance. In those problems, we learned how to tackle problems that ask you to measure items against one another.

Today, we will look at some puzzles that require the use of a traditional weighing scale. When we put an object on this scale, it shows us the weight of the object.

This is what such a scale looks like:

Puzzles involving a weighing scale can be quite tricky! Let’s take a look at a couple of examples:

You have 10 bags with 1000 coins in each. In one of the bags, all of the coins are forgeries. A true coin weighs 1 gram; each counterfeit coin weighs 1.1 grams. 

If you have an accurate weighing scale, which you can use only once, how can you identify the bag with the forgeries?

We are allowed only a single weighing, so we cannot weigh all 10 bags on the scale individually to measure which one has counterfeit coins. We need to find the bag in only one weighing, so we need to somehow make the coins in the bags distinctive.

How do we do that? We can take out one coin from the first bag, two coins from the second bag, three coins from the third bag and so on. Finally, we will have 1 + 2 + 3 + … + 10 = 10*11/2 = 55 coins.

Let’s weigh these 55 coins now.

If all coins were true, the total weight would have been 55 grams. But since some coins are counterfeit, the total weight will be more. Say, the total weight comes out to be 55.2 grams. What can we deduce from this? We can deduce that there must be two counterfeit coins (because each counterfeit coin weighs 0.1 gram extra). So the second bag must be the bag of counterfeit coins.

Let’s try one more:

A genuine gummy bear has a mass of 10 grams, while an imitation gummy bear has a mass of 9 grams. You have 7 cartons of gummy bears, 4 of which contain real gummy bears while the others contain imitation bears. 

Using a scale only once and the minimum number of gummy bears, how can you determine which cartons contain real gummy bears?

Now this has become a little complicated! There are three bags with imitation gummy bears. Taking a cue from the previous question, we know that we should take out a fixed number of gummy bears from each bag, but now we have to ensure that the sum of any three numbers is unique. Also, we have to keep in mind that we need to use the minimum number of gummy bears.

So from the first bag, take out no gummy bears.

From the second bag, take out 1 gummy bear.

From the third bag, take out 2 gummy bears (if we take out 1 gummy bear, the sum will be the same in case the second bag has imitation gummy bears or in case third bag has imitation gummy bears.

From the fourth bag, take out 4 gummy bears. We will not take out 3 because otherwise 0 + 3 and 1 + 2 will give us the same sum. So we won’t know whether the first and fourth bags have imitation gummy bears or whether second and third bags have imitation gummy bears.

From the fifth bag, take out 7 gummy bears. We have obtained this number by adding the highest triplet: 1 + 2 + 4 = 7. Note that anything less than 7 will give us a sum that can be made in multiple ways, such as:

0 + 1 + 6 = 7 and 1 + 2 + 4 = 7
or
0 + 1 + 5 = 6 and 0 + 2 + 4 = 6

But we need the sum to be obtainable in only one way so that we can find out which three bags contain the imitation gummy bears.

At this point, we have taken out 0, 1, 2, 4, and 7 gummy bears.

From the sixth bag, take out 13 gummy bears. We have obtained this number by adding the highest triplet: 2 + 4 + 7 = 13. Note that anything less than 13 will, again, give us a sum that can be made in multiple ways, such as:

12 + 1 + 0 = 13 and 2 + 4 + 7 = 13
or
0 + 1 + 9 = 10 and 1 + 2 + 7 = 10
…etc.

Note that this way, we are also ensuring that we measure only the minimum number of gummy bears, which is what the question asks us to do.

From the seventh bag, take out 24 gummy bears. We have obtained this number by adding the highest triplet again: 4 + 7 + 13 = 24. Again, anything less than 24 will give us a sum that can be made in multiple ways, such as:

0 + 1 + 15 = 16 and 1 + 2 + 13 = 16
or
0 + 1 + 19 = 20 and 0 + 7 + 13 = 20
or
0 + 1 + 23 = 24 and 4 + 7 + 13 = 24
…etc.

Thus, this is the way we will pick the gummy bears from the 7 bags: 0, 1, 2, 4, 7, 13, 24.

In all, 51 gummy bears will be weighed. Their total weight should be 510 grams (51*10 = 510) but because three bags have imitation gummy bears, the weight obtained will be less.

Say the weight is less by 8 grams. This means that the first bag (which we pulled 0 gummy bears from), the second bag (which we pulled 1 gummy bear from) and the fifth bag (which we pulled 7 gummy bears from) contain the imitation gummy bears. This is because 0 + 1 + 7 = 8 – note that we will not be able to make 8 with any other combination.

We hope this tricky little problem got you thinking. Work those grey cells and the GMAT will not seem hard at all!

Getting ready to take the GMAT? We have free online GMAT seminars running all the time. And, be sure to follow us on FacebookYouTubeGoogle+, and Twitter!

Karishma, a Computer Engineer with a keen interest in alternative Mathematical approaches, has mentored students in the continents of Asia, Europe and North America. She teaches the GMAT for Veritas Prep and regularly participates in content development projects such as this blog!

Quarter Wit, Quarter Wisdom: Solving the Hourglass Puzzle

Quarter Wit, Quarter WisdomLet’s continue our puzzles discussion today with another puzzle type – time measurement using an hourglass. (Before you continue reading this article, check out our posts on how to solve pouring water puzzles and weighing and balancing puzzles)

First, understand what an hourglass is – it is a mechanical device used to measure the passage of time. It is comprised of two glass bulbs connected vertically by a narrow neck that allows a regulated trickle of sand from the upper bulb to fall into the lower one. The sand also takes a fixed amount of time to fall from the upper bulb to the lower bulb. Hourglasses may be reused indefinitely by inverting the bulbs once the upper bulb is empty.

This is what they look like:

Say a 10-minute hourglass will let us measure time in intervals of 10 minutes. This means all of the sand will flow from the upper bulb to the lower bulb in exactly 10 minutes. We can then flip the hourglass over – now sand will start flowing again for the next 10 minutes, and so on. We cannot measure, say, 12 minutes using just a 10-minute hourglass, but we can measure more time intervals when we have two hourglasses of different times. Let’s look at this practice problem to see how this can be done:

A teacher of mathematics used an unconventional method to measure a 15-minute time limit for a test. He used a 7-minute and an 11-minute hourglass. During the whole time, he turned the hourglasses only 3 times (turning both hourglasses at once counts as one flip). Explain how the teacher measured out 15 minutes.

Here, we have a 7-minute hourglass and an 11-minute hourglass. This means we can measure time in intervals of 7 minutes as well as in intervals of 11 minutes. But consider this: if both hourglasses start together, at the end of 7 minutes, we will have 4 minutes of sand leftover in the top bulb of the 11-minute hourglass. So we can also measure out 4 minutes of time.

Furthermore, if we flip the 7-minute hourglass over at this time and let it flow for that 4 minutes (until the sand runs out of the top bulb of the 11-minute hourglass), we will have 3 minutes’ worth of sand leftover in the 7-minute hourglass. Hence, we can measure a 3 minute time interval, too, and so on…

Now, let’s see how we can measure out 15 minutes of time using our 7-minute and 11-minute hourglasses.

First, start both hourglasses at the same time. After the top bulb of the 7-minute hourglass is empty, flip it over again. At this time, we have 4 minutes’ worth of sand still in the top bulb of the 11-minute hourglass. When the top bulb of the 11-minute hourglass is empty, the bottom bulb of 7-minute hourglass will have 4 minutes’ worth of sand in it. At this point, 11 minutes have passed

Now simply flip the 7-minute hourglass over again and wait until the sand runs to the bottom bulb, which will be in 4 minutes.

This is how we measure out 11 + 4 = 15 minutes of time using a 7-minute hourglass and an 11-minute hourglass.

Let’s look at another problem:

Having two hourglasses, a 7-minute one and a 4-minute one, how can you correctly time out 9 minutes?

Now we need to measure out 9 minutes using a 7-minute hourglass and a 4-minute hourglass. Like we did for the last problem, begin by starting both hourglasses at the same time. After 4 minutes pass, all of the sand in the 4-minute hourglass will be in the lower bulb. Now flip this 4-minute hourglass back over again. In the 7-minute hourglass, there will be 3 minutes’ worth of sand still in the upper bulb.

After 3 minutes, all of the sand from the 7-minute hourglass will be in the lower bulb and 1 minute’s worth of sand will be in the upper bulb of the 4-minute hourglass.

This is when we will start our 9-minute interval.

The 1 minute’s worth of sand will flow to the bottom bulb of the 4-minute hourglass. Then we just need to flip the 4-minute hourglass over and let all of the sand flow out (which will take 4 minutes), and then flip the hourglass over to let all of the sand flow out again (which will take another 4 minutes).

In all, we have measured out a 1 + 4 + 4 = 9-minute interval, which is what the problem has asked us to find.

Getting ready to take the GMAT? We have free online GMAT seminars running all the time. And, be sure to follow us on FacebookYouTubeGoogle+, and Twitter!

Karishma, a Computer Engineer with a keen interest in alternative Mathematical approaches, has mentored students in the continents of Asia, Europe and North America. She teaches the GMAT for Veritas Prep and regularly participates in content development projects such as this blog!

Quarter Wit, Quarter Wisdom: Solving the Weighing and Balancing Puzzle

Quarter Wit, Quarter WisdomLet’s continue the discussion on puzzles that we began last week. Today we look at another kind of puzzle – weighing multiple objects using a two-pan balance while we are given a limited number of times to weight the objects against each other.

First of all, do we understand what a two-pan balance looks like?

Here is a picture.

Law School Images

 

 

 

 

 

As you can see, it has two pans that will be even if the weights in them are equal. If one pan has heavier objects in it, that pan will go down due to the weight. With this in mind, let’s try our first puzzle:

One of twelve coins is a bit lighter than the other 11 (which have the same weight). How would you identify this lighter coin if you could use a two-pan balance scale only 3 times? (You can only balance one set of coins against another, i.e. you have no weight measurements.)

There are various ways in which we can solve this.

We are given 12 coins, all of same weight, except one which is a bit lighter.

Let’s split the coins into two groups of 6 coins each and put them in the two pans. Since there is one lighter coin, one pan will be lighter than the other and will rise higher. So now we know that one of these 6 coins is the lighter coin.

Now split these 6 coins into another two groups of 3 coins each. Again, one pan will rise higher since it will have the lighter coin on it. Now we know that one of these three coins is the lighter coin.

Now what do we do? We have 3 coins and we cannot split them equally. What we can do is put one coin in each pan. What happens if the pans are not balanced? Then we know the pan that rises higher has the lighter coin on it (and thus, we have identified our coin). But what if both pans are balanced? The catch is that then the leftover coin is the lighter one! In any case, we would be able to identify the lighter coin using this strategy.

We hope you understand the logic here. Now let’s try another puzzle:

One of 9 coins is a bit lighter than the other 8. How would you identify this lighter coin if you could use a two-pan balance scale only 2 times?

Now we can use the balance only twice, and we are given an odd number of coins so we cannot split them evenly. Recall what we did in the first puzzle when we had an odd number of coins – we put one coin aside. What should we do here? Can we try putting 1 coin aside and splitting the rest of the 8 coins into two groups of 4 each? We can but once we have a set of 4 coins that contain the lighter coin, we will still need 2 more weighings to isolate the light coin, and we only have a total of 2 weighings to use.

Instead, we should split the 9 coins into 3 groups of 3 coins each. If we put one group aside and put the other two groups into the two pans of the scale, we will be able to identify the group which has the lighter coin. If one pan rises up, then that pan is holding the lighter coin; if the pans weight the same, then the group put aside has the lighter coin in it.

Now the question circles back to the strategy we used in the first puzzle. We have 3 coins, out of which one is lighter than the others, and we have only one chance left to weigh the coins. Just like in the first puzzle, we can put one coin aside and weigh the other two against each other – if one pan rises, it is holding the lighter coin, otherwise the coin put aside is the lighter coin! Thus, we were able to identify the lighter coin in just two weighings. Can you use the same method to answer the first puzzle now?

We will leave you with a final puzzle:

On a Christmas tree there were two blue, two red, and two white balls. All seemed the same, however in each color pair, one ball was heavier. All three lighter balls weighed the same, just like all three heavier balls weighed the same. Using a 2-pan balance scale only twice, identify the lighter balls.

Can you solve this problem using the strategies above? Let us know in the comments!

Getting ready to take the GMAT? We have free online GMAT seminars running all the time. And, be sure to follow us on FacebookYouTubeGoogle+, and Twitter!

Karishma, a Computer Engineer with a keen interest in alternative Mathematical approaches, has mentored students in the continents of Asia, Europe and North America. She teaches the GMAT for Veritas Prep and regularly participates in content development projects such as this blog!

Quarter Wit, Quarter Wisdom: Solving the Pouring Water Puzzle

Quarter Wit, Quarter WisdomSome time back, we came across a GMAT Data Sufficiency word problem question based on the “pouring water puzzle”. That made us think that it is probably a good idea to be comfortable with the various standard puzzle types. From this week on, we will look at some fundamental puzzles to acquaint ourselves with these mind benders in case we encounter them on test day.

Today, we will look at the popular “pouring water puzzle”. You may remember a similar puzzle from the movie Die Hard with a Vengeance, where Bruce Willis and Samuel L. Jackson had to diffuse a bomb by placing a 4 gallon jug of water on a set of scales.

Here is the puzzle:

You have a 3- and a 5-liter water container – each container has no markings except for that which gives us its total volume. We also have a running tap. We must use the containers and the tap in such a way that we measure out exactly 4 liters of water. How can this be done?

Don’t worry that this question is not written in a traditional GMAT format! We need to worry only about the logic behind the puzzle – we can then answer any question about it that is given in any GMAT format.

Let’s break down what we are given. We have only two containers – one of 3-liter and the other of 5-liter capacity. The containers have absolutely no markings on them other than those which give us the total volumes, i.e. the markings for 3 liters and 5 liters respectively. There is no other container. We also have a tap/faucet of running water, so basically, we have an unlimited supply of water. Environmentalists may not like my saying this, but this fact means we can throw out water when we need to and just refill again.

Now think about it:

STEP 1: Let’s fill up the 5-liter container with water from the tap. Now we are at (5, 0), with 5 being the liters of water in the 5-liter container, and 0 being the liters of water in the 3-liter container.

STEP 2: Now, there is nothing we can do with this water except transfer it to the 3-liter container (there is no other container and throwing out the water will bring us back to where we started). After we fill up the 3-liter container, we are left with 2 liters of water in the 5-liter container. This brings us to (2, 3).

STEP 3: We gain nothing from transferring the 3 liters of water back to 5-liter container, so let’s throw out the 3 liters that are in the 3-liter container. Because we just threw out the water from the 3-liter container, we will gain nothing by simply refilling it with 3 liters of water again. So now we are at (2, 0).

STEP 4: The next logical step is to transfer the 2 liters of water we have from the 5-liter container to the 3-liter container. This means the 3-liter container has space for 1 liter more until it reaches its maximum volume mark. This brings us to (0, 2).

STEP 5: Now fill up the 5-liter container with water from the tap and transfer 1 liter to the 3-liter container (which previously had 2 liters of water in it). This means we are left with 4 liters of water in the 5-liter container. Now we are at (4, 3).

This is how we are able to separate out exactly 4 liters of water without having any markings on the two containers. We hope you understand the logic behind solving this puzzle. Let’s take a look at another question to help us practice:

We are given three bowls of 7-, 4- and 3-liter capacity. Only the 7-liter bowl is full of water. Pouring the water the fewest number of times, separate out the 7 liters into 2, 2, and 3 liters (in the three bowls).

This question is a little different in that we are not given an unlimited supply of water. We have only 7 liters of water and we need to split it into 2, 2 and 3 liters. This means we can neither throw away any water, nor can we add any water. We just need to work with what we have.

We start off with (7, 0, 0) – with 7 being the liters of water in the 7-liter bowl, the first 0 being the liters of water in the 4-liter bowl, and the second 0 being the liters of water in the 3-liter bowl – and we need to go to (2, 2, 3). Let’s break this down:

STEP 1: The first step would obviously be to pour water from the 7-liter bowl into the 4-liter bowl. Now you will have 3 liters of water left in the 7-liter bowl. We are now at (3, 4, 0).

STEP 2: From the 4-liter bowl, we can now pour water into the 3-liter bowl. Now we have 1 liter in the 4-liter bowl, bringing us to (3, 1, 3).

STEP 3: Empty out the 3-liter bowl, which is full, into the 7-liter bowl for a total of 6 liters – no other transfer makes sense [if we transfer 1 liter of water to the 7-liter bowl, we will be back at the (4, 0, 3) split, which gives us nothing new]. This brings us to (6, 1, 0).

STEP 4: Shift the 1 liter of water from the 4-liter bowl to the 3-liter bowl. We are now at (6, 0, 1).

STEP 5: From the 7-liter bowl, we can now shift 4 liters of water into the 4-liter bowl. This leaves us with with 2 liters of water in the 7-liter bowl. Again, no other transfer makes sense – pouring 1 liter of water into some other bowl takes us back to a previous step. This gives us (2, 4, 1).

STEP 6: Finally, pour water from the 4-liter bowl into the 3-liter bowl to fill it up. 2 liters will be shifted, bringing us to (2, 2, 3). This is what we wanted.

We took a total of 6 steps to solve this problem. At each step, the point is to look for what helps us advance forward. If our next step takes us back to a place at which we have already been, then we shouldn’t take it.

Keeping these tips in mind, we should be able to solve most of these pouring water puzzles in the future!

Getting ready to take the GMAT? We have free online GMAT seminars running all the time. And, be sure to follow us on FacebookYouTubeGoogle+, and Twitter!

Karishma, a Computer Engineer with a keen interest in alternative Mathematical approaches, has mentored students in the continents of Asia, Europe and North America. She teaches the GMAT for Veritas Prep and regularly participates in content development projects such as this blog!

How to Answer GMAT Questions That are About an Unfamiliar Topic

Quarter Wit, Quarter WisdomUsually, GMAT questions that are based on your field of work or interests are simple to comprehend and relatively easy to answer correctly. But what happens when, say, an engineer gets a question based on psychiatry? Is he or she bound to fail? No.

Remember that the GMAT offers a level playing field for test takers from different backgrounds – it doesn’t matter whether your major was literature or physics. If you feel lost on a question about renaissance painters, remember that the guy next to you is lost on the problem involving planetary systems.

So how can you successfully handle GMAT questions on any topic? By sticking to the basics. The logic and reasoning required to answer these questions will stay the same no matter which field the information in the question stem comes from.

To give an example of this, let’s today take a look at a GMAT question involving psychoanalysis:

Studies in restaurants show that the tips left by customers who pay their bill in cash tend to be larger when the bill is presented on a tray that bears a credit-card logo. Consumer psychologists hypothesize that simply seeing a credit-card logo makes many credit-card holders willing to spend more because it reminds them that their spending power exceeds the cash they have immediately available.

Which of the following, if true, most strongly supports the psychologists’ interpretation of the studies? 

(A) The effect noted in the studies is not limited to patrons who have credit cards. 
(B) Patrons who are under financial pressure from their credit-card obligations tend to tip less when presented with a restaurant bill on a tray with credit-card logo than when the tray has no logo.
(C) In virtually all of the cases in the studies, the patrons who paid bills in cash did not possess credit cards.
(D) In general, restaurant patrons who pay their bills in cash leave larger tips than do those who pay by credit card.
(E) The percentage of restaurant bills paid with given brand of credit card increases when that credit card’s logo is displayed on the tray with which the bill is prepared.

Let’s break down the argument:

Argument: Studies show that cash tips left by customers are larger when the bill is presented on a tray that bears a credit-card logo.

Why would that be? Why would there be a difference in customer behavior when the tray has no logo from when the tray has a credit card logo? Psychologists’ hypothesize that seeing a credit-card logo reminds people of the spending power given by the credit card they carry (and that their spending power exceeds the actual cash they have right now).

The question asks us to support the psychologists’ interpretation. And what is the psychologists’ interpretation of the studies? It is that seeing a logo reminds people of their own credit card status. Say we change the argument a little by adding a line:

Argument: Studies show that cash tips left by customers are larger when the bill is presented on a tray that bears a credit-card logo. Patrons under financial pressure from credit-card obligations tend to tip less when presented with a restaurant bill on a tray with credit-card logo than when the tray has no logo.

Now, does the psychologists’ interpretation make even more sense? The psychologists’ interpretation is only that “seeing a logo reminds people of their own credit card status.” The fact “that their spending power exceeds the cash they have right now” explains the higher tips. If we are given that some customers tip more upon seeing that card logo and some tip less upon seeing it, it makes sense, right? Different people have different credit card obligation status, hence, people are reminded of their own card obligation status and they tip accordingly.

Answer choice B increases the probability that the psychologists’ interpretation is true because it tells you that in the cases of very high credit card obligations, customers tip less. This is what you would expect if the psychologists’ interpretation were correct.

In simpler terms, the logic here is similar to the following situation:

A: After 12 hours of night time sleep, I can’t study.
B: Yeah, because your sleep pattern is linked to your level of concentration. After a long sleep, your mind is still muddled and lazy so you can’t study.
A: After only 4 hrs of night time sleep, I can’t study either.

Does B’s theory make sense? Sure! B’s theory is that “sleep pattern is linked to level of concentration.” If A sleeps too much, her concentration is affected. If she sleeps too little, again her concentration is affected. So B’s theory certainly makes more sense.

Let’s now review answer choice E since it tends to confuse people:

(E) The percentage of restaurant bills paid with given brand of credit card increases when that credit card’s logo is displayed on the tray with which the bill is prepared.

This option supports the hypothesis that credit card logos remind people of their own card – not of their card obligations. The psychologists’ interpretation talks about the logo reminding people of their card status (high spending power or high obligations). Hence, this option is not correct.

Now let’s examine the rest of the answer choices to see why they are also incorrect:

(A) The effect noted in the studies is not limited to patrons who have credit cards.

This argument is focused only on credit cards, not on credit cards and their logos, so this is irrelevant.

(C) In virtually all of the cases in the studies, the patrons who paid bills in cash did not possess credit cards.

This option questions the validity of the psychologists’ interpretation. Hence, this is also not correct.

(D) In general, restaurant patrons who pay their bills in cash leave larger tips than do those who pay by credit card.

This argument deals with people who have credit cards but are tipping by cash, hence this is also irrelevant.

Therefore, our answer is B.

We hope you see that if you approach GMAT questions logically and stick to the basics, it is not hard to interpret and solve them, even if they include information from an unfamiliar field.

Getting ready to take the GMAT? We have free online GMAT seminars running all the time. And, be sure to follow us on FacebookYouTubeGoogle+, and Twitter!

Karishma, a Computer Engineer with a keen interest in alternative Mathematical approaches, has mentored students in the continents of Asia, Europe and North America. She teaches the GMAT for Veritas Prep and regularly participates in content development projects such as this blog!

Solving GMAT Geometry Problems That Involve Infinite Figures

Quarter Wit, Quarter WisdomSometimes, we come across GMAT geometry questions that involve figures inscribed inside other figures. One shape inside of another shape may not be difficult to work with, but how do we handle problems that involve infinite figures inscribed inside one another? Such questions can unsettle even the most seasoned test takers. Let’s take a look at one of them today:

A square is drawn by joining the midpoints of the sides of a given square. A third square is drawn inside the second square in this way and this process is continued indefinitely. If a side of the first square is 4 cm. Determine the sum of areas of all squares?

(A) 18
(B) 32
(C) 36
(D) 64
(E) None

Now the first thing that might come to our mind is this – how do we mathematically, in the time limit of approximately 2 minutes, calculate areas of infinite squares?

There has to be a formula for this. Recall that we do, in fact, have a formula that calculates the sum of infinite terms – the geometric progression formula! Let’s see if we can use that to find the areas of the squares mentioned in this problem.

ConcentricSquares

First, we’ll see if we can find a pattern in the areas of the squares:

Say the side of the outermost square is “s“. The area of the outermost square will be s^2 and half of the side will be s/2. The side of the next square inside this outermost square (the second square) forms the hypotenuse of a right triangle with legs of length s/2 each. Using the Pythagorean Theorem:

Hypotenuse^2 = (s/2)^2 + (s/2)^2 = s^2/2
Hypotenuse = s/√(2)

So now we know the sides of the second square will each equal s/√(2), and the area of the second square will be s^2/2.

Our calculations will be far easier if we note that the diagonal of the second square will be the same length as the side of the outer square. We know that area of a square given diagonal d is d^2/2, so that would directly bring us to s^2/2 as the area of the second square.

The second square and the square inscribed further inside it (the third square) will have the same relation. The area of the third square will be (s^2/2)*(1/2) = s^2/4.

Now we know the area of every subsequent square will be half the area of the outside square. So the total area of all squares = s^2 + s^2/2 + s^2/4 + s^2/8 + …Each term is half the previous term.

Therefore, the sum of an infinite Geometric Progression where the common ratio is less than 1 is:

Total Sum = a/(1-r)
a: First Term
r: Common Ratio

Sum of areas of all squares = s^2 + s^2/2 + s^2/4 + s^2/8 + …
Sum of areas of all squares = s^2/(1 – 1/2)
Sum of areas of all squares = 2s^2

Since s is the length of the side of the outermost square, and s = 4 (this fact is given to us by the questions stem), the sum of the areas of all the squares = 2*4^2 = 32 cm^2. Therefore, our answer is B.

We hope you understand how we have used the geometric progression formula to get to our answer. To recap, the sum of an infinite geometric progression is a/(1 – r).

Getting ready to take the GMAT? We have free online GMAT seminars running all the time. And, be sure to follow us on FacebookYouTubeGoogle+, and Twitter!

Karishma, a Computer Engineer with a keen interest in alternative Mathematical approaches, has mentored students in the continents of Asia, Europe and North America. She teaches the GMAT for Veritas Prep and regularly participates in content development projects such as this blog!

How to Find the Maximum Distance Between Points on a 3D Object

Quarter Wit, Quarter WisdomHow do we find the the two farthest points on a 3D object? For example, we know that on a circle, any two points that are diametrically opposite will be the farthest from each other than from any other points on the circle. Which two points will be the farthest from each other on a square? The diagonally opposite vertices. Now here is a trickier question – which two points are farthest from each other on a rectangular solid? Again, they will be diagonally opposite, but the question is, which diagonal?

A rectangular box is 10 inches wide, 10 inches long, and 5 inches high. What is the greatest possible (straight-line) distance, in inches, between any two points on the box?

(A) 15
(B) 20
(C) 25
(D) 10 * √(2)
(E) 10 * √(3)

There are various different diagonals in a rectangular solid. Look at the given figure:
MaxDistRectangularSolid

 

 

 

 

 

BE is a diagonal, BG is a diagonal, GE is a diagonal, and BH is a diagonal. So which two points are farthest from each other? B and E, G and E, B and G, or B and H?

The inside diagonal BH can be seen as the hypotenuse of the right triangle BEH. So both BE and EH will be shorter in length than BH.

The inside diagonal BH can also be seen as the hypotenuse of the right triangle BHG. So both HG and BG will also be shorter in length than BH.

The inside diagonal BH can also be seen as the hypotenuse of the right triangle BDH. So both BD and DH will also be shorter in length than BH.

Thus, we see that BH will be longer than all other diagonals, meaning B and H are the points that are the farthest from each other. Solving for the exact value of BH then should not be difficult.

In our question we know that:

l = 10 inches
w = 10 inches
h = 5 inches

Let’s consider the right triangle DHB. DH is the length, so it is 10 inches.

DB is the diagonal of the right triangle DBC. If DC = w = 10 and BC = h = 5, then we can solve for DB^2 using the Pythagorian Theorem:

DB^2 = DC^2 + BC^2
DB^2 = 10^2 + 5^2 = 125

Going back to triangle DHB, we can now say that:

BH^2 = HD^2 + DB^2
BH^2 = 10^2 + 125
BH = √(225) = 15

Thus, our answer to this question is A.

Similarly, which two points on a cylinder will be the farthest from each other? Let’s examine the following practice GMAT question to find out:

The radius of cylinder C is 5 inches, and the height of cylinder C is 5 inches. What is the greatest possible straight line distance, in inches, between any two points on a cylinder C?

(A) 5 * √2
(B) 5 * √3
(C) 5 * √5
(D) 10
(E) 15

Look at where the farthest points will lie – diametrically opposite from each other and also at the opposite sides of the length of the cylinder:
MaxDistanceCylinder

 

 

 

 

 

 

The diameter, the height and the distance between the points forms a right triangle. Using the given measurements, we can now solve for the distance between the two points:

Diameter^2 + Height^2 = Distance^2
10^2 + 5^2 = Distance^2
Distance = 5 * √5

Thus, our answer is C.

In both cases, if we start from one extreme point and traverse every length once, we reach the farthest point. For example, in case of the rectangular solid, say we start from H, cover length l and reach D – from D, we cover length w and reach C, and from C, we cover length h and reach B. These two are the farthest points.

Getting ready to take the GMAT? We have free online GMAT seminars running all the time. And, be sure to follow us on FacebookYouTubeGoogle+, and Twitter!

Karishma, a Computer Engineer with a keen interest in alternative Mathematical approaches, has mentored students in the continents of Asia, Europe and North America. She teaches the GMAT for Veritas Prep and regularly participates in content development projects such as this blog!

Advanced Number Properties on the GMAT – Part VII

Quarter Wit, Quarter WisdomWe have seen a number of posts on divisibility, odd-even concepts and perfect squares. Individually, each topic has very simple concepts but when they all come together in one GMAT question, it can be difficult to wrap one’s head around so many ideas. The GMAT excels at giving questions where multiple concepts are tested. Let’s take a look at one such Data Sufficiency question today:

If p, x, and y are positive integers, y is odd, and p = x^2 + y^2, is x divisible by 4?

1) When p is divided by 8, the remainder is 5.
2) x – y = 3

This Data Sufficiency question has a lot of information in the question stem.  First, we need to sort through this information before we move on to the statements.

We know that p, x and y are positive integers. y is an unknown odd number, so it can be written in the form 2n + 1. We also know that p = x^2 + y^2.

Because y is written in the form 2n + 1, y^2 can be written as:

y^2 =(2n + 1)^2

y^2 = 4n^2 + 4n + 1

y^2 = 4n(n + 1) + 1

An interesting thing to note here is that one case of n and (n+1) will be odd and the other will be even. In every case, n(n + 1) is even. Therefore, y^2 is 1 more than a multiple of 8. In other words, we can write it as y^2 = 8m + 1.

Now we can say p = x^2 + 8m + 1.

With this in mind, is x divisible by 4? Let’s examine the statements to find out:

Statement 1: When p is divided by 8, the remainder is 5.

Because y^2 = 8m + 1, we can see that when y^2 is divided by 8, the remainder will be 1. Therefore, to get a remainder of 5 when p is divided by 8, when x^2 is divided by 8, we should get a remainder of 4.

Now we know that x^2 can be written in the form 8a + 4 (i.e. we can make “a’” groups of 8 each and have 4 leftover).

x^2 = 4*(2a + 1)

So x = 2 * √(an odd number)

Note that square root of an odd number will be an odd number only. If there is no 2 in the perfect square, obviously there was no 2 in the number, too.

So, x = 2 * some other odd number, which means x will be a multiple of 2, but not of 4 definitely. This statement alone is sufficient.

Now let’s look at the next statement:

Statement 2: x – y = 3

Since y is odd, we can say that x will be even (an even – an odd = an odd). But whether x is divisible by 2 only or by 4 as well, we cannot say since we have no constraints on p.

This statement alone is not sufficient to answer the question. Therefore, our answer is A.

Test takers might feel that not every step in this solution is instinctive. For example, how do we know that we should put y^2 in the form 4n(n+1) + 1? Keep the target in mind – we know that we need to find whether x is divisible by 4. Hence, try to get everything in terms of multiples of 4 + a remainder.

See you next week!

(For more advanced number properties on the GMAT, check out Parts IIIIIIIV, V and VI of this series.)

Getting ready to take the GMAT? We have free online GMAT seminars running all the time. And, be sure to follow us on FacebookYouTubeGoogle+, and Twitter!

Karishma, a Computer Engineer with a keen interest in alternative Mathematical approaches, has mentored students in the continents of Asia, Europe and North America. She teaches the GMAT for Veritas Prep and regularly participates in content development projects such as this blog!

How NOT to Write the Equation of a Line on the GMAT

Quarter Wit, Quarter WisdomA question brought an interesting situation to our notice. Let’s start by asking a question: How do we write the equation of a line? There are two formulas:

y = mx + c (where m is the slope and c is the y-intercept)
and
yy1 = m * (xx1) [where m is the slope and (x1,y1) is a point on the line]

We also know that m = (y2y1)/(x2x1) – this is how we find the slope given two points that lie on a line. The variables are x1, y1 and x2, y2, and they represent specific values.

But think about it, is m = (y2y)/(xx1) really the equation of a line? Let’s further clarify this idea using a GMAT practice question:

In the coordinate plane, line k passes through the origin and has slope 2. If points (3,y) and (x,4) are on line k, then x + y =

(A) 3.5
(B) 7 
(C) 8
(D) 10
(E) 14

We have been given that the line passes through (0, 0) and has a slope of 2. We can find the equation of the line from this information.

y = mx + c
y = 2x + 0 (Since the line passes through (0, 0), its y-intercept is 0 – when x is 0, y is also 0.)
y = 2x

Since we are given two other points, (3, y) and (x, 4), on the line and we have a slope of 2, many test-takers will be tempted to make another equation for the line using this information.

(4 – y)/(x – 3) = 2
(4 – y) = 2*(x – 3)
Thus, 2+ y = 10

Here, test-takers will use the two equations to solve for x and y and get x = 5/2 and y = 5.

After adding x and y together, they then wonder why 7.5 is not one of the answer choices. If this were an actual GMAT question, it is quite likely that 7.5 would have been one of the options. So all in all, the test-taker would not even have realized that he or she made a mistake, and would choose 7.5 as the (incorrect) answer.

The error is conceptual here. Note that the equation of the line, 2x + y = 10, is not the same as the equation we obtained above, y = 2x. They represent two different lines, but we have only a single line in the question. So which is the actual equation of that line?

To get the second equation, we have used m = (y2y)/(xx1). But is this really the equation of a line? No. This formula doesn’t have y and x, the generic variables for the x– and y-coordinates in the equation of a line.

To further clarify, instead of x and y, try using the variables a and b in the question stem and see if it makes sense:

“In the coordinate plane, line k passes through the origin and has slope 2. If points (3, a) and (b, 4) are on line k, then a + b =”

You can write (4 – a)/(b – 3) = 2 and this would be correct. But can we solve for both a and b here? No – we can write one of them in terms of the other, but we can’t get their exact values.

We know a and b must have specific values. (3, a) is a point on the line y = 2x. For x = 3, the value of of the y-coordinate, a, will be y = 2*3 = 6. Therefore, a = 6.

(b, 4) is also on the line y = 2x. So if the y-coordinate is 4, the x-coordinate, b, will be 4 = 2b, i.e. b = 2. Thus, a + b = 6 + 2 = 8, and our answer is C.

This logic remains the same even if the variables used are x and y, although test-takers often get confused because of it. Let’s solve the question in another way using the variables as given in the original question.

Recall what we have learned about slope in the past. If the slope of the line is 2 and the point (0, 0) lies on the line, the value of y – if point (3, y) also lies on the line – will be 6 (a slope of 2 means a 1-unit increase in x will lead to a 2-unit increase in y).

Again, if point (x, 4) lies on the line too, an increase of 4 in the y-coordinate implies an increase of 2 in the x-coordinate. So x will be 2, and again, x + y = 2 + 6 = 8.

Getting ready to take the GMAT? We have free online GMAT seminars running all the time. And, be sure to follow us on FacebookYouTubeGoogle+, and Twitter!

Karishma, a Computer Engineer with a keen interest in alternative Mathematical approaches, has mentored students in the continents of Asia, Europe and North America. She teaches the GMAT for Veritas Prep and regularly participates in content development projects such as this blog!

3 Formats for GMAT Inequalities Questions You Need to Know

Quarter Wit, Quarter WisdomAs if solving inequalities wasn’t already hard enough, sometimes the way a GMAT question is framed will make us wonder which answer option to choose, even after we have already solved solved the problem.

Let’s look at three different question formats today to understand the difference between them:

  1. Must Be True
  2. Could Be True
  3. Complete Range

Case 1: Must Be True
If |-x/3 + 1| < 2, which of the following must be true?
(A) x > 0
(B) x < 8
(C) x > -4
(D) 0 < x < 3
(E) None of the above

We have two linked inequalities here. One is |-x/3 + 1| < 2 and the other is the correct answer choice. We need to think about how the two are related.

We are given that |-x/3 + 1| < 2. So we know that x satisfies this inequality. That will give us the universe which is relevant to us. x will take one of those values only. So let’s solve this inequality. (We will not focus on how to solve the inequality in this post – it has already been discussed here. We will just quickly show the steps.)

|x/3 – 1| < 2
(1/3) * |x – 3| < 2
|x – 3| < 6

The distance of x from 3 is less than 6, so -3 < x < 9. Now we know that every value that x can take will lie within this range.

The question now becomes: what must be true for each of these values of x? Let’s assess each of our answer options with this question:

(A) x > 0
Will each of the values of x be positive? No – x could be a negative number greater than -3, such as -2.

(B) x < 8
Will each of the values of x be less than 8? No – x could be a number between 8 and 9, such as 8.5

(C) x > -4
Will each of the values of x be more than -4? Yes! x will take values ranging from -3 to 9, and each of the values within that range will be greater than -4. So this must be true.

(D) 0 < x < 3
Will each of these values be between 0 and 3. No – since x can take any of the values between -3 and 9, not all of these will be just between 0 and 3.

Therefore, the answer is C (we don’t even need to evaluate answer choice E since C is true).

Case 2: Could Be True
If −1 < x < 5, which is the following could be true?
(A) 2x > 10
(B) x > 17/3
(C) x^2 > 27
(D) 3x + x^2 < −2
(E) 2x – x^2 < 0

Again, we have two linked inequalities, but here the relation between them will be a bit different. One of the inequalities is  −1 < x < 5 and the other will be the correct answer choice.

We are given that -1 < x < 5, so x lies between -1 and 5. We need an answer choice that “could be true”. This means only some of the values between -1 and 5 should satisfy the condition set by the correct answer choice – all of the values need not satisfy. Let’s evaluate our answer options:

(A) 2x > 10
x > 5
No values between -1 and 5 will be greater than 5, so this cannot be true.

(B) x > 17/3
x > 5.67
No values between -1 and 5 will be greater than 5.67, so this cannot be true.

(C) x^2 > 27
x^2 – 27 > 0
x > 3*√(3) or x < -3*√(3)
√(3) is about 1.73 so 3*1.73 = 5.19. No value of x will be greater than 5.19. Also, -3*1.73 will be -5.19 and no value of x will be less than that. So this cannot be true.

(Details on how to solve such inequalities are discussed here.)

(D) 3x + x^2 < −2
x^2 + 3x + 2 < 0
(x + 1)(x + 2) < 0
-2 < x < -1
No values of x will lie between -2 and -1, so this also cannot be true.

(E) 2x – x^2 < 0
x * (x – 2) > 0
x > 2 or x < 0
If -1 < x < 5, then x could lie between -1 and 0 (x < 0 is possible) or between 2 and 5 (x > 2 is possible). Therefore, the correct answer is E.

Case 3: Complete Range
Which of the following represents the complete range of x over which x^3 – 4x^5 < 0?
(A) 0 < |x| < ½
(B) |x| > ½
(C) -½ < x < 0 or ½ < x
(D) x < -½ or 0 < x < ½
(E) x < -½ or x > 0

We have two linked inequalities, but the relation between them will be a bit different again. One of the inequalities is  x^3 – 4x^5 < 0 and the other will be the correct answer choice.

We are given that x^3 – 4x^5 < 0. This inequality can be solved to:

x^3 ( 1 – 4x^2) < 0
x^3*(2x + 1)*(2x – 1) > 0
> 1/2 or -1/2 < x < 0

This is our universe of the values of x. It is given that all values of x lie in this range.

Here, the question asks us the complete range of x. So we need to look for exactly this range. This is given in answer choice C, and therefore C is our answer.

We hope these practice problems will help you become able to distinguish between the three cases now.

Getting ready to take the GMAT? We have free online GMAT seminars running all the time. And, be sure to follow us on FacebookYouTubeGoogle+, and Twitter!

Karishma, a Computer Engineer with a keen interest in alternative Mathematical approaches, has mentored students in the continents of Asia, Europe and North America. She teaches the GMAT for Veritas Prep and regularly participates in content development projects such as this blog!

Using Special Formats on GMAT Variable Problems

Quarter Wit, Quarter WisdomIn today’s post, we will discuss some special formats when we assume variables on the GMAT. These will allow us to minimize the amount of manipulations and calculations that are required to solve certain Quant problems.

Here are some examples:

An even number: 2a
Logic: It must be a multiple of 2.

An odd number: (2a + 1) or (2a – 1)
Logic: It will not be a multiple of 2. Instead, it will be 1 more (or we can say 1 less) than a multiple of 2.

Two consecutive integers: 2a, (2a + 1) or (2a – 1), 2a
Logic: One number will be even and the other will be the next odd number (or the other way around).

Four consecutive odd numbers: (2a – 3), (2a – 1), (2a + 1), (2a + 3)
In this case, the sum of the numbers comes out to be a clean 8a. This can be very useful in many cases.

Five consecutive even numbers: (2a – 4), (2a – 2), 2a, (2a + 2), (2a + 4)
In this case, the sum of the numbers comes out to be a clean 10a. This can also be very useful in many cases.

A prime number: (6a+1) / (6a – 1)
Every prime number greater than 3 is of the form (6a + 1) or (6a – 1). Note, however, that every number of this form is not prime.

Three consecutive numbers:
If we know one number is even and the other two are odd, we will have: (2a – 1), 2a, (2a + 1).
Logic: They add up to give 6a.
In a more generic case, we will have: 3a, (3a+1), (3a+2).
This gives us some important information. It tells us that one of the numbers will definitely be a multiple of 3 and the other two numbers will not be. Note that the numbers can be in a different order such as (3a + 1), (3a + 2) and (3a + 3). (3a + 3) can be written as 3b, so the three numbers will still have the same properties.

Basically, try to pick numbers in a way that will make it easy for you to manage them. Remember, three numbers do not need to be a, b and c – there could be, and in fact often are, several other hints which will give you the relations among the numbers.

Now, let’s see how picking the right format of these numbers can be helpful using a 700-level GMAT question:

The sum of four consecutive odd numbers is equal to the sum of 3 consecutive even numbers. Given that the middle term of the even numbers is greater than 101 and lesser than 200, how many such sequences can be formed?

(A) 12
(B) 17
(C) 25
(D) 33
(E) 50

Let’s have the four consecutive odd numbers be the following, where “a” is any integer: (2a – 3), (2a – 1), (2a + 1), (2a + 3)

The sum of these numbers is: (2a – 3) + (2a – 1) + (2a + 1) + (2a + 3) = 8a

Now let’s have the three consecutive even numbers be the following, where “b” is any integer: (2b – 2), 2b, (2b + 2)

The sum of these numbers is: (2b – 2) + 2b + (2b + 2) = 6b

Note here that instead of 2a, we used 2b. There is no reason that the even numbers would be right next to the odd numbers, hence we used different variables so that we don’t establish relations that don’t exist between these seven numbers.

We are given that the sum 8a is equal to the sum 6b.

8a = 6b, or a/b = 3/4, where a and b can be any integers. So “a” has to be a multiple of 3 and “b” has to be a multiple of 4.

With this in mind, possible solutions for a and b are:

a = 3, b = 4;
a = 6, b = 8;
a = 9, b = 12
etc.

We are also given that the middle term of the even numbers is greater than 101 and less than 200.

So 101 < 2b < 200, i.e. 50.5 < b < 100.

B must be an integer, hence, 51 ≤ b ≤ 99.

Also, b has to be a multiple of 4, so the values that b can take are 52, 56, 60, 64 … 96

The number of values b can take = (Last term – First term)/Common Difference + 1 = (96 – 52)/4 + 1 = 12

For each of these 12 values of b, there will be a corresponding value of a and, hence, we will get 12 such sequences. Therefore, the answer to our question is A.

Getting ready to take the GMAT? We have free online GMAT seminars running all the time. And, be sure to follow us on FacebookYouTubeGoogle+, and Twitter!

Karishma, a Computer Engineer with a keen interest in alternative Mathematical approaches, has mentored students in the continents of Asia, Europe and North America. She teaches the GMAT for Veritas Prep and regularly participates in content development projects such as this blog!

Quarter Wit, Quarter Wisdom: Beware of Sneaky Answer Choices on the GMAT!

Quarter Wit, Quarter WisdomTest-takers often ask for tips and short cuts to cut down the amount of work necessary to solve a GMAT problem. As such, the Testmaker might want to award the test-taker who pays attention to detail and puts in the required effort.

Today, we will look at an example of this concept – if it seems to be too easy, it is a trap!

QWQW_11_21

 

 

 

 

 

In the figure given above, the area of the equilateral triangle is 48. If the other three figures are squares, what is the perimeter, approximately, of the nine-sided shape they form?

(A) 8√(2)
(B) 24√(3)
(C) 72√(2)
(D) 144√(2)
(E) 384

The first thing I notice about this question is that we have an equilateral triangle. So I am thinking, the area = s^2 * √(3)/4 and/or the altitude = s*√(3)/2.

The irrational number in play is √(3). There is only one answer choice with √(3) in it, so will this be the answer?

Now, it actually makes me uncomfortable that  there is only one option with √(3). At first glance, it seems that the answer has been served to us on a silver plate. But the question format doesn’t seem very easy – it links two geometrical figures together. So I doubt very much that the correct answer would be that obvious.

The next step will be to think a bit harder:

The area of the triangle has √(3) in it, so the side would be a further square root of √(3). This means the actual irrational number would be the fourth root of 3, but we don’t have any answer choice that has the fourth root of 3 in it.

Let’s go deeper now and actually solve the question.

The area of the equilateral triangle = Side^2 * (√(3)/4) = 48

Side^2 = 48*4/√(3)
Side^2 = 4*4*4*3/√(3)
Side = 8*FourthRoot(3)

Now note that the side of the equilateral triangle is the same length as the sides of the squares, too. Hence, all sides of the three squares will be of length 8*FourthRoot(3).

All nine sides of the figure are the sides of squares. Hence:

The perimeter of the nine sided figure = 9*8*FourthRoot(3)
The perimeter of the nine sided figure =72*FourthRoot(3)

Now look at the answer choices. We have an option that is 72√(2). The other answer choices are either much smaller or much greater than that.

Think about it – the fourth root of 3 = √(√(3)) = √(1.732), which is actually very similar to √(2). Number properties will help you figure this out. Squares of smaller numbers (that are still greater than 1) are only a bit larger than the numbers themselves. For example:

(1.1)^2 = 1.21
(1.2)^2 = 1.44
(1.3)^2 = 1.69
(1.414)^2 = 2

Since 1.732 is close to 1.69, the √(1.732) will be close to the √(1.69), i.e. 1.3. Also, √(2) = 1.414. The two values are quite close, therefore, the perimeter is approximately 72√(2). This is the reason the question specifically requests the “approximate” perimeter.

We hope you see how the Testmaker could sneak in a tempting answer choice – beware the “easiest” option!

Getting ready to take the GMAT? We have free online GMAT seminars running all the time. And, be sure to follow us on FacebookYouTubeGoogle+, and Twitter!

Karishma, a Computer Engineer with a keen interest in alternative Mathematical approaches, has mentored students in the continents of Asia, Europe and North America. She teaches the GMAT for Veritas Prep and regularly participates in content development projects such as this blog!

The Holistic Approach to Absolute Values – Part V

Quarter Wit, Quarter WisdomWe will continue our holistic approach to absolute values and add more complications to these types of questions. This article should set you up for any question of this kind. Note that this is a 750+ level concept, so if you are targeting a lower score, it may not be necessary for you to know.

(Before you continue reading, be sure to check out Part I, Part II, Part III and Part IV of this lesson.)

Let’s look at the following GMAT question:

For how many integer values of x, is |x – 6| > |3x + 6|?

(A) 1
(B) 3
(C) 5
(D) 7
(E) Infinite

In this question, we are given the inequality |x – 6| > 3*|x + 2|

Using the same logic as we did in the previous two posts, we will word the inequality like this: the distance from 6 should be more than three times the distance from -2.
QWQW image 2

 

At x = -2, the distance from 6 is 8 and the distance from -2 is 0. This means the distance from 6 is more than three times the distance from -2.

At x = -1, the distance from 6 is 7 and the distance from -2 is 1. Three times the distance from -2 is 3. This means the distance from 6 is more than three times the distance from -2.

At some point on the right of -1, the distance from 6 will be equal to three times the distance from -2. The distance between -2 and 6 is 8. If we split this 8 into 4 equal parts to get to x = 0, the distance from 6 will be equal to three times the distance from -2.

Now for every point to the right of 0, the distance from 6 will be less than three times the distance from -2.

Let’s try to go to the left of -2 instead. Will there be a point to the left of -2 where the distance from 6 will be equal to three times the distance from -2? Say that point is “a” units away from -2. -2 must then be 2a units away from 6 to ensure that 6 is a total of 3a units away from that point.

The distance between -2 and 6 is 8 – this 8 needs to be equal to 2a, so “a” must be 4 units.

The point where the distance from 6 will be equal to three times the distance from -2 will be 4 units to the left of -2, i.e. at -6. So at points to the right of -6 (but left of 0), the distance from 6 will be more than three times the distance from -2.

Note that for all values to the left of -6, the distance from 6 will be less than three times the distance from -2.

Hence, our x will lie in the range from -6 to 0.

-6 < x < 0

With these parameters, we will have 5 integer solutions: -5, -4, -3, -2 and -1. Hence, our answer is C.

Let’s look at a second question:

For how many integer values of x, is |x – 8| + |5 – x| > |x + 7|?

(A) 1
(B) 3
(C) 5
(D) 7
(E) Infinite

Now the true value of this method is visible, as we have three or more terms. The arduous algebra involved in this given inequality makes our logical approach much more attractive.

First note that we have the term |5 – x|. This is the same as |x – 5| because |x| = |-x|.

We will word the inequality like this: the distance from 5 + the distance from 8 should be greater than the distance from -7.

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Let’s find the point where the sum of the distance from 5 and the distance from 8 is equal to distance from -7. Say that point is “a” units to the left of 5.

a + a + 3 = 12 – a
a = 3

So the point is 3 units to the left of 5, which means it is at 2. For all points to the left of 2, the sum of the distance from 5 and the distance from 8 will be greater than the distance from -7.

How about the points that are to the right of 8? Say there is a point “b” units away from 8 where the sum of the distance from 5 and the distance from 8 is equal to the distance from -7.

3 + b + b = 15 + b
b = 12

So if we go 12 units to the right of 8, i.e. at x = 20, the sum of the distance from 5 and the distance from 8 is equal to the distance from -7.

For all points to the right of 20, the sum of the distance from 5 and the distance from 8 is greater than the distance from -7, so there will be infinite points for which the sum of the distance from 5 and the distance from 8 is greater than the distance from -7. Therefore, our answer is E.

Using this concept, try to answer the following question on your own: For how many integer values of x, is |x – 6| – |3x + 6| > 0?

Getting ready to take the GMAT? We have free online GMAT seminars running all the time. And, be sure to follow us on FacebookYouTubeGoogle+, and Twitter!

Karishma, a Computer Engineer with a keen interest in alternative Mathematical approaches, has mentored students in the continents of Asia, Europe and North America. She teaches the GMAT for Veritas Prep and regularly participates in content development projects such as this blog!

The Holistic Approach to Absolute Values – Part IV

Quarter Wit, Quarter WisdomLast week, we looked at some absolute value questions involving inequalities. Today, we’ll continue this discussion by adding some more complications to our questions. Consider the question: What is the minimum value of the expression |x – 3| + |x + 1| + |x|? Technically, |x – 3| + |x + 1| + |x| is the sum of “the distance of x from 3,” “the distance of x from -1” and “the distance of x from 0.” To make solving such questions simpler, we’ll often use a parallel situation:

Imagine that there are 3 friends with houses at points -1, 0 and 3 in a straight line. They decide to meet at the point x.

  • |x – 3| will be the distance covered by the friend at 3 to reach x.
  • |x + 1| will be the distance covered by the friend at -1 to reach x.
  • |x| will be the distance covered by the friend at 0 to reach x.

So, the total distance the friends will cover to meet at x will be |x – 3| + |x + 1| + |x|.

Now we can choose to minimize this total distance, bring it to some particular value or make it more or less than some particular value.

If we want to minimize the total distance, we just make the friends meet at the second guy’s house, i.e. at the point 0. The friend at 3 and the friend at -1 need to travel 4 units total to meet anyway, so there’s no point in making the guy at 0 travel any distance at all. So the minimum total distance would be 4, which would then be the minimum value of |x – 3| + |x + 1| + |x|. This minimum value is given by the expression at x = 0.

With this in mind, when we move to the right or to the left of x = 0, the total distance will increase and, hence, the value of the expression |x – 3| + |x + 1| + |x| will also increase.

Thereafter, it is easy to solve for |x – 3| + |x + 1| + |x| = 10 or |x – 3| + |x + 1| + |x| < 10, etc., as seen in our previous post.

Today, let’s look at how to solve a more advanced GMAT question using the same logic:

For how many integer values of x, is |x – 5| + |x + 1| + |x|  + |x – 7| < 15?

(A) 0
(B) 2
(C) 4
(D) 6
(E) Infinite

In our parallel situation of friends and houses, we now have 4 friends with houses at points -1, 0, 5 and 7.

The friends at -1 and 7 are 8 units apart, so they will need to cover at least this total distance together to meet. It doesn’t matter where they meet between -1 and 7 (inclusive), they will need to cover exactly 8 units.

The friends at 0 and 5 will need to travel a minimum distance of 5 to meet. They can meet anywhere between 0 and 5 (inclusive) and the distance they will cover will still be 5.

So, all four friends can meet anywhere between 0 and 5 (inclusive) and the total distance covered will be 8 + 5 = 13. This would be the minimum total distance, and hence, the minimum value of the expression |x – 5| + |x + 1| + |x|  + |x – 7|.

When we move to the left of 0 or to the right of 5, the total distance covered will be more than 13. At any point between -1 and 7, the total distance covered by the friends at -1 and 7 will be only 8. When we move 1 unit to the left of 0 and reach -1, the total distance covered by the friends at 0 and 5 will be 1 + 6 = 7. So to meet at -1, the total distance traveled by all friends together will be 8 + 7 = 15.

Similarly, when we move 1 unit to the right of 5 and reach 6, the total distance covered by the four friends will be again 8 + 7 = 15. So at points x = -1 and x = 6, the value of the expression will be 15. Between these two points (excluding the points themselves), the value of the expression will be less than 15.

So now we know -1 < x < 6. With these parameters, x can take 6 integer values: 0, 1, 2, 3, 4, 5. Therefore, the answer is D.

Note that when we had 3 points on the number line, the minimum total distance was found at the second point. Now when we have 4 points on the number line, the minimum total distance has been found to be in the range between second and third points.

Let’s look at another question:

For how many integer values of x, is |2x – 5| + |x + 1| + |x| < 10?

(A) 1
(B) 2
(C) 4
(D) 5
(E) Infinite

|2x – 5| + |x + 1| + |x| < 10

2*|x – 5/2| + |x + 1| + |x| < 10

In this sum, now the distance from 5/2 is added twice.

In our parallel situation, this is equivalent to two friends living at 5/2, one living at 0 and one living at -1. Now note that the expression may not take the minimum value of x = 0 because there are 2 people who will need to travel from 5/2.

We have four friends in all, so we can expect to get a range in which we will get the minimum value of the expression. The second and third people are at 0 and 5/2, respectively.

The total distance at x = 0 will be 1 + 2*(5/2) = 6. The total distance at x = 5/2 will be 7/2 + 5/2 = 6.

So if we move to the left of 0 or to the right of 5/2, the total distance will increase. If we move 1 unit to the right of 5/2 and reach 7/2, the total distance covered by the four friends will be 9/2 + 7/2 + 2 = 10. If we move 1 unit to the left of 0 and reach -1, the total distance covered by the four friends will be 0 + 1 + 2*(7/2) = 8. Now all four friends are at -1. To cover a distance of another 2, they should move another 0.5 units to the left of -1 to reach -1.5.

Now the total distance covered by the four friends will be 0.5 + 1.5 + 2*4 = 10, so the total distance when x lies between the points -1.5 and 3.5 (excluding the points themselves) will be less than 10.

Now we know -1.5 < x < 3.5. With these parameters, x can take 5 integer values: -1, 0, 1, 2 and 3. Therefore, the answer is D.

Now use these concepts to solve the following question: For how many integer values of x, is |3x – 3| + |2x + 8| < 15?

Getting ready to take the GMAT? We have free online GMAT seminars running all the time. And, be sure to follow us on FacebookYouTubeGoogle+, and Twitter!

Karishma, a Computer Engineer with a keen interest in alternative Mathematical approaches, has mentored students in the continents of Asia, Europe and North America. She teaches the GMAT for Veritas Prep and regularly participates in content development projects such as this blog!

The Holistic Approach to Absolute Values – Part III

Quarter Wit, Quarter WisdomA while back, we discussed some holistic approaches to answering absolute value questions. Today, we will enhance our understanding of absolute values with some variations that you might see on the GMAT.

Instead of looking at how to solve equations, like we did in our previous post, we will look at how to solve inequalities using the same concept.

A quick review:

  • |x| = The distance of x from 0 on the number line. For example, if |x| = 4, x is 4 away from 0. So x can be 4 or -4.
  • |x – 1| = The distance of x from 1 on the number line. For example, if |x – 1| = 4, x is 4 away from 1. so x can be 5 or  -3.
  • |x| + |x – 1| = The sum of distance of x from 0 and distance of x from 1 on the number line. for example, if x = 5, the distance of x from 0 is 5 and the distance of x from 1 is 4. The sum of the distances is 5 + 4 = 9. So |x| + |x – 1| = 5 + 4 = 9.

Let’s move ahead now and see how we can use these concepts to solve inequalities:

For how many integer values of x, is |x – 3| + |x + 1| + |x| < 10?

(A) 0
(B) 2
(C) 4
(D) 6
(E) Infinite

In the previous post, we saw the a similar question, except it involved an equation rather than an inequality. For that problem, we found that the two points where the total distance is equal to 10 are -2.667 and 4:

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What will be the total distance at any value of x between these two points?

Say, x = 0
|x – 3| + |x + 1| + |x|
= 3 + 1 + 0
= 4

Say, x = 3
|x – 3| + |x + 1| + |x|
0 + 4 + 3
= 7

In both cases, we see that the total distance covered is less than 10. Note that the minimum distance covered will be 4 at x = 0 (discussed in the previous post) so by moving to the right of 0 or to the left of 0 on the number line, we get to the points where the distance increases to 10. So for every point in between, the total distance will be less than 10 (the entire red region).

Hence, at integer points x = -2, -1, 0, 1, 2 and 3 (which are all between -2.667 and 4), the total distance will be less than 10. The total distance will be less than 10 for all non-integer points lying between -2.667 and 4 too, but the question only asks for the integer values, so that is all we need to focus on. (Of course, there are infinite non-integer points between any two distinct points on the number line.) Hence, the answer will be 6 points, or D.

Along the same lines, consider a slight variation of this question:

For how many integer values of x, is |x – 3| + |x + 1| + |x| > 10?

(A) 0
(B) 2
(C) 4
(D) 6
(E) Infinite

What will the answer be here? We hope you immediately jumped to answer choice E – for every integer value of x to the right of 4 or to the left of -2.667, the total distance will be more than 10 (the blue regions). So there will be infinite such integer points (all integers greater than 4 or less than -2.667). Thus, the answer is E.

We hope this logic is clear. We will look at some other variations of this concept next week!

Getting ready to take the GMAT? We have free online GMAT seminars running all the time. And, be sure to follow us on FacebookYouTubeGoogle+, and Twitter!

Karishma, a Computer Engineer with a keen interest in alternative Mathematical approaches, has mentored students in the continents of Asia, Europe and North America. She teaches the GMAT for Veritas Prep and regularly participates in content development projects such as this blog!

Data Sufficiency Questions: How to Know When Both Statements Together Are Not Sufficient

Quarter Wit, Quarter WisdomToday we will discuss a problem we sometimes face while attempting to solve Data Sufficiency questions for which the answer is actually E (when both statements together are not sufficient to answer the question). Ideally, we would like to find two possible answers to the question asked so that we know that the data of both statements is not sufficient to give us a unique answer. But what happens when it is not very intuitive or easy to get these two distinct cases?

Let’s try to answer these questions in today’s post using using one of our own Data Sufficiency questions.

A certain car rental agency rented 25 vehicles yesterday, each of which was either a compact car or a luxury car. How many compact cars did the agency rent yesterday?

(1) The daily rental rate for a luxury car was $15 higher than the rate for a compact car.
(2) The total rental rates for luxury cars was $105 higher than the total rental rates for compact cars yesterday

We know from the question stem that the total number of cars rented is 25. Now we must find how many compact cars were rented.

There are four variables to consider here:

  1. Number of compact cars rented (this is what we need to find)
  2. Number of luxury cars rented
  3. Daily rental rate of compact cars
  4. Daily rental rate of luxury cars

Let’s examine the information given to us by the statements:

Statement 1: The daily rental rate for a luxury car was $15 higher than the rate for a compact car.

This statement gives us the difference in the daily rental rates of a luxury car vs. a compact car. Other than that, we still only know that a total of 25 cars were rented. We have no data points to calculate the number of compact cars rented, thus, this statement alone is not sufficient. Let’s look at Statement 2:

Statement 2: The total rental rates for luxury cars was $105 higher than the total rental rates for compact cars yesterday.

This statement gives us the difference in the total rental rates of luxury cars vs. compact cars (we do not know the daily rental rates). Again, we have no data points to calculate the number of compact cars rented, thus, this statement alone is also not sufficient.

Now, let’s try to tackle both statements together:

The daily rate for luxury cars is $15 higher than it is for compact cars, and the total rental rates for luxury cars is $105 higher than it is for compact cars. What constitutes this $105? It is the higher rental cost of each luxury car (the extra $15) plus adjustments for the rent of extra/fewer luxury cars hired. That is, if n compact cars were rented and n luxury cars were rented, the extra total rental will be 15n. But if more  luxury cars were rented, 105 would account for the $15 higher rent of each luxury car and also for the rent of the extra luxury cars.

Event with this information, we still should not be able to find the number of compact cars rented. Let’s find 2 cases to ensure that answer to this question is indeed E – the first one is quite easy.

We start with what we know:

The total extra money collected by renting luxury cars is $105.

105/15 = 7

Say out of 25 cars, 7 are luxury cars and 18 are compact cars. If the rent of compact cars is $0 (theoretically), the rent of luxury cars is $15 and the extra rent charged will be $105 (7*15 = 105) – this is a valid case.

Now how do we get the second case? Think about it before you read on – it will help you realize why the second case is more of a challenge.

Let’s make a slight change to our current numbers to see if they still fit:

Say out of 25 cars, 8 are luxury cars and 17 are compact cars. If the rent of compact cars is $0 and the rent of luxury cars is $15, the extra rent charged should be $15*8 = $120, but notice, 9 morecompact cars were rented than luxury cars. In reality, the extra total rent collected is $105 – the $15 reduction is because of the 9 additional compact cars. Hence, the daily rental rate of each compact car would be $15/9 = $5/3.

This would mean that the daily rental rate of each luxury car is $5/3 + $15 = $50/3

The total rental cost of luxury cars in this case would be 8 * $50/3 = $400/3

The total rental cost of compact cars in this case would be 17 * $5/3 = $85/3

The difference between the two total rental costs is $400/3 – $85/3 = 315/3 = $105

Everything checks out, so we know that there is no unique answer to this question – for any number of compact cars you use, you will come up with the same answer. Thus, Statements 1 and 2 together are not sufficient.

The strategy we used to find this second case to test is that we tweaked the numbers we were given a little and then looked for a solution. Another strategy is to try plugging in some easy numbers. For example:

Instead of using such difficult numbers, we could have tried an easier split of the cars. Say out of 25 cars, 10 are luxury and 15 are compact. If the rent of compact cars is $0 and the rent of luxury cars is $15, the extra rent charged should be 10*$15 = $150 extra, but it is actually only $105 extra, a difference of $45, due to the 5 additional compact cars. The daily rental rent of 5 extra compact cars would be $45/5 = $9. Using these numbers in the calculations above, you will see that the difference between the rental costs is, again, $105. This is a valid case, too.

Hence, there are two strategies we saw in action today:

  • Tweak the numbers slightly to see if you will get the same results
  • Go for the easy split when choosing numbers to plug in

Getting ready to take the GMAT? We have free online GMAT seminars running all the time. And, be sure to follow us on FacebookYouTubeGoogle+, and Twitter!

Karishma, a Computer Engineer with a keen interest in alternative Mathematical approaches, has mentored students in the continents of Asia, Europe and North America. She teaches the GMAT for Veritas Prep and regularly participates in content development projects such as this blog!

How to Solve “Unsolvable” Equations on the GMAT

Quarter Wit, Quarter WisdomThe moment we see an equation involving the variable x, we have a habit of jumping right into attempting to solve it. But what happens when we are not able to solve it? Let’s say, for example, we have an equation such as x^2 + 1 = 0. How would we solve for x here? We can’t because x has no real value. Note that x^2 is non-negative – it would be either 0 or positive. 1, we know, is positive. So together, a positive number and a non-negative number cannot add up to 0.

In this example, it relatively easy to see that the equation has no real solution. In others, it may not be so obvious, so we will need to use other strategies.

We know how to solve third degree equations. The first solution is found by trial and error – we try simple values such as -2, -1, 0, 1, 2, etc. and are usually able to find the first solution. Then the equation of third degree is split into two factors, including a quadratic. We know how to solve a quadratic, and that is how we get all three solutions, if it has any.

But what if we are unable to find the first solution to a third degree equation by trial and error? Then we should force ourselves to wonder if we even need to solve the equation at all. Let’s take a look at a sample question to better understand this idea:

Is x < 0?
(1) x^3 + x^2 + x + 2 = 0
(2) x^2 – x – 2 < 0

In this problem, x can be any real number – we have no constraints on it. Now, is x negative?

Statement 1: x^3 + x^2 + x + 2 = 0

If we try to solve this equation as we are used to doing, look at what happens:

If you plug in x = 2, you get 16 = 0
If you plug in x = 1, you get 5 = 0
If you plug in x = 0, you get 2 = 0
If you plug in x = -1, you get 1 = 0
If you plug in x = -2, you get -4 = 0

We did not find any root for the equation. What should we do now? Note that when x goes from -1 to -2, the value on the left hand side changes from 1 to -4, i.e. from a positive to a negative. So, in between -1 and -2 there will be some value of x for which the left hand side will become 0. That value of x will not be an integer, but some decimal value such as -1.3 or -1.4, etc.

Even after we find the first root, making the quadratic will be very tricky and then solving it will be another uphill task. So we should ask ourselves whether we even need to solve this equation.

Think about it – can x be positive? If x is indeed positive, x^3, x^2 and x all will be positive. Then, if we add four positive numbers (x^3, x^2, x and 2) we will get a positive sum – we cannot get 0. Obviously x cannot be 0 since that will give us 2 = 0.

This means the value of x must be negative, but what it is exactly doesn’t matter. We know that x has to be negative, and that is sufficient to answer the question.

Statement 2: x^2 – x – 2 < 0

This, we can easily solve:

x^2 – 2x + x – 2 < 0
(x – 2)*(x + 1) < 0

We know how to solve this inequality using the method discussed here.

This this will give us -1 < x < 2.

Since x can be a non-integer value too, x can be negative, 0, or positive. This statement alone is not sufficient,and therefore, the answer is A.

To evaluate Statement 1, we didn’t need to solve the equation at all. We figured out everything we wanted to know by simply using some logic.

Getting ready to take the GMAT? We have free online GMAT seminars running all the time. And, be sure to follow us on FacebookYouTubeGoogle+, and Twitter!

Karishma, a Computer Engineer with a keen interest in alternative Mathematical approaches, has mentored students in the continents of Asia, Europe and North America. She teaches the GMAT for Veritas Prep and regularly participates in content development projects such as this blog!

Quarter Wit, Quarter Wisdom: A GMAT Quant Question That Troubles Many!

Quarter Wit, Quarter WisdomWhat determines whether or not a question can be considered a GMAT question? We know that GMAT questions that are based on seemingly basic concepts can be camouflaged such that they may “appear” to be very hard. Is it true that a question requiring a lot of intricate calculations will not be tested in GMAT? Yes, however it is certainly possible that a question may “appear” to involve a lot of calculations, but can actually be solved without any!

In the same way, it is possible that a question may appear to be testing very obscure concepts, while it is really solvable by using only basic ones.

This happens with one of our own practice questions – we have often heard students exclaim that this problem isn’t relevant to the GMAT since it “tests an obscure number property”. It is a question that troubles many people, so we decided to tackle it in today’s post.

We can easily solve this problem with just some algebraic manipulation, without needing to know any obscure properties! Let’s take a look:

† and ¥ represent non-zero digits, and (†¥)² – (¥†)² is a perfect square. What is that perfect square?

(A) 121
(B) 361
(C) 576
(D) 961
(E) 1,089

The symbols † and ¥ are confusing to work with, so the first thing we will do is replace them with the variables A and B.

The question then becomes: A and B represent non-zero digits, and (AB)² – (BA)² is a perfect square. What is that perfect square?

As I mentioned before, we have heard students complain that this question isn’t relevant to the GMAT because it “uses an obscure number property”.  Now here’s the thing – most advanced number property questions CAN be solved in a jiffy using some obscure number property such as, “If you multiply a positive integer by its 22nd multiple, the product will be divisible by …” etc. However, those questions are not actually about recalling these so-called “properties” – they are about figuring out the properties using some generic technique, such as pattern recognition.

For this question, the complaint is often that is that the question tests the property, “(x + y)*(x – y) (where x and y are two digit mirror image positive integers) is a multiple of 11 and 9.” It doesn’t! Here is how we should solve this problem, instead:

Given the term (AB)^2, where A and B are digits, how will you square this while keeping the variables A and B?

Let’s convert (AB)^2 to (10A + B)^2, because A is simply the placeholder for the tens digit of the number. If you are not sure about this, consider the following:

58 = 50 + 8 = 10*5 + 8
27 = 20 + 7 = 10*2 + 7
…etc.

Along those same lines:

AB = 10A + B
BA = 10B + A

Going back to our original question:

(AB)^2 – (BA)^2
= (10A + B)^2 – (10B + A)^2
= (10A)^2 + B^2 + 2*10A*B – (10B)^2 – A^2 – 2*10B*A
= 99A^2 – 99B^2
= 9*11*(A^2 – B^2)

We know now that the expression is a multiple of 9 and 11. We would not have known this beforehand. Now we’ll just use the answer choices to figure out the solution. Only 1,089 is a multiple of both 9 and 11, so the answer must be E.

We hope you see that this question is not as hard as it seems. Don’t get bogged down by unknown symbols – just focus on the next logical step at each stage of the problem.

Getting ready to take the GMAT? We have free online GMAT seminars running all the time. And, be sure to follow us on FacebookYouTubeGoogle+, and Twitter!

Karishma, a Computer Engineer with a keen interest in alternative Mathematical approaches, has mentored students in the continents of Asia, Europe and North America. She teaches the GMAT for Veritas Prep and regularly participates in content development projects such as this blog!

Quarter Wit, Quarter Wisdom: Try to Answer This GMAT Challenge Question!

Quarter Wit, Quarter WisdomToday, we will give you a GMAT challenge question. The challenge of reviewing this question is not that the question is hard to understand – it is that you will need to solve this official question within a minute using minimum calculations.

Let’s take a look at the question stem:

Date of Transaction

Type of Transaction

June 11

Withdrawal of $350

June 16

Withdrawal of $500

June 21

Deposit of x dollars

For a certain savings account, the table shows the three transactions for the month of June. The daily balance for the account was recorded at the end of each of the 30 days in June. If the daily balance was $1,000 on June 1 and if the average (arithmetic mean) of the daily balances for June was $1,000, what was the amount of the deposit on June 21?

(A) $1,000
(B) $1,150
(C) $1,200
(D) $1,450
(E) $1,600

Think about how you might answer this question:

The average of daily balances = (Balance at the end of June 1 + Balance at the end of June 2 + … + Balance at the end of June 30) / 30 = 1000

Now we have been given the only three transactions that took place:

  • A withdrawal of $350 on June 11 – so on June 11, the account balance goes down to $650.
  • A withdrawal of $500 on June 16 – so on June 16, the account balance goes down to $150.
  • A deposit of $x on June 21 – So on June 21, the account balance goes up to 150 + x.

Now we can plug in these numbers to say the average of daily balances = [1000 + 1000 + …(for 10 days, from June 1 to June 10) + 650 + 650 + … (for 5 days, from June 11 to June 15) +  150 + … (for 5 days, from June 16 to June 20) + (150 + x) + (150 + x) + … (for 10 days, from June 21 to June 30)] / 30 = 1000

One might then end up doing this calculation to find the value of x:

[(1000 * 10) + (650 * 5) + (150 * 5) + ((150 + x) * 10)] / 30 = 1000
x = $1,450
The answer is D.

But this calculation is rather tedious and time consuming. Can’t we use the deviation method we discussed for averages and weighted averages, instead? After all, we are dealing with large values here! How?

Note that we are talking about the average of certain data values. Also, we know the deviations from those data values:

  • The amount from June 11 to June 30 is 350 less.
  • The amount from June 16 to June 30 is another 500 less.
  • The amount from June 21 to June 30 is x in excess.

Through the deviation method, we can see the shortfall = the excess:

350 * 20 + 500 * 15 = x * 10
x = 1,450 (D)

This simplifies our calculation dramatically! Though saving only one minute on a question like this may not seem like a very big deal, saving a minute on every question by using a more efficient method could be the difference between a good Quant score and a great Quant score!

Getting ready to take the GMAT? We have free online GMAT seminars running all the time. And, be sure to follow us on FacebookYouTubeGoogle+, and Twitter!

Karishma, a Computer Engineer with a keen interest in alternative Mathematical approaches, has mentored students in the continents of Asia, Europe and North America. She teaches the GMAT for Veritas Prep and regularly participates in content development projects such as this blog!

Quarter Wit, Quarter Wisdom: Evaluating Nasty GMAT Answer Choices

Quarter Wit, Quarter WisdomIn some Quant questions, we are given big nasty numbers in the answer choices and little else in the question stem. Often in such cases, the starting point is difficult for the test-taker to find, so today, we will discuss how to handle such questions.

The first and only rule with these types of problems is that familiarity helps. Evaluate the answer choices that make sense to you first.

Let’s look at a few questions to understand how to do that:

Which of the following is NOT prime?

(A) 1,556,551
(B) 2,442,113
(C) 3,893,257
(D) 3,999,991
(E) 9,999,991

The first thing that comes to mind when we consider how to find prime numbers should be to “check the number N for divisibility by all prime factors until we get to the √N.” But note that here, we have four numbers that are prime and one number that is not. Also, the numbers are absolutely enormous and, hence, will be very difficult to work with. So, let’s slide down to a number that seems a bit more sane: 3,999,991 (it is very close to a number with lots of 0’s).

3,999,991 = 4,000,000 – 9
= (2000)^2 – 3^2

This is something we recognise! It’s a difference of squares, which can be written as:

= (2000 + 3) * (2000 – 3)
= 2003 * 1997

Hence, we see that 3,999,991 is a product of two factors other than 1 and itself, so it is not a prime number. We have our answer! The answer is D.

Let’s try another problem:

Which of the following is a perfect square?

 (A) 649
 (B) 961
 (C) 1,664
 (D) 2,509
 (E) 100,000

Here, start by looking at the answer choices. The first one that should stand out is option E, 100,000, since multiples of 10 are always easy to handle. However, we have an odd number of zeroes here, so we know this cannot be a perfect square.

Next, we look at the answer choices that are close to the perfect squares that we intuitively know, such as 30^2 = 900, 40^2 = 1600, 50^2 = 2500. The only possible number whose perfect square could be 961 is 31 – 31^2 will end with a 1 and will be a bit greater than 900 (32^2 will end with a 4, so that cannot be the square root of 961, and the perfect squares of other greater numbers will be much greater than 900).

31^2 = (30 + 1)^2 = 900 + 1 + 2*30*1 = 961

So, we found that 961 is a perfect square and is, hence, the answer!

In case 961 were not a perfect square, we would have tried 1,664 since it is just 64 greater than 1,600. It could be the perfect square of 42, as the perfect square of 42 will end in a 4.

If 1,664 were also not a perfect square (it is not), we would have looked at 2,509. We would have known immediately that 2,509 cannot be a perfect square because it is too close to 2,500. 2,509 ends in a 9, so we may have considered 53 to be its square root, but the difference between consecutive perfect squares increases as we get to greater numbers.

(4^2 is 16 while 5^2 is 25 – the difference between them is 9. The difference between 5^2 and 6^2 will be greater than 9, and so will the difference between the perfect squares of any pair of consecutive integers greater than 6. Hence, the difference between the squares of 50 and 53 certainly cannot be 9.)

Therefore, our answer is B. Let’s try one more question:

When a certain perfect square is increased by 148, the result is another perfect square. What is the value of the original perfect square?

(A) 1,296
(B) 1,369
(C) 1,681
(D) 1,764
(E) 2,500

This question is, again, on perfect squares. We can use the same concepts here, too.

30^2 = 900
31^2 = 961 (=(30+1)^2 = 900 + 1 + 2*30)

40^2= 1,600
41^2 = 1,681 (=(40+1)^2 = 1,600 + 1 + 2*40)

50^2 = 2,500
51^2 = 2,601 (=(50+1)^2 = 2,500 + 1 + 2*50)

We know that the difference between consecutive squares increases as we go to greater numbers: going from 30^2 to 31^2 is a difference of 61, while jumping from 40^2 to 41^2 is a difference of 81.

All the answer choices lie in the range from 900 to 2500. In this range, the difference between consecutive squares is between 60 and 100. So, when you add 148 to a perfect square to get another perfect square in this range, we can say that the numbers must be 2 apart, such as 33 and 35 or 42 and 44, etc. Also, the numbers must lie between 30 and 40 because twice 61 is 122 and twice 81 is 162 – 148 lies somewhere in between 122 and 162.

A and B are the only two possible options.

Consider option A – it ends in a 6, so the square root must end in a 6, too. If you add 148, then it will end with a 4 (the perfect square of a number ending in 8 will end in 4). So this answer choice works.

Consider option B – it ends in a 9, so the square root must end in a 3 or a 7. When you add 148, it ends in 7. No perfect square ends in 7, so this option is out. Our answer is, therefore, A.

We hope you see how a close evaluation of the answer choices can help you solve questions of this type. Go get ’em!

Getting ready to take the GMAT? We have free online GMAT seminars running all the time. And, be sure to follow us on FacebookYouTubeGoogle+, and Twitter!

Karishma, a Computer Engineer with a keen interest in alternative Mathematical approaches, has mentored students in the continents of Asia, Europe and North America. She teaches the GMAT for Veritas Prep and regularly participates in content development projects such as this blog!

Evaluating “Useful to Evaluate” Critical Reasoning Questions – Part II

Quarter Wit, Quarter WisdomLast week we looked at how to handle “useful to evaluate” questions in the Verbal section, and we left you with a tricky “useful to evaluate except” question. Let’s take a look at that problem today. “Except” questions are usually more difficult to deal with since we need to find four “correct” options (which we are not as used to). So, let’s take a look at this question:

Following several years of declining advertising sales, the Greenville Times reorganized its advertising sales force two years ago. Before the reorganization, the sales force was organised geographically, with some sales representatives concentrating on city center businesses and others concentrating on different outlying regions. The reorganization attempted to increase the sales representatives’ knowledge of clients’ businesses by having each sales representative deal with only one type of industry or of retailing. After the reorganization, advertising sales increased. 

In assessing whether the improvement in advertising sales can properly be attributed to the reorganization, it would be helpful to find out each of the following EXCEPT:

(A) Two years ago, what proportion of the Greenville Times’ total revenue was generated by advertising sales?
(B) Has the circulation of the Greenville Times increased substantially in the last two years?
(C) Has there been a substantial turnover in personnel in the advertising sales force over the last two years?
(D) Before the reorganization, had sales representatives found it difficult to keep up with relevant developments in all types of businesses to which they are assigned?
(E) Has the economy in Greenville and the surrounding regions been growing rapidly over the last two years?

Let’s first break down what the argument says:

  • Advertising sales were declining.
  • The paper reorganized the advertising sales team two years back.
  • Advertising sales increased after reorganisation.

Now, we want to figure out whether the increase actually happened due to the reorganization; in other words, we need to evaluate what else could have caused the increase in sales, if not the reorganization. Say the lead of the sales team changed two years back – it is possible that he is responsible for the increase in revenue. Four of the five answer choices will raise similar questions, while the leftover option (which will be our answer) will not. Let’s take a look at each of the answer choices:

(A) Two years ago, what proportion of the Greenville Times’ total revenue was generated by the advertising sales?

The proportion of advertising sales as a part of the total revenue is immaterial to us – we only need to evaluate why the advertising sales have increased. It is possible that the revenue from other sources has increased much more than the revenue from advertising sales, and hence, advertising sales could be a smaller proportion of the overall revenue now, however this doesn’t matter at all. This option has nothing to do with the increase in advertising sales, and hence, is the correct answer.

Let’s take a look at all the other options too, just to be safe:

(B) Has the circulation of the Greenville Times increased substantially in the last two years?

This answer choice can be evaluated in two ways:

  1. Yes, it has increased – If the circulation increased substantially in the last two years, that could have led to the increase in advertising sales.
  2. No, it has not increased – If the circulation hasn’t increased substantially, then there must be another reason for the increase in advertising sales. In that case, the reorganization could be the reason.

These two answers affect the argument differently, and hence, this option will be useful in evaluating the argument.

(C) Has there been a substantial turnover in personnel in the advertising sales force over the last two years?

Again, the answer choice can be evaluated in two ways:

  1. Yes, there has – If there has been a substantial turnover in personnel, it is possible that more capable people have been hired, which could have led to higher advertising sales.
  2. No, there hasn’t – If there hasn’t been a substantial turnover in personnel, then there would need to be another reason for the increased advertising sales. In that case, the reorganization could be the reason.

The two answers affect the argument differently, so this option will also be useful in evaluating the argument.

(D) Before the reorganization, had sales representatives found it difficult to keep up with relevant developments in all types of businesses to which they are assigned?

This option can also be evaluated in two ways:

  1. Yes, they did find it difficult – Did reorganization make it easier to keep track of relevant developments? If yes, then the reorganization could be responsible for the increase in sales.
  2. No, they did not find it difficult – If they did not find it difficult to keep up with relevant developments, then we cannot say whether the reorganization was responsible for the increase in sales or not.

These two responses affect the argument differently. Hence, this option will be useful in evaluating the argument.

(E) Has the economy in Greenville and the surrounding regions been growing rapidly over the last two years?

Answer choice E can also be evaluated in two ways:

  1. Yes, it has – If the economy has been growing rapidly over the past two years, it could be the reason for higher advertising sales. Then we may not be able to attribute the improvement in advertising sales to the reorganization.
  2. No it has not – If there has been no such growth in the economy, then reorganization could be the reason for higher advertising sales.

Again, the two responses affect the argument differently, so this option will also be useful in evaluating the argument.

We see that B, C, D and E are all useful in evaluating the argument. Therefore, our answer is A. We hope you will find it easier to handle such questions in the future!

Getting ready to take the GMAT? We have free online GMAT seminars running all the time. And, be sure to follow us on FacebookYouTubeGoogle+, and Twitter!

Karishma, a Computer Engineer with a keen interest in alternative Mathematical approaches, has mentored students in the continents of Asia, Europe and North America. She teaches the GMAT for Veritas Prep and regularly participates in content development projects such as this blog!

Evaluating “Useful to Evaluate” Critical Reasoning Questions on the GMAT

Quarter Wit, Quarter WisdomIn today’s post, we will look at how to answer “useful to evaluate” Critical Reasoning questions in the Verbal section of the GMAT. Arguably, this is one of the toughest question types for test-takers to tackle (perhaps right after boldfaced questions).

To answer this type of question, all you will need to do is follow these six simple steps:

1) Identify the conclusion.
2) Ask yourself the question raised by answer choice A.
3) Answer it with a “yes” and figure out whether it affects the conclusion.
4) Answer it with a “no” and figure out whether it affects the conclusion.
5) Repeat this for all other answer choices.
6) Only one option will affect the conclusion differently in the two cases – that is your answer.

Let’s illustrate this concept with a problem:

In a certain wildlife park, park rangers are able to track the movements of many rhinoceroses because those animals wear radio collars. When, as often happens, a collar slips off, it is put back on. Putting a collar on a rhinoceros involves immobilizing the animal by shooting it with a tranquilizer dart. Female rhinoceroses that have been frequently re-collared have significantly lower fertility rates than uncollared females. Probably, therefore, some substance in the tranquilizer inhibits fertility. 

In evaluating the argument, it would be most useful to determine which of the following? 

(A) Whether there are more collared female rhinoceroses than uncollared female rhinoceroses in the park. 
(B) How the tranquilizer that is used for immobilizing rhinoceroses differs, if at all, from tranquilizers used in working with other large mammals 
(C) How often park rangers need to use tranquilizer darts to immobilize rhinoceroses for reasons other than attaching radio collars 
(D) Whether male rhinoceroses in the wildlife park lose their collars any more often than the park’s female rhinoceroses do 
(E) Whether radio collars are the only practical means that park rangers have for tracking the movements of rhinoceroses in the park

First, we need to break down the argument to find the premises and the conclusion:

  • Many rhinoceroses wear radio collars.
  • Often, collars slip.
  • When a collar slips, the animal is shot with a tranquilizer to re-collar.
  • The fertility of frequently re-collared females is less than the fertility of uncollared females.
  • Conclusion: Some substance in the tranquilizer inhibits fertility.

Let’s take a look at each answer choice:

(A) Whether there are more collared female rhinoceroses than uncollared female rhinoceroses in the park.

Even if there are more collared female rhinoceroses than uncollared females, this does not affect the argument’s conclusion. This answer choice talks about collared females vs. uncollared females; we are comparing the fertility of re-collared females with that of uncollared females. Anyway, how many of either type there are doesn’t matter. So, whether you answer “yes” or “no” to this question, it is immaterial.

(B) How the tranquilizer that is used for immobilizing rhinoceroses differs, if at all, from tranquilizers used in working with other large mammals.

This option is comparing the tranquilizers used for rhinoceroses with the tranquilizers used for other large mammals. What the conclusion does, however, is compare collared female rhinoceroses with uncollared female rhinoceroses. Hence, whether you answer “very different” or “not different at all” to this question, in the end, it doesn’t matter.

(C) How often park rangers need to use tranquilizer darts to immobilize rhinoceroses for reasons other than attaching radio collars.

This answer choice can be evaluated in two ways:

  • Very Often – Tranquilizers are used very often for uncollared females, too. In this case, can we still say that “tranquilizers inhibit fertility”? No! If they did, fertility in uncollared females would have been low, too.
  • Rarely – This would strengthen our conclusion. If tranquilizers are not used on uncollared females, it is possible that something in these tranquilizers inhibits fertility.

(D) Whether male rhinoceroses in the wildlife park lose their collars any more often than the park’s female rhinoceroses do.

This answer choice is comparing the frequency of tranquilizers used on male rhinoceroses with the frequency of tranquilizers used on female rhinoceroses. What the conclusion actually does is compare collared female rhinoceroses with uncollared female rhinoceroses. Hence, whether you answer this question with “more frequently” or “not more frequently,” it doesn’t matter.

(E) Whether radio collars are the only practical means that park rangers have for tracking the movements of rhinoceroses in the park.

This option is comparing radio collars with other means of tracking. What the conclusion does is compare collared female rhinoceroses with uncollared female rhinoceroses. Hence, whether you answer this question with “there are other means” or “there are no other means,” again, it does not matter.

Note that only answer choice C affects the conclusion – if you answer the question it raises differently, it affects the conclusion differently. Option C would be good to know to evaluate the conclusion of the argument, therefore, the answer must be C.

Now try this question on your own:

Following several years of declining advertising sales, the Greenville Times reorganized its advertising sales force two years ago. Before the reorganization, the sales force was organized geographically, with some sales representatives concentrating on city-center businesses and others concentrating on different outlying regions. The reorganization attempted to increase the sales representatives’ knowledge of clients’ businesses by having each sales representative deal with only one type of industry or of retailing. After the reorganization, advertising sales increased. 

In assessing whether the improvement in advertising sales can properly be attributed to the reorganization, it would be helpful to find out each of the following EXCEPT:

(A) Two years ago, what proportion of the Greenville Times’ total revenue was generated by advertising sales?
(B) Has the circulation of the Greenville Times increased substantially in the last two years?
(C) Has there been a substantial turnover in personnel in the advertising sales force over the last two years?
(D) Before the reorganization, had sales representatives found it difficult to keep up with relevant developments in all types of businesses to which they are assigned?
(E) Has the economy in Greenville and the surrounding regions been growing rapidly over the last two years?

We hope you will find this post useful to evaluate the “useful to evaluate” questions!

Getting ready to take the GMAT? We have free online GMAT seminars running all the time. And, be sure to follow us on FacebookYouTubeGoogle+, and Twitter!

Karishma, a Computer Engineer with a keen interest in alternative Mathematical approaches, has mentored students in the continents of Asia, Europe and North America. She teaches the GMAT for Veritas Prep and regularly participates in content development projects such as this blog!

Quarter Wit, Quarter Wisdom: Using the Deviation Method for Weighted Averages

Quarter Wit, Quarter WisdomWe have discussed how to use the deviation method to find the arithmetic mean of numbers. It is very useful in cases where the numbers are huge, as it considerably brings down the calculation time.

The same method can be applied to weighted averages, as well. Let’s look at an example very similar to the one we examined when we were working on deviations in the case of arithmetic means:

What is the average of 452, 452, 453, 460, 467, 480, 499,  499, 504?

What would you say the average is here? Perhaps, around 470?

Shortfall:
We have two 452s – 452 is 18 less than 470.
453 is 17 less than 470.
460 is 10 less than 470.
467 is 3 less than 470.

Overall, the numbers less than 470 are (2*18) + 17 + 10 + 3 = 66 less than 470.

Excess:
480 is 10 more than 470.
We have two 499s – 499 is 29 more than 470.
504 is 34 more than 470.

Overall, the numbers more than 470 are 10 + (2*29) + 34 = 102 more than 470.

The shortfall is not balanced by the excess; there is an excess of 102-66 = 36.

So what is the average? If we assume that the average of these 9 numbers is 470, there will be an excess of 36. We need to distribute this excess evenly among all of the numbers, and hence, the average will increase by 36/9 = 4.

Therefore, the required mean is 470 + 4 = 474. (If we had assumed the mean to be 474, the shortfall would have balanced the excess.)

This method is used in exactly the same way when we have a simple average as when we have a weighted average. The reason we are reviewing it is that it can be very handy in weighted average questions involving more than two quantities.

We often deal with questions on weighted averages involving two quantities using the scale method. Let’s see how to use the deviation method for more than 2 quantities on an official GMAT question:

Three grades of milk are 1 percent, 2 percent and 3 percent fat by volume. If x gallons of the 1 percent grade, y gallons of the 2 percent grade, and z gallons of the 3 percent grade are mixed to give x+y+z gallons of a 1.5 percent grade, what is x in terms of y and z?

(A) y + 3z
(B) (y +z) / 4
(C) 2y + 3z
(D) 3y + z
(E) 3y + 4.5z

Grade 1 milk contains 1% fat. Grade 2  milk contains 2% fat. Grade 3 milk contains 3% fat. The mixture of all three contains 1.5% fat. So, grade 1 milk provides the shortfall and grades 2 and 3 milk provide the excess.

Shortfall = x*(1.5 – 1)
Excess = y*(2 – 1.5) + z*(3 – 1.5)

Since 1.5 is the actual average, the shortfall = the excess.

x*(1.5 – 1) = y*(2 – 1.5) + z*(3 – 1.5)
x/2 = y/2 + 3z/2
x = y + 3z

And there you have it – the answer is A.

We easily used deviations here to arrive at the relation. It’s good to have this method – useful for both simple averages and weighted averages – in your GMAT toolkit.

Getting ready to take the GMAT? We have free online GMAT seminars running all the time. And, be sure to follow us on FacebookYouTubeGoogle+, and Twitter!

Karishma, a Computer Engineer with a keen interest in alternative Mathematical approaches, has mentored students in the continents of Asia, Europe and North America. She teaches the GMAT for Veritas Prep and regularly participates in content development projects such as this blog!

Quarter Wit, Quarter Wisdom: How to Negate Assumption Answer Choices on the GMAT

Quarter Wit, Quarter WisdomMost GMAT test-takers come across the Assumption Negation Technique at some point in their preparation. It is one of the most effective techniques for assumption questions (which are usually fairly difficult) if you learn to apply it successfully.

We already know that many sentences are invalidated by negating the verb of the dominant clause. For example:

There has been a corresponding increase in the number of professional companies devoted to other performing arts.

becomes

There has not been a corresponding increase in the number of professional companies devoted to other performing arts.

Recently, we got a query on how to negate various modifiers such as “most” and “a majority”. So today, we will examine how to negate the most popular modifiers we come across:

  • All -> Not all
  • Everything -> Not everything
  • Always -> Not always
  • Some -> None
  • Most -> Half or less than half
  • Majority -> Half or less than half
  • Many -> Not many
  • Less than -> Equal to or more than
  • Element A -> Not element A
  • None ->  Some
  • Never ->  Sometimes

Let’s take a look at some examples with these determiners:

1) “All of the 70 professional opera companies are commercially viable options.”
This becomes, “Not all of the 70 professional opera companies are commercially viable options.”

2) “There were fewer than 45 professional opera companies that had been active 30 years ago and that ceased operations during the last 30 years.”
This becomes, “There were 45 or more professional opera companies that had been active 30 years ago and that ceased operations during the last 30 years.”

3) “No one who is feeling isolated can feel happy.”
This becomes, “Some who are feeling isolated can feel happy.”

4) “Anyone who is able to trust other people has a meaningful emotional connection to at least one other human being.”
This becomes, “Not everyone who is able to trust other people has a meaningful emotional connection to at least one other human being.”

5) “The 45 most recently founded opera companies were all established as a result of enthusiasm on the part of a potential audience.”
This becomes, “The 45 most recently founded opera companies were not all established as a result of enthusiasm on the part of a potential audience.”

6) “Many of the vehicles that were ticketed for exceeding the speed limit were ticketed more than once in the time period covered by the report.”
This becomes, “Not many of the vehicles that were ticketed for exceeding the speed limit were ticketed more than once in the time period covered by the report.”

7) “The birds of prey capture and kill every single Spotted Mole that comes above ground.”
This becomes, “Not every single Spotted Mole that comes above ground is captured and killed by the birds of prey.”

8) “At least some people who do not feel isolated are happy.”
This becomes, “No people who do not feel isolated are happy.”

9) “Some land-based mammals active in this region, such as fox, will also hunt and eat the Spotted Mole on a regular basis.”
This becomes, “None of the land-based mammals active in this region, such as fox, will also hunt and eat the Spotted Mole on a regular basis.”

10) “No other animal could pose as significant a threat to the above-ground fruits as could the Spotted Mole.”
This becomes, “Some other animals could pose as significant a threat to the above-ground fruits as could the Spotted Mole.”

We hope the next time you come across an assumption question, you will not face any trouble negating the answer choices!

Getting ready to take the GMAT? We have free online GMAT seminars running all the time. And, be sure to follow us on FacebookYouTubeGoogle+, and Twitter!

Karishma, a Computer Engineer with a keen interest in alternative Mathematical approaches, has mentored students in the continents of Asia, Europe and North America. She teaches the GMAT for Veritas Prep and regularly participates in content development projects such as this blog!

Quarter Wit, Quarter Wisdom: Know Your Subtraction for the GMAT!

Quarter Wit, Quarter WisdomYour first reaction to the title of this post is probably, “I already know my subtraction!” No surprise there. But what is surprising is that our statistics tell us that the following GMAT question – which is nothing extraordinary, but does involve some tricky subtraction – is a 700-level question. That made us decide to write this post. We will discuss this concept along with the question:

The last digit of 12^12 + 13^13 – 14^14 × 15^15 =

(A) 0
(B) 1
(C) 5
(D) 8
(E) 9

This is a simple question based on the cyclicity of units digits. There are 3 terms here: 12^12, 13^13 and (14^14)*(15^15). Let’s find the last digit of each of these terms:

12^12
The units digit of 12 is 2.
2 has a cyclicity of 2 – 4 – 8 – 6.
The cycles end at the powers 4, 8, 12 … etc. So, twelve 2’s will end in a units digit of 6.

13^13
The units digit of 13 is 3.
3 has a cyclicity of 3 – 9 – 7 – 1.
A new cycle starts at the powers 1, 5, 9, 13 … etc. So, thirteen 3’s will end in a units digit of 3.

(14^14)*(15^15)
This term is actually the most simple to manage in the case of its units digit – an even number multiplied by a multiple of 5 will end in 0. Also, note that this will be a huge term compared to the other two terms.

This is what our expression looks like when we consider just the units digits of these terms:

(A number ending in 6) + (A number ending in 3) – (A much greater number ending in 0)

Looking at our most basic options, a number ending in 6 added to a number ending in 3 will give us a number ending in 9 (as 3 + 6 = 9). So, the expression now looks like this:

(A number ending in 9) – (A much greater number ending in 0)

It is at this point that many people mess up. They deduce that 9-0 will end in a 9, and hence, the answer will be E. All their effort goes to waste when they do this. Let’s see why:

How do you subtract one number out of another? Take, for example, 10-7 = 3

This can also be written as 7-10 = -3. (Here, you are still subtracting the number with a lower absolute value from the number with a greater absolute value, but giving it a negative sign.)

Let’s try to look at this in tabular form. The number with the greater absolute value goes on the top and the number with the smaller absolute value goes under it. You then subtract and the result gets the sign of the number with the greater absolute value.

(i) 100-29
100
-29
071

(ii) 29-100
100
-29
071
(But since the sign of 100 is negative, your answer is actually -71.)

So, the number with greater absolute value is always on top. Going back to our original question now, (A number ending in 9) – (A much greater number ending in 0) will look like:

abcd0
–  pq9
ghjk1

Ignoring the letter variables (these are simply placeholders), note that the greater number ending in 0 will be on the top and the smaller one ending in 9 will be below it. This means the answer will be a negative number ending in a units digit of 1. Therefore, our answer is B.

As we learn more advanced concepts, make sure you are not taking your basic principles for granted!

Getting ready to take the GMAT? We have free online GMAT seminars running all the time. And, be sure to follow us on FacebookYouTube, Google+, and Twitter!

Karishma, a Computer Engineer with a keen interest in alternative Mathematical approaches, has mentored students in the continents of Asia, Europe and North America. She teaches the GMAT for Veritas Prep and regularly participates in content development projects such as this blog!

Quarter Wit, Quarter Wisdom: Linear Relations in GMAT Questions

Quarter Wit, Quarter WisdomWe have covered the concepts of direct, inverse and joint variation in previous posts. Today, we will discuss what we mean by “linearly related”. A linear relation is one which, when plotted on a graph, is a straight line. In linear relationships, any given change in an independent variable will produce a corresponding change in the dependent variable, just like a change in the x-coordinate produces a corresponding change in the y-coordinate on a line.

We know the equation of a line: it is y = mx + c, where m is the slope and c is a constant.

Let’s illustrate this concept with a GMAT question. This question may not seem like a geometry question, but using the concept of linear relations can make it easy to find the answer:

A certain quantity is measured on two different scales, the R-scale and the S-scale, that are related linearly. Measurements on the R-scale of 6 and 24 correspond to measurements on the S-scale of 30 and 60, respectively. What measurement on the R-scale corresponds to a measurement of 100 on the S-scale?

(A) 20
(B) 36
(C) 48
(D) 60
(E) 84

Let’s think of the two scales R and S as x- and y-coordinates. We can get two equations for the line that depicts their relationship:

30 = 6m + c ……. (I)
60 = 24m + c ……(II)

(II) – (I)
30 = 18m
m = 30/18 = 5/3

Plugging m = 5/3 in (I), we get:

30 = 6*(5/3) + c
c = 20

Therefore, the equation is S = (5/3)R + 20. Let’s plug in S = 100 to get the value of R:

100 = (5/3)R + 20
R = 48

48 (answer choice C) is our answer.

Alternatively, we have discussed the concept of slope and how to deal with it without any equations in this post. Think of each corresponding pair of R and S as points lying on a line – (6, 30) and (24, 60) are points on a line, so what will (r, 100) be on the same line?

We see that an increase of 18 in the x-coordinate (from 6 to 24) causes an increase of 30 in the y-coordinate (from 30 to 60).

So, the y-coordinate increases by 30/18 = 5/3 for every 1 point increase in the x-coordinate (this is the concept of slope).

From 60 to 100, the increase in the y-coordinate is 40, so the x-coordinate will also increase from 24 to 24 + 40*(3/5) = 48. Again, C is our answer.

Getting ready to take the GMAT? We have free online GMAT seminars running all the time. And, be sure to follow us on FacebookYouTubeGoogle+, and Twitter!

Karishma, a Computer Engineer with a keen interest in alternative Mathematical approaches, has mentored students in the continents of Asia, Europe and North America. She teaches the GMAT for Veritas Prep and regularly participates in content development projects such as this blog!

Advanced Number Properties on the GMAT – Part VI

Quarter Wit, Quarter WisdomMost people feel that the topic of number properties is hard or at least a little tricky. The reason is that no matter how much effort you put into it, you will still come across new concepts every time you sit with some 700+ level problems of this topic. There will be some concepts you don’t know and will need to “figure out” during the actual test. I came across one such question the other day. It brought forth a concept I hadn’t thought about before so I decided to share it today:

Say you have N consecutive integers (starting from any integer). What can you say about their sum? What can you say about their product?

Say N = 3
The numbers are 5, 6, 7 (any three consecutive numbers)
Their sum is 5 + 6 + 7 = 18
Their product is 5*6*7 = 210
Note that both the sum and the product are divisible by 3 (i.e. N).

Say N = 5
The numbers are 2, 3, 4, 5, 6 (any five consecutive numbers)
Their sum is 2 + 3 + 4 + 5 + 6 = 20
Their product is 2*3*4*5*6 = 720
Again, note that both the sum and the product are divisible by 5 (i.e. N)

Say N = 4
The numbers are 3, 4, 5, 6 (any five consecutive numbers)
Their sum is 3 + 4 + 5 + 6 = 18
Their product is 3*4*5*6 = 360
Now note that the sum is not divisible by 4, but the product is divisible by 4.

If N is odd then the sum of N consecutive integers is divisible by N, but this is not so if N is even.
Why is this so? Let’s try to generalize – if we have N consecutive numbers, they will be written in the form:

(Multiple of N),
(Multiple of N) +1,
(Multiple of N) + 2,
… ,
(Multiple of N) + (N-2),
(Multiple of N) + (N-1)

In our examples above, when N = 3, the numbers we picked were 5, 6, 7. They would be written in the form:

(Multiple of 3) + 2 = 5
(Multiple of 3)       = 6
(Multiple of 3) + 1 = 7

In our examples above, when N = 4, the numbers we picked were 3, 4, 5, 6. They would be written in the form:

(Multiple of 4) + 3 = 3
(Multiple of 4)        = 4
(Multiple of 4) + 1 = 5
(Multiple of 4) + 2 = 6
etc.

What happens in case of odd integers? We have a multiple of N and an even number of other integers. The other integers are 1, 2, 3, … (N-2) and (N-1) more than a multiple of N.

Note that these extras will always add up in pairs to give the sum of N:

1 + (N – 1) = N
2 + (N – 2) = N
3 + (N – 3) = N

So when you add up all the integers, you will get a multiple of N.

What happens in case of even integers? You have a multiple of N and an odd number of other integers. The other integers are 1, 2, 3, … (N-2) and (N-1) more than a multiple of N.

Note that these extras will add up to give integers of N but one will be leftover:

1 + (N – 1) = N
2 + (N – 2) = N
3 + (N – 3) = N

The middle number will not have a pair to add up with to give N. So when you add up all the integers, the sum will not be a multiple of N.

For example, let’s reconsider the previous example in which we had four consecutive integers:

(Multiple of 4)      = 4
(Multiple of 4) + 1 = 5
(Multiple of 4) + 2 = 6
(Multiple of 4) + 3 = 3

1 and 3 add up to give 4 but we still have a 2 extra. So the sum of four consecutive integers will not be a multiple of 4.

Let’s now consider the product of N consecutive integers.

In any N consecutive integers, there will be a multiple of N. Hence, the product will always be a multiple of N.

Now take a quick look at the GMAT question that brought this concept into focus:

Which of the following must be true?
1) The sum of N consecutive integers is always divisible by N.
2) If N is even then the sum of N consecutive integers is divisible by N.
3) If N is odd then the sum of N consecutive integers is divisible by N.
4) The Product of K consecutive integers is divisible by K.
5) The product of K consecutive integers is divisible by K!

(A) 1, 4, 5
(B) 3, 4, 5
(C) 4 and 5
(D) 1, 2, 3, 4
(E) only 4

Let’s start with the first three statements this question gives us. We can see that out of Statements 1, 2 and 3, only Statement 3 will be true for all acceptable values of N. Therefore, all the answer choices that include Statements 1 and 2 are out, i.e. options A and D are out. The answer choices that don’t have Statement 3 are also out, i.e. options C and E are out. This leaves us with only answer choice B, and therefore, B is our answer.

This question is a direct application of what we learned above so it doesn’t add much value to our learning as such, but it does have an interesting point. By establishing that B is the answer, we are saying that Statement 5 must be true.

5) The product of K consecutive integers is divisible by K!

We will leave it to you to try to prove this!

(For more advanced number properties on the GMAT, check out Parts I, II, III, IV and V of this series.)

Getting ready to take the GMAT? We have free online GMAT seminars running all the time. And, be sure to follow us on FacebookYouTubeGoogle+, and Twitter!

Karishma, a Computer Engineer with a keen interest in alternative Mathematical approaches, has mentored students in the continents of Asia, Europe and North America. She teaches the GMAT for Veritas Prep and regularly participates in content development projects such as this blog!

Quarter Wit, Quarter Wisdom: Divisibility by Powers of 2

Quarter Wit, Quarter WisdomWe know the divisibility rules of 2, 4 and 8:

For 2 – If the last digit of the number is divisible by 2 (is even), then the number is divisible by 2.

For 4 – If the number formed by last two digits of the number is divisible by 4, then the number is divisible by 4.

For 8 – If the number formed by last three digits of the number is divisible by 8, then the number is divisible by 8.

A similar rule applies to all powers of 2:

For 16 – If the number formed by last four digits of the number is divisible by 16, then the number is divisible by 16.

For 32 – If the number formed by last five digits of the number is divisible by 32, then the number is divisible by 32.

and so on…

Let’s figure out why:

The generic rule can be written like this: A number M is divisible by 2^n if the last n digits of M are divisible by 2^n.

Take, for example, a division by 8 (= 2^3), where M = 65748048 and n = 3.

Our digits of interest are the last three digits, 048.

48 is completely divisible by 8, so we conclude that 65748048 is also divisible by 8.

A valid question here is, “What about the remaining five digits? Why do we ignore them?”

Breaking down M, we can see that 65748048 = 65748000 + 048 (we’ve separated the last three digits).

Now note that 65748000 = 65748 * 1000. Since 1000 has three 0s, it is made up of three 2s and three 5s. Because 1000 it has three 2s as factor, it also has 8 as a factor. This means 65748000 has 8 as a factor by virtue of its three 0s.

All we need to worry about now is the last three digits, 048. If this is divisible by 8, 65748048 will also be divisible by 8. If it is not, 65748048 will not be divisible by 8.

In case the last three digits are not divisible by 8, you can still find the remainder of the number. Whatever remainder you get after dividing the last three digits by 8 will be the remainder when you divide the entire number by 8. This should not be a surprise to you now – 65748000 won’t have a remainder when divided by 8 since it is divisible by 8, so whatever the remainder is when the last 3 digits are divided by 8 will be the remainder when the entire number is divided by 8.

In the generic case, the number M will be split into a number with n zeroes and another number with n digits. The number with n zeroes will be divisible by 2^n because it has n 2s as factors. We just need to see the divisibility of the number with n digits.

We hope you have understood this concept. Let’s take look at a quick GMAT question to see this in action:

What is the remainder when 1990990900034 is divided by 32 ?

(A) 16
(B) 8
(C) 4
(D) 2
(E) 0

Breaking down our given number, 1990990900034 = 1990990900000 + 00034.

1990990900000 ends in five 0’s so it is divisible by 32. 34, when divided by 32, gives us a remainder of 2. Hence, when 1990990900034 is divided by 32, the remainder will be 2. Our answer is D.

Getting ready to take the GMAT? We have free online GMAT seminars running all the time. And, be sure to follow us on FacebookYouTubeGoogle+, and Twitter!

Karishma, a Computer Engineer with a keen interest in alternative Mathematical approaches, has mentored students in the continents of Asia, Europe and North America. She teaches the GMAT for Veritas Prep and regularly participates in content development projects such as this blog!

Quarter Wit Quarter Wisdom: What is Your Favorite Number?

Quarter Wit, Quarter WisdomFans of The Big Bang Theory will remember Sheldon Cooper’s quote from an old episode on his favorite number:

The best number is 73. Why? 73 is the 21st prime number. Its mirror, 37, is the 12th and its mirror, 21, is the product of multiplying 7 and 3… and in binary 73 is a palindrome, 1001001, which backwards is 1001001.”

Though Sheldon’s logic is infallible, my favorite number is 1001 because it has a special role in standardized tests.

1001 is 1 more than 1000 and hence, is sometimes split as (1000 + 1). It sometimes appears in the a^2 – b^2 format such as 1001^2 – 1, and its factors are 7, 11 and 13 (not the factors we usually work with).

Due to its unusual factors and its convenient location (right next to 1000), it could be a part of some tough-looking GMAT questions and should be remembered as a “special” number. Let’s look at a question to understand how to work with this  number.

Which of the following is a factor of 1001^(32) – 1 ?

(A) 768
(B) 819
(C) 826
(D) 858
(E) 924

Note that 1001 is raised to the power 32. This is not an exponent we can easily handle. If  we try to use a binomial here and split 1001 into (1000 + 1), all we will achieve is that upon expanding the given expression, 1 will be cancelled out by -1 and all other terms will have 1000 in common. None of the answer choices are factors of 1000, however, so we must look for some other factor of 1001^(32) – 1.

Without a calculator, it is not possible for us to find the factors of 1001^(32) – 1, but we do know the prime factors of 1001 and hence, the prime factors of 1001^32. We may not be able to say which numbers are factors of 1001^(32) – 1, but we will be able to say which numbers are certainly not factors of this!

Let me explain:

1001 = 7 * 11 * 13 (Try dividing 1001 by 7 and you’ll get 143. 143 is divisible by 11, giving you 13.)

1001^32 = 7^32 * 11^32 * 13^32

Now, what can we say about the prime factors of 1001^(32) – 1? Whatever they are, they are certainly not 7, 11 or 13 – two consecutive integers cannot have any common prime factor (discussed here and continued here).

Now look at the answer choices and try dividing each by 7:

(A) 768 – Not divisible by 7

(B) 819 – Divisible by 7

(C) 826 – Divisible by 7

(D) 858 – Not divisible by 7

(E) 924 – Divisible by 7

Options B, C and E are eliminated. They certainly cannot be factors of 1001^(32) – 1 since they have 7 as a prime factor, and we know 1001^(32) – 1 cannot have 7 as a prime factor.

Now try dividing the remaining options by 11:

(A) 768 – Not divisible by 11

(D) 858 – Divisible by 11

D can also be eliminated now because it has 11 as a factor. By process of elimination, the answer is A; it must be a factor of 1001^(32) – 1.

I hope you see how easily we used the factors of 1001 to help us solve this difficult-looking question. And yes, another attractive feature of 1001 – it is a palindrome in the decimal representation itself!

Getting ready to take the GMAT? We have free online GMAT seminars running all the time. And, be sure to follow us on FacebookYouTubeGoogle+, and Twitter!

Karishma, a Computer Engineer with a keen interest in alternative Mathematical approaches, has mentored students in the continents of Asia, Europe and North America. She teaches the GMAT for Veritas Prep and regularly participates in content development projects such as this blog!

Quarter Wit, Quarter Wisdom: Attacking Gerunds on the GMAT!

Quarter Wit, Quarter WisdomA few weeks back, we talked about participles and how they are used on the GMAT. In that post, we had promised to discuss gerunds more in depth at another time. So today, as promised, we’ll be looking at gerunds. Before we do that, however, let’s examine Verbals.

A Verbal is a verb that acts as a different part of speech – not as a verb.

There are three types of verbals:

  • Infinitives – these take the form of “to + verb”
  • Gerunds – these are the “-ing” form of the verb
  • Participles – these can take the “-ing,” “-ed,” “-en” etc. forms

Gerunds end in “-ing” and act as nouns in the sentence. They can act as a subject, direct object, subject complement or object of a preposition. For example:

Running a marathon is very difficult. – Subject
I love swimming. – Direct object
The activity I enjoy the most is swimming. – Subject complement
She thanked me for helping her. – Object of a preposition

You don’t have to identify the part of speech the gerund represents in a sentence; you just need to identify whether a verb’s “-ing” form is being used as a gerund and evaluate whether it is being used correctly.

A sentence could also use a gerund phrase that begins with a gerund, such as, “Swimming in the morning is exhilarating.”

Let’s take a look at a couple of official questions now:

A recent study has found that within the past few years, many doctors had elected early retirement rather than face the threats of lawsuits and the rising costs of malpractice insurance.

(A) had elected early retirement rather than face
(B) had elected early retirement instead of facing
(C) have elected retiring early instead of facing
(D) have elected to retire early rather than facing
(E) have elected to retire early rather than face

Upon reading the original sentence, we see that there is a gerund phrase here – “rising costs of malpractice insurance” – which is parallel to the noun “threat of lawsuits.”

The two are logically parallel too, since there are two aspects that the doctors do not want to face: rising costs and the threat of lawsuits.

Note, however, that they are not logically parallel to “face.” Hence, the use of the form “facing” would not be correct, since it would put “facing” and “rising” in parallel. So answer choices B, C and D are incorrect.

Actually, “retire” and “face” are logically parallel so they should be grammatically parallel, too. Answer choice E has the two in parallel in infinitive form – to retire and (to is implied here) face are in parallel.

Obviously, there are other decision points to take note of here, mainly the question of “had elected” vs. “have elected.” The use of “had elected” will not be correct here, since we are not discussing two actions in the past occurring at different times. Therefore, the correct answer is E.

Take a look at one more:

In virtually all types of tissue in every animal species dioxin induces the production of enzymes that are the organism’s trying to metabolize, or render harmless, the chemical that is irritating it.

(A) trying to metabolize, or render harmless, the chemical that is irritating it
(B) trying that it metabolize, or render harmless, the chemical irritant
(C) attempt to try to metabolize, or render harmless, such a chemical irritant
(D) attempt to try and metabolize, or render harmless, the chemical irritating it
(E) attempt to metabolize, or render harmless, the chemical irritant

Notice the use of the gerund “trying” in answer choice A. “Organism’s” is in possessive form and acts as an adjective for the noun verbal “trying.” Usually, with possessives, a gerund does not work. We need to use a noun only. With this in mind, answer choices A and B will not work.

The other three options replace “trying” with “attempt” and hence correct this error, however options C and D use the redundant “attempt to try.” The use of “attempt” means “try,” so there is no need to use both. Option E corrects this problem, so it is our correct answer.

Unlike participles, which can be a bit confusing, gerunds are relatively easy to understand and use. Feeling more confident about them now?

Getting ready to take the GMAT? We have free online GMAT seminars running all the time. And, be sure to follow us on FacebookYouTubeGoogle+, and Twitter!

Karishma, a Computer Engineer with a keen interest in alternative Mathematical approaches, has mentored students in the continents of Asia, Europe and North America. She teaches the GMAT for Veritas Prep and regularly participates in content development projects such as this blog!

How to Go From a 48 to 51 in GMAT Quant – Part VII

Quarter Wit, Quarter WisdomBoth a test-taker at the 48 level and one at the 51 level in the GMAT Quant section, are conceptually strong – given an unlimited time frame, both will be able to solve most GMAT questions correctly. The difference lies in the two things a test-taker at the 51 level does skillfully:

  1. Uses holistic, big-picture methods to solve Quant questions.
  2. Handles questions he or she finds difficult in a timely manner.

We have been discussing holistic methods on this blog for a long time now and will continue discussing them. (Before you continue reading, be sure to check out parts I, II, III, IVV and VI of this series.)

Today we will focus on “handling the hard questions in a timely manner.” Note that we do not say “solving the hard questions in a timely manner.” Occasionally, one might be required to make a quick call and choose to guess and move on – but again, that is not the focus of this post. We are actually going to talk about the “lightbulb” moment that helps us save on time. There are many such moments for the 51 level test-taker – in fact, the 51 scorers often have time left over after attempting all these questions.

Test takers at the 48 level will also eventually reach the same conclusions but might need much more time. That will put pressure on them the next time they look at the ticking clock, and once their cool is lost, “silly errors” will start creeping in. So it isn’t about just that one question – one can end up botching many other questions too.

There are many steps that can be easily avoided by a lightbulb moment early on. This is especially true for Data Sufficiency questions.

Let’s take an official example:

Pam owns an inventory of unopened packages of corn and rice, which she has purchased for $17 and $13 per package, respectively. How many packages of corn does she have ?

Statement 1: She has $282 worth of packages.

Statement 2: She has twice as many packages of corn as of rice.

A high scorer will easily recognize that this question is based on the concept of “integral solutions to an equation in two variables.” Since, in such real world examples, x and y cannot be negative or fractional, these equations usually have a finite number of solutions.

After we find one solution, we will quickly know how many solutions the equation has, but getting the first set of values that satisfy the equation requires a little bit of brute force.

The good thing here is that this is a Data Sufficiency question – you don’t need to find the actual solution. The only thing we need is to establish that there is a single solution only. (Obviously, there has to be a solution since Pam does own $282 worth of packages.)

So, the test-taker will start working on finding the first solution (using the method discussed in this post). We are told:

Price of a packet of corn = $17
Price of a packet of rice = $13

Say Pam has “x” packets of corn and “y” packets of rice.

Statement 1: She has $282 worth of packages

Using Statement 1, we know that 17x + 13y = 282.

We are looking for the integer values of x and y.

If x = 0, y will be 21.something (not an integer)
If x = 1, y = 20.something
If x = 2, y = 19.something
If x = 3, y = 17.something

This is where the 51 level scorer stops because they never lose sight of the big picture. The “lightbulb” switches on, and now he or she knows that there will be only one set of values that can satisfy this equation. Why? Because y will be less than 17 in the first set of values that satisfies this equation. So if we want to get the next set that satisfies, we will need to subtract y by 17 (and add 13 to x), which will make y negative.

So in any case, there will be a unique solution to this equation. We don’t actually need to find the solution and hence, nothing will be gained by continuing these calculations. Statement 1 is sufficient.

Statement 2: She has twice as many packages of corn as of rice.

Statement 2 gives us no information on the total number of packages or the total amount spent. Hence, we cannot find the total number of packages of corn using this information alone. Therefore, our answer is A.

I hope you see how you can be alert to what you want to handle these Quant questions in a timely manner.

Getting ready to take the GMAT? We have free online GMAT seminars running all the time. And, be sure to follow us on FacebookYouTubeGoogle+, and Twitter!

Karishma, a Computer Engineer with a keen interest in alternative Mathematical approaches, has mentored students in the continents of Asia, Europe and North America. She teaches the GMAT for Veritas Prep and regularly participates in content development projects such as this blog!

Quarter Wit, Quarter Wisdom: Using Prepositional Phrases on the GMAT

Quarter Wit, Quarter WisdomIn previous posts, we have already discussed participles as well as absolute phrases. Today, let’s take a look at another type of modifier – the prepositional phrase.

A prepositional phrase will begin with a preposition and end with a noun, pronoun, gerund, or clause – the “object” of the preposition. The object of the preposition might have one or more modifiers to describe it.

Here are some examples of prepositional phrases (with prepositions underlined):

  • along the ten mile highway…
  • with a cozy blanket…
  • without worrying…
  • about what he likes…

A prepositional phrase can function as an adjective or an adverb. As an adjective, it answers the question, “Which one?” while as an adverb it can answer the questions, “How?” “When?” or “Where?”.

For example:

  • The book under the table belongs to my mom. Here, the prepositional phrase acts as an adjective and tells us “which one” of the books belongs to my mom.
  • We tried the double cheeseburger at the new burger joint. Here, the prepositional phrase acts as an adverb and tells us “where” we tried the cheeseburger.

Like other modifiers, a prepositional modifier should be placed as close as possible to the thing it is modifying.

Let’s take a look at a couple of official GMAT questions to see how understanding prepositional phrases can help us on this exam:

The nephew of Pliny the Elder wrote the only eyewitness account of the great eruption of Vesuvius in two letters to the historian Tacitus.

(A) The nephew of Pliny the Elder wrote the only eyewitness account of the great eruption of Vesuvius in two letters to the historian Tacitus.
(B) To the historian Tacitus, the nephew of Pliny the Elder wrote two letters, being the only eyewitness accounts of the great eruption of Vesuvius.
(C) The only eyewitness account is in two letters by the nephew of Pliny the Elder writing to the historian Tacitus an account of the great eruption of Vesuvius.
(D) Writing the only eyewitness account, Pliny the Elder’s nephew accounted for the great eruption of Vesuvius in two letters to the historian Tacitus.
(E) In two letters to the historian Tacitus, the nephew of Pliny the Elder wrote the only eyewitness account of the great eruption of Vesuvius.

There are multiple prepositional phrases here:

  • of the great eruption of Vesuvius (answers “Which eruption?”)
  • in two letters (tells us “where” he wrote his account)
  • to the historian Tacitus (answers “Which letters?”)

Therefore, the phrase “to the historian Tacitus” should be close to what it is describing, “letters,” which makes answer choices B and C incorrect.

Also, “in two letters to the historian Tacitus” should modify the verb “wrote.” In options A and D, “in two letters to the historian Tacitus” seems to be modifying “eruption,” which is incorrect. (There are other errors in answer choices B, C and D as well, but we will stick to the topic at hand.)

Option E corrects the prepositional phrase errors by putting the modifier close to the verb “wrote,” so therefore, E is our answer.

Let’s try one more:

Defense attorneys have occasionally argued that their clients’ misconduct stemmed from a reaction to something ingested, but in attributing criminal or delinquent behavior to some food allergy, the perpetrators are in effect told that they are not responsible for their actions.

(A) in attributing criminal or delinquent behavior to some food allergy
(B) if criminal or delinquent behavior is attributed to an allergy to some food
(C) in attributing behavior that is criminal or delinquent to an allergy to some food
(D) if some food allergy is attributed as the cause of criminal or delinquent behavior
(E) in attributing a food allergy as the cause of criminal or delinquent behavior

This sentence has two clauses:

Clause 1: Defense attorneys have occasionally argued that their clients’ misconduct stemmed from a reaction to something ingested,

Clause 2: in attributing criminal or delinquent behavior to some food allergy, the perpetrators are in effect told that they are not responsible for their actions.

These two clauses are joined by the conjunction “but,” and the underlined part is a prepositional phrase in the second clause.

Answer choices A, C and E imply that the perpetrators are attributing their own behaviors to food allergies. That is not correct – their defense attorneys are attributing their behavior to food allergies, and hence, all three of these options have modifier errors.

This leaves us with B and D. Answer choice D uses the phrase “attributed as,” which is grammatically incorrect – the correct usage should be “X is attributed to Y,” rather than “X attributed as Y.” Therefore, option B is our answer.

As you can see, the proper placement of prepositional phrases is instrumental in creating a sentence with a clear, logical meaning.  Since that type of clear, logical meaning is a primary emphasis of correct Sentence Correction answers, you should be prepared to look for prepositional phrases (here we go…) *on the GMAT*.

Getting ready to take the GMAT? We have free online GMAT seminars running all the time. And, be sure to follow us on FacebookYouTubeGoogle+, and Twitter!

Karishma, a Computer Engineer with a keen interest in alternative Mathematical approaches, has mentored students in the continents of Asia, Europe and North America. She teaches the GMAT for Veritas Prep and regularly participates in content development projects such as this blog!

Quarter Wit, Quarter Wisdom: Some GMAT Questions Using the “Like” vs. “As” Concept

Quarter Wit, Quarter WisdomToday we will look at some official GMAT questions testing the “like” vs. “as” concept we discussed last week.

(Review last week’s post – if you haven’t read it already – before you read this one for greater insight on this concept.)

Take a look at the following GMAT Sentence Correction question:

As with those of humans, the DNA of grape plants contains sites where certain unique sequences of nucleotides are repeated over and over.

(A) As with those of humans, the DNA of grape plants contains sites where
(B) As human DNA, the DNA of grape plants contain sites in which
(C) As it is with human DNA, the DNA of grape plants, containing sites in which
(D) Like human, the DNA of grape plants contain sites where
(E) Like human DNA, the DNA of grape plants contains sites in which

Should we use “as” or “like”? Well, what are we comparing? We’re comparing the DNA of humans to the DNA of grape plants. Answer choice E compares these two properly – “Like human DNA, the DNA of grape plants…” DNA is singular, so it uses the singular verb “contains”.

All other options are incorrect. Answer choice A uses “those of” for DNA, but DNA is singular, so this cannot be right. B uses “as” to compare the two nouns, which is also incorrect. C is a sentence fragment without a main verb. And D compares “human” to “DNA”, which is not the “apples-to-apples” comparison we need to make this sentence correct. Therefore, our answer must be E.

Let’s try another one:

Like Auden, the language of James Merrill is chatty, arch, and conversational — given to complex syntactic flights as well as to prosaic free-verse strolls.

(A) Like Auden, the language of James Merrill
(B) Like Auden, James Merrill’s language
(C) Like Auden’s, James Merrill’s language
(D) As with Auden, James Merrill’s language
(E) As is Auden’s the language of James Merrill

Here, we’re comparing Auden’s language to James Merrill’s language. Answer choice C correctly uses the possessive “Auden’s” to show that language is implied. “Like Auden’s language, James Merrill’s language …” contains both parallel structure and a correct comparison.

Answer choices A, B and D incorrectly compare “Auden” to “language,” rather than “Auden’s language” to “language,” so those options are out. The structure of answer choice E is not parallel – “Auden’s” vs. “the language of James Merrill”. Therefore, the answer must be C.

Let’s try something more difficult:

More than thirty years ago Dr. Barbara McClintock, the Nobel Prize winner, reported that genes can “jump,” as pearls moving mysteriously from one necklace to another.

(A) as pearls moving mysteriously from one necklace to another
(B) like pearls moving mysteriously from one necklace to another
(C) as pearls do that move mysteriously from one necklace to others
(D) like pearls do that move mysteriously from one necklace to others
(E) as do pearls that move mysteriously from one necklace to some other one

This is a tricky question – it’s perfect for us to re-iterate how important it is to focus on the meaning of the given sentence. Do not try to follow grammar rules blindly on the GMAT!

Is the comparison between “genes jumping” and “pearls moving”? Do pearls really move mysteriously from one necklace to another? No! This is a hypothetical situation, so we must use “like” – genes are like pearls. Answer choices B and D are the only ones that use “like,” so we can eliminate our other options. D uses a clause with “like,” which is incorrect. In answer choice B, “moving from …” is a modifier – “moving” doesn’t act as a verb here, so it doesn’t need a clause. Hence, answer choice B is correct.

Here’s another one:

According to a recent poll, owning and living in a freestanding house on its own land is still a goal of a majority of young adults, like that of earlier generations.

(A) like that of earlier generations
(B) as that for earlier generations
(C) just as earlier generations did
(D) as have earlier generations
(E) as it was of earlier generations

Note the parallel structure of the comparison in answer choice E – “Owning … a house… is still a goal of young adults, as it was of earlier generations.” It correctly uses “as” with a clause.

Answer choice A uses “that” but its antecedent is not very clear; there are other nouns between “goal” and “like,” and hence, confusion arises. None of the other answer choices give us a clear, parallel comparison, so our answer is E.

Alright, last one:

In Hungary, as in much of Eastern Europe, an overwhelming proportion of women work, many of which are in middle management and light industry.

(A) as in much of Eastern Europe, an overwhelming proportion of women work, many of which are in
(B) as with much of Eastern Europe, an overwhelming proportion of women works, many in
(C) as in much of Eastern Europe, an overwhelming proportion of women work, many of them in.
(D) like much of Eastern Europe, an overwhelming proportion of women works, and many are.
(E) like much of Eastern Europe, an overwhelming proportion of women work, many are in.

Another tricky question. The comparison here is between “what happens in Hungary” and “what happens in much of Eastern Europe,” not between “Hungary” and “much of Eastern Europe.” A different sentence structure would be required to compare “Hungary” to “much of Eastern Europe” such as “Hungary, like much of Eastern Europe, has an overwhelming …”

With prepositional phrases, as with clauses, “as” is used. So, we have two relevant options – A and C. Answer choice A uses “which” for “women,” and hence, is incorrect. Therefore, our answer is C.

Here are some takeaways to keep in mind:

  • You should be comparing “apples” to “apples”.
  • Parallel structure is important.
  • Use “as” with prepositional phrases.

Getting ready to take the GMAT? We have free online GMAT seminars running all the time. And, be sure to follow us on FacebookYouTubeGoogle+, and Twitter!

Karishma, a Computer Engineer with a keen interest in alternative Mathematical approaches, has mentored students in the continents of Asia, Europe and North America. She teaches the GMAT for Veritas Prep and regularly participates in content development projects such as this blog!

Quarter Wit, Quarter Wisdom: Using “Like” vs. “As” on the GMAT Verbal Section

Quarter Wit, Quarter WisdomIf you have seen the Veritas Prep curriculum, then you know we frequently highlight the strategy of “Think like the Testmaker” to answer GMAT questions. Recently, we had a student question the grammatical validity of this construct – this brought the “like” vs. “as” debate to mind, so we decided to tackle it this week.

When should you use “like” and when should you use “as” in a sentence?

Both words can be used in comparisons, however the structure of the sentence will be different in the two cases. This is because traditionally, “like” is a preposition and “as” is a conjunction – a preposition takes the form of an object while a conjunction takes the form of a clause. Therefore:

Using “like,” we compare nouns/pronouns (including gerunds). Usually, a single verb will be used.

Using “as,” we compare actual actions. There will be two verbs used when we compare using “as.”

So, this is how we are going to compare “like” and “as”:

  • He runs like a madman. – A single verb, “runs.”
  • He runs as a madman does. – Two verbs, “runs” and “does” (which is equivalent to “does run”).

In the same way, both of the following sentences are correct:

  • Think like the Testmaker.
  • Think as the Testmaker does.

But beware – “as” used with a noun or pronoun alone does not mean that this usage is incorrect. “As” can also be used to show a role or capacity. For example, in the sentence, “She works as a consultant,” the word “as” means that she works in the capacity of a consultant. There is no comparison here, but the sentence is still grammatically correct.

Also, we usually use “like” in the case of hypothetical comparisons. Take, for instance, the sentence, “She screams like a banshee.” Here, it would be odd to say, “She screams as a banshee does,” because we don’t really know how a banshee screams.

Let’s look at a few GMAT Sentence Correction questions now:

Like many self-taught artists, Perle Hessing did not begin to paint until she was well into middle age.

(A) Like
(B) As have
(C) Just as with
(D) Just like
(E) As did

In this sentence, the word “like” is correctly comparing “Perle Hessing” to “many self taught artists.” There is no clause after “like” and we are using a single verb. Hence, the use of “like” is correct and our answer is A.

Not too bad, right? Let’s try another question:

Based on recent box office receipts, the public’s appetite for documentary films, like nonfiction books, seems to be on the rise. 

(A) like nonfiction books 
(B) as nonfiction books 
(C) as its interest in nonfiction books 
(D) like their interest in nonfiction books 
(E) like its interest in nonfiction books

This sentence also has a comparison, and it is between “appetite” and “interest” and how they are both are on a rise. Answer choice E compares “appetite” to “interest” using “like” as a single verb. None of the answer choices have “as” with a clause so the answer must be E.

These were two simple examples of “like” vs. “as.” Now let’s look at a higher-level GMAT question:

During an ice age, the buildup of ice at the poles and the drop in water levels near the equator speed up the Earth’s rotation, like a spinning figure skater whose speed increases when her arms are drawn in

(A) like a spinning figure skater whose speed increases when her arms are drawn in 
(B) like the increased speed of a figure skater when her arms are drawn in 
(C) like a figure skater who increases speed while spinning with her arms drawn in 
(D) just as a spinning figure skater who increases speed by drawing in her arms 
(E) just as a spinning figure skater increases speed by drawing in her arms

There is a comparison here, but between which two things? Answer choice A seems to be comparing “Earth’s rotation” to “spinning figure skater,” but these two things are not comparable. Option E is the correct choice here – it compares “speed up Earth’s rotation” to “skater increases speed.” Therefore, our answer is E.

Getting ready to take the GMAT? We have free online GMAT seminars running all the time. And, be sure to follow us on FacebookYouTubeGoogle+, and Twitter!

Karishma, a Computer Engineer with a keen interest in alternative Mathematical approaches, has mentored students in the continents of Asia, Europe and North America. She teaches the GMAT for Veritas Prep and regularly participates in content development projects such as this blog!

Quarter Wit, Quarter Wisdom: Are Official Answers Debatable on the GMAT?

Quarter Wit, Quarter WisdomLet’s begin with the bottom line: no, they are not. If you are thinking along the lines of, “This official answer cannot be correct! How can the answer be A? It must be C, or C is at least just as valid as A,” then you are wasting your time. The answer given is never debatable. What you should be thinking instead is, “The answer given is A, but  I thought it was C. I must find out where I made a mistake.”

The point is that since you are going to take GMAT, you must learn to think like the GMAT testmakers. The answers they give for these questions are the correct answers, so need to accept that – this way, the next step of figuring out the gap in your understanding will be far easier. Today, let’s take a look at an official question that is often debated:

The average hourly wage of television assemblers in Vernland has long been significantly lower than that in neighboring Borodia. Since Borodia dropped all tariffs on Vernlandian televisions three years ago, the number of televisions sold annually in Borodia has not changed. However, recent statistics show a drop in the number of television assemblers in Borodia. Therefore, updated trade statistics will probably indicate that the number of televisions Borodia imports annually from Vernland has increased.

Which of the following is an assumption on which the argument depends?

(A) The number of television assemblers in Vernland has increased by at least as much as the number of television assemblers in Borodia has decreased.
(B) Televisions assembled in Vernland have features that televisions assembled in Borodia do not have.
(C) The average number of hours it takes a Borodian television assembler to assemble a television has not decreased significantly during the past three years.
(D) The number of televisions assembled annually in Vernland has increased significantly during the past three years.
(E) The difference between the hourly wage of television assemblers in Vernland and the hourly wage of television assemblers in Borodia is likely to decrease in the next few years.

First, let’s look at the premises of the argument:

  • The hourly wage of assemblers in Vernland is much lower than that in Borodia.
  • 3 years ago, Borodia dropped all tariffs on TVs imported from Vernland.
  • The number of TVs sold annually in Borodia is same.
  • However, the number of assemblers in Borodia has decreased.

The conclusion is that the trade statistics will probably indicate that the number of televisions Borodia imports annually from Vernland has increased.

This conclusion might look logical, but it is full of assumptions.

Why does this conclusion seem so logical? Wages in Vernland are lower, so it would seem like TVs should be cheaper here. Borodia dropped all tariffs on imported TVs, which means there will be no artificial inflation of Vernland TV prices. Finally, the number of TVs sold in Borodia has not dropped, but number of assemblers in Borodia has dropped, which makes it look like fewer TVs are getting made in Borodia.

An onlooker might conclude that Borodia is importing more TVs from Vernland because they are cheaper, but here are some assumptions that come to mind:

  • The cost of a TV in Vernland is lower because assembler’s wage is lower. What if the raw material cost is higher in Vernland? Or other costs are higher? The cost to produce a Vernland TV could actually be higher than the cost to produce a Borodia TV.
  • Fewer TVs are getting made in Borodia, but that does not mean that Borodian assemblers have not become more productive. What if fewer assemblers are needed because they can actually complete the assembly process much faster? The number of TVs sold is the same, however, if each assembler is doing more work, fewer assemblers will be needed. In this case, the number of TVs made in Borodia might not have changed even though the number of producers dropped.

Coming to our question now: Which of the following is an assumption on which the argument depends?

We are looking for an assumption, i.e. a NECESSARY premise. We have already identified some assumptions, so let’s see if any of the answer choices gives us one of those:

(A) The number of television assemblers in Vernland has increased by at least as much as the number of television assemblers in Borodia has decreased.

This is the most popular incorrect answer choice. Test takers keep trying to justify why it makes perfect sense, but actually, it is not required for the conclusion to hold true.

The logic of test takers that pick this answer choice is often on the lines of, “If the number of workers from Borodia decreased, in order for Borodia to show an increased number of imports from Vernland, Vernland must have increased their number of workers by at least as much as the number of workers that left Borodia.”

Note that although this may sound logical, it is not necessary to the argument. There are lots of possible situations where this may not be the case:

Perhaps number of TVs being manufactured in Vernland is the same and, hence, the number of assemblers is the same, too. It is possible that out of the fixed number of TVs manufactured, fewer are getting locally bought and more are getting exported to Borodia. So, it is not necessarily true that number of TV assemblers in Vernland has increased.

(B) Televisions assembled in Vernland have features that televisions assembled in Borodia do not have.

This is also not required for the conclusion to hold – the TVs could actually be exactly the same, but the TVs assembled in Vernland could still be cheaper than the TVs assembled in Borodia due to a potentially lower cost of assembly in Vernland.

(C) The average number of hours it takes a Borodian television assembler to assemble a television has not decreased significantly during the past three years.

This is one of the assumptions we discussed above – we are assuming that the reduction in the number of assemblers must not be due to an increase in the productivity of the assemblers because if the assemblers have got more productive, then the number of TVs produced could be the same and, hence, the number of TVs imported would not have increased.

(D) The number of televisions assembled annually in Vernland has increased significantly during the past three years.

This is not required for the conclusion to hold. Perhaps the number of TVs being sold in Vernland has actually reduced while more are getting exported to Borodia, so the overall number of TVs being made is the same.

(E) The difference between the hourly wage of television assemblers in Vernland and the hourly wage of television assemblers in Borodia is likely to decrease in the next few years.

This is also not required for the conclusion to hold. What happens to the hourly wages of assemblers in Vernland and Borodia in the future doesn’t concern this argument – we are only concerned about what has been happening in the last 3 years.

Therefore, our answer is C.

Getting ready to take the GMAT? We have free online GMAT seminars running all the time. And, be sure to follow us on FacebookYouTubeGoogle+, and Twitter!

Karishma, a Computer Engineer with a keen interest in alternative Mathematical approaches, has mentored students in the continents of Asia, Europe and North America. She teaches the GMAT for Veritas Prep and regularly participates in content development projects such as this blog!

How to Go From a 48 to 51 in GMAT Quant – Part VI

Quarter Wit, Quarter WisdomToday’s post the next part in our “How to Go From 48 to 51 in Quant” series. Again, we will learn a technique that can be employed by the test-taker at an advanced stage of preparation – requiring one to understand the situations in which one can use this simplifying technique.

(Before you continue reading, be sure to check out parts I, II, III, IV, and V of this series.)

We all love to use the plug-in method on GMAT Quant questions. We have an equation given, and if the answer choices are the possible values of x, we just plug in these values to find the one that satisfies the equation.

But what if the answer choices are all complicated values? What if it seems that five times the calculation (in the worst case) will be far more time consuming than actually solving the given equation? Then one is torn between using the favorite plug-in method and using algebra. Let’s take an example to review the methods we can use to solve the question and learn how to simplify the plug-in process by approximating the five available options:

If |4x−4|=|2x+30|, which of the following could be a value of x?

(A) –35/3
(B) −21/2
(C) −13/3
(D) 11/5 
(E) 47/5

This question is an ideal candidate for the “plug-in” method. Here, you have the absolute value equation with the potential values of x given in the answer choices. The problem is that the values of x given are fractional. Of course, if we do plan to solve the equation rather than “plug-in”, we can still solve it using our holistic approach rather than pure algebra. Let’s take a look at that now, and later we will discuss the trick to making the answer choices easier for us to plug in.

Method 1:
|4x – 4| = |2x + 30|

4 * |x – 1| = 2 * |x + 15|

2 * |x – 1| = |x + 15|

This is how we rephrase the equation in our words: twice the distance of x from 1 should be equal to the distance of x from -15.

——————(-15) —————————————————(0)——(1)——————

There are two ways to find the value of x:

Case 1: x could be between -15 and 1 such that the distance between them is split in the ratio 2:1.

or

Case 2: x could be to the right of 1 such that the distance between x and -15 is twice the distance between x and 1.

Let’s examine both of these cases in further detail:

Case 1: The distance from -15 to 1 is of 16 units – this can be split into 3 sections of 16/3 units each. So, the distance of x from 1 should be 16/3, which would make the distance of x from -15 two times 16/3, i.e. 32/3.

So, x should be at a point 16/3 away from 1 toward the left.

x = 1 – 16/3 = -13/3

This is one of our answer choices and, hence, the correct answer. Normally, we would just move on to the next question at this point, but had we not found -13/3 in the answer options, we would have moved on to Case 2:

Case 2: The distance between -15 and 1 is 16 units. x should be an additional 16 units to the right of 1, so the distance between x and 1 is 16 and the distance between x and -15 is two times 16, i.e. 32. This means that x should be 16 units to the right of 1, i.e. x = 17. If you would not have found -13/3 in the answer choices, then you would have found 17.

Now let’s move on to see how we can make the plug-in method work for us in this case by examining each answer choice we are given:

Method 2:
|4x – 4| = |2x + 30|

2 * |x – 1| = |x + 15|

(A) -35/3

It is difficult to solve for x = -35/3 to see if both sides match. Instead, let’s solve for the closest integer, -12.

2 * |-12 – 1| = |-12 + 15|

On the left-hand side, you will get 26, but on the right-hand side, you will get 3.

These values are far away from each other, so x cannot be -35/3.  As the value of x approaches the point where the equation holds – i.e. where the two sides are equal to each other – the gap between the value of the two sides keeps reducing. With such a huge gap between the value of the two sides in this case, it is unlikely that a small adjustment of -35/3 from -12 will bring the two sides to be equal.

(B) -21/2

For this answer choice, let’s solve for the nearest integer, x = -10.

2 * |-10 – 1| = |-10 + 15|

On the left-hand side, you will get 22; on the right-hand side, you will get 5.

Once again, these values are far away from each other and, hence, x will not be -21/2.

(C) -13/3

For this answer choice, let’s solve for x = -4.

2 * |-4 -1| = |-4 + 15|

On the left-hand side, you will get 10; on the right-hand side, you will get 11.

Here, there is a possibility that x can equal -13/3, as the two sides are so close to one another – plug in the actual value of -13/3 and you will see that the left-hand side of the equation does, in fact, equal the right-hand side. Therefore, C is the correct answer.

Basically, we approximated the answer choices we were given and shortlisted the one that gave us very close values. We checked for that and found that it is the answer.

We can also solve this question using pure algebra (taking positive and negative signs of absolute values) but in my opinion, the holistic Method 1 is almost always better than that. Out of the two methods discussed above, you can pick the one you like better, but note that Method 2 does have limited applications – only if you are given the actual values of x, can you use it. Method 1 is far more generic for absolute value questions.

Getting ready to take the GMAT? We have free online GMAT seminars running all the time. And, be sure to follow us on FacebookYouTubeGoogle+, and Twitter!

Karishma, a Computer Engineer with a keen interest in alternative Mathematical approaches, has mentored students in the continents of Asia, Europe and North America. She teaches the GMAT for Veritas Prep and regularly participates in content development projects such as this blog!

Quarter Wit, Quarter Wisdom: Squares and Square Roots on the GMAT

Quarter Wit, Quarter WisdomIn today’s post, we will try to clear up your doubts regarding positive and negative solutions in the case of squares and square roots. We will explain the reasons behind each case, which will help you recall the fundamentals when you need to use them. While preparing for the GMAT, you have probably come across a discussion that says x^2 = 4 has two roots, 2 and -2, while √4 has only one value, 2.

Now, let’s try to understand why this is so:

1) x^2 = 4
Basic algebra tells us that quadratics have two roots. Here, x can be either 2 or -2; each, when squared, will give you 4.

x^2 – 4 = 0 and (x + 2)*(x – 2) = 0 when x equals -2 or 2.

2) √x is positive, only
Now this is odd, right? √4 must be 2. Why is that? Shouldn’t it be 2 or -2. After all, when we square both 2 and -2, we get 4 (as discussed above). So, √4 should be 2 or -2.

Here is the concept: √x denotes only the principal square root. x has two square roots – the positive square root (or principal square root) written as √x and the negative square root written as -√x. Therefore, when you take the square root of 4, you get two roots: √4 and -√4, which  is 2 and -2 respectively.

On a GMAT question, when you see √x, this is specifically referring to the positive square root of the number. So √4 is 2, only.

3) (√x)^2 = x
This is fairly straightforward – since x has a square root, it must be non-negative. When you square it, just the square root sign vanishes and you are left with x.

4) √(x^2) = |x|
Now this isn’t intuitive either. √(x^2) should simply be x – why do we have absolute value of x, then? Again, this has to do with the principal square root concept. First you will square x, and then when you write √, it is by default just the principal square root. The negative square root will be written as -√(x^2). So, irrespective of whether x was positive or negative initially, √(x^2) will definitely be positive x. Therefore, we will need to take the absolute value of x.

Here’s a quick recap with some examples:

  • √9 = 3
  • x^2 = 16 means x is either 4 or -4
  • √(5^2) = 5
  • √(-5^2) = 5
  • (√16)^2 = 16
  • √100 = 10

To see this concept in action, let’s take a look at a very simple official problem:

If x is not 0, then √(x^2)/x =

(A) -1
(B) 0
(C) 1
(D) x
(E) |x|/x

We know that √(x^2) is not simply x, but rather |x|. So, √(x^2)/x = |x|/x.

Depending on whether x is positive or negative, |x|/x will be 1 or -1 – we can’t say which one. Hence, there is no further simplification that we can do, and our answer must be E.

Now that you are all warmed up, let’s examine a higher-level question:

Is √[(x – 3)^2] = (3 – x)?

Statement 1: x is not 3
Statement 2: -x * |x| > 0

We know that √(x^2) = |x|, so √[(x – 3)^2] = |x – 3|.

This means that our question is basically:

Is |x – 3| = 3 – x?

Note that 3 – x can also be written as -(x – 3).

Is |x – 3| = -(x – 3)?

Recall the definition of absolute values: |a| = a if a is greater than or equal to 0, and -a if a < 0.

So, “Is |x – 3| = -(x – 3)?” depends on whether (x – 3) is positive or negative. If (x – 3) is negative (or 0), then |x – 3| is equal to -(x – 3).

So our question now boils down to:

Is (x – 3) negative (or 0)?

Statement 1: x is not 3

This means we know that (x – 3) is not 0, but we still don’t know whether it is negative or positive. This statement is not sufficient.

Statement 2: -x * |x| > 0

|x| is always non-negative, so for the product to be positive, “-x” must also be positive. This means x must be negative. If x is negative, x – 3 must be negative, too.

If (x – 3) is negative, |x – 3| is equal to -(x – 3). Hence, this statement alone is sufficient, and our answer is B.

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Karishma, a Computer Engineer with a keen interest in alternative Mathematical approaches, has mentored students in the continents of Asia, Europe and North America. She teaches the GMAT for Veritas Prep and regularly participates in content development projects such as this blog!

Quarter Wit, Quarter Wisdom: Why Critical Reasoning Needs Your Complete Attention on the GMAT!

Quarter Wit, Quarter WisdomLet’s look at a tricky and time consuming official Critical Reasoning question today. We will learn how to focus on the important aspects of the question and quickly evaluate our answer choices:

Tiger beetles are such fast runners that they can capture virtually any nonflying insect. However, when running toward an insect, a tiger beetle will intermittently stop and then, a moment later, resume its attack. Perhaps the beetles cannot maintain their pace and must pause for a moment’s rest; but an alternative hypothesis is that while running, tiger beetles are unable to adequately process the resulting rapidly changing visual information and so quickly go blind and stop. 

Which of the following, if discovered in experiments using artificially moved prey insects, would support one of the two hypotheses and undermine the other? 

(A) When a prey insect is moved directly toward a beetle that has been chasing it, the beetle immediately stops and runs away without its usual intermittent stopping. 
(B) In pursuing a swerving insect, a beetle alters its course while running and its pauses become more frequent as the chase progresses.
(C) In pursuing a moving insect, a beetle usually responds immediately to changes in the insect’s direction, and it pauses equally frequently whether the chase is up or down an incline. 
(D) If, when a beetle pauses, it has not gained on the insect it is pursuing, the beetle generally ends its pursuit. 
(E) The faster a beetle pursues an insect fleeing directly away from it, the more frequently the beetle stops.

First, take a look at the argument:

  • Tiger beetles are very fast runners.
  • When running toward an insect, a tiger beetle will intermittently stop and then, a moment later, resume its attack.

There are two hypotheses presented for this behavior:

  1. The beetles cannot maintain their pace and must pause for a moment’s rest.
  2. While running, tiger beetles are unable to adequately process the resulting rapidly changing visual information and so quickly go blind and stop.

We need to support one of the two hypotheses and undermine the other. We don’t know which one will be supported and which will be undermined. How will we support/undermine a hypothesis?

The beetles cannot maintain their pace and must pause for a moment’s rest.

Support: Something that tells us that they do get tired. e.g. going uphill they pause more.

Undermine: Something that says that fatigue plays no role e.g. the frequency of pauses do not increase as the chase continues.

While running, tiger beetles are unable to adequately process the resulting rapidly changing visual information and so quickly go blind and stop.

Support: Something that says that they are not able to process changing visual information e.g. as speed increases, frequency of pauses increases.

Undermine: Something that says that they are able to process changing visual information e.g. it doesn’t pause on turns.

Now, we need to look at each answer choice to see which one supports one hypothesis and undermines the other. Focus on the impact each option has on our two hypotheses:

(A) When a prey insect is moved directly toward a beetle that has been chasing it, the beetle immediately stops and runs away without its usual intermittent stopping.

This undermines both hypotheses. If the beetle is able to run without stopping in some situations, it means that it is not a physical ailment that makes him take pauses. He is not trying to catch his breath – so to say – nor is he adjusting his field of vision.

(B) In pursuing a swerving insect, a beetle alters its course while running and its pauses become more frequent as the chase progresses.

If the beetle alters its course while running, it is obviously processing changing visual information and changing its course accordingly while running. This undermines the hypothesis “it cannot process rapidly changing visual information”. However, if the beetle pauses more frequently as the chase progresses, it is tiring out more and more due to the long chase and, hence, is taking more frequent breaks. This supports the hypothesis, “it cannot maintain its speed and pauses for rest”.

Answer choice B strengthens one hypothesis and undermines the other. This must be the answer, but let’s check our other options, just to be sure:

(C) In pursuing a moving insect, a beetle usually responds immediately to changes in the insect’s direction, and it pauses equally frequently whether the chase is up or down an incline.

This answer choice undermines both hypotheses. If the beetle responds immediately to changes in direction, it is able to process changing visual information. In addition, if the beetle takes similar pauses going up or down, it is not the effort of running that is making it take the pauses (otherwise, going up, it would have taken more pauses since it takes more effort going up).

(D) If, when a beetle pauses, it has not gained on the insect it is pursuing, the beetle generally ends its pursuit.

This answer choice might strengthen the hypothesis that the beetle is not able to respond to changing visual information since it decides whether it is giving up or not after pausing (in case there is a certain stance that tells us that it has paused), but it doesn’t actually undermine the hypothesis that the beetle pauses to rest. It is very possible that it pauses to rest, and at that time assesses the situation and decides whether it wants to continue the chase. Hence, this option doesn’t undermine either hypothesis and cannot be our answer.

(E) The faster a beetle pursues an insect fleeing directly away from it, the more frequently the beetle stops.

This answer choice strengthens both of the hypotheses. The faster the beetle runs, the more rest it would need, and the more rapidly visual information would change causing the beetle to pause. Because this option does not undermine either hypothesis, it also cannot be our answer.

Only answer choice B strengthens one hypothesis and undermines the other, therefore, our answer must be B.

Getting ready to take the GMAT? We have free online GMAT seminars running all the time. And, be sure to follow us on FacebookYouTubeGoogle+, and Twitter!

Karishma, a Computer Engineer with a keen interest in alternative Mathematical approaches, has mentored students in the continents of Asia, Europe and North America. She teaches the GMAT for Veritas Prep and regularly participates in content development projects such as this blog!