The post Quarter Wit, Quarter Wisdom: Solving the Fuel-Up Puzzles appeared first on Veritas Prep Blog.

]]>(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 Facebook, YouTube, 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!*

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]]>The post Quarter Wit Quarter Wisdom: Solving the Weighing Puzzle (Part 2) appeared first on Veritas Prep Blog.

]]>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 Facebook, YouTube, 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!*

The post Quarter Wit Quarter Wisdom: Solving the Weighing Puzzle (Part 2) appeared first on Veritas Prep Blog.

]]>The post GMAT Tip of the Week: Keep Your GMAT Score Safe from the Bowling Green Massacre appeared first on Veritas Prep Blog.

]]>Whatever Ms. Conway’s intentions (or lack thereof; again we’ll let you decide) with the quote, she is certainly guilty of inadvertently doing one thing: she didn’t likely intend to help you avoid a disaster on the GMAT, but if you’re paying attention she did.

Your GMAT test day does not have to be a Bowling Green Massacre!

Here’s the thing about the Bowling Green Massacre: it never happened. But by now, it’s lodged deeply enough in the psyche of millions of Americans that, to them, it did. And the same thing happens to GMAT test-takers all the time. They think they’ve seen something on the test that isn’t there, and then they act on something that never happened in the first place. And then, sadly, their GMAT hopes and dreams suffer the same fate as those poor souls at Bowling Green (#thoughtsandprayers).

Here’s how it works:

**The Quant Section’s Bowling Green Massacre**

On the Quant section, particularly with Data Sufficiency, your mind will quickly leap to conclusions or jump to use a rule that seems relevant. Consider the example:

*What is the perimeter of isosceles triangle LMN?*

*(1) Side LM = 4*

* (2) Side LN = 4√2*

*A. Statement (1) ALONE is sufficient, but statement (2) alone is insufficient*

* B. Statement (2) ALONE is sufficient, but statement (1) alone is insufficient*

* C. BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient*

* D. EACH statement ALONE is sufficient*

* E. Statements (1) and (2) TOGETHER are NOT sufficient*

When people see that square root of 2, their minds quickly drift back to all those flash cards they studied – flash cards that include the side ratio for an isosceles right triangle: x, x, x√2. And so they then leap to use that rule, inferring that if one side is 4 and the other is 4√2, the other side must also be 4 to fit the ratio and they can then calculate the perimeter. With both statements together, they figure, they can derive that perimeter and select choice C.

But think about where that side ratio comes from: an isosceles **right** triangle. You’re told in the given information that this triangle is, indeed, isosceles. But you’re never told that it’s a right triangle. Much like the Bowling Green Massacre, “right” never happened. But the mere suggestion of it – the appearance of the √2 term that is directly associated with an isosceles, right triangle – baits approximately half of all test-takers to choose C here instead of the correct E (explanation: “isosceles” means only that two sides match, so the third side could be either 4, matching side LM, or 4√2, matching side LN).

Your mind does this to you often on Data Sufficiency problems: you’ll limit the realm of possible numbers to integers, when that wasn’t defined, or to positive numbers, when that wasn’t defined either. You’ll see symptoms of a rule or concept (like √2 leads to the isosceles right triangle side ratio) and assume that the entire rule is in play. The GMAT preys on your mind’s propensity for creating its own story when in reality, only part of that story really exists.

**The Verbal Section’s Bowling Green Massacre**

This same phenomenon appears on the Verbal section, too – most notably in Critical Reasoning. Much like what many allege that Kellyanne Conway did, your mind wants to ascribe particular significance to events or declarations, and it will often exaggerate on you. Consider the example:

*About two million years ago, lava dammed up a river in western Asia and caused a small lake to form. The lake existed for about half a million years. Bones of an early human ancestor were recently found in the ancient lake-bottom sediments that lie on top of the layer of lava. Therefore, ancestors of modern humans lived in Western Asia between two million and one-and-a-half million years ago.*

*Which one of the following is an assumption required by the argument?*

*A. There were not other lakes in the immediate area before the lava dammed up the river.*

* B. The lake contained fish that the human ancestors could have used for food.*

* C. The lava that lay under the lake-bottom sediments did not contain any human fossil remains.*

* D. The lake was deep enough that a person could drown in it.*

* E. The bones were already in the sediments by the time the lake disappeared.*

The key to most Critical Reasoning problems is finding the conclusion and knowing EXACTLY what the conclusion says – nothing more and nothing less. Here the conclusion is the last sentence, that “ancestors of modern humans lived” in this region at this time. When people answer this problem incorrectly, however, it’s almost always for the same reason. They read the conclusion as “the FIRST/EARLIEST ancestors of modern humans lived…” And in doing so, they choose choice C, which protects against humans having come before the ones related to the bones we have.

“First/earliest” is a classic Bowling Green Massacre – it’s a much more noteworthy event (“scientists have discovered human ancestors” is pretty tame, but “scientists have discovered the FIRST human ancestors” is a big deal) that your brain wants to see. But it’s not actually there! It’s just that, in day to day life, you’d rarely ever read about a run-of-the-mill archaeological discovery; it would only pop up in your social media stream if it were particularly noteworthy, so your mind may very well assume that that notoriety is present even when it’s not.

In order to succeed on the GMAT, you need to become aware of those leaps that your mind likes to take. We’re all susceptible to:

- Assuming that variables represent integers, and that they represent positive numbers
- Seeing the symptoms of a rule and then jumping to apply it
- Applying our own extra superlatives or limits to conclusions

So when you make these mistakes, commit them to memory – they’re not one-off, silly mistakes. Our minds are vulnerable to Bowling Green Massacres, so on test day #staywoke so that your score isn’t among those that are, sadly, massacred.

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

*By Brian Galvin.*

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]]>The post Quarter Wit, Quarter Wisdom: Solving the Hourglass Puzzle appeared first on Veritas Prep Blog.

]]>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 Facebook, YouTube, 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!*

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]]>The post Quarter Wit, Quarter Wisdom: Solving the Weighing and Balancing Puzzle appeared first on Veritas Prep Blog.

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

Here is a picture.

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!

**free online GMAT seminars** running all the time. And, be sure to follow us on **Facebook**, **YouTube**, **Google+**, and **Twitter**!

*GMAT** for Veritas Prep and regularly participates in content development projects such as this blog!*

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]]>The post Quarter Wit, Quarter Wisdom: Solving the Pouring Water Puzzle appeared first on Veritas Prep Blog.

]]>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!

**free online GMAT seminars** running all the time. And, be sure to follow us on **Facebook**, **YouTube**, **Google+**, and **Twitter**!

*GMAT** for Veritas Prep and regularly participates in content development projects such as this blog!*

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]]>The post How to Answer GMAT Questions That are About an Unfamiliar Topic appeared first on Veritas Prep Blog.

]]>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.

**free online GMAT seminars** running all the time. And, be sure to follow us on **Facebook**, **YouTube**, **Google+**, and **Twitter**!

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]]>The post GMAT Tip of the Week: Taking the Least Amount of Time to Solve “At Least” Probability Problems appeared first on Veritas Prep Blog.

]]>Fortunately, and contrary to popular belief, the GMAT isn’t “pure evil.” Wherever it provides opportunities for less-savvy examinees to waste their time, it also provides a shortcut for those who have put in the study time to learn it or who have the patience to look for the elevator, so to speak, before slogging up the stairs. And one classic example of that comes with the “at least one” type of probability question.

To illustrate, let’s consider an example:

*In a bowl of marbles, 8 are yellow, 6 are blue, and 4 are black. If Michelle picks 2 marbles out of the bowl at random and at the same time, what is the probability that at least one of the marbles will be yellow?*

*(A) 5/17*

* (B) 12/17*

* (C) 25/81*

* (D) 56/81*

* (E) 4/9*

Here, you can first streamline the process along the lines of one of those “There are two types of people in the world: those who _______ and those who don’t _______” memes. Your goal is to determine whether you get a yellow marble, so you don’t care as much about “blue” and “black”…those can be grouped into “not yellow,” thereby giving you only two groups: 8 yellow marbles and 10 not-yellow marbles. Fewer groups means less ugly math!

But even so, trying to calculate the probability of every sequence that gives you one or two yellow marbles is labor intensive. You could accomplish that “not yellow” goal several ways:

First marble: Yellow; Second: Not Yellow

First: Not Yellow; Second: Yellow

First: Yellow; Second: Yellow

That’s three different math problems each involving fractions and requiring attention to detail. There ought to be an easier way…and there is. When a probability problem asks you for the probability of “at least one,” consider the only situation in which you WOULDN’T get at least one: if you got none. That’s a single calculation, and helpful because if the probability of drawing two marbles is 100% (that’s what the problem says you’re doing), then 100% minus the probability of the unfavorable outcome (no yellow) has to equal the probability of the favorable outcome. So if you determine “the probability of no yellow” and subtract from 1, you’re finished. That means that your problem should actually look like:

PROBABILITY OF NO YELLOW, FIRST DRAW: 10 non-yellow / 18 total

PROBABILITY OF NO YELLOW, SECOND DRAW: 9 remaining non-yellow / 17 remaining total

10/18 * 9/17 reduces to 10/2 * 1/17 = 5/17. Now here’s the only tricky part of using this technique: 5/17 is the probability of what you DON’T want, so you need to subtract that from 1 to get the probability you do want. So the answer then is 12/17, or B.

More important than this problem is the lesson: when you see an “at least one” probability problem, recognize that the probability of “at least one” equals 100% minus the probability of “none.” Since “none” is always a single calculation, you’ll always be able to save time with this technique. Had the question asked about three marbles, the number of favorable sequences for “at least one yellow” would be:

Yellow Yellow Yellow

Yellow Not-Yellow Not-Yellow

Yellow Not-Yellow Yellow

Yellow Yellow Not-Yellow

Not-Yellow Yellow Yellow

…

(And note here – this list is not yet exhaustive, so under time pressure you may very well forget one sequence entirely and then still get the problem wrong even if you’ve done the math right.)

Whereas the probability of No Yellow is much more straightforward: Not-Yellow, Not-Yellow, Not-Yellow would be 10/18 * 9/17 * 8/16 (and look how nicely that last fraction slots in, reducing quickly to 1/2). What would otherwise be a terrifying slog, the “long way” becomes quite quick the shorter way.

So, remember, when you see “at least one” probability on the GMAT, employ the “100% minus probability of none” strategy and you’ll save valuable time on at least one Quant problem on test day.

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

*By Brian Galvin.*

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]]>The post Solving GMAT Geometry Problems That Involve Infinite Figures appeared first on Veritas Prep Blog.

]]>*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.

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 = 2*s*^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).

**free online GMAT seminars** running all the time. And, be sure to follow us on **Facebook**, **YouTube**, **Google+**, and **Twitter**!

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]]>The post How to Find the Maximum Distance Between Points on a 3D Object appeared first on Veritas Prep Blog.

]]>*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:

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:

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.

**free online GMAT seminars** running all the time. And, be sure to follow us on **Facebook**, **YouTube**, **Google+**, and **Twitter**!

*GMAT** for Veritas Prep and regularly participates in content development projects such as this blog!*

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