Integrated Reasoning – Multi Source Reasoning

Quarter Wit, Quarter WisdomFor the past couple of weeks, we have been talking about integrated reasoning. Today we will continue with that and take up a multi-source reasoning question. These questions often include substantial data and require you to make inferences based on it. They test your logical and reasoning aptitude so don’t get lost in the data. Review the given information and then jump on to the questions. Then come back to the relevant part of the given information and peruse it in detail.

Given Data:

Minerals possess a number of properties that are used as an aid in their identification. These are listed below with a brief description:

Color – The color of the specimen as it appears to the naked eye under normal lighting conditions. Some minerals such as gold will only appear as one color, but due to impurities and crystal light distortion, many minerals can appear in multiple colors. Therefore, observable specimen color is the least effective property in identification.

Streak – The color of a mineral in powdered form. A streak test is performed by dragging a freshly cleaved mineral surface across an unglazed porcelain (Mohs hardness 7) surface. If the mineral is less hard than the porcelain, it will leave a stripe of color (the mineral in a powdered state). This is the true color of a mineral specimen as it lessens the impurity impact and eliminates the light distortion from the crystal. Although a mineral may have multiple observable specimen colors, it will only have one streak color.

Hardness – Minerals are identified roughly by their hardness based on the Mohs scale of mineral hardness, a list of ten minerals from #1 (softest) to #10 (hardest). All minerals will fall somewhere along the scale, based on their ability to scratch any mineral with a number lower than theirs and their inability to scratch any mineral with a number higher than theirs.

Mohs Scale of Mineral Hardness

1         Talc

2         Gypsum

3         Calcite

4         Fluorite

5         Apatite

6         Orthoclase

7         Quartz

8         Topaz

9         Corundum

10     Diamond

Specific Gravity – It is the relative weight of a mineral as compared to the weight of an equal volume of water. The specific gravity is also referred to as density and is expressed normally as an average of a small range of densities.

Common Minerals and Their Specific Gravity

Halite – 2.1

Diamond – 2.26

Gypsum – 2.3

Quartz – 2.7

Talc – 2.8

Muscovite Mica – 2.8

Corundum – 4.0

Cinnabar – 8.0

Gold – 19.3

Optical Properties – Used mainly by scientists, X-rays are sent through thin slices of mineral, producing identifying patterns of light which measure their index of refraction which is distinct for each mineral.

Properties of 3 unidentified minerals:

  1. Mineral A was not able to scratch any of the top 9 minerals on Mohs scale. Its specific gravity is 2.3 (rounded to one decimal place).
  2. The streak color of Mineral B is white. Its specific gravity is 2.3 (rounded to one decimal place)
  3. Mineral C scratches Calcite and Topaz. It is pink in color and its index of refraction is 2.417.

Questions:

Question 1: For each of the following, select ‘Yes’ if the mineral can be uniquely identified based on the information provided. Otherwise, select No.

Question 2. State True/False: It is possible that mineral A and mineral B are the same mineral.
Solutions:

Solution 1. Mineral A – No

We know that the mineral lies between 9  and 10 (excluding 9 but including 10) on the Mohs scale. Also its specific gravity could be anything from 2.25 to 2.35 (excluding 2.35). There could be many minerals with these two properties. From the given data, we see that diamond is one such mineral but it may not be the only one.

Mineral B – No

Again, there can be many minerals with streak color white. Note that every mineral has a single streak color but every streak color may not belong to a single mineral. But we can say that its hardness must be less than 7 (hardness of unglazed porcelain) on Mohs scale since it has a streak color. Also its specific gravity could be anything from 2.25 to 2.35 (excluding 2.35).

Mineral C – Yes

Index of refraction is distinct for each mineral hence given the index of refraction, we can uniquely identify the mineral.

Solution 2. We need to compare mineral A with mineral B. The known properties of the two should not clash if they are to be the same mineral. Note from above that mineral A has a hardness of more than 9 on Mohs scale while mineral B has a hardness of less than 7. So it is not possible that mineral A and B are the same.

Hope the example gave you some idea about the multi source reasoning questions.

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!

Integrated Reasoning – Grasping the Graphs

Quarter Wit, Quarter WisdomContinuing our discussion on IR, let’s look at a graph today and learn how to infer from the data given in it. You may not need to do too many calculations because the options in the drop down menu may allow you to approximate i.e. the options may be quite far apart. Also, you will need to segment the graph into regions using imaginary vertical lines e.g. number of household spending less than 2 hrs at the mall in the graph given below.

The segmentation may not be perfect and hence you may not get the exact answer. The options need to have some scope of error and that is another reason why the options may be far apart. Anyway, the purpose of the graphs is to depict relative trends so exact values are not important.

Question: In a survey of 60 households, the following data was obtained. It gives the relation between ‘annual household income’ and ‘number of hours spent in shopping malls per week’.

1. No household with an annual income of greater than $135,000 spends more than ___ hours in the shopping malls per week.

(A) 3                (B) 5               (C) 6

The households with income above $135,000 are the ones lying above the solid blue line. None of them spends more than 6 hrs at the mall. Note that if you are unsure of where the red point lies (less than or greater than 5), it doesn’t matter. You know for sure that all points lie to the left of 6 and you can have only one correct answer. Hence the red point must lie to the right of 5.

Answer (C)

2. From the households surveyed in the graph above, the pool that spends maximum time on average at the mall is the one with the annual household income between____

(A)   75,000 – 90,000                (B) 90,000 – 105,000                      (C) 105,000 – 120,000

The average time spent by households in the $75,000 – 90,000  range will be to the left of the red dot. Note that there is a point to the extreme left of the red dot and one to the extreme right. Both these points are almost equidistant from the red dot. There is another dot slightly to the left of the red dot. Hence the average will be very near the red dot but to its left.

The average time spent by households in the $90,000 – 105,000  range will be around the green dot. Look at the two dots to the far left of the green dot. The three dots to the right of the green dot are much closer to the green dot. Hence, if anything, the average will be to the left of the green dot, not to its right.

The average time spent by households in the $105,000 – 120,000  range will be close to the orange dot. It will be to the right of the orange dot, not to the left.

The orange dot is the rightmost and hence the average in the range 105,000 – 120,000 will be the highest.

Answer (C)

3. Number of households spending more than 6 hrs per week in a mall is around ____% of the number of household spending less than 2 hrs per week at the mall.

(A)   10%                    (B) 30%                   (C) 50%

We need to segment the graph into three sections – less than 2 hrs, between 2 to 6 hrs and more than 6 hrs.

Number of households that spend less than 2 hrs per week in a shopping mall – around 21

Number of households that spend more than 6 hrs per week in a shopping mall – 7

Number of households that spend more than 6 hrs per week is about 30% of the number of households that spend less than 2 hrs per week.

Answer (B)

4. Number of households earning more than $135,000 annually and spending less than 3 hrs per week at the mall is around ____% of the total number of households surveyed.

(A) 10%                   (B) 30%                 (C) 40%

Again we segment the graph such that the top left corner comprises of households with annual income more than $135,000 but spending less than 3 hrs per week at the mall. On counting them, we find that there are 18 such households. We also know that total there are 60 households surveyed. Hence the required percentage is 18/60 i.e. 30%.

Answer (B)

Hope you have a better grasp of graphs now. More to come shortly!

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!

Integrated Reasoning – Turn the Tables

Quarter Wit, Quarter WisdomStarting today, for the next few weeks we would like to focus on the ‘Integrated Reasoning’ section of the GMAT. The 1.5 yr old section of the GMAT has been giving jitters to many people. We have come across people with 48+ Quant scores but a 2/3 on the IR section. In my opinion, that’s a little strange. If you have strong reasoning skills, there is no reason you cannot apply those to this section as well.

If you are comfortable with the rest of the GMAT but get nightmares thinking of the IR section, I think the problem is psychological – that’s not to say that there is no problem – there is, but it can be easily remedied. In the next few posts we will see how the IR section is also based on the same fundamentals that we have been playing with all along.

There are different types of questions you get in the IR section. We will take Table Analysis today. The good thing about table analysis is that it requires close to 0.0 calculations to arrive at the answer. We know we keep talking about how GMAT Quant questions don’t require any calculations but IR table analysis honestly requires almost none. The question may seem to ask you to sum 10 big numbers but actually you don’t need to do that. Estimates will be more than sufficient in any case. We will discuss a complicated table analysis question today and if you are comfortable with this, you will be comfortable with most table questions.

Table Analysis Question:

Which of the following conclusions can be drawn from the table?

1. The state with the highest ‘number of candidates appeared’ in 2009 also had the lowest pass percentage in 2009.

Let’s sort the data by ‘2009-Appeared’. When put in increasing order, we get Wyoming at the bottom with number of candidates who appeared in 2009 as 496958.

The pass percentage of Wyoming is approximately 200/500 = 40%. Now we need to find whether there is any state with a lower pass percentage. Obviously we cannot and will not calculate the pass percentage of every state. Let’s analyze the 2009 data and group the pass percentage into 2 camps – greater than 50% and less than 50%. If we come across a pass percentage much much lower than 50%, we will give extra attention to it. Note that it is very easy to group them into >50% or <50%. E.g. In Iowa 203587 people appeared and 146084 passed. Since number of candidates who passed is more than 100000, the pass percentage is obviously more than 50%. Ignore it. Similarly, carry on till you reach Florida. 67036 is much less than 50% of 235451. In fact, it will be less than 30% since 10% of 235451 will be approximately 23000 and 30% will be 69000. Hence pass percentage of Florida is less than pass percentage of Wyoming. This means the given statement is false.

2. In 2010, all states experienced an increase in the ‘number of candidates appeared’ but the increase was not more than 16.2% for any state.

This might look ominous. Will we need to calculate the percentage increase for some states accurately up to one decimal place? We will follow the proverbial ‘cross the bridge when you come to it’. First let’s figure out whether all states experienced an increase in ‘the number of candidates appeared’ in 2010. Compare the 2010 values with 2009 values. We see an increase till we reach Michigan. The number of candidates appeared has decreased slightly in Michigan in 2010. This means the statement is false. No further analysis is required! We didn’t have to go to the bridge at all, much less cross it!

3. The state with the maximum number of passed candidates in 2011 is also the state with the minimum number of passed candidates in 2008.

First we sort the data by ‘2011-Passed’. We get Indiana at the bottom with 281565 passed candidates in 2011. Next, sort by ‘2008-Passed’. If Indiana is the state with the minimum number of passed candidates in 2008, it should appear at the top of the table now. It does! So the given statement is true.

4. The same state had the maximum pass percentage in every year.

Now this is tricky. We have to believe that the data must be such that we are able to get the answer in 2 mins. What kind of sorting could help us here? To get an idea of relative values of pass percentages, let’s sort the data by 2008-Appeared. The reason for that is that similar values will come together. States with 200k – 300k candidates will be together, those with 300k-400k candidates will be together etc.

Georgia appears at the top. We see that its pass percentage in 2008 was around 75% (15/20) which is quite high. Let’s look at the pass percentages of other states in 2008. Note that ‘the number of passed candidates’ doesn’t increase much by states but ‘the number of candidates appeared’ keeps increasing. Kansas, Michigan and Wyoming come closest with about 60%. No other state comes even close. So Georgia does have the highest pass percentage in 2008. Let’s focus on Georgia then. In all years thereafter it seems to have very high pass percentage except in 2012 when its pass percentage drops to about 50%. Now all we have to do is find a state with pass percentage more than 50% in 2012 to prove this statement to be wrong. We can easily see that Idaho (13/20) has a pass percentage much higher than 50%. Hence this statement is false.

Hope you learnt some tricks of the trade of tackling tables. More to come shortly!

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!

Facing Too-Much-Knowledge Problems?

Quarter Wit, Quarter WisdomContinuing our scrutiny of interesting standalone questions with important takeaways, let’s discuss today how too much knowledge can actually let you down. We often come across people wondering whether they should learn up the many formulas in permutation/combination, co-ordinate geometry etc. Our take on the question is a flat ‘No’. Formulas won’t take you far in GMAT, perhaps up to 600 but certainly not further. In fact, until and unless you have an eidetic memory or a Math PHD, chances are that knowing too many formulas will be a disadvantage. Let me show you why:

Say you see this question: What is the area of a triangle with vertices at (1, 4), (7, 1) and (4, 7)?

What is the first thing that comes to your mind? I am fairly certain that my fellow engineering grads will think of either Matrices or Heron’s formula.  With Matrices, the confusion will be: was it (y2 – y1) or (y1 – y2)? Or was it –x2 or +x2? With Heron’s formula, the problem will be too many calculations. So we need to get the length of the sides first, then find s, then plug it all in the formula which was ummm… ?s(s-a)… or s?(s-a)… Hope you get my point. Until and unless you spend a fair bit of time everyday with all the formulas you intend to remember and the exact cases in which they can be used, it’s a waste of time and effort. In fact, if you are too attuned to the use of formulas, you will find it hard to think of a non-formula method to solve the problem. It will be hard for you to think beyond the formula and you will spend the two minutes you get in recalling the exact formula to be used in this particular situation, that is assuming there are formulas for most situations.

Now, try to think of a non-formula method and in this, I am sure the non Math background people will do better because they are used to figuring out innovative methods of solving problems. They do not come to the floor with pre-conceived notions on ‘methods to be used while solving particular question types’ and hence can keep their minds open. I can think of and have come across at least 3 different methods of solving this problem without using any exotic formulas. We just have to think in terms of right angles since we know how to find the area of figures with right angles. Let’s discuss each of those methods:

Question: What is the area of a triangle with vertices at (1, 4), (7, 1) and (4, 7)?

(A) 9/2

(B) 9

(C) 27/2

(D) 18

(E) 27

Solution: First of all, follow the golden rule of coordinate geometry – draw the triangle.

We don’t see any right triangles so let’s make some. We know how to find the area when we have right angles around.

Method 1: Use Trapezoids

Area of PQR = Area of APQB + Area of QBCR – Area of APRC

Area of PQR = (1/2)*3*(4 + 7) + (1/2)*3*(7 + 1) – (1/2)*6*(4+1) = 27/2

Method 2: Use a Rectangle

Area of PQR = Area of ABRC – Area of AQP – Area of BQR – Area of PCR

Area of PQR = 6*6 – (1/2)*3*3 – (1/2)*3*6 – (1/2)*3*6 = 36 – (1/2)*45 = 27/2

Method 3: Use right triangles

Area of PQR = Area of PAQ + Area of QAC + Area of PRC

There is one complication here – we don’t know the coordinates of point C. It is easy to figure out. Just find the equation of line QR and find the value of x when y = 4.

Equation of a line is given by: y – y1 = (y2 – y1)/(x2 – x1) * (x – x1)

Equation of QR: y – 1 = (7 – 1)/(4 – 7) * (x – 7)

2x + y = 15

When y = 4, x = 11/2. So C (11/2, 4)

Area of PQR = Area of PAQ + Area of QAC + Area of PRC

Area of PQR = (1/2)*3*3 + (1/2) *(11/2 – 4)*3 + (1/2) *(11/2 -1)*3 = 27/2

Answer (C)

I am sure you can come up with some other methods if you try!

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 Play of Words

Quarter Wit, Quarter WisdomSome days back I came across a question which was a slight twist on a regular question type. The usual active voice of the sentence had been changed to passive but in such a way that the meaning had been altered. It was a lesson in DS as well as SC – read every word carefully. One word could change a 600 level to a 750 level one, a mundane everyday question to a smart question. We often see this interesting transformation in P&C questions but for that to happen in algebra was quite a delight. Let’s discuss that particular question today.

First let’s look at the mundane version.

Question: If 9 notebooks and 3 pencils cost 20 Swiss Francs, do 12 notebooks and 12 pencils cost 40 Swiss Francs?

Statement 1: 7 notebooks and 5 pencils cost 20 Swiss Francs.
Statement 2: 4 notebooks and 8 pencils cost 20 Swiss Francs.

Solution: Both statements give very similar information. It looks like the answer will be (D). That is, if one statement is enough to answer the question alone, the other will probably be enough to answer alone too. Also it seems that we will have two simultaneous equations in two variables so we will be able to solve for the variables.

Let’s quickly review how we actually solve this

Given: If 9 notebooks and 3 pencils cost 20 Swiss Francs –> 9N + 3P = 20 (assuming N is the cost of each notebook and P is the cost of each pencil)……..(I)

Question: do 12 notebooks and 12 pencils cost 40 Swiss Francs – > Is 12N + 12P = 40? OR Is 6N + 6P = 20?

Statement 1: 7 notebooks and 5 pencils cost 20 Swiss Francs.

7N + 5P = 20 ……… (II)

Equating (I) and (II), we get N = P = 20/12. This is sufficient to answer whether 6N + 6P is equal to 20.

Statement 2: 4 notebooks and 8 pencils cost 20 Swiss Francs.

4N + 8P = 20 ……… (III)

Equating (I) and (III), we get N = P = 20/12. This is sufficient to answer whether 6N + 6P is equal to 20.

So as expected, answer is (D) in this case.

The problem arises when the question is changed a bit.

Question 2: If 20 Swiss Francs is enough to buy 9 notebooks and 3 pencils, is 40 Swiss Francs enough to buy 12 notebooks and 12 pencils?

Statement 1: 20 Swiss Francs is enough to buy 7 notebooks and 5 pencils.
Statement 2: 20 Swiss Francs is enough to buy 4 notebooks and 8 pencils.

Solution: Note that the numbers are unchanged. Does the changed wording convey the same meaning? At first glance, you may think so but that is not true. Now, 9 notebooks and 3 pencils may cost less than 20 SF too. All that the statement tells us is that 20 SF is enough – whether it is just enough or comfortably enough, we don’t know. So we don’t have the actual cost of 9N and 3 pencils. We just know that 9N + 3P <= 20. So here we will have to solve inequalities.

But it still seems that both statements give very similar information and so if one alone is sufficient, the other alone should be sufficient too.

Given: If 20 Swiss Francs is enough to buy 9 notebooks and 3 pencils –> 9N + 3P <= 20 ……. (I)

Question: is 40 Swiss Francs enough to buy 12 notebooks and 12 pencils – > “Is 12N + 12P <= 40?” OR “Is 6N + 6P <= 20?” OR “Is N + P < 10/3?”

Statement 1: 20 Swiss Francs is enough to buy 7 notebooks and 5 pencils.

7N + 5P <= 20 ……..(II)

Adding (I) and (II), we get 2N + P <= 5. We want the coefficients of N and P to be the same in the resulting inequality. Since coefficient of N is greater in both inequalities, we cannot have the same coefficient of N and P in the resultant inequality. So we cannot say whether N + P < 10/3 so this statement alone is not sufficient.

Statement 2: 20 Swiss Francs is enough to buy 4 notebooks and 8 pencils.

4N + 8P <= 20 …………(III)

Again, we need the coefficient of N and P to be the same in the resultant inequality. You can get this as

2N + 4P <= 10 (from III)

3N + P <= 20/3 (from I)

Adding them, we get 5N + 5P <= 50/3 OR N + P <= 10/3

This statement alone is sufficient to answer the question.

Answer (B)

As opposed to our instinct, we find that the second statement alone is sufficient while the first is not.

Let’s try to understand why this is so using a logical solution.

Given: 9N + 3P <= 20

Question: Is 12N + 12P <= 40? OR Is 6N + 6P <= 20?

We know that 9 notebooks and 3 pencils cost 20 SF or less. We want to find whether 6 notebooks and 6 pencils will cost 20 SF or less i.e. if you drop 3 notebooks but take another 3 pencils, will your total cost still not exceed 20 SF? That depends on the relative cost of notebooks and pencils. If pencils are cheaper than (or have same cost as) notebooks, then obviously the total cost will stay less than or equal to 20. If pencils are more expensive than notebooks, we need to know how much more expensive they are to be able to judge whether the cost of 6 notebooks and 6 pencils will exceed 20 SF.

Statement 1: 20 Swiss Francs is enough to buy 7 notebooks and 5 pencils.

Now we know that we can substitute 2 notebooks with 2 pencils i.e. pencils may be cheaper than notebooks, may have the same cost or may be a little more expensive but there is enough leeway in our total cost for us to bear the extra cost of 2 pencils in place of 2 notebooks. But do we have enough leeway in our total cost to replace 3 notebooks with 3 pencils, we don’t know. Let me explain this using an example:

Say 1 notebook costs 1 SF and 1 pencil costs 1 SF too. So 9 notebooks and 3 pencils costs 12 SF. 7 notebooks and 5 pencils cost 12 SF. 6 notebooks and 6 pencils will cost 12 SF.

Take a different case – say 1 notebook costs 1.5 SF and 1 pencil costs 1.85 SF
Then 9 notebooks and 3 pencils cost 19.05 SF (which is less than 20 SF). 7 notebooks and 5 pencils cost 19.75 SF. (So even though 1 pencil costs more than 1 notebook, 2 pencils can substitute 2 notebooks because total cost is less than 20 SF. Obviously 1 pencil can substitute 1 notebook since there was enough leeway for even 2 pencils in place of 2 notebooks)
But 6 notebooks and 6 pencils cost 20.1 SF. (Now we see that 3 pencils cannot substitute 3 notebooks because there isn’t enough leeway. This time it crossed 20 SF)

Hence this statement alone is not sufficient to answer the question.

Statement 2: 20 Swiss Francs is enough to buy 4 notebooks and 8 pencils.

Now we know that we can drop 5 notebooks and buy 5 extra pencils in their place and the total cost will still stay below 20 SF. Hence there is enough leeway for 5 replacements. This obviously means that there is enough leeway for 3 replacements and hence the cost of 6 notebooks and 6 pencils will stay below 20 SF. This statement alone is sufficient to answer the question.

Answer (B)

We hope all this made sense. If you are reeling after reading all these numbers, give it another try. The question is a good 750 level question and certainly not easy.

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 Curious Case of the Incorrect Answer

Quarter Wit, Quarter WisdomMany of us are hooked on to using algebra in Quant questions. The thought probably is that how can it be a Quant question if one did not need to take a couple of variables and make a couple of equations/inequalities. We, at Veritas Prep, love to harp on about how algebra is time consuming and unnecessary in most cases. But today we will go one step further and discuss how indiscriminate use of algebra can actually result in incorrect answers. Surprised, eh?

Of course if you make correct equations and solve them correctly, there is no reason you shouldn’t get the correct answer. The problem that arises is in making correct equations/inequalities. Now I am sure you are thinking that you know how to make equations and so probably this post is a waste of your time. Hold on to that thought – Let me give you a statement:

The total number of apples is more than 20.

How will you convert it in terms of algebra? Will you write it as ‘Na > 20’? If this is what you did, please do go through the post. I am sure there will be a couple of things you will find interesting.

Let me take an example to show you why something like this may not be enough.

Question: Harry bought some red books that cost $8 each and some blue books that cost $25 each. If Harry bought more than 10 red books, how many blue books did he buy?

Statement 1: The total cost of blue books that Harry bought was at least $150.
Statement 2: The total cost of all books that Harry bought was less than $260.

Solution: The first thing we will do is look at the most common algebra solution.

Let the number of blue books be B and red books be R.

He bought more than 10 red books so R > 10.

Statement 1: The total cost of blue books that Harry bought was at least $150
25B >= 150

So B >= 6
This statement alone is not sufficient to give the actual value of B.

Statement 2: The total cost of all books that Harry bought was less than $260

8R + 25B < 260 …………. (I)
Also, 10 < R (from above) which gives 80 < 8R …………..(II)
Adding (I) and (II), we get (note that the two inequalities have the same sign ‘<‘ so they can be added)
8R + 25B + 80 < 260 + 8R

B < 7.2

This statement alone is not sufficient to give the actual value of B.
Using both statements together, we get that B >= 6 and B < 7.2

So B could be 6 or 7 (since number of blue books must be an integer value)

Answer (E)

This is incorrect and actually, answer is (C). The question is how? There is no calculation mistake in the above given solution. Then why do we get the incorrect answer? Let me give you the logical solution and prove that answer is actually (C). Then we will discuss why algebra fails us here.

Logical Solution:

Red Books – $8 each

Blue Books – $25 each

No of Red books is more than 10.

Statement 1: The total cost of blue books that Harry bought was at least $150

Blue books cost $25 each so he bought at least 6 books. He could have bought more too. This statement alone is not sufficient.

Statement 2: The total cost of all books that Harry bought was less than $260

He bought more than 10 red books so he bought at least 11 red books. He spent at least $8*11 = $88 on the red books. Out of the total 260, he is left with 260 – 88 = $172 for  the blue books. Since each blue book costs $25, he could have bought at most 6 blue books. He could have bought fewer too. This statement alone is not sufficient.

Using both statements together: He bought at least 6 and at most 6 blue books. So he must have bought 6 blue books.

Answer (C)

I think you would have figured out the problem with the algebra solution by now. In the algebra solution, the inequality 10 < R does not include all the information you have available. You know that R cannot be 10.5 or 10.8 i.e. a decimal. It must be an integer since it represents the number of red books. So you might want to use 11 <= R to get a tighter value for B. Mind you, it is true that R is greater than 10. Important is that it is equal to or greater than 11 too.

Hence the analysis of statement 2 changes a little:

8R + 25B < 260 ………(I)
11 <= R which gives us 88 <= 8R …………. (II)

When you add (I) and (II) now, you get 25B < 172 i.e. B < 6.9. So B must be 6 or less.

This gives us enough information such that when considering both statements together, we get B = 6.

So when you use algebra, but be mindful of the hidden constraints.

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!

Dealing with the Third Degree

Quarter Wit, Quarter WisdomOne of the basic things you need to know before you start your GMAT preparation is how to solve quadratic equations i.e. factorize the quadratic  and equate each factor to 0 to get the possible values that x can take. Today we will discuss how you can solve a third degree equation.

Say an equation such as x^3 – 6x^2 + 11x – 6 = 0.

How do we get the values of x which satisfy this equation?

If you do get a third degree equation, it will have one very easy root such as 0 or 1 or -1 or 2 or -2 etc. Try a few of these values to get the first root. Here x = 1 works. It is easy to see since you have two 6s, a 1 and an 11 as the coefficients.

Putting x = 1: (1)^3 – 6(1)^2 + 11(1) – 6 = 0

So you know that (x – 1) is a factor. Now figure out the quadratic which when multiplied by (x – 1) gives the third degree expression
(x – 1)(ax^2 + bx + c) = x^3 – 6x^2 + 11x – 6

How do we figure out the values of a, b and c? Let’s see.

Coefficient of x^3 on right hand side is 1. So you know that all you need is x^2 so that it multiplies with x to give x^3 on left hand side too. So ‘a’ must be 1.

(x – 1)(x^2 + bx + c) = x^3 – 6x^2 + 11x – 6

The constant term, ‘c’, is easy to figure out too. It should multiply with -1 to give -6 on right hand side. Hence c must be 6.

(x – 1)(x^2 + bx + 6) = x^3 – 6x^2 + 11x – 6

Getting the middle term is slightly more complicated. bx multiplies with x to give x^2 term and you also get the x^2 term by multiplying -1 with x^2. You have -6x^2 on right hand side so you need the same on the left hand side too. You already have -x^2 (by multiplying -1 with x^2) so you need another -5x^2 from bx^2. So b must be -5.

(x – 1)(x^2 – 5x + 6) = x^3 – 6x^2 + 11x – 6
Now you just factorize the quadratic in the usual way. Let’s see how exactly we would do it using a question.

Question: Is x^3 + 2x^2 – 5x – 6 < 0

Statement 1: -3 < x <= -1
Statement 2: -1 <= x < 2

Solution:

We know how to deal with inequalities with multiple factors (discussed here). But the inequality is not split into factors here so we will have to do it on our own.

Let’s first try to find the simple root that this expression must have. Try x = 1, -1 etc. We see that when we put x = -1, the expression becomes 0.

x^3 + 2x^2 – 5x – 6 = (-1)^3 + 2(-1)^2 – 5(-1) – 6 = 0

So the first factor we get is (x + 1).

(x + 1)(ax^2 + bx + c) = x^3 + 2x^2 – 5x – 6

a must be 1 since we have x^3 on the right hand side.

c must be -6 since we have -6 on the right hand side.

(x + 1)(x^2 + bx – 6) = x^3 + 2x^2 – 5x – 6

bx^2 + x^2 = 2x^2

So b must be 1.

We get: (x + 1)(x^2 + x – 6) which is equal to (x + 1)(x + 3)(x – 2) (after splitting the quadratic)

So the question becomes:

Is (x + 1)(x + 3)(x – 2) < 0?

We already know how to deal with inequalities with multiple factors. The transition points here will be -3, -1 and 2. The expression will be negative in the ranges -1 < x < 2 and x < -3

Statement 1: -3 < x <= -1

In this region, the expression is positive (when -3 < x < -1) or 0 (when x = -1). Hence it will certainly not be negative. This is sufficient to answer the question with ‘No’. Hence statement 1 is sufficient to answer the question.

Statement 2: -1 <= x < 2

In this region, the expression is negative (when -1 < x < 2) or 0 (when x = -1). We cannot say for certain whether it will be negative or not. Hence statement 2 alone is not sufficient to answer the question.

Answer (A)

It’s not hard to deal with third degree equations. All you have to do is bring it down to second degree by figuring out one root and then the problem is in a format you already know.

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!

When the Question is Harder than the Solution!

Quarter Wit, Quarter WisdomLast week we looked at a question whose solution was quite hard to explain. This week we will look at a question in which the question itself is hard to explain (so no point worrying about the difficulty in explaining the solution as of now!) So why are we discussing such a question? Because it is certainly not out of GMAT-scope. It uses the concepts of relative speed and GMAT could give you some pretty intimidating questions at higher levels. So what should be your strategy when you come across a question which takes a minute or more to sink in? After you understand the question, first of all you should congratulate yourself that the toughest part is already over. If the question is hard to understand, the solution would be cake walk (well, at least it will feel like it).

Of course, another approach is to skip such a question within 20 secs and move on but in the interest of this post, we will assume that you will not do that. Also, if you get such a question, chances are that you are ‘good’ at quant and that you would have performed quite well in the test till then. In that case, you would have plenty of extra time to challenge your intellect with such a question.

Let’s look at this question now:

Question: On a straight road, a biker noticed that every 12 minutes a bus overtakes him and every 4 minutes he meets an oncoming bus. If all buses leave the station at the same fixed time intervals and run at the same constant speed and the biker moves at a constant speed, what is the time interval between consecutive buses?

(A) 5 minutes
(B) 6 minutes
(C) 8 minutes
(D) 9 minutes
(E) 10 minutes

Solution:

First let’s review the information given in the question:

– All buses run at the same constant speed.

– They leave the station at fixed time intervals, say every t mins. (We have to find the value of t) Had the biker been stationary, he would have met a bus every t mins from both directions.

–  The biker is moving at a constant speed which is less than the speed of the bus. We can infer that the biker’s speed must be less than the speed of the bus because buses overtake him from behind every 12 mins. Had his speed been equal to or more than the speed of the buses, the buses could not have overtaken him.

– Since the biker is moving too, (say going due east) he meets buses coming from one direction (say going from east to west) more frequently and buses coming from the other direction less frequently.

Imagine a scenario where the biker is stationary:

Bus –>         Bus –>         Bus –>         Bus –>         Bus –>         Bus –>

Biker

Bus <–         Bus <–         Bus <–         Bus <–         Bus <–         Bus <–

He will meet a bus coming from either direction every t mins. Note that the distance between consecutive buses will stay the same. Why? Let me explain this using an example:

Assume that starting from a bus station, all buses run at the same speed of 50 mph.
Say a bus starts at 12:00 noon. Another starts at 1:00 pm i.e. exactly one hr later on the same route. Can we say that the previous bus is 50 miles away at 1:00 pm? Yes, so the distance between the two buses initially will be 50 miles. The 1 o clock bus also runs at 50 mph. Will the distance between these two buses always stay the same i.e. the initial 50 miles? Since both buses are moving at the same speed of 50 mph, relative to each other, they are not moving at all and the distance between them remains constant. The exact same concept is used in this question.

Now imagine what happens when the biker starts moving too. Say, he is traveling due east.

Bus –>        Bus –>         Bus –>         Bus –>         Bus –>         Bus –>

…………………………..Biker ->

Bus <–         Bus <–        Bus <–         Bus <–         Bus <–         Bus <–

Say, he just met two buses, one from each direction. Now the Bus (in bold) is a fixed distance away from him. The biker and the Bus are traveling toward each other so they will cover the distance between them faster. Their relative speed is the sum of the speed of the bus and the speed of the biker. They take only 4 mins to meet up. t must be more than 4.

On the other hand, the biker is moving away from the Bus (in italics) so the effective speed of Bus is only the difference between the speed of the bus and the speed of the biker. So Bus takes 12 mins to catch up with the biker. t must be less than 12.

Now that we have understood the question, solving it is relatively easy.

Say the speed of the biker is K and  the speed of the bus is B. The ratio of the relative speeds in the two cases will be the inverse of the ratio of time taken (Ratio of speed and time is covered in a previous post).

When the relative speed is (B + K), the time taken is 4 mins.

When the relative speed is (B – K), the time taken is 12 mins.
(B + K) /(B –K) = 12/4 (inverse of 4/12)

This gives us K = (1/2)B

This means that the bus travelling at a relative speed which is half its usual speed (B-K = B/2) takes 12 minutes to meet the man. If it were travelling at its usual speed (i.e. if the biker were stationary), it would have taken half the time i.e. 12/2 = 6 mins to meet the biker. Had the biker been stationary, the time taken by the bus to cover the distance between them would be the same as the time interval between consecutive buses i.e. t mins.

Hence the value of t is 6 mins.

Answer (B)

Hope you see that once you understand the question well, solving it becomes quite easy.

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!

When Does Order Matter?

Quarter Wit, Quarter WisdomI have to admit that probability is confusing. The problem is not so much that students find it hard to understand as that teachers find it hard to explain. There are subtle points in a probability question that make all the difference in the world and it takes a ton of ingenuity to explain them in a manner that others understand. You either get the point right away, or you don’t.

Today, I will try to explain a probability concept I have always found very difficult to explain in person so the fact that I am attempting to explain it in a post is making me queasy. Nevertheless, the concept is important and I think it deserves a post.

Let me give you two questions:

Question 1: First 15 positive integers are written on a board.  If two numbers are selected one by one from the board at random (the numbers are not necessarily different), what is the probability that the sum of these numbers is odd?

Question 2: There is a group of people consisting of 10 men and 6 women. Among these 16 people, there are 4 married couples (man-woman couples). If one man and one woman are selected at random, what is the probability that a married couple gets selected?

Now, let me give you the solutions to these questions. Note that the two solutions are different. We will discuss the reasons behind the difference today.

Solution 1:

Numbers: 1, 2, 3, 4, …, 13, 14, 15

When will the sum of two of these numbers be odd? When one number is odd and the other is even.

P(Sum is Odd) = P(First number is Odd)*P(Second number is even) + P(First number is Even)*P(Second number is Odd)

P(Sum is Odd) = (8/15)*(7/15) + (7/15)*(8/15) = 112/225

Solution 2:

P(Selecting a Married Man) = 4/10

P(Selecting the Wife of that Man) = 1/6

P(Married Couple is Selected) = (4/10)*(1/6) = 4/60

The question I come across here is this: Why is the second question not solved the way we solved the first question? After all, selecting two things together is the same as selecting them one after another (explained in your Combinatorics book) i.e. why don’t we solve the second question in this way:

P(Selecting a Married Couple) = P(Selecting a Married Man)*P(Selecting the Wife of that Man) + P(Selecting a Married Woman)*P(Selecting the Husband of the Woman)

= (4/10)*(1/6) + (4/6)*(1/10)

Other than the fact that it gives the wrong answer, why can’t we solve it like this? Because the order doesn’t matter here. It doesn’t matter whether we pick the husband first or the wife first. The end result is the same. After we pick either one, the probability of picking the other one stays the same. The two selections have to be made from two different groups. They cannot be made from the same group (contrary to the first question). It doesn’t matter whether you catch hold of the man first or the woman first.

In the first question, the probability of picking the correct second number depends on what you picked in the first selection. Hence we consider the order. I will explain this by trying to solve the first question the way we solved the second question:

On first selection, we can pick any number so the probability is 1. The second selection depends on what you selected in the first pick. If you selected an odd number in the first pick, the probability of selecting an even number is 7/15. If you selected an even number in the first pick, the probability of selecting an odd number is 8/15. So what do you do? Do you use 7/15 or 8/15 with 1? You cannot say so you must take individual cases.

Case 1: Select an odd number and then an even number: (8/15) * (7/15)

Case 2: Select an even number and then an odd number: (7/15) * (8/15)

The total cases considered here are 15*15 (select first number in 15 ways and select the second number in 15 ways since the second number can be the same as the first number). In 8*7 ways, you will select an odd number and then an even number. In 7*8 ways, you will select an even number and then an odd number. In both the cases, the sum will be odd. This gives us a probability of (56+56)/225 = 112/225

The total probability of 1 is obtained as follows:

1 = P(first number odd, second number even) + P(first number even, second number odd) + P(first number odd, second number odd) + P(first number even, second number even)

= 56/225 + 56/225 + 64/225 + 49/225 = 1

We are only interested in the 56/225 + 56/225 part.

In the second question, we need to select a couple. Here, it doesn’t matter whether you select the man first or the woman; the two member types are different and there is only one way in which you can select the corresponding partner. You cannot select two members of the same type e.g. two men or two women. Hence we don’t need to bother with calculating the different cases of selecting the man first or the woman first.

Of course, even if we do it, we will get the correct answer. Let me show you the calculation.

The total number of ways of selecting a man and a woman are ‘select a man in 10 ways’ and ‘a woman in 6 ways’. Then ‘select a woman in 6 ways’ and ‘then a man in 10 ways’ i.e. total 120 ways. To select a couple, you can select a man in 4 ways and the woman in 1 way. You can select a woman in 4 ways and the man in 1 way. So total 4 + 4 = 8 ways.

Probability of selecting a couple = 8/120 = 4/60 (same as before).

To sum it, the two questions are quite different.

In the first question, you have two groups of numbers: Even Numbers and Odd Numbers

You can select the two numbers from different groups or from the same group. Hence the total number of cases is 15*15. Also, you can select the same number again.

In the second question, you have two groups of members: Men and Women

You must select the two members from different groups. You cannot select two men or two women. Hence the total number of cases is only 10*6 (and not 16*15). You cannot select the same member again.

In case of confusion, just use the combinations approach rather than probability. You will invariably get the correct answer.

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 the Number Line

Quarter Wit, Quarter WisdomBy now, you know that we like to discuss visual approaches to problems.  A visual tool that we have used before for solving inequality and modulus questions is the number line. The number line is also useful in helping us solve many number properties questions.

A few things to keep in mind when dealing with number line:

  1. x < y (in other words, x is less than y) implies x is to the left of y on the number line. x and y could be in any region i.e. negative or positive but x must be to the left of y in any case.
  2. ‘x – y > 0’ (in other words, x – y is positive) implies x is to the right of y on the number line. Again, x and y could be in any region of the number line but x will be to the right of y i.e. x will be greater than y in any case.

The importance of these points is not apparent without a couple of questions.

Question 1: If a, b, and c are positive integers, is b between a and c?

Statement 1: b is 3 greater than a, and b is 5 less than c.

Statement 2: c is 5 greater than b, and c is 8 greater than a.

Solution: You might be tempted to use algebra with equations such as b = a + 3, b = c – 5 etc. But the question ‘is b between a and c’ should remind you of the number line. If we can figure out the relative position of ‘a’, ‘b’ and ‘c’ on the number line, we can say whether ‘b’ is between ‘a’ and ‘c’. Many of these ‘is this number less than that number’ questions can be easily done using the number line.

The question ‘Is b between a and c?’ essentially means ‘does b lay between a and c on the number line?’

Statement 1: b is 3 greater than a, and b is 5 less than c.

This means ‘b’ is 3 steps to the right of ‘a’ but 5 steps to the left of ‘c’ on the number line. It must lay between ‘a’ and ‘c’.

This statement alone is sufficient to answer the question.

Statement 2: c is 5 greater than b, and c is 8 greater than a.

‘c’ is 5 steps to the right of ‘b’ which means ‘b’ is 5 steps to the left of ‘c’. ‘c’ is 8 steps to the right of ‘a’ which means ‘a’ is 8 steps to the left of ‘c’. ‘a’ is further to the left of ‘c’ than ‘b’. So ‘b’ must be between ‘a’ and ‘c’.

This statement alone is sufficient to answer the question too.

Hence the answer is (D).

Working with equations would have been far too cumbersome. Don’t take my word for it; try it on your own.

Let’s look at another question based on the same concepts.

Question 2: The points A, B, C and D are on a number line, not necessarily in this order. If the distance between A and B is 18 and the distance between C and D is 8, what is the distance between B and D?

Statement 1: The distance between C and A is the same as the distance between C and B.

Statement 2: A is to the left of D on the number line.

Solution: This question specifically mentions number line.

We are given that distance between A and B is 18. We don’t know how to place A and B on the number line yet:

We don’t know in which region they lay. We can make a similar diagram for C and D. Note that we don’t know how to place these points. All we know is the relative distance between them. We also don’t know which one lays to the left and which one lays to the right.

Statement 1: The distance between C and A is the same as the distance between C and B.

Since distance between C and A is the same as distance between C and B, C must lay in the center of A and B. There are still many different ways of placing B and D so the distance between B and D is not known yet.

This statement alone is not sufficient.

Statement 2: A is to the left of D on the number line.

If the only constraint is that A is to the left of D, there are many different ways of placing A relative to D.

The distance between B and D will be different in different cases. This statement alone is not sufficient.

Let’s consider both the statements together.  C is in the middle of A and B and A is to the left of D. There are still two different cases possible.

The distance between B and D will be different in the two cases. Hence, we still cannot say what the distance between the two points is.

Answer (E)

Number line is a very simple yet powerful visualization tool. Try to use it in various questions for a simpler, more intuitive solution.

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!

Plugging Numbers Without Using Transition Points

Quarter Wit, Quarter WisdomA few months back, one of our posts talked about knowing which numbers to plug-in in case you want to use the number–plugging method. To be more exact, we discussed that you need to find the transition points i.e. the points where the two sides of the inequality become equal. The transition points tend to reverse the relation between the two sides. For a detailed discussion of this concept, revisit this post.

A question that arises here is what if the transition points are not apparent? What do we do in that case? First of all, let us say that the use of logic is preferable in every question. There are few questions (but there sure are!) where the number plugging method is the only decent option. But that’s beside the point. Let’s take up a question and see what to do in case the transition point is hard to see.

Question: If x is an integer, is 4^x < 3^(x+1)?

Statement I: x is positive

Statement II: |x – 1| < 2

Solution:

Again, our moral duty is to first give you the logical solution since we would like you to think in those terms as far as possible. (Though in this question, plugging in numbers might seem easier.) We will discuss how to get the answer by plugging in appropriate numbers in this case.

Method 1: Logical Solution

The question stem only tells us that x can take any integral value. We need to find whether 4^x is less than 3^(x+1).

Consider statement I: x is positive. Given that x is positive, is 4^x less than 3^(x+1)? If we put x = 1, it is easy to see that 4^x is less than 3^(x+1). So the inequality holds in this case. The point is how do we prove that this will be true for all positive values of x? It’s tough to prove that something holds for a lot of numbers. It’s easier to show that it doesn’t hold for at least one number since we need only one suitable value in that case.

Question Stem: Is 4^x < 3^(x+1)?

Reframe it as: Is 4^x < 3^x * 3?

Or Is (4/3)^x < 3?

Note that 4/3 (= 1.33) is greater than 1. When you raise it to a very high power, it will take a very large value. There is no reason it should stay less than 3. Hence the inequality will not hold for large values of x. Hence this statement alone is not sufficient.

Some properties to note:

  • When you raise a positive number greater than 1 to a large positive integer power, it takes a large value.
  • When you raise 1 to a large positive integer power, it stays 1.
  • When you raise a positive number less than 1 to a large positive integer power, the number becomes even smaller than the original value.

These are some number properties you need to work through and be comfortable with.

Consider statement 2: |x – 1| < 2

Hopefully, you understand modulus well now. We can say that this inequality implies that x is a point at a distance less than 2 from the point 1 on the number line i.e. -1 < x < 3.

Is (4/3)^x < 3?

For  small values of x e.g. x = 0, we know the inequality holds. Let’s check for only the largest value x can take i.e. 2 since x must be an integer. Even if x were 2, (4/3)^x = 16/9 i.e. less than 2. (4/3)^x would still be less than 3. Hence the inequality will hold in this case. This statement alone is sufficient to answer the question.

Note that we did use some basic number plugging here too but that number plugging helps us get a clear picture and makes us ask the right questions. There is nothing wrong with plugging in some numbers here and there to understand the logic. If you know why you are plugging in a particular number, it means you are on the right track. Blindly plugging in is the problem.

2. Putting in Numbers

Let’s look at this method too now.

Is 4^x < 3^(x+1)?

Here it’s not easy to find the transition point. We would have to plug in numbers again to find where the two sides of the inequality are equal! So let’s ignore the transition points and directly start plugging in numbers.

Statement 1: x is positive.

Put x = 1, you get 4 < 9 (Holds)

Put x = 2, you get 16 < 27 (Holds)

Put x = 3, you get 64 < 81 (Holds)

What do we do now? How long are we supposed to keep putting in numbers? We cannot do it for all positive integers. How do we decide when to stop? Note that the relative difference between the left hand side and the right hand side is reducing. 9 is more than twice of 4. 27 is more than 16 but not quite twice. It is more than 1.5 times of 16. 81 is somewhat more than 64 but not more than 1.5 times of 64. What this means is that soon enough, the difference will go to 0 and left hand side will become more than the right hand side. If you want to check and you are comfortable with higher powers of numbers,

Put x = 4, you get 256 < 243 (Does not hold)

What you need to do in case the transition point is not apparent is focus on the pattern of the numbers. Is the difference between them narrowing or widening?

This statement alone is not sufficient to answer the question.

Statement II: |x – 1| < 2

As above, we get -1 < x < 3 so this is simply a matter of putting in x = 0, 1 and 2 to see that the inequality holds in each case. Sufficient.

Answer (B)

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!

 

Bewildered by your Verbal Score?

Quarter Wit, Quarter WisdomMost people who plan to take GMAT seriously take a few prep tests, practice tests or mock tests, whatever you may like to call them. Usually, the tests are taken to gauge one’s current level i.e. to get an approximate idea of what one would score if one were to take GMAT that day. Of course, they have other uses too – practice in timed environment, build stamina, identify strengths and weaknesses etc. Usually, these tests are fairly accurate (with an error of up to 40-50 points in the total score). A recent phenomena has been much lower score (especially verbal) compared to the prep test scores (not among Veritas Prep students though – I will explain the reason for this soon).

Today I would like to discuss possible reasons for this unfortunate phenomena and measures you can take to avoid this situation.

I cannot take credit for this discussion since the offered reasoning was Brian’s brain child. Nevertheless, I am putting up some pointers so that people are aware of these issues and can take remedial actions.

Today’s Issues  

  • First let’s pick up the problem with RC passages. A lot of test prep RC passages are old style i.e.  long-winded 400+ word count passages using simple language. The newer passages are shorter and denser so they are more convenient but also more convoluted. It takes more time to read them through and everyone seems to get 4 passages (I have also heard about some people getting five but I cannot be sure.  It seems less likely since it will increase the time pressure manifold. More passages will mean fewer questions from each passage which means your return on time investment in reading the passage will be lower).
  • The official SC is getting tougher. If you are a memorizer, you might have gotten by on previous “memorize between vs. among” or “memorize such as vs. like” questions but now the questions are more reasoning-based (more questions based on accurate meaning). Not only do you need to catch grammatical mistakes, you also need to catch meaning errors e.g. which sentence conveys the correct meaning: “I will try and call you” or “I will try to call you”? I doubt some of the popular tests have caught up to this changing trend, so people are still practicing for older questions. Overall, the GMAT seems to be evolving faster than some popular practice tests.
  • Another major reason is the new IR section. Many people skip the AWA and IR sections while taking mock tests (or they don’t take these sections seriously). But these sections, which account for one hour, sap people’s energy more than they realize and hence they lose focus and will toward the end of the actual test. Since the verbal section appears at the end, it suffers the most.
  • A downside of being net savvy – if people spend a lot of time online discussing questions and solutions, they invariably come across lots of prep test questions (or direct rip-offs of prep test questions) and their variations. This inflates their prep test scores when they take them later. The actual GMAT questions are all new i.e. they test the same concept but in innovative styles. It takes some effort to uncover the concept they are testing and hence the actual GMAT score could be lower than the prep test score.

On the whole, the point is that actual GMAT is tougher than some popular practice tests. These tests have probably been the same for the past 3-4 yrs though the pattern especially of Verbal has changed drastically.

Remedial Measures:

My suggestion going forward will be to adapt yourself to the new GMAT – shorter but denser RCs and reasoning based SCs. Try to get hold of some latest tests for two reasons – the questions would be better suited to the current GMAT and the questions would have novelty value. Also, web based tests are better than downloadable onse since web based tests are easy to update. Keeping these things in mind, Veritas Prep has created tons of new relevant questions in the past few months (they appear in our Next Gen Tests). Also, they have novelty value since they have not been discussed online yet and are web based so we keep updating them with new questions. Also, I have noticed that the scores given by these Veritas Prep practice tests are quite indicative of the actual GMAT scores. So if you are one of our students, we have already taken care of these issues for you (You may want to close this window now and call it a day.)

If you are not our student, you may get in touch with the practice test providers you use and ask them about the age and relevance of their tests in the current format. Take the score of only their latest tests (which are according to the changing GMAT pattern) seriously. Their older tests can be used for practice. Let me add an important point here – it is common knowledge that officially released GMAT questions are old questions (which appeared in GMAT a few years back), but they are a must do to acquaint yourself with the GMAT style so don’t ignore them. However, if you try out enough official questions before taking the test, keep in mind that your score in the official test will be inflated.

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 Matter of Squares

Quarter Wit, Quarter WisdomLet’s look at a question today which encompasses most of what we have discussed in this topic. This will be the last post on this topic for a while now. We assume that after going through these posts thoroughly, if you come across any question on ‘this inscribed in that’, you should be able to handle it. Just a reminder, keep in mind the symmetry of the figures you are handling.

Question: Two identical squares EFGH and JKLM are inscribed in a square ABCD such that AJ:JE:EB = 1:?2:1. What is the area of the octagon obtained by joining points E, K, F, L, G, M, H and J if AB = (2 + ?2) cm?

(A)   8 cm^2

(B)   4 cm^2

(C)   4?2 + 2 cm^2

(D)   4(?2+1) cm^2

(E)    2(?2 + 1) cm^2

Solution:

We are given the length of side AB = (2 + ?2) cm

Also AJ:JE:EB = 1: ?2:1

AJ + JE + EB = (2 + ?2)  = a + ?2a + a

a = 1 cm

AJ = 1 cm

JE = ?2 cm

EB = 1 cm

Now let’s make the octagon as required.

Since AJ = 1 cm and AH = 1 cm, JH = ?(1^2 + 1^2) = ?2 cm

Notice that the octagon is a regular octagon: JE = KF = LG = MH = ?2 cm. Also, HJ = EK = FL = GM = ?2 cm

The area of the octagon = Area of trapezoid MHJG + area of rectangle JELG + Area of trapezoid KFLE

Area of trapezoid MHJG = (1/2) *(Sum of parallel sides)*Altitude = (1/2)*( ?2 + 2 + ?2)*(1) = (?2+1) cm^2

Area of trapezoid KFLE = (?2 +1) cm^2 (by symmetry)

Area of rectangle JELG = ?2*(2 + ?2) = 2(?2 + 1) cm^2

Area of the octagon = (?2 + 1) + 2(?2 + 1) + (?2 + 1) = 4(?2 + 1) cm^2

Answer (D)

Hope you see that it doesn’t matter how the question setter twists the concepts, they are still easy to apply if you understand them well!

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!

Questions on Circles Inscribed in Polygons

Quarter Wit, Quarter WisdomLast week we looked at questions on polygons inscribed in a circle. This week, let’s look at questions on circles inscribed in regular polygons. As noted earlier, it’s important to keep in mind that regular polygons are symmetrical figures. You need very little information to solve for anything in a symmetrical figure.

Question 1: A circle is inscribed in a regular hexagon. A regular hexagon is inscribed in this circle. Another circle is inscribed in the inner regular hexagon and so on. What is the area of the tenth such circle?

Statement I: The length of the side of the outermost regular hexagon is 6 cm.

Statement II: The length of a diagonal of the outermost regular hexagon is 12 cm.

Solution: Thankfully, in DS questions, we don’t need to calculate the answer. We just need to establish the sufficiency of the given data. Note that we have found that there is a defined relation between the sides of a regular hexagon and the radius of an inscribed circle and there is also a defined relation between the radius of a circle and the side of an inscribed regular hexagon.

When the circle is inscribed in a regular hexagon,

Radius of the inscribed circle = (?3/2)* Side of the hexagon

When a regular hexagon is inscribed in a circle,

Side of the inscribed regular hexagon = Radius of the circle

So all we need is the side of any one regular hexagon or the radius of any one circle and we will know the length of the sides of all hexagons and the radii of all circles.

Statement I: The length of the side of the outermost regular hexagon is 6 cm.

If length of the side of the outermost regular hexagon is 6 cm, the radius of the inscribed circle is (?3/2)*6 = 3?3 cm

In that case, the side of the regular hexagon inscribed in this circle is also 3?3 cm. Now we can get the radius of the circle inscribed in this second hexagon and go on the same lines till we reach the tenth circle. This statement alone is sufficient.

Statement II: The length of a diagonal of the outermost regular hexagon is 12 cm.

Note that a hexagon has diagonals of two different lengths. The diagonals that connect vertices with one vertex between them are smaller than the diagonals that connect vertices with two vertices between them. Length of AC will be shorter than length of AD. Given the length of a diagonal, we do not know which diagonal it is. Is AC = 12 or is AD = 12? The length of the side will be different in the two cases. So this statement alone is not sufficient.

Answer (A)

Keep in mind that you don’t actually need to solve for an answer is DS; in fact, in some questions you will not be able to solve for the answer under the given time constraints. All you need to do is ensure that given unlimited time, you will get a unique answer.

Question 2: Four identical circles are drawn in a square such that each circle touches two sides of the square and two other circles (as shown in the figure below). If the side of the square is of length 20 cm, what is the area of the shaded region?

(A) 400 – 100?

(B) 200 – 50?

(C) 100 – 25?

(D) 8?

(E) 4?

Solution: First let’s recall that squares and circles are symmetrical figures. The given figure is symmetrical.

We don’t know any formula that will help us get the area of the curved shaded grey shape in the center. In such cases, very often what you need is to find the area of one region and subtract the area of another out of it. Here, if we subtract the area of the four circles out of the area of the square, the leftover area includes the shaded region but it includes other regions (around the corners etc) too. This is where symmetry helps us.

Notice that we can split the figure into four equal regions to get four smaller squares. Now focus on the diagram give below which shows you one such smaller square. The area around the four corners of the smaller squares is equal i.e. the area of the red region = area of the blue region = area of the yellow region = area of the green region.

Our shaded grey region has four such equal areas so

Area of the shaded grey region = Area of the smaller square – Area of one circle

Area of the shaded grey region = (10)^2 – ?(5)^2 = 100 – 25?

Answer (C)

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!

 

Questions on Polygons Inscribed in Circles

Quarter Wit, Quarter WisdomFor today’s post, I have two questions for you – both on polygons inscribed in a circle. You must go through the previous post based on this topic before trying these questions.

Question 1: Four points that form a polygon lie on the circumference of the circle. What is the area of the polygon ABCD?

Statement I: The radius of the circle is 3 cm.
Statement II: ABCD is square.

Solution:

Notice that you have been given that angles B and D are right angles. Does that imply that the polygon is a square? No. You haven’t been given that the polygon is a regular polygon. The diagonal AC is the diameter since arc ADC subtends a right angle ABC. Hence arc ADC and arc ABC are semi-circles. But the sides of the polygon (AB, BC, CD, DA) may not be equal. Look at the diagram given below:

Statement I: The radius of the circle is 3 cm.

This statement alone is not sufficient. Look at the two figures given above. The area in the two cases will be different depending on the length of the sides. Just knowing the diagonal AC is not enough. Hence this statement alone is not sufficient.

Statement II: ABCD is square.

This tells us that the first figure is valid i.e. the polygon is actually a square. But this statement alone doesn’t give us the measure of any side/diagonal. Hence this statement alone is not sufficient.

Using both statements together, we know that ABCD is a square with a diagonal of length 6 cm. This means that the side of the square is 6/?2 cm giving us an area of (6/?2)^2 = 18 cm^2.

Answer (C)

Let’s look at a more complicated question now.

Question 2: A regular polygon is inscribed in a circle. How many sides does the polygon have?

Statement I: The length of the diagonal of the polygon is equal to the length of the diameter of the circle.
Statement II: The ratio of area of the polygon to the area of the circle is less than 2:3.

Solution:

In this question, we know that the polygon is a regular polygon i.e. all sides are equal in length. As the number of sides keeps increasing, the area of the circle enclosed in the regular polygon keeps increasing till the number of sides is infinite (i.e. we get a circle) and it overlaps with the original circle. The diagram given below will make this clearer.

Let’s look at each statement:

Statement I: The length of one of the diagonals of the polygon is equal to the length of the diameter of the circle.

Do we get the number of sides of the polygon using this statement? No. The diagram below tells you why.

Regular polygons with even number of sides will be symmetrical around their middle diagonal and hence the diagonal will be the diameter. Hence the polygon could have 4/6/8/10 etc sides. Hence this statement alone is not sufficient.

Statement II: The ratio of area of the polygon to the area of the circle is less than 2:3.

Let’s find the fraction of area enclosed by a square.

In the previous post we saw that

Side of the square = ?2 * Radius of the circle

Area of the square = Side^2 = 2*Radius^2

Area of the circle = ?*Radius^2 = 3.14*Radius^2

Ratio of area of the square to area of the circle is 2/3.14 i.e. slightly less than 2/3.

So a square encloses less than 2/3 of the area of the circle. This means a triangle will enclose even less area. Hence, we see that already the number of sides of the regular polygon could be 3 or 4. Hence this statement alone is not sufficient.

Using both statements together, we see that the polygon has 4/6/8 etc sides but the area enclosed should be less than 2/3 of the area of the circle. Hence the regular polygon must have 4 sides. Since the area of a square is a little less than 2/3rd the area of the circle, we can say with fair amount of certainty that the area of a regular hexagon will be more than 2/3rd the area of the circle. But just to be sure, you can do this:

Side of the regular hexagon = Radius of the circle

Area of a regular hexagon = 6*Area of each of the 6 equilateral triangles = 6*(?3/4)*Radius^2 = 2.6*Radius^2

2.6/3.14 is certainly more than 2/3 so the regular polygon cannot be a hexagon. The regular polygon must have 4 sides only.

Answer (C)

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!

And Now, the Other Way

Quarter Wit, Quarter WisdomToday we will work with circles inscribed in regular polygons.

We begin by considering an equilateral triangle whose each side is of length ‘a’. Recall that every triangle has an incircle i.e. a circle can be inscribed in every triangle. The diagram given below shows the circle of radius ‘r’ inscribed in an equilateral triangle.

How can we find the relation between ‘r’ and ‘a’? Every angle of an equilateral triangle is 60 degrees. Since it is an equilateral triangle, due to the symmetry, angle OBD = angle OBA = 30 degrees. So we see that triangle BOD is a 30-60-90 triangle. So the ratio of the sides OD:BD:OB = 1: ?3 = r : a/2.

Therefore, a = 2?3 * r

Side of the triangle = 2?3 * Radius of the circle

As discussed last week, there are many other methods of getting this result. We can use the altitude method.

Altitude of an equilateral triangle is given by (?3/2)*a. The incenter is at a distance 2/3rd of the altitude so OD (radius) = (1/3)* (?3/2)*a = a/2?3

Or Side of the triangle = 2?3 * Radius of the circle

Now we will look at a square.

The figure itself shows us that r = a/2

Side of the square = 2 * Radius of the circle

There is no need to delve deeper into it. Though, here is something for you to think about: Can you have a circle inscribed in a rectangle?

Now let’s consider a circle inscribed in a regular hexagon.

We know that the interior angle of a regular hexagon is 120 degrees. OA will bisect that angle making angle OAD = 60 degrees. Since AB is tangent to the circle, OD will be perpendicular to AB. Hence OAD is a 30-60-90 triangle. Therefore, a/2 : r = 1: ?3

Hence, a = 2r/?3

Side of the hexagon = (2/?3) * Radius of the circle

Again, remember, you are not expected to ‘know’ these results so don’t try to learn them up. You can always derive any relation you want once you know some basic tricks. The intent of these posts is to familiarize you with those tricks.

Next week, we will look at some interesting Geometry questions based on these concepts!

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!

Circle and Inscribed Regular Polygon Relations

Quarter Wit, Quarter WisdomAs promised last week, let’s figure out the relations between the sides of various inscribed regular polygons and the radius of the circle.

We will start with the simplest regular polygon – an equilateral triangle. We will use what we already know about triangles to arrive at the required relations.

Look at the figure given below. AB, BC and AC are sides (of length ‘a’) of the equilateral triangle. OA, OB and OC are radii (of length ‘r’) of the circle.

The interior angles of an equilateral triangle are 60 degrees each. Therefore, angle OBD is 30 degrees (since ABC is an equilateral triangle, BO will bisect angle ABD). So, triangle BOD is a 30-60-90 triangle.

As discussed in your geometry book, the ratio of sides in a 30-60-90 triangle is 1:?3:2 therefore, a/2 : r = ?3:2 or a:r = ?3:1

Side of the triangle = ?3 * Radius of the circle

You don’t have to learn up this result. You can derive it if needed. Note that you can derive it using many other methods. Another method that easily comes to mind is using the altitude AD. Altitude AD of an equilateral triangle is given by (?3/2)*a. The circum center is at a distance 2/3rd of the altitude so AO (radius) = (2/3)* (?3/2)*a = a/?3

Or side of the triangle = ?3 * radius of the circle

Let’s look at a square now.

AB is the side of the square and AO and BO are the radii of the circle. Each interior angle of a square is 90 degrees so half of that angle will be 45 degrees. Therefore, ABO is a 45-45-90 triangle. We know that the ratio of sides in a 45-45-90 triangle is 1:1:?2.

r:a = 1: ?2

Side of the square = ?2*Radius of the circle

Again, no need to learn up the result. Also, there are many methods of arriving at the relation. Another one is using the diagonal of the square. The diagonal of a square is ?2 times the side of the square. The radius of the circle is half the diagonal. So the side is the square is ?2*radius of the circle.

The case of a pentagon is more complicated since it needs the working knowledge of trigonometry which is beyond GMAT scope so we will not delve into it.

We will look at a hexagon though.

Notice that the interior angle of a regular hexagon is 120 degrees so half of that will be 60 degrees. Therefore, both angles OAB and OBA will be 60 degrees each. This means that triangle OAB is an equilateral triangle with all angles 60 degrees and all sides equal. Hence,

Side of the regular hexagon = Radius of the circle.

The higher order regular polygons and more complicated and we will not take them up. We will discuss a circle inscribed in a polygon next week.

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!

Inscribing Polygons and Circles

Quarter Wit, Quarter WisdomLast week we looked at regular and irregular polygons.  Today, let’s try to understand how questions involving one figure inscribed in another are done.  The most common example of a figure inscribed in another is a polygon inscribed in a circle or a circle inscribed in a polygon. Let’s see the various ways in which this can be done.

To inscribe a polygon in a circle, the polygon is placed inside the circle so that all the vertices of the polygon lie on the circumference of the circle.

There are a few points about inscribing a polygon in a circle that you need to keep in mind:

–  Every triangle has a circumcircle so all triangles can be inscribed in a circle.

–  All regular polygons can also be inscribed in a circle.

–  Also, all convex quadrilaterals whose opposite angles sum up to 180 degrees can be inscribed in a circle.

There are also a few points about inscribing a circle in a polygon that you need to keep in mind:

–  All triangles have an inscribed circle (called incircle). When a circle is inscribed in a triangle, all sides of the triangle must be tangent to the circle.

–  All regular polygons have an inscribed circle.

–  Most other polygons do not have an inscribed circle

A simple official question will help us see the relevance of these points:

Question: Which of the figures below can be inscribed in a circle?

(A) I only
(B) III only
(C) I & III only
(D) II & III only
(E) I, II & III

Solution:

I think it will suffice to say that the answer is (C).

Next week, we will look at the relations between the sides of these polygons and the radii of the circles.

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!

Regular Polygons and the Irregular Ones

Quarter Wit, Quarter WisdomContinuing our Geometry journey, let’s discuss polygons today. Some years back, I used to often get confused in the polygon sum-of-the-interior-angles formula if I had to recall it after a gap of some months because I had seen two variations of it:

Sum of interior angles of a polygon = (n – 2)*180

Sum of interior angles of a polygon = (2n – 4)*90

Now, I don’t want you to judge me. Of course, in the second formula, 2 has been removed from 180 and multiplied to the first factor. It is quite simple so why would anyone get tricked here, you wonder? The problem was that after a few months, I would somehow remember (2n – 4) and 180. So I was mixing up the two and I wasn’t sure of the logic behind this formula. That is until I came across the simple explanation of this formula in our Veritas Prep Geometry book (the one which explains how you can divide every polygon with n sides into (n – 2) triangles and hence get the sum of (n – 2)*180). Now it made perfect sense! I couldn’t believe that I had not come across that explanation before and had just learned up (well, tried to!) the formula blindly. So now I ensure that all my students understand every formula that I teach them.

Usually, we are given a regular polygon and we need to find the measure of interior angles or the number of sides. But what if we are given a polygon instead, not a regular polygon. Does this formula still apply? We wouldn’t know if we didn’t understand how the formula came into being. But since we know that we obtain the formula by dividing the polygon into (n-2) triangles, we know that the sum of all interior angles of a triangle is 180 irrespective of the kind of triangle. So it doesn’t matter whether the polygon is regular or not. The sum of all interior angles will still be (n-2)*180.

Let’s look at a question to see the application of this formula in irregular polygon scenario.

Question: The measures of the interior angles in a polygon are consecutive odd integers. The largest angle measures 153 degrees. How many sides does this polygon have?

A) 8
B) 9
C) 10
D) 11
E) 12

Solution:

The interior angles are: 153, 151, 149, 147 … and so on.

Now there are two ways to approach this question – one which is straight forward but uses algebra so is time consuming, another which makes you think but doesn’t take much time. You can guess which one we are going to focus on! But before we do that let’s take a quick look at the algebraic solution too.

Method 1: Algebra

Sum of interior angles of this polygon = 153 + 151 + 149 + … (153 – 2(n-1)) = (n – 2)*180

If there are n sides, there are n interior angles. The second largest angle will be 153 – 2*1. The third largest will be 153 – 2*2. The smallest will be 153 – 2*(n-1). This is an arithmetic progression.

Sum of all terms = [(First term + Last term)/2] * n = [(153 + 153 – 2(n-1))/2] * n

Equating, we get [(153 + 153 – 2(n-1))/2] * n = (n – 2)*180

Solving this you get, n = 10

But let’s figure out a solution without going through this painful calculation.

Method 2: Capitalize on what you know

Angles of the polygon: 153, 151, 149, 147, 145, 143, 141, … , (153 – 2(n-1))

The average of these angles must be equal to the measure of each interior angle of a regular polygon with n sides since the sum of all angles is the same in both the cases.

Measure of each interior angle of n sided regular polygon = Sum of all angles / n = (n-2)*180/n

Using the options:

Measure of each interior angle of 8 sided regular polygon = 180*6/8 = 135 degrees

Measure of each interior angle of 9 sided regular polygon = 180*7/9 = 140 degrees

Measure of each interior angle of 10 sided regular polygon = 180*8/10 = 144 degrees

Measure of each interior angle of 11 sided regular polygon = 180*9/11 = 147 degrees apprx
and so on…

Notice that the average of the given angles can be 144 if there are 10 angles.

The average cannot be higher than 144  i.e. 147 since that will give us only 7 sides (153, 151, 149, 147, 145, 143, 141 – the average is 147 is this case). But the regular polygon with interior angle measure of 147 has 11 sides. Similarly, the average cannot be less than 144 i.e. 140 either because that will give us many more sides than the required 9.

Hence, the polygon must have 10 sides.

Answer (C).

Interesting, eh? Well, it will be when you understand method 2 well and can do it intuitively!

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!

Geometry Diagrams for DS Questions

Quarter Wit, Quarter WisdomLet’s go back to geometry now. We will discuss how to use diagrams to solve DS questions today. Though we discussed a DS question in a previous geometry post, we didn’t discuss how the thought process used for a DS question is different from the thought process used for a PS question. To find whether a statement is sufficient to answer the question, you should try to prove that it is not sufficient. Try to make two cases which answer the question differently using the give information. If there are two or more different answers possible, it means the given information is not enough. Let’s discuss this with the help of an official question.

Question: A circle and a line lie in the XY plane. The circle is centered at the origin and has a radius 1. Does the line intersect the circle?

Statement I: The x-Intercept of the line is greater than 2
Statement II: The slope of the line is -1/5

Solution:

We are given that there is a circle and a line on the XY plane. The line can lie anywhere – it may or may not intersect the circle. The circle has radius 1 so it intersects the x axis at (1, 0) and (-1, 0).

Let’s look at the information given in the two statements:

Statement I: The x-Intercept of the line is greater than 2.

If x intercept > 2, the line can be any of the following (and can be drawn in many more ways)

We found two cases – one in which the line intersects the circle and another in which it doesn’t. The line could have different slopes and different x intercepts (as long as it is greater than 2) to get different cases. Hence we see that this information alone is not sufficient to answer the question.

Statement II: The slope of the line is -1/5.

If slope of the line is -1/5, the line can be drawn in any of the following ways (and many more).

Again, we found two cases – one in which the line intersects the circle and another in which it doesn’t. The slope of the line stays the same but you can move it up or down to get the two different cases (and different x intercepts). Hence we see that this information alone is not sufficient to answer the question.

Using both together: Now the line has a defined slope = -1/5 but it has no defined x intercept. To get x intercept greater than 2, all we need is that y intercept must be greater than 2/5. If you are wondering how we arrived at this, recall from an earlier post:

Slope of a line = – (y intercept)/(x intercept) = – 1/5

y intercept = (1/5) * x intercept

Since x intercept must be greater than 2, y intercept will be greater than 2/5.

If the y intercept is much more than 2/5, it will not intersect the circle. If the y intercept is a little more than 2/5, the line will intersect the circle (as shown by the two diagrams in Fig 2). In one case, it will intersect the circle, in the other case, it will not. So both statements together are not sufficient.

Answer (E)

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!

 

Or Just Use Inequalities!

Quarter Wit, Quarter WisdomIf you are wondering about the absurd title of this post, just take a look at last week’s title. It will make much more sense thereafter. This post is a continuation of last week’s post where we discussed number plugging. Today, as per students’ request, we will look at the inequalities approach to the same official question. You will need to go through our inequalities post to understand the method we will use here.

Recall that, given a < b, (x – a)(x – b) < 0 gives us the range a < x < b and (x – a)(x – b) > 0 gives us the range x < a or x > b.

Question: If x is positive, which of the following could be the correct ordering of 1/x, 2x and x^2?

(I) x^2 < 2x < 1/x

(II) x^2 < 1/x < 2x

(III) 2x < x^2 < 1/x

(A) none
(B) I only
(C) III only
(D) I and II
(E) I, II and III

Solution:  The question has three complex inequalities. We will take each in turn. Note that each inequality consists of two more inequalities. We will split the complex inequality into two simpler inequalities e.g. x^2 < 2x < 1/x gives us x^2 < 2x and 2x < 1/x. Next we will find the range of values of x which satisfy each of these two inequalities and we will see if the two ranges have an overlap i.e. whether there are any values of x which satisfy both these simpler inequalities. If there are, it means there are values of x which satisfy the entire complex inequality too. Things will become clearer once we start working on it so hold on.

Let’s look at each inequality in turn. We start with the first one:

(I) x^2 < 2x < 1/x

We split it into two inequalities:

(i) x^2 < 2x

We can rewrite x^2 < 2x as x^2 – 2x < 0 or x(x – 2) < 0.

We know the range of x for such inequalities can be easily found using the curve on the number line. This will give us 0 < x < 2.

(ii) 2x < 1/x

It can be rewritten as x^2 – 1/2 < 0 (Note that since x must be positive, we can easily multiply both sides of the inequality with x)

This gives us the range -1/?2 < x < 1/?2 (which is 0 < x < 1/?2 since x must be positive).

Is there a region of overlap in these two ranges i.e. can both inequalities hold simultaneously for some values of x? Yes, they can hold for 0 < x < 1/?2. Hence, x^2 < 2x < 1/x will be true for the range 0 < x < 1/?2. So this could be the correct ordering. Let’s go on to the next complex inequality.

(II) x^2 < 1/x < 2x

Again, let’s break up the inequality into two parts:

(i) x^2 < 1/x

x^1 < 1/x is rewritten as x^3 – 1 < 0 which gives us x < 1.

(ii) 1/x < 2x

1/x < 2x is rewritten as x^2 – 1/2 > 0 which gives us x < -1/?2 (not possible since x must be positive) or x > 1/?2

Can both x < 1 and x > 1/?2 hold simultaneously? Sure! For 1/?2 < x < 1, both inequalities will hold and hence x^2 < 1/x < 2x will be true. So this could be the correct ordering too.

(III) 2x < x^2 < 1/x

The inequalities here are:

(i)  2x < x^2

2x < x^2 can be rewritten as x(x – 2) > 0 which gives us x < 0 (not possible) or x > 2.

(ii) x^2 < 1/x

x^2 < 1/x gives us x^3 – 1 < 0 i.e. x < 1

Can x be less than 1 and greater than 2 simultaneously? No. Therefore, 2x < x^2 < 1/x cannot be the correct ordering.

Answer (D)

Is this method simpler?

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!

Plug Using Transition Points

Quarter Wit, Quarter WisdomLet’s take a break from Geometry today and discuss the concept of transition points. This is especially useful in questions where you are tempted to plug in values. A question often asked is: how do I know which values to plug and how do I know that I have covered the entire range in the 3-4 values I have tried? What transition points do is that they give you the ranges in which the relationships differ. All you have to do is try one value from each range. If you do, you would have figured out all the different relationships that can hold. We will discuss this concept using a GMAT Prep question. You can solve it using our discussion on inequalities too. But if number plugging is what comes first to your mind in this question, then it will be a good idea to get the transition points.

Let’s begin:

If x is positive, which of the following could be correct ordering of 1/x, 2x and x^2?

(I) x^2 < 2x < 1/x

(II) x^2 < 1/x < 2x

(III) 2x < x^2 < 1/x

(A) none
(B) I only
(C) III only
(D) I and II
(E) I, II and III

Solution:

Notice that the question says “could be correct ordering”. This means that for different values of x, different orderings could hold. We need to find the one (or two or three) which will not hold in any case. So what do we do? We cannot try every value that x can take so how do we know for sure that one or more of these relations cannot hold? What if we try 4-5 values and only one relation holds for all of those values? Can we say for sure that the other two relations will not hold for any value of x? No, we cannot since we haven’t tried all values of x. So there are two options you have in this case:

1. Use logic to figure out which relations can hold and which cannot. This you can do using inequalities (but we will not discuss that today).

2. You can figure out the ranges in which the relationships are different and then try one value from each range. This is our transition points concept which we will discuss today.

Let’s discuss the second option in more detail.

First of all, we are just dealing with positives so there is less to worry about. That’s good.

To picture the relationship between two functions, we first need to figure out the points where they are equal.

x^2 = 2x

x^2 – 2x = 0

x = 0 or 2

x cannot be 0 since x must be positive so this equation holds when x = 2

So x =2 is the transition point of their relation. x^2 is less than 2x when x is less than 2 and it will be greater than 2x when x is greater than 2.

Let’s try to figure out the relation between 1/x and 2x now.

1/x = 2x

x = 1/sqrt(2)

Since 1/x is less than 2x when x is greater than 1/?2, it will be more than 2x when x is less than 1/sqrt(2).

Move on to the relation between 1/x and x^2.

1/x = x^2

x^3 = 1 (notice that since x must be positive, we can easily multiply/divide by x without any complications)

x = 1

So you have got three transition points: 1/sqrt(2), 1 and 2.

Now all you need to do is try a number from each of these ranges:

(i) x < 1/sqrt(2)

(ii) 1/sqrt(2) < x < 1

(iii) 1 < x < 2

(iv) x > 2

If a relation doesn’t hold in any of these ranges, it will not hold for any value of x.

(i) For x < 1/sqrt(2), put x = a little more than 0 (e.g. 0.01)

1/x = 100, 2x = 0.02, x^2 = 0.0001

We get x^2 < 2x < 1/x is possible. So (I) is possible

(ii) For 1/sqrt(2) < x < 1, put x = a little less than 1 (e.g. 0.99)

1/x = slightly more than 1, 2x = slightly less than 2, x^2 = slightly less than 1

We get x^2 < 1/x < 2x is possible. So (II) is also possible.

(iii) For 1 < x < 2, put x = 3/2

1/x = 2/3, 2x = 3, x^2 = 9/4 = 2.25

We get 1/x < x^2 < 2x is possible.

(iv) For x > 2, put x = 3

1/x = 1/3, 2x = 6, x^2 = 9

We get 1/x < 2x < x^2 is possible.

We see that for no positive value of x is the third relation possible. We have covered all different ranges of values of x.

Answer (D)

Try using inequalities instead of number plugging to see if solving the question becomes easier in that case.

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!

Diagrams of Geometry – Part II

Quarter Wit, Quarter WisdomLast week, we discussed how drawing extreme diagrams can help solve Geometry questions. Today we will see how to solve another Geometry question by making diagrams. The diagram can help you understand exactly what it is that you need to do; doing it will be quite straightforward.

Question: If 10, 12 and ‘x’ are sides of an acute angled triangle, how many integer values of ‘x’ are possible?

 

(A) 7
(B) 12
(C) 9
(D) 13
(E) 11

Solution: The question is very interesting. It asks you for an acute triangle i.e. a triangle with all angles less than 90 degrees. It’s a little hard to wrap your head around it, isn’t it? We know that the third side of a triangle can take many values. Right from a little more than the difference of the other two sides to a little less than the sum of the other two sides (Since we know that the sum of any two sides of a triangle is always greater than the third side). So x can be anything from a little more than 2 to a little less than 22. But how do we find out the values for which all the angles will be less than 90?

We want no obtuse or right angles. An obtuse angled triangle has one angle more than 90. So the thought here is that before one of the angles reaches 90, find out all the values that x can take.

Look at the figure given above. The value of x in the first figure is very small – slightly more than 2 – minimum required to make a triangle. There is an obtuse angle in that triangle. We keep making x bigger and bigger and the angle keeps becoming smaller till it reaches 90 (Fig III). We use Pythagorean theorem to get the value of x in that case:

x = ?(12^2 – 10^2)
x = ?44 which is 6.something
x should be greater than 6.something because the angle cannot be 90.

We further keep increasing x and all the angles are acute now. We reach Fig V where we hit another right triangle. We use Pythagorean theorem again to get the value of x (the hypotenuse) in this case:

x = ?(12^2 + 10^2)
x = ?244 which is 15.something
x should be less than 15.something so that the angle is not 90.

Further on, in Fig VI, we obtain an obtuse angle again.

We only need integral values of x so values that x can take range from 7 to 15 which is 9 values.

Answer (C).

Note: We made two angles 90 and found the values of x in between those two angles. The third angle cannot be 90 because that will make 10 the hypotenuse but hypotenuse is always the greatest side.

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!

Diagrams of Geometry – Part I

Quarter Wit, Quarter WisdomLet’s continue with geometry today. We would like to discuss how drawing extreme diagrams can help you solve questions. Most GMAT questions are quite intuitive and hence our non-traditional methods are perfect for them. They are not typical MATH problems per se; instead, they are logical puzzles. If you can prove why some things will not work, it means whatever is left will work.

Let me explain with the help of an official Data Sufficiency question.

 

Question:

In the figure above, is the area of the triangle ABC equal to the area of the triangle ADB?

Statement 1: (AC)^2=2(AD)^2

Statement 2: ?ABC is isosceles.

Solution:

When presented with this question, people see right triangles and jump to Pythagorean theorem, isosceles triangles and then wage a war on AC, AB, CB and AD relations. Well, that is our traditional approach. But what do we do if making equations and solving for relations isn’t our style?

We make diagrams and figure out the relations! One thing that is apparent the moment we read statement 1 is that the figure is not to scale. From the figure it looks as if AD is greater than or at best, equal to AC. That itself is an indication that if you draw the figure on your own, you could see something that will make this question very simple. The question setter doesn’t want to show you that and hence he made the distorted figure.

Anyway, let’s first analyze the question. Then we will look at the statements.

We need to compare areas of ABC and ADB. Both are right angled triangles.

Area of ABC = (1/2) * AC * BC

Area of ADB = (1/2) * AD * AB

We need to figure out whether these two are the same.

Think about it this way – we are given a triangle ABC with a particular area. So the length of AD must be defined. If AD is very small, (shown by the dotted lines in the diagram given below) the area of ADB will be very close to 0. If AD is very large, the area will be much larger than the area of ABC. So for only one value of AD, the area of DAB will be equal to the area of ABC.

We need to figure out whether for the given relations, the triangles have equal area.

Statement 1: (AC)^2=2(AD)^2
This gives us AD = AC/?2. Let’s draw AC and AD such that AD is somewhat shorter than AC. Now can we say that the areas of the two triangles are the same? No. The area of ABC is decided by AC and BC both not just AC. We can vary the length of BC to see that the relation between AC and AD is not enough to say whether the areas will be the same (see the diagrams given below).

So this statement alone is not sufficient.

(2) ?ABC is isosceles.
This means that AB = BC. Notice that the triangle is right angled so the hypotenuse must be the largest side. If ABC is isosceles, it means that the two legs of the triangle must be equal. Hence sides of ABC must be in the ratio 1:1:?2 = AC:BC:AB. Since we only need to consider relative length of the sides, let’s say that AC = 1, BC = 1 and AB = ?2 or some multiple thereof.

We have no idea about the length of AD so this statement alone is also not sufficient.

Let’s consider both statements together now:

AD = AC/?2 = 1/?2 (Since AC = 1)

Area of ABC = (1/2) * AC * BC = (1/2) * 1 * 1 = 1/2

Area of ADB = (1/2) * AD * AB = (1/2) *  (1/?2 ) * ?2 = 1/2

Both triangles have the same area. Sufficient!

Answer (C)

Now compare this approach with your Pythagorean approach. Is this simpler?

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!

Graphs of Geometry – Part III

Quarter Wit, Quarter WisdomThis week, we will further build up on what we have discussed in the past two weeks. You will need to sum up everything we discussed last week in a few seconds and arrive at a conclusion and then, move on and solve the question on the basis of that conclusion. We will take you through the ‘summing up’ and ‘getting a feel for it’ process step by step so that it’s intuitive to you next time you come across this concept.

Question: A certain square is to be drawn on a coordinate plane such that all the coordinates of its vertices are integers. One of the vertices must be at the origin, and the area of the square must be 25. How many different squares can be drawn?

(A) 4
(B) 6
(C) 8
(D) 10
(E) 12

Solution:

Since the question tells us that there are many different ways to draw the square, let’s draw it in one particular way. We will have something with which to proceed.

One vertex must be at (0, 0). Area must be 25 which means the side of the square must be 5. The easiest way to draw such a square would be the blue square shown in the figure.

How will we get many different squares? By turning the square around (0, 0) as shown (Push one side of the square – notice the vertex with the green dot at (0, 5) moving clockwise). Now the problem is that we need the coordinates of all the vertices to be integers. How do we assure that? One vertex is at (0, 0) and will always stay at (0, 0) so we don’t need to worry about that. What about the rest of the three vertices? When you turn the square, at any one position, they may or may not have integer coordinates – and we have three of them to worry about!

This is where our last two posts come in. First let’s establish that we only need to focus on one vertex – the other two will follow. Look at the blue square. Each one of its vertices is at integer coordinates: (0, 0), (0, 5) – the green dot vertex, (5, 5) and (5, 0). Let’s say, when you push the side joining the center and the green dot, the green dot moves by 1 unit down and 3 units to the right. What happens to the side perpendicular to this side (the one joining the green and the yellow dots)? Let’s think about it.

Pay attention to the diagram. When we turn the square, the coordinates of all three vertices change. The change is similar in all three vertices.

Hope it’s intuitive that if one vertex takes integral steps, all vertices will move in a similar fashion. The diagram shown above is just to reinforce this point. So if our first square has all integral coordinates and we move one vertex such that its coordinates remain integers, all the other vertices will follow suit.

Let’s catch hold of the green dot vertex with coordinates (0, 5). We want the x and y coordinates to stay integers such that

x^2 + y^2 = 5^2 (since the length of the side must stay 5)

As discussed above, x = 0 and y = 5 satisfies this. When x = 1, will y be integral? No.

What about when x = 2? No.

When x = 3? Yes, y will be 4.

When x = 4? Yes, y will be 3.

When x = 5? Yes, y will be 0.

So you have two pairs of numbers 0, 5 and 3, 4 (and their variations) which satisfy this equation. (Hope it reminded you of Pythagorean theorem and you jumped to this conclusion right away!)

Notice that coordinates can be negative too so 0, -5 and -3, -4 will also work. So how many total squares do we get?

So the green dot can take any of the following coordinates:

(0, 5), (0, -5)

(5, 0), (-5, 0)

(3, 4), (3, -4)

(4, 3), (4, -3)

(-3, 4), (-3, -4)

(-4, 3), (-4, -3)

A total of 12 values.

Alternatively, you have 3 different squares keeping the green dot in the first quadrant and when rotated, they will give you one square in each of the three other quadrants too.

 

Total 12 such squares.

The correct answer is (E).

If it seems difficult, that is because it is – 750 level.

But keep practicing. Most of it will become intuitive once you get the hang of it.

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!

Graphs of Geometry – Part II

Quarter Wit, Quarter WisdomLet’s pick up from where we left last week. We had discussed a coordinate geometry concept using clock faces and had left you with a tough question. Today we will see how you can solve that question using the concepts discussed last week.

You might wonder whether we can expect such a question in actual GMAT. The question we discussed in the last post was an official question and we could solve it easily using this concept. Of course there are many other ways of solving it but this is simplest (or trickiest depending on how you look at it), and it certainly is the fastest, no two ways about it! It is a very logical big-picture approach and people who get Q50-51 often use such methods. The question we will discuss today can also be solved in other ways but we will use the last week’s ‘turning minute hand 90 degrees’ approach.

Question: On the coordinate plane (6, 2) and (0, 6) are the endpoints of the diagonal of a square. What are the coordinates of the point on the corner of the square which is closest to the origin?

(A) (0, 1)
(B) (1, 0)
(C) (1, 1)
(D) (2, 0)
(E) (2, 2)

Solution:

First rule of coordinate geometry – draw what you can.

So we make the xy axis and plot the given points, (6, 2) and (0, 6) on it. Let’s say the square is denoted by points ABCD. Say, A is (6, 2) and C is (0, 6). We see that AC is a sloping line. Its two end points are two vertices of the square. We need to find the other two vertices of the square. One of them will lie closest to the origin. The other two vertices will be the end points of the diagonal BD. BD will be perpendicular to AC at the mid-point of AC since a square’s diagonals bisect each other and are perpendicular. So the question is, how do we obtain the end points, B and D? Let’s try to figure out what information we need to draw BD. BD must pass through the mid-point of AC.

How will we obtain the mid-point of AC? By averaging x and y co-ordinates of the points A and C:

x coordinate of mid-point is (0 + 6)/2 = 3

y coordinate of mid-point is (6 + 2)/2 = 4

So BD must pass through (3, 4). When AC turns by 90 degrees, with point (3, 4) as the axis, we get the diagonal BD. So how do the coordinates of AC change when it turns by 90 degrees? Go back to last week’s post and look at the clock face again.

Think of a horizontal line PQ passing through (3, 4).

P coordinate will be given by (0, 4) and Q coordinate will be given by (6, 4) since length of P to mid point is 3 and length of mid-point to Q will also be 3. P shifts up by 2 units to give the point A and Q shifts down by 2 units to give the point C.

Now rotate PQ by 90 degrees and you get RS. We know the coordinates of a line perpendicular to PQ. R will be (3, 1) and S will be (3, 7). This is because R and S will have the same x coordinates as the mid-point (3, 4) and S is 3 units above the mid-point and R is 3 units below the mid-point. We are assuming that you can intuitively see these values on the graph. If not, it may be too soon to spend time on this post.

Now, can we obtain the diagonal BD using RS as reference? If you move S two units to the right, you will get point B (just like A was obtained by moving P two units up) and if you move R two units to the left, you will get point D. Notice that we are using PQ and RS as reference lines. It is easy to calculate vertical/horizontal distances. So B will be given by (5, 7) and D will be given by (1, 1).

The closest co-ordinate to (0, 0) is (1, 1).

Answer (C).

Take a few minutes to review the logic discussed here. The ability to ‘see’ such symmetry makes GMAT Quant very simple.

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!

 

Graphs of Geometry – Part I

Let’s start with geometry today. It has some very interesting and intuitive concepts. We will discuss one of them today. It’s surprising how a little bit of imagination can go a long way in helping you solve questions. Let’s discuss the concept first. We will look at a question later.

Imagine a clock face. Think of the minute hand on 10. Ignore the hour hand for our discussion today. Say, the length of the minute hand is 2 cm. Its distance from the vertical and horizontal axis is shown in the diagram below (using the green and the red dotted lines). Let’s say the minute hand moves to 1. Can you say something about the lengths of the dotted black and dotted blue lines?

Isn’t it apparent that when the minute hand moved by 90 degrees, the dotted green line became the dotted black line and the dotted red line became the dotted blue line. So can we say that the dotted black line is ?3 cm in length and the dotted blue line is 1 cm in length? The same thing will happen when the minute hand goes to 4 and to 7. We don’t think there is much explanation needed here, right? The diagram makes it all clear.

Let’s look at the clock from coordinate geometry perspective. Let’s say the center of the clock is the origin (0, 0). What are the x and y coordinates of the “10’o clock point” i.e. the tip of the minute hand before it moves to 1? Notice that the x coordinate will be -?3 (since the point is in the second quadrant, x coordinate will be negative) and y coordinate will be 1.

What are the x and y coordinates of the “1’o clock point” i.e. the tip of the minute hand after it moves to 1. Notice that the absolute values of x coordinate and y coordinate have switched because the hand has turned 90 degrees. The x coordinate is 1 now and the y coordinate is ?3. Since it is the first quadrant, both the coordinates will be positive.

Now, think, what will be the coordinates of the “4’o clock point”, “7’o clock point”? What about the “11’o clock point”, “12’o clock point” etc?

Using this concept, we can solve a very tough GMAT Prep question in a few seconds.

Question 1:

In the figure above, points P and Q lie on the circle with center O. What is the value of s?

(A) 1/2

(B) 1

(C) ?2

(D) ?3

(E) 1/?2

Solution: You might be tempted to think on the lines of ‘slope of a line’ or ’30-60-90’ triangle (because of the presence of ?3) etc. But we should be able to arrive at the answer without using any of those.

Point P is (-?3, 1). O is the center of the circle at (0, 0). When OP is turned 90 degrees to give OQ, the x and y co-ordinates get interchanged. Also both x and y co-ordinates will be positive in the first quadrant. Hence s, the x co-ordinate of Q will be 1 (and y co-ordinate of Q will be ?3).

Answer (B)

The question doesn’t seem difficult now (after understanding the concept); actually, it is a 700+ level.

Try another question using the same concept:

Question 2: On the coordinate plane (6, 2) and (0, 6) are the endpoints of the diagonal of a square. What are the coordinates of the point on the corner of the square which is closest to the origin?

(A) (0, 1)

(B) (1, 0)

(C) (1, 1)

(D) (2, 0)

(E) (2, 2)

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: Stuck in Assumptions Again

Quarter Wit, Quarter WisdomThere is a particular issue in assumption questions that I would like to discuss today. We discussed in our previous posts that assumptions are ‘necessary missing premises’. Many students get stuck between two options in assumption questions. The correct option is the necessary premise. The incorrect one is often a sufficient premise. Due to the sufficiency, they believe that that particular option is a stronger assumption. But the point to remember is that an assumption is only necessary for the conclusion to be true. It may not actually lead to the conclusion beyond a reasonable doubt. You only have to answer what has been asked (which is an assumption), not what you think is better to make the conclusion true.

Let me explain this with an example:

Question: Exports of United States cotton will rise considerably during this year. The reason for the rise is that the falling value of the dollar will make it cheaper for cloth manufacturers in Japan and Western Europe to buy American cotton than to get it from any other source.

Which of the following is an assumption made in drawing the conclusion above?

(A) Factory output of cloth products in Japan and Western Europe will increase sharply during this year.
(B) The quality of the cotton produced in the United States would be adequate for the purposes of Japanese and Western European cloth manufacturers.
(C) Cloth manufacturers in Japan and Western Europe would prefer to use cotton produced in the United States if cost were not a factor.
(D) Demand for cloth products made in Japan and Western Europe will not increase sharply during this year.
(E) Production of cotton by United States companies will not increase sharply during this year.

Solution:

First, let’s analyze the given argument:

Premises:
– Dollar is falling.
– It will be cheaper for cloth manufacturers in Japan and Western Europe to buy American cotton than to get it from any other source.

Conclusion:
– Exports of United States cotton will rise considerably during this year.

The conclusion links ‘sale of cotton’ to ‘cost of cotton’. It says that since the cost of American cotton will be lower than the cost of cotton from any other source, American cotton will sell. We are assuming here that the American cotton is adequate in all other qualities that the cloth manufacturers look for while buying cotton or that lower cost is all that matters. We are assuming that lowest cost will automatically lead to sale.

Let’s look at each of the options now:

(A)   Factory output of cloth products in Japan and Western Europe will increase sharply during this year.

Notice that we don’t NEED the factory output to increase. Even if it stays the same or in fact, even if it falls, as long as the manufacturers find American cotton suitable, the cotton exports of US could rise. Hence this is not the assumption.

(B)   The quality of the cotton produced in the United States would be adequate for the purposes of Japanese and Western European cloth manufacturers.

This option says that the quality is adequate and hence this is an assumption. Notice that it is necessary for our conclusion. If the quality is not adequate, no matter what the cost, US cotton sale may not increase. Hence, answer is (B).

(C)   Cloth manufacturers in Japan and Western Europe would prefer to use cotton produced in the United States if cost were not a factor.

This is the tricky non-correct option! Many people will swing between (B) and (C) for a while and then choose (C).  This option says that Japanese and Europeans prefer to use US cotton if cost does not matter. Do we NEED this to be true? No. It is good if it is true because it means that if cost of US cotton goes down, US cotton will sell more (hence, it is sufficient for the conclusion to be true – assuming all else stays constant). But do we NEED them to prefer US cotton? No. It is not necessary for our conclusion to be true. Even if the manufacturers don’t particularly prefer US cotton, US cotton exports could still increase if the price is the lowest.

Beware of this difference between ‘necessary’ and ‘sufficient’ conditions. Remember that assumptions are NECESSARY conditions; they don’t need to be sufficient. We end up incorrectly choosing sufficient conditions because they seem to be all encompassing and hence more attractive for our conclusion. If the sufficient condition is satisfied, then the conclusion has to be true. But mind you, that is not what the question is asking you. The question is looking for only a necessary condition, not a sufficient one. Also notice that sufficient conditions may not be necessary.

(D)   Demand for cloth made in Japan and Western Europe will not increase sharply during this year.

This is incorrect. We are not assuming that the demand for their cloth will not increase.

(E)    Production of cotton by United States companies will not increase sharply during this year.

It doesn’t matter what happens to the production of cotton in US. All we care about is that exports should rise.

The correct answer is (B). 

Hope you will be careful next time when you come across such a question.

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: And Now, Evading Formulas!

Quarter Wit, Quarter WisdomToday, we again pay homage to the lazy bum within each one of us in our QWQW series. If you are wondering what we mean by ‘again’, check out our last two posts of the QWQW series. We have been discussing how to avoid calculations. Today let’s learn why it is advisable to avoid learning formulas too!

You really don’t need to know many formulas for GMAT – just the basic ones e.g. Distance = Speed*Time, Work = Rate*Time (which are actually the same if you look at them closely) etc. If a Time-Distance-Speed question pertains to GMAT, rest assured it can be solved using just the formula given above and that too, within 1-2 mins. Then, do you need to learn the many formulas that people claim speed up question solving? No! In fact, the more specific the formula, the more constraints it has. It can be used in only particular circumstances and hence when the situation differs even a little bit from the ideal, you could end up using the formula incorrectly. Therefore, we recommend our students to stay away from the umpteen, less generic formulas until and unless they have already used them extensively. Let’s discuss this point with an example:

Question: A and B start from Opladen and Cologne respectively at the same time and travel towards each other at constant speeds along the same route. After meeting at a point between Opladen and Cologne, A and B proceed to their destinations of Cologne and Opladen respectively. A reaches Cologne 40 minutes after the two meet and B reaches Opladen 90 minutes after their meeting. How long did A take to cover the distance between Opladen and Cologne?

(A) 1 hour
(B) 1 hour 10 minutes
(C) 2 hours 30 minutes
(D) 1 hour 40 minutes
(E) 2 hours 10 minutes

Solution: People often like to use a formula for this situation. Let’s quickly discuss that first.

If two objects A and B start simultaneously from opposite points and, after meeting, reach their destinations in ‘a’ and ‘b’ hours respectively (i.e. A takes ‘a hrs’ to travel from the meeting point to his destination and B takes ‘b hrs’ to travel from the meeting point to his destination), then the ratio of their speeds is given by:

Sa/Sb = sqrt(b/a)

i.e. Ratio of speeds is given by the square root of the inverse ratio of time taken.

Sa/Sb = sqrt(90/40) = 3/2

This gives us that the ratio of the speed of A : speed of B as 3:2. We know that time taken is inversely proportional to speed. If ratio of speed of A and B is 3:2, the time taken to travel the same distance will be in the ratio 2:3. Therefore, since B takes 90 mins to travel from the meeting point to Opladen, A must have taken 60 (= 90*2/3) mins to travel from Opladen to the meeting point

So time taken by A to travel from Opladen to Cologne must be 60 + 40 mins = 1 hr 40 mins

Now let’s see how we can solve the question without using the formula.

Think of the point in time when they meet:

A starts from Opladen and B from Cologne simultaneously. After some time, say t mins of travel, they meet. Since A covers the entire distance of Opladen to Cologne in (t + 40) mins and B covers it in (t + 90) mins, A is certainly faster than B and hence the Meeting point is closer to Cologne.

Now think, what information do we have? We know the time taken by A and B to reach their respective destinations from the meeting point. We also know that they both traveled the same distance i.e. the distance between Opladen and Cologne. So let’s try to link distance with time taken. We know that ‘Distance’ varies directly with ‘Time taken’. (Check out this post if you don’t know what we are talking about here.)

Distance between Opladen and Meeting point /Distance between Meeting point and Cologne = Time taken to go from Opladen to Meeting point/Time taken to go from Meeting point to Cologne = t/40 (in case of A) = 90/t (in case of B)
t = 60 mins

So A takes 60 mins + 40 mins = 1 hr 40 mins to cover the entire distance.

Answer (D)

We could easily solve the question without using any specific formula. So stick to your basics and kick those little grey cells to get to the answer!

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: Evading Calculations Part II

Quarter Wit, Quarter WisdomLast week we discussed how to solve equations with the variable in the denominator. We also said that the technique generally works for PS questions but you need to be careful while working on DS questions. Today, let’s look at the reason behind the caveat.

Say, the question stem of a DS question asks you to find the value of n, the number of people in the room. Statement 1 of the question gives you the following equation:

60/(n – 5) – 60/n = 2

We can easily figure out that a value of n that satisfies this equation is 15. Now, is that enough to say that statement 1 is sufficient alone? No! It could be a trap! The equation, when manipulated, gives us a quadratic. It is important to find out whether the second solution of the quadratic works for us. When n is the number of people, it must be positive. So one extra step that we should take is re-arrange the equation to get the quadratic. If the constant term i.e. the product of the roots is negative, it means one root is positive and one is negative. Since we have already found the positive root, it is the only answer and hence we can say that the statement 1 is sufficient alone.

60/(n – 5) – 60/n = 2

60*n – 60*(n – 5) = 2*n*(n – 5)

n^2 – 5n – 150 = 0

The constant term, -150, is negative so the product of the roots must be negative. This means one root must be negative and the other must be positive. Since we have already found the positive root i.e. the number of people in the room, we can say that statement 1 is sufficient alone.

Let’s look at an example where we could fall in the trap.

Say statement 1 gives us an equation which looks like this:

60/(n +5) – 10/(n – 5) = 2

As discussed last week, we will easily see that n = 10 satisfies this equation. So should we move on now and say that statement 1 is sufficient alone? No, not so fast! Let’s try to manipulate the equation to get the quadratic.

60/(n +5) – 10/(n – 5) = 2

60*(n – 5) – 10*(n + 5) = 2*(n – 5)(n + 5)

n^2 – 25n + 150 = 0

n = 10 or 15

So actually, there are two values of n that satisfy this equation. In PS questions, since we have a single answer, there would be only one solution so once you get one, you are done. In DS questions, you need to be certain that only one value satisfies. There is a possibility that both values satisfy your constraints in which case your answer would change.

Therefore, it may not be necessary to solve the equation for the PS question, but it is certainly necessary to solve it for DS. That’s counter intuitive, isn’t it? We hope you understand the reason.

Another related trap in DS questions: Statement 1 gives you a quadratic and asks you for the value of x (no constraints that x must be an integer or positive number etc). You know that it is a quadratic and it will give you two values of x so you say that statement 1 is not sufficient alone and move on. But hold it! What if both the roots of the equation are same? It may not apparent to you when you look at the equation. When you solve it, you realize that the roots are the same. Hence, ensure that you solve the equation in DS questions before you decide on the sufficiency.

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 in Detroit, Michigan, and regularly participates in content development projects such as this blog!

Quarter Wit, Quarter Wisdom: Evading Calculations!

Quarter Wit, Quarter WisdomWe have discussed before how GMAT is not a calculation intensive exam. Whenever you land on an equation which looks something like this: 60/(n – 5) – 60/n = 2, you probably think that we don’t know what we are talking about! You obviously need to cross multiply, make a quadratic and finally, solve the quadratic to get the value of n. Actually, you usually don’t need to do any of that for GMAT questions. You have an important leverage – the options. Even if the options don’t directly give you the values of n or n-5, you can use the knowledge that every GMAT question is do-able in 2 mins and that the numbers fit in beautifully well.

Let’ see whether we can get a value of n which satisfies this equation without going the whole nine yards. We will not use any options and will try to rely on our knowledge that GMAT questions don’t take much time.

60/(n – 5) – 60/n = 2

So, the difference between the two terms of the left hand side is 2. Try to look for values of n which give us simple numbers i.e. try to plug in values which are factors of the numerator.
Say, if n = 10, you get 60/5 – 60/10 = 12 – 6 = 6. The difference between them is much more than 2. 60/n and 60/(n – 5) need to be much closer to each other so that the difference between them is 2. The two terms should be smaller to bring them closer together. So increase the value of n.

Put n = 15 since it is the next number such that (15 – 5 =) 10 as well as 15 divide 60 completely. You get 60/10 – 60/15 = 6 – 4 = 2. It satisfies and you know that a value that n can take is 15. Usually, you will get a solution within 2-3 iterations. This is enough for a PS question. Notice that this equation gives us a quadratic so be careful while working on DS questions. You might need to manipulate the equation a little to figure out whether the other root is a possible solution as well. Anyway, today we will focus on the application of such equations in PS questions only. Let’s take a question now to understand the concept properly:

Question: Machine A takes 2 more hours than machine B to make 20 widgets. If working together, the machines can make 25 widgets in 3 hours, how long will it take machine A to make 40 widgets?

(A) 5
(B) 6
(C) 8
(D) 10
(E) 12

Solution: We need to find the time taken by machine A to make 40 widgets. It will be best to take the time taken by machine A to make 40 widgets as the variable x. Then, when we get the value of x, we will not need to perform any other calculations on it and hence the scope of making an error will reduce. Also, value of x will be one of the options and hence plugging in to check will be easy.

Machine A takes x hrs to make 40 widgets.

Rate of work done by machine A = Work done/Time taken = 40/x

Machine B take 2 hrs less than machine A to make 20 widgets hence it will take 4 hrs less than machine B to make 40 widgets. Think of it this way: Break down the 40 widgets job into two 20 widget jobs. For each job, machine B will take 2 hrs less than machine A so it will take 4 hrs less than machine A for both the jobs together.

Time taken by machine B to make 40 widgets = x – 4

Rate of work done by machine B = Work done/Time taken = 40/(x – 4).

We know the combined rate of the machines is 25/3

So here is the equation:

40/x + 40/(x – 4) = 25/3

The steps till here are not complicated. Getting the value of x poses a bit of a problem.

Notice here that that the right hand side is not an integer. This will make the question a little harder for us, right? Wrong! Everything has its pros and cons. The 3 of the denominator gives us ideas for the values of x (as do the options).  To get a 3 in the denominator, we need a 3 in the denominator on the left hand side too.

x cannot be 3 but it can be 6. If x = 6, 40/(6 – 4) = 20 i.e. the sum will certainly not be 20 or more since we have 25/3 = 8.33 on the right hand side.

The only other option that makes sense is x = 12 since it has 3 in it.

40/12 + 40/(12 – 4) = 10/3 + 5 = 25/3

Answer (E)

If we did not have the options, we might have tried x = 9 too before landing on x = 12. Nevertheless, these calculations are not time consuming at all since you can get rid of the incorrect numbers orally. Making a quadratic and solving it is certainly much more time consuming.

Another method could be to bring 3 to the left hand side to get the following equation:

120/x + 120/(x – 4) = 25

This step doesn’t change anything but it helps if you face a mental block while working with fractions. Try to practice such questions using these techniques – they will save you a lot of time.

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: An Official Assumption Question

Quarter Wit, Quarter WisdomToday we will look at an OG question of critical reasoning (as promised last week). We will use the concept discussed last week – remember what an assumption is. An assumption is a missing necessary premise. It will bring in new information essential to the conclusion.

Now let’s jump on to the OG question.

Question: A recent report determined that although only three percent of drivers on Maryland highways equipped their vehicles with radar detectors, thirty-three percent of all vehicles ticketed for exceeding the speed limit were equipped with them. Clearly, drivers who equip their vehicles with radar detectors are more likely to exceed the speed limit regularly than are drivers who do not.

The conclusion drawn above depends on which of the following assumptions?

(A) Drivers who equip their vehicles with radar detectors are less likely to be ticketed for exceeding the speed limit than are drivers who do not.

(B) Drivers who are ticketed for exceeding the speed limit are more likely to exceed the speed limit regularly than are drivers who are not ticketed.

(C) The number of vehicles that were ticketed for exceeding the speed limit was greater than the number of vehicles that were equipped with radar detectors.

(D) 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.

(E) Drivers on Maryland highways exceeded the speed limit more often than did drivers on other state highways not covered in the report.

Solution:

Let’s look at the question stem first. We need to find an assumption. An assumption is a missing necessary premise. Something that will not only strengthen the conclusion but also be essential to the argument.
An assumption is a statement that needs to be added to the premises for the conclusion to be true. Let’s first find the premises and the conclusion of this argument.

Premises:

– Only 3% of drivers on Maryland highways had radar detectors.

– 33% of vehicles that got speeding tickets had radar detectors.

Conclusion:

Drivers with radar detectors are more likely to exceed the speed limit regularly than other drivers.

There must be some disconnect between the premises and conclusion since there is an assumption in the argument. Look carefully. Premises give you the connection between ‘vehicles that have radar detectors’ and ‘vehicles that get speeding tickets’. The conclusion, on the other hand, concludes a relation between ‘vehicles that have radar detectors’ and ‘vehicles that exceed the speed limit’. The assumption must then give a connection between ‘vehicles that get speeding tickets’ and ‘vehicles that exceed speed limit’.

To clarify it further,

A – vehicles that have radar detectors

B – vehicles that get speeding tickets/vehicles that were ticketed for speeding

C – vehicles that exceed the speed limit

Premises:

– Only 3% of all vehicles are A

– 33% of B are A

Conclusion:

– A are  more likely to be C

The assumption needs to be something that links B to C i.e. that links ‘vehicles that get speeding tickets’

to ‘vehicles that exceed the speed limit’. Option (B) gives us that relation. It says ‘B are more likely to be C’.

Lets add it to premises and see if the conclusion makes more sense now:

– Only 3% of drivers on Maryland highways had radar detectors.

– 33% of vehicles that got speeding tickets had radar detectors.

– Drivers who get speeding tickets are more likely to exceed the speed limit regularly than others.

Conclusion: Drivers with radar detectors are more likely to exceed the speed limit regularly than other drivers.

Now it makes sense!

Let’s take a quick look at the other options and see why they don’t work. We will retain the A, B, C structure given above.

Option (A) says ‘A are less likely to be B’ – Cannot be our assumption

Option (C) only tells us that number of B are greater than number of A.

Option (D) tells us that many vehicles were ticketed multiple times.

Option (E) compares drivers on Maryland highways with drivers on other state highways. This is out of scope.

It is clear that option (B) is the outright winner. This question is one of the tougher questions. You can easily handle it using this technique. We hope you will be able to put this technique to good use.

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: Making Sense of Assumptions

Quarter Wit, Quarter WisdomToday we would like to discuss a technique which is very useful in solving assumption questions. No, I am not talking about the ‘Assumption Negation Technique’ (ANT), which, by the way, is extremely useful, no doubt. The point is that ANT is explained beautifully and in detail in your book so there is no point of re-doing it here. You already know how to use it.

What we are going to discuss today is not so much a technique as a revelation of how you can identify the assumption in a particular argument by just fully understanding this – ‘what is an assumption?’. Actually, this is also already discussed in your CR book but I would like to draw your attention to it. We usually end up ignoring the finer points here and hence get stuck on something that is supposed to be quite obvious.

Ok so, what is an assumption?

An assumption is a missing necessary premise. (Doesn’t seem like much a revelation, right? You already knew that! Right! Focus on every word now.)

An assumption is a premise – it gives you some new fact/information.

It is also necessary – necessary for the conclusion to be true. The conclusion cannot be true if the assumption doesn’t hold. Our ANT is based on this premise.

To add, it is also missing – it is not something already mentioned in the argument.

Let’s take a very simplistic example to understand the implication of a missing necessary premise.

Argument: A implies B. B implies C. Hence, A implies D.

Premises given in the argument:
– A implies B
– B implies C

Conclusion given in the argument:
– A implies D

Is it apparent that something is missing here? Sure! The premises give us the relations between A, B and C. They do not mention D. But while drawing the conclusion, we are concluding about the relation between A and D. We can’t do that. We must know something about D too to be able to conclude a relation between A and D. Hence, there is a necessary premise that is missing here. What we are looking for is something that says ‘C implies D’.

When we add this to our premises, our argument makes sense.

Argument: A implies B. B implies C. C implies D. Hence, A implies D.

This little point will help us in solving the trickiest of questions. We get so lost in the n number of things mentioned in the argument that we forget to consider this aspect.

We will discuss an LSAT question today because it seems to be created just to exemplify this concept! Many people falter on this question. After going through it with us here, you will wonder why.

Question:

Therapist: The ability to trust other people is essential to happiness, for without trust there can be no meaningful emotional connection to another human being, and without meaningful emotional connections to others we feel isolated.
Which one of the following, if assumed, allows the conclusion of the therapist’s argument to be properly inferred?
(A) No one who is feeling isolated can feel happy.
(B) Anyone who has a meaningful emotional connection to another human being can be happy.
(C) To avoid feeling isolated, it is essential to trust other people.
(D) At least some people who do not feel isolated are happy.
(E) Anyone who is able to trust other people has a meaningful emotional connection to at least one other human being.

Solution:

First, we break down the argument into premises and conclusion.

Premises:

– Without trust there can be no meaningful emotional connection.

– Without meaningful emotional connections, we feel isolated.

Conclusion:

Ability to trust is essential to happiness.

Do you see something missing here? We are concluding about trust and happiness but in the premises, the link between ‘feeling isolated’ and ‘happiness’ is missing. The premises do not talk about happiness at all. So we need a premise which says, ‘feeling isolated’ means ‘not happy’ for the conclusion to make sense.
Look at the premises now:

Premises:

– Without trust there can be no meaningful emotional connection.

– Without meaningful emotional connections, we feel isolated.

– When we feel isolated, we cannot be happy. (The assumption)

Conclusion:

Ability to trust is essential to happiness.

Now it all makes sense, doesn’t it?

Look at the options now.

Option (A) says – ‘No one who is feeling isolated can feel happy.’ – exactly what we needed.

Hence (A) will be your answer.

Hope this makes sense to you. Next week, we will see how you can easily solve OG questions using this concept.

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: The Efficiency of Using Variation

Quarter Wit, Quarter WisdomToday, we would like to discuss one of our own work questions. The intent is to show you how simple your calculations can get when you use the methods we discussed in the last few weeks. I couldn’t say it enough – develop a love for ratios. You will save a huge amount of time in lots of questions. If you haven’t been following the last few weeks’ posts, take a look at this link before checking out the question. Otherwise the method may not make sense to you.

Question: 16 horses can haul a load of lumber in 24 minutes. 12 horses started hauling a load and after 14 minutes, 12 mules joined the horses. Will it take less than a quarter-hour for all of them together to finish hauling the load?

(1) Mules work more slowly than horses.
(2) 48 mules can haul the same load of lumber in 16 minutes.

Solution: First do this question on your own and see the calculations involved. Thereafter, check out the solution given below to know how we can solve the question using our joint variation method.

We are given that 16 horses can complete the work in 24 mins. Let’s find out how much work is done by 12 horses in 14 mins (before the mules join in)

16 horses ……… 24 mins ………. 1 work
12 horses ……… 14 mins ………. ?? work

Work done = 1*(14/24)*(12/16) = 7/16 work (if you don’t know how we arrived at this, seriously, check out last week’s post first)

So in 14 mins, the 12 horses can complete 7/16 of the work i.e. they do 1/16 of the work every 2 mins.

How much work is leftover for the mules and horses to do together? 1 – 7/16 = 9/16

Leftover work = 9/16

This makes us think that 12 horses alone will take 9*2 = 18 mins to finish the work. When 12 mules join in, depending on the rate of work of mules, time taken to complete this work could be less than or more than 15 mins.

Statement 1: Mules work more slowly than horses.

This statement doesn’t give us enough information. It just tells us that mules work slower than horses. Say if they work very slowly so that, effectively, they are not adding much to the work done, the work will get done in approximately 18 mins. If they work faster, time taken will keep decreasing. If they work as fast as the horses, the rate at which the work will be done will double (because we already have 12 horses and we will add 12 mules which will be equivalent to 12 horses) and time taken will become half i.e. it will be 9 mins. So the time taken will vary in the range 9 mins to 18 mins depending on the rate of work of mules. This statement alone is not sufficient.

Statement 2: 48 mules can haul the same load of lumber in 16 minutes.

We now know the rate of work of mules. The point is that now we can easily calculate the exact time taken by 12 horses and 12 mules to complete 9/16 of the work. Once we calculate the exact time, we will be able to say whether the time taken will be less than or more than 15 mins. Hence this statement alone is sufficient to answer the question. We don’t really need to find out exactly how much is taken by the 24 animals together since it is a DS question. Ideally, we should mark the answer as (B) and move on.

Nevertheless, let’s do the calculations if only to practice application of work concepts.

Let’s try to find the equivalence of mules and horses (the way we did with cars in the previous post)

We know that 16 horses can haul a load of lumber in 24 minutes. Let’s find out the number of mules that are needed to complete the work in 24 mins.

48 mules …….16 mins.

?? mules ……..24 mins

No. of mules required = 48*16/24 = 32 mules

So, 32 mules do the same work in the same time as done by 16 horses. Or we can say that 2 mules are equivalent to 1 horse. Hence, 12 mules are equivalent to 6 horses. When 12 mules join the 12 horses, equivalently we get 12+6 = 18 horses.

16 horses ……… 24 mins ………. 1 work

18 horses ……….. ?? mins ……….. 9/16 work

Time taken by 18 horses (i.e. 12 horses and 12 mules) = 24*(9/16)*(16/18) = 12 mins

Yes, the horses and mules together will take less than a quarter-hour to finish hauling the load.

Answer (B)

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: Work-Rate Using Joint Variation

Quarter Wit, Quarter WisdomThis week, let’s look at some work-rate questions which use joint variation. Check out the last three posts of QWQW series if you are not comfortable with joint variation.

Question 1: A contractor undertakes to do a job within 100 days and hires 10 people to do it. After 20 days, he realizes that one fourth of the work is done so he fires 2 people. In how many more days will the work get over?

(A) 60

(B) 70

(C) 75

(D) 80

(E) 100

Solution: Can we say that 10 people can finish the work in 100 days? No. If that were the case, after 20 days, only 1/5th of the work would have been over. But actually 1/4th of the work is over. This means that ‘10 people can complete the work in 100 days’ was just the contractor’s estimate (which turned out to be incorrect). Actually 10 people can do 1/4th of the work in 20 days. The contractor fires 2 people. So the question is how many days are needed to complete 3/4th of the work if 8 people are working?

We need to find the number of days. How is ‘no. of days’ related to ‘no. of people’ and ‘work done’?

If we have more ‘no. of days’ available, we need fewer people. So ‘no. of days’ varies inversely with ‘no. of people’.

If we have more ‘no. of days’ available, ‘work done’ will be more too. So ‘no. of days’ varies directly with ‘work done’.

Therefore,

‘no. of days’ * ‘no. of people’/’work done’ = constant

20*10/(1/4) = ‘no. of days’*8/(3/4)

No. of days = 75

So, the work will get done in 75 days if 8 people are working.

We can also do this question using simpler logic. The concept used is joint variation only. Just the thought process is simpler.

10 people can do 1/4th of the work in 20 days.

8 people can do 3/4th of the work in x days.

Start with the no. of days since you want to find the no of days:

x = 20*(10/8)*(3/1) = 75

From where do we get 10/8? No. of people decreases from 10 to 8. If no. of people is lower, the no of days taken to do the work will be more. So 20 (the initial no. of days) is multiplied by 10/8, a number greater than 1, to increase the number of days.

From where do we get (3/1)? Amount of work increases from 1/4 to 3/4. If more work has to be done, no. of days required will be more. So we further multiply by (3/4)/(1/4) i.e. 3/1, a number greater than 1 to further increase the number of days.

This gives us the expression 20*(10/8)*(3/1)

We get that the work will be complete in another 75 days.

Answer (C)

Let’s take another question to ensure we understand the logic.

Question 2: A company’s four cars running 10 hrs a day consume 1200 lts of fuel in 10 days. In the next 6 days, the company will need to run 9 cars for 12 hrs each so it rents 5 more cars which consume 20% less fuel than the company’s four cars. How many lts of fuel will be consumed in the next 6 days?

(A) 1200 lt

(B) 1555 lt

(C) 1664 lt

(D) 1728 lt

(E) 4800 lt

Solution: First let’s try to figure out what is meant by ‘consume 20% less fuel than the company’s cars’. It means that if company’s each car consumes 1 lt per hour, the hired cars consume only 4/5 lt per hour. So renting 5 more cars is equivalent to renting 4 cars which are same as the company’s cars. Hence, the total number of cars that will be run for the next 6 days is 8 company-equivalent cars.

4 cars running 10 hrs for 10 days consume 1200 lt of fuel

8 cars running 12 hrs for 6 days consume x lt of fuel

x = 1200*(8/4)*(12/10)*(6/10) = 1728 lt

We multiply by 8/4 because more cars implies more fuel so we multiply by a number greater than 1.

We multiply by 12/10 because more hours implies more fuel so we multiply by a number greater than 1.

We multiply by 6/10 because fewer days implies less fuel so we multiply by a number smaller than 1.

Answer (D)

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: Varying Jointly

Quarter Wit, Quarter WisdomNow that we have discussed direct and inverse variation, joint variation will be quite intuitive. We use joint variation when a variable varies with (is proportional to) two or more variables.

Say, x varies directly with y and inversely with z. If y doubles and z becomes half, what happens to x?

“x varies directly with y” implies x/y = k (keeping z constant)

If y doubles, x doubles too.

“x varies inversely with z” implies xz = k (now keeping y constant)

If z becomes half, x doubles.

So the overall effect is that x becomes four times of its initial value.

The joint variation expression in this case will be xz/y = k. Notice that when z is constant, x/y = k and when y is constant, xz = k; hence both conditions are being met. Once you get the expression, it’s very simple to solve for any given conditions.

x1*z1/y1 = x2*z2/y2 = k (In any two instances, xz/y must remain the same)

x1*z1/y1 = x2*(1/2)z1/2*y1

x2 = 4*x1

Let’s look at some more examples. How will you write the joint variation expression in the following cases?

1. x varies directly with y and directly with z.

2. x varies directly with y and y varies inversely with z.

3. x varies inversely with y^2 and inversely with z^3.

4. x varies directly with y^2 and y varies directly with z.

5. x varies directly with y^2, y varies inversely with z and z varies directly with p^3.

Solution: Note that the expression has to satisfy all the conditions.

1. x varies directly with y and directly with z.

x/y = k

x/z = k

Joint variation: x/yz = k

2. x varies directly with y and y varies inversely with z.

x/y = k

yz = k

Joint variation: x/yz = k

3. x varies inversely with y^2 and inversely with z^3.

x*y^2 = k

x*z^3 = k

Joint variation: x*y^2*z^3 = k

4. x varies directly with y^2 and y varies directly with z.

x/y^2 = k

y/z = k which implies that y^2/z^2 = k

Joint variation: x*z^2/y^2 = k

5. x varies directly with y^2, y varies inversely with z and z varies directly with p^3.

x/y^2 = k

yz = k which implies y^2*z^2 = k

z/p^3 = k which implies z^2/p^6 = k

Joint variation: (x*p^6)/(y^2*z^2) = k

Let’s take a GMAT prep question now to see these concepts in action:

Question 1: The rate of a certain chemical reaction is directly proportional to the square of the concentration of chemical M present and inversely proportional to the concentration of chemical N present. If the concentration of chemical N is increased by 100 percent, which of the following is closest to the percent change in the concentration of chemical M required to keep the reaction rate unchanged?

(A) 100% decrease
(B) 50% decrease
(C) 40% decrease
(D) 40% increase
(E) 50% increase

Solution:

Rate/M^2 = k

Rate*N = k

Rate*N/M^2 = k

If Rate has to remain constant, N/M^2 must remain the same too.

If N is doubled, M^2 must be doubled too i.e. M must become ?2 times. Since ?2 = 1.4 (approximately),

M must increase by 40%.

Answer (D)

Simple enough?

Quarter Wit, Quarter Wisdom: Varying Inversely

Quarter Wit, Quarter WisdomAs promised, we will discuss inverse variation today. The concept of inverse variation is very simple – two quantities x and y vary inversely if increasing one decreases the other proportionally.

If x takes values x1, x2, x3… and y takes values y1, y2, y3 … correspondingly, then x1*y1 = x2*y2 = x3*y3 = some constant value

This means that if x doubles, y becomes half; if x becomes 1/3, y becomes 3 times etc. In other words, product of x and y stays the same in different instances.

Notice that x1/x2 = y2/y1; The ratio of x is inverse of the ratio of y.

The concept will become clearer after working on a few examples.

Question 1: The price of a diamond varies inversely with the square of the percentage of impurities. The cost of a diamond with 0.02% impurities is $2500. What is the cost of a diamond with 0.05% impurities (keeping everything else constant)?

(A)   $400

(B)   $500

(C)   $1000

(D)   $4000

(E)    $8000

Solution:

Price1*(% of Impurities1)^2 = Price2*(% of Impurities2)^2

2500*(.02)^2 = Price2*(.05)^2

Price = $400

Answer (A)

The answer is quite intuitive in the sense that if % of impurities in the diamond increases, the price of the diamond decreases.

There is an important question type related to inverse variation. It often uses the formula:

Total Price = Number of units*Price per unit

If, due to budgetary constraints, we need to keep the total money spent on a commodity constant, number of units consumed varies inversely with price per unit. If price per unit increases, we need to reduce the consumption proportionally.

Question 2: The cost of fuel increases by 10%. By what % must the consumption of fuel decrease to keep the overall amount spent on the fuel same?

(A)   5%

(B)   9%

(C)   10%

(D)   11%

(E)    20%

Solution: Do you think the answer is 10%? Think again.

Total Cost = Number of units*Price per unit

If the price per unit increases by 10%, it becomes 11/10 of its original value. To keep the total cost same, you need to multiply number of units by 10/11. i.e. you need to decrease the number of units by 1/11 i.e. 9.09%. In that case,

New Total Cost = (10/11)*Number of units*(11/10)*Price per unit

This new total cost will be the same as the previous total cost.

Answer (B)

Let’s look at one more example of the same concept but this one is a little trickier.

Question 3: Recently, fuel price has seen a hike of 20%. Mr X is planning to buy a new car with better mileage as compared to his current car. By what % should the new mileage be more than the previous mileage to ensure that Mr X’s total fuel cost stays the same for the month? (assuming the distance traveled every month stays the same)

(A)   10%

(B)   17%

(C)   20%

(D)   21%

(E)    25%

Solution:  The problem here is ‘how is mileage related to fuel price?’

Total fuel cost = Fuel price * Quantity of fuel used

Since the ‘total fuel cost’ needs to stay the same, ‘fuel price’ varies inversely with ‘quantity of fuel used’.

Quantity of fuel used = Distance traveled/Mileage

Distance traveled = Quantity of fuel used*Mileage

Since the same distance needs to be traveled, ‘quantity of fuel used’ varies inversely with the ‘mileage’.

We see that ‘fuel price’ varies inversely with ‘quantity of fuel used’ and ‘quantity of fuel used’ varies inversely with ‘mileage’. So, if fuel price increases, quantity of fuel used decreases proportionally and if quantity of fuel used decreases, mileage increases proportionally. Hence, if fuel price increases, mileage increases proportionally or we can say that fuel price varies directly with mileage.

If fuel price becomes 6/5 (20% increase) of previous fuel price, we need the mileage to become 6/5 of the previous mileage too i.e. mileage should increase by 20% too.

Another method is that you can directly plug in the expression for ‘Quantity of fuel used’ in the original equation.

Total fuel cost = Fuel price * Distance traveled/Mileage

Since ‘total fuel cost’ and ‘distance traveled’ need to stay the same, ‘fuel price’ is directly proportional to ‘mileage’.

Answer (C)

We hope you are comfortable with fundamentals of direct and inverse variation now. More next week!

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: Varying Directly

Quarter Wit, Quarter WisdomWe can keep working on ‘pattern recognition’ questions for a long time and not run out of questions of different types on which it can be used. We hope you have understood the basic concepts involved. So let’s move on to another topic now: Variation.

Basically, variation describes the relation between two or more quantities. e.g. workers and work done, children and noise, entrepreneurs and start ups. More workers means more work done; more children means more noise; more entrepreneurs means more start ups and so on… These are examples of direct variation i.e. if one quantity increases, the other increases proportionally. Then there are quantities that have inverse variation between them e.g. workers and time taken. If there are more workers, time taken to complete a work will be less.

Let’s discuss direct variation today.

Formally, let’s say x varies directly with y. If x takes values x1, x2, x3… and y takes values y1, y2, y3 … correspondingly, then x1/y1 = x2/y2 = x3/y3 = Some constant value

In other words, ratio of x and y stays the same in different instances.

(Notice that this is the same as x1/x2 = y1/y2)

It might seem a little cumbersome when put this way but the truth is that direct variation is quite intuitive. A couple of questions will make it clear.

Question 1: 20 workmen can make 35 widgets in 5 days. How many workmen are needed to make 105 widgets in 5 days?

(A)   7

(B)   20

(C)   25

(D)   30

(E)    60

Solution: Notice that the number of days stays the same so we can ignore it. Now think, how are workmen and widgets related? If the number of workmen increases, the number of widget made also increases proportionally. You need to find the new number of workmen required. The number of widgets has become thrice (105/35 = 3) so number of workmen needed will become thrice as well (remember, the number of workmen will increase in the same proportion).

We need 20*3 = 60 workmen

Answer (E)

The concept of variation is very intuitive. If the number of widgets required doubles, the number of workmen required to make them in the same amount of time will double too. If the number of widgets required becomes one fourth, the number of workmen required to make them in the same amount of time will become one fourth too.

A quantity can directly vary with some power of another quantity. Let’s take an example of this scenario too.

Question 2: If the ratio of the volumes of two right circular cylinders is given by 64/9, what is the ratio of their radii? (Both the cylinders have the same height)

(A)   4/3

(B)   8/3

(C)   16/9

(D)   4/1

(E)    16/3

Solution: This question involves a little bit of geometry too. The volume of a right circular cylinder is given by Area of base * height i.e.

Volume of a right circular cylinder = pi*radius^2 * height

So volume varies directly with the square of radius.

Va/Vb = 64/9 = Ra^2/Rb^2

Ra/Rb = 8/3

Answer (B)

We hope this little concept is not hard to understand. We will work on inverse proportion next week and then work on problems involving both (that’s where the good questions are!).

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: Pattern Recognition or Number Properties?

Quarter Wit, Quarter WisdomContinuing our quest to master ‘pattern recognition’, let’s discuss a tricky little question today. It is best done using divisibility and remainders logic we discussed in some previous posts. We suggest you check out these divisibility posts if you haven’t yet.

We are first going to see how to solve the question conceptually. The interesting thing is – what do you do if you are under immense pressure during the test and are unable to remember anything you ever read on divisibility? It is a fairly common phenomenon – students have reported that they had blanked out during the test and couldn’t think of an appropriate approach. Our suggestion is that in that case, you should lean on trying to figure out the pattern. Try out a couple of values and see what you get. You may not understand why you are getting what you are getting but that doesn’t stop you from getting the correct answer. Let’s jump on to the question – we will first discuss the ideal approach and then go on to what happens if you don’t have a clue of what to do in the question.

Question: If p, m, and n are positive integers, n is odd, and p = m^2 + n^2, is m divisible by 4?

Statement 1: When p is divided by 8, the remainder is 5.
Statement 2: m – n = 3

Solution:  Given that p, m and n are positive integers. If n is odd, n^2 must also be odd. How do you represent an odd integer? As (2k + 1)

n^2 = (2k + 1)^2 = 4k^2 + 4k + 1 = 4k(k + 1) + 1
Since one of k and (k+1) will definitely be even (out of any two consecutive integers, one is always even, the other is always odd – discussed in detail here), 4k(k+1) will be divisible by 8. Therefore, when n^2 is divided by 8, it will leave a remainder of 1.

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

When p ( i.e. m^2 + n^2) is divided by 8, we get a remainder of 5. When n^2 is divided by 8, the remainder will be 1. To get a remainder of 5, when m^2 is divided by 8, we should get a remainder of 4.
m^2 = 8a + 4 (i.e. we can make ‘a’ groups of 8 and 4 will be leftover)
m^2 = 4(2a + 1)

This implies m = 2*?(Odd Number) because (2a+1) is an odd number. Square root of an odd number will also be odd.
Therefore, we can say that m is not divisible by 4.

This statement alone is sufficient.

Statement 2: m – n = 3

The difference between m and n is 3 i.e. an odd number. Since n is odd, we can say that m will be even (Even – Odd = Odd). But whether m is divisible by 2 only or by 4 as well, we cannot say since here we have no constraints on p.

This statement alone is not sufficient.

Answer (A)

In this question, analyzing the question stem and statement 1 is a little complicated. Say we don’t analyze the question stem and jump to statement 1. Let’s see how we can use pattern recognition to make it easier.

Question: If p, m, and n are positive integers, n is odd, and p = m^2 + n^2, is m divisible by 4?

Statement 1: When p is divided by 8, the remainder is 5.
Statement 2: m – n = 3

Solution:

Statement 1: When p is divided by 8, the remainder is 5.
When p ( i.e. m^2 + n^2) is divided by 8, we get a remainder of 5.  We need a remainder of 5 when m^2 + n^2 is divided by 8. Let’s try to find the remainders when m^2 and n^2 are divided by 8.

We are given that n is odd. Let’s try to figure out what this implies.

n = 1; When n^2 ( = 1) is divided by 8, the remainder is 1.
n = 3; When n^2 ( = 9) is divided by 8, the remainder is 1.
n = 5; When n^2 ( = 25) is divided by 8, the remainder is 1.
n = 7; When n^2 (= 49) is divided by 8, the remainder is 1.

There is a pattern here! Whenever you divide the square of an odd number by 8, you get the remainder 1. (We have discussed ‘why’ above in the logical explanation of statement 1)

This implies that when we divide m^2 by 8, we get 4 as remainder. If m^2 gives 4 as remainder, it means it is of the form m^2 = 8a + 4 = 4(2a + 1). So m must be of the form 2?(Odd Number). Hence m is not divisible by 4.

This statement is sufficient alone.

Through this example, you can see that pattern recognition is a very important tool (in easy as well as difficult questios)

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!