Properties of Absolute Values on the GMAT

Quarter Wit, Quarter WisdomWe have talked about quite a few concepts involving absolute value of x in our previous posts. But some absolute value questions involve two variables. Then do we need to consider the positive and negative values of both x and y? Certainly! But there are some properties of absolute value that could come in handy in such questions. Let’s take a look at them:

(I)    For all real x and y, |x + y| <= |x| + |y|

(II)   For all real x and y, |x – y| >= |x| – |y|

We don’t need to learn them of course and there is no need to look at how to prove them either. All we need to do is understand them – why do they hold, when is the equality sign applicable and when can they be useful. Let’s look at both the properties one by one.

(I)     For all real x and y, |x + y| <= |x| + |y|

The result of both the left hand side and the right hand side will be positive or zero. On the right hand side, the absolute values of x and y will always get added irrespective of the signs of x and y. On the left hand side, the absolute values of x and y might get added or subtracted depending on whether they have the same sign or different signs. Hence the result of the left hand side might be smaller than or equal to that of the right hand side.

For which values of x and y will the equality hold and for which values will the inequality hold? Let’s think logically about it.

The absolute values of x and y get added on the right hand side. We want the absolute values of x and y to get added on the left hand side too for the equality to hold. This will happen when x and y have the same sign. So the equality should hold when they have the same signs.

For example, x = 4, y = 8:

|4 + 8| = |4| + |8| = 12

OR x = -3, y = -4:

|-3 -4| = |-3| + |-4| = 7

Also, when at least one of x and y is 0, the equality will hold.

For example, x = 0, y = 8:

|0 + 8| = |0| + |8| = 8

OR x = -3, y = 0:

|-3 + 0| = |-3| + |0| = 3

What happens when x and y have opposite signs? On the left hand side, the absolute values of x and y get subtracted hence the left hand side will be smaller than the right hand side (where they still get added). That is when the inequality holds i.e. |x + y| < |x| + |y|

For example, x = -4, y = 8:

|-4 + 8| < |-4| + |8|

4 < 12

OR x = 3, y = -4:

|3 -4| < |3| + |-4|

1 < 7

Let’s look at our second property now:

(II) For all real x and y, |x – y| >= |x| – |y|

Thinking on similar lines as above, we see that the right hand side of the inequality will always lead to subtraction of the absolute values of x and y whereas the left hand side could lead to addition or subtraction depending on the signs of x and y. The left hand side will always be positive whereas the right hand side could be negative too. So in any case, the left hand side will be either greater than or equal to the right hand side.

When will the equality hold?

When x and y have the same sign and x has greater (or equal) absolute value than y, both sides will yield a positive result which will be the difference between their absolute values

For example, x = 9, y = 2;

|9 – 2| = |9| – |2| = 7

OR x = -7, y = -3

|-7 – (-3)| = |-7| – |-3| = 4

Also when y is 0, the equality will hold.

For example, x = 8, y = 0:

|8 – 0| = |8| – |0| = 8

OR x = -3, y = 0:

|-3 – 0| = |-3| – |0| = 3

What happens when x and y have the same sign but absolute value of y is greater than that of x?

It is easy to see that in that case both sides have the same absolute value but the right hand side becomes negative.

For example, x = -4, y = -9

|x – y| = |-4 – (-9)| = 5

|x| – |y| = |-4| – |-9| = -5

So even though the absolute values will be the same since we will get the difference of the absolute values of x and y on both sides, the right hand side will be negative. If we were to take further absolute value of the right hand side, the two will become equal i.e. the right hand side will become |(|x| – |y|)| = |-5| = 5 in our example above. In that case, the equality will hold again.

Similarly, what happens when only x = 0? The right hand side becomes negative again so taking further absolute value will make both sides equal.

For example, x = 0, y = -5

|x – y| = |0 – (-5)| = 5

|x| – |y| = |0| – |5| = -5

Taking further absolute value, |(|x| – |y|)| = |-5| = 5

So when we take further absolute value of the right hand side, this property becomes similar to property 1 above: |x – y| = |(|x| – |y|)| when x and y have the same sign or at least one of x and y is 0.

Now let’s look at the inequality part of property 2.

Whenever x and y have opposite signs, |x – y| > |x| – |y|

On the left hand side, the absolute values will get added while on the right hand side, the absolute values will get subtracted. So the absolute value of the left hand side will always be greater than the absolute value of the right hand side. The left hand side will always be positive while the right hand side could be negative too. Hence even if we take the further absolute value of the right hand side, the inequality will hold: |x – y| > |(|x| – |y|)| when x and y have opposite signs

For example, x = -4, y = 8:

|-4 – 8| > |-4| – |8|

12 > -4

Taking further absolute value of the right hand side, we get |(|x| – |y|)| = |-4| = 4

Still, 12 > 4 i.e. |x – y| > |(|x| – |y|)|

OR x = 3, y = -4:

|3 –(-4)| > |3| – |-4|

7 > -1

Taking further absolute value of the right hand side, we get |(|x| – |y|)| = |-1| = 1

Still, 7 > 1 i.e. |x – y| > |(|x| – |y|)|

Note that the inequality of the original property 2 also holds when x and y have the same sign but absolute value of y is greater than the absolute value of x since the right hand side becomes negative. It also holds when x is 0 but y is not.

To sum it all neatly,

(I) For all real x and y, |x + y| <= |x| + |y|

|x + y| = |x| + |y| when (1) x and y have the same sign (2) at least one of x and y is 0.

|x + y| < |x| + |y| when (1) x and y have opposite signs

(II) For all real x and y, |x – y| >= |x| – |y|

|x – y| = |x| – |y|when (1) x and y have the same sign and x has greater (or equal) absolute value than y (2) y is 0

|x – y| > |x| – |y| in all other cases

(III) For all real x and y, |x – y| >= |(|x| – |y|)|

|x – y| = |(|x| – |y|)| when (1) x and y have the same sign (2) at least one of x and y is 0.

|x – y| > |(|x| – |y|)| when (1) x and y have opposite signs

Note that property (III) matches property (I).

There is another property we would like to discuss but let’s take it up next week along with some GMAT questions where we put these properties to 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!

How Well Do You Know Your Factors?

Quarter Wit, Quarter WisdomIn the last three weeks, we discussed a couple of strategies we can use to solve max-min questions: ‘Establishing Base Case’ and ‘Focus on Extremes’. Now try to use those to solve this question:

Question: A carpenter has to build 71 wooden boxes in one week. He can build as many per day as he wants but he has decided that the number of boxes he builds on any one day should be within 4 off the number he builds on any other day.
(A) What is the least number of boxes that he could have build on Saturday?
(B) What is the greatest number of boxes that he could have build on Saturday?

Meanwhile, let’s move on to something else today. What we will discuss today is a very simple concept but it seems odd to us when we first confront it even if we are very comfortable with factors and divisibility. If we tell you the concept right away, you will probably not believe us when we say that many people are unable to come up with it on their own. Hence, we will first give you a question which you need to answer in 30 seconds. If you are unable to do so, then we will discuss the concept with you!

Question: A, B, C and D are positive integers such that A/B = C/D. Is C divisible by 5?

Statement 1: A is divisible by 210

Statement 2: B = 7^x, where x is a positive integer

Solution: Let’s discuss the solution till the point I assume you will be quite comfortable.

We need to find whether C is divisible by 5. So let’s separate the C out of the variables.

C = AD/B

Since C is an integer, AD will be divisible by B but what we don’t know is that after the division, is the quotient divisible by 5?

Statement 1: A is divisible by 210

We still have no idea what B is so this statement alone is not sufficient. Let’s take an example of how the value of B could change our answer. Assume A is 210.

If B is 3, AD/B will be divisible by 5.

If B is 10, AD/B may not be divisible by 5 (depending on the value of D).

Statement 2: B = 7^x, where x is a positive integer

We have no idea what A and D are hence this statement alone is not sufficient.

Using both together: Now, this is where the trick comes in. Using both statements together, we see that C = (210*a*D)/(7^x)

Now we can say for sure that C will be divisible by 5. If you are not sure why, read on.

The Concept:

As you know, factors (also called divisors) of a number N are those positive integers which completely divide number N i.e. they do not leave a remainder on dividing N. If F is a factor of N, N/F leaves no remainder. This also means that N can be written as F*m where m is an integer. Sure you feel this is elementary but this concept is not as internalized in your conscience as you believe. To prove it, let me give you a question.

Example 1: Is 3^5 * 5^9 * 7 divisible by 18?

Did you take more than 2 seconds to say ‘No’ confidently?

For N to be divisible by F, you should be able to write N as F*m i.e. N must have F as a factor. F here is 18 (= 2*3^2) but we have no 2 in N (which is 3^5 * 5^9 * 7) though we do have a couple of 3s. Hence this huge product is not divisible by 18.

This helps us deduce that odd numbers are never divisible by even numbers.

Example 2: Is 3^5*7^6*11^3 divisible by 13?
The answer is simply ‘No’.

For the numerator to be divisible by the denominator, the denominator MUST BE a factor of the numerator. In the entire numerator, there is no 13 so the numerator is not divisible by 13.

Example 3: On the other hand, is 3^5*7^6*11^3*13 divisible by 13?

Yes, it is. 13 gets cancelled and the quotient will be 3^5*7^6*11^3.

Example 4: Is 2^X divisible by 3?

No. No matter what X is, you will only have X number of 2s in the numerator and will never have a 3. So this will not be divisible by 3.

Let’s come back to the original question now:

Given that C = (210*a*D)/(7^x)

Whatever x is, 7^x will get cancelled out by the numerator and we will be left with something. That something will include 5 (obtained from 210) since only 7s will be cancelled out from the numerator. Hence C is divisible by 5.

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!

Max-Min Strategies: Focus on Extremes

Quarter Wit, Quarter WisdomIn the last two weeks, we discussed some max min strategies. Today, let’s look at another max-min question in which we apply the strategy of focusing on the extremes. The largest or the smallest values are often found at the extremes of a given range.

Question: If x and y are integers such that (x+1)^2 is less than or equal to 36 and (y-1)^2 is less than 64, what is the sum of the maximum possible value of xy and the minimum possible value of xy?

(A) -16

(B) -14

(C) 0

(D) 14

(E) 16

Solution: To get the sum of the maximum and minimum possible values of xy, we need to know the maximum and minimum values of xy. For those, we need to find the values that x and y can take. So first, we should review the information given:

(x + 1)^2 <= 36

(y – 1)^2 < 64

We need to find the values that x and y can take. There are many ways of doing that. We can solve the inequality using the wave method discussed in this post or using the concept of absolute values. Let’s discuss both the methods.

Wave method to solve inequalities:

Solve for x: (x + 1)^2 <= 36

(x + 1)^2 – 6^2 <= 0

(x + 1 + 6)(x + 1 – 6) <= 0

(x + 7)(x – 5) <= 0

-7 <= x <= 5 (Using the wave method)

Solve for y: (y – 1)^2 < 64

(y – 1)^2 – 8^2 < 0

(y – 1 + 8)(y – 1 – 8) < 0

(y + 7)(y – 9) < 0

-7 < y < 9 (Using the wave method)

Or you can solve taking the square root on both sides

Solve for x: (x + 1)^2 <= 36

|x + 1| <= 6

-6 <= x + 1 <= 6 (discussed in your Veritas Algebra book)

-7 <= x <= 5

So x can take values: -7, -6, -5, -4, … 3, 4, 5

Solve for y: (y – 1)^2 < 64

|y – 1| < 8

-8 < y – 1 < 8 (discussed in your Veritas Algebra book)

-7 < y < 9

So y can take values: -6, -5, -4, -3, … 6, 7, 8.

Now that we have the values of x and y, we should try to find the minimum and maximum values of xy.

Note that the values of xy can be positive as well as negative. The minimum value will be the negative value with largest absolute value (largest negative) and the maximum value will be the positive value with the largest absolute value.

Minimum value – For the value to be negative, one and only one of x and y should be negative. Focus on the extreme values: if x is -7 and y is 8, we get xy = -56. This is the negative value with largest absolute value.

Maximum value – For the value to be positive, both x and y should have the same signs. If x = -7 and y = -6, we get xy = 42. This is the largest positive value.

The sum of the maximum value of xy and minimum value of xy is -56 + 42 = -14

Answer (B)

Try to think of it in terms of a number line. x lies in the range -7 to 5 and y lies in the  range -6 to 8. The range is linear so the end points give us the maximum/minimum values. Think of what happens when you plot a quadratic – the minimum/maximum could lie anywhere.

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!

 

Max-Min Strategies: Establishing Base Case

Quarter Wit, Quarter WisdomContinuing our discussion on maximizing/minimizing strategies, let’s look at another question today. Today we discuss the strategy of establishing a base case, a strategy which often comes in handy in DS questions. The base case gives us a starting point and direction to our thoughts. Otherwise, with the number of possible cases in any given scenario, we may find our mind wandering from one direction to another without reaching any conclusions. That is a huge waste of time, a precious commodity.

Question: Four friends go to Macy’s for shopping and buy a top each. Three of them buy a pillow case each too. The prices of the seven items were all different integers, and every top cost more than every pillow case. What was the price, in dollars, of the most expensive pillow case if the total price of the seven items was $89?

Statement 1: The most expensive top cost $16.

Statement 2: The least expensive pillow case cost $9.

Solution: The first problem here is figuring out the starting point. There must be many ways in which you can price the seven items such that the total cost is $89. So we need to establish a base case (which conforms to all the conditions given in the question stem) first and then we will tweak it around according to the additional information obtained from our statements.

‘Seven items for $89’ means the average price for each item is approximately $12. But 12 is not the exact average. 12*7 = 84 which means another $5 were spent.

A sequence with an average of 12 and different integers is $9, $10, $11, $12, $13, $14, $15.

But actually another $5 were spent so the prices could be any one of the following variations (and many others):

$9, $10, $11, $12, $13, $14, $20 (Add $5 to the highest price)

$9, $10, $11, $12, $13, $16, $18 (Split $5 into two and add to the two highest prices)

$9, $10, $12, $13, $14, $15, $16 (Split $5 into five parts of $1 each and add to the top 5 prices)

$7, $9, $13, $14, $15, $16, $17 (Take away some dollars from the lower prices and add them to the higher prices along with the $5)

etc

Let’s focus on another piece of information given in the question stem: “every top cost more than every pillow case.”

This means that when we arrange all the prices in the increasing order (as done above), the last four are the prices of the four tops and the first three are the prices of the three pillow cases. The most expensive pillow case is the third one.

Now that we have accounted for all the information given in the question stem, let’s focus on the statements.

Statement 1: The most expensive top cost $16.

We have already seen a case above where the maximum price was $16. Is this the only case possible? Let’s look at our base case again:

$9, $10, $11, $12, $13, $14, $15

(a further $5 needs to be added to bring the total price up to $89)

Since the prices need to be all unique, if we add 1 to any one price, we also need to add at least $1 to each subsequent price. E.g. if we increase the price of the least expensive pillow case by $1 and make it $10, we will need to increase the price of every subsequent item by $1 too. But we have only $5 more to give.

If the maximum price is $16, it means the rightmost price can increase by only $1. So all prices before it can also only increase by $1 only and except the first two prices, they must increase by $1 to adjust the extra $5.

Hence the only possible case is $9, $10, $12, $13, $14, $15, $16.

So the cost of the most expensive pillow case must have been $12.

Statement 1 is sufficient alone.

Statement 2: The least expensive pillow case cost $9.

A restriction on the lowest price is much less restrictive. Starting from our base case

$9, $10, $11, $12, $13, $14, $15,

we can distribute the extra $5 in various ways. We can do what we did above in statement 1 i.e. give $1 to each of the 5 highest prices: $9, $10, $12, $13, $14, $15, $16

We can also give the entire $5 to the highest price: $9, $10, $11, $12, $13, $14, $20

So the price of the most expensive pillow case could take various values. Hence, statement 2 alone is not sufficient.

Answer (A)

Note that the answer is a little unexpected, isn’t it? If we were to read the question and guess within 20 secs, we would probably guess that the answer is (C), (D) or (E). The two statements give similar but complementary information. It would be hard to guess that one will be sufficient alone while other will not be. This is what makes this question interesting and hard too.

Our strategy here was to establish a base case and tweak it according to the information given in the statements. This strategy is often useful in DS – not just in max-min questions but others too.

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 series!

How to Deal with Maximizing/Minimizing Strategies on the GMAT

Quarter Wit, Quarter WisdomWe haven’t dealt with maximizing/minimizing strategies in our QWQW series yet (except in sets). The reason for this is that the strategy to be used varies from question to question. What works in one question may not work in another. You might have to think up on what to do in a question from scratch and you have only 2 mins to do it in. The saving grace is that once you know what you have to do, the actual work involved to arrive at the answer is very little.

Let’s look at some maximizing minimizing strategies in the next few weeks. We start with an OG question today with a convoluted question stem.

Question: List T consists of 30 positive decimals, none of which is an integer, and the sum of the 30 decimals is S. The estimated sum of the 30 decimals, E, is defined as follows. Each decimal in T whose tenths digit is even is rounded up to the nearest integer, and each decimal in T whose tenths digits is odd is rounded down to the nearest integer. If 1/3 of the decimals in T have a tenths digit that is even, which of the following is a possible value of E – S?

I. -16
II. 6
III. 10

A. I only
B. I and II only
C. I and III only
D. II and III only
E. I, II, and III

Solution:

There is a lot of information in the question stem and a lot of variables are explained. Let’s review the given data in our own words first.

T has 30 decimals. The sum of all the decimals is S.

10 decimals have even tenths digit. They will be rounded up.

20 decimals have odd tenths digit. They will be rounded down.

The sum of rounded numbers is E.

E – S can take many values so how do we figure which ones it cannot take? We need to find the minimum value E – S can take and the maximum value it can take. That will help us figure out the values that E – S cannot take. Note that E could be greater than S and it could be less than S. So E – S could be positive or negative.

Step 1: Getting Minimum Value of E – S

Let’s try to make E as small as possible. For that, we need to do two things:

1. When we round up the decimals (even tenths digit), the difference between actual and estimate should be very small. The estimate should add a very small number to round it up so that E is not much greater than S. Say the numbers are something similar to 3.8999999 (the tenths digit is the largest even digit) and they will be rounded up to 4 i.e. the estimate gains about 0.1 per number. Since there are 10 even tenths digit numbers, the estimate will be approximately .1*10 = 1 more than actual.

2. When we round down the decimals (odd tenths digit), the difference between actual and estimate should be as large as possible. Say the numbers are something similar to 3.999999 (tenths digit is the largest odd digit) and they will be rounded down to 3 i.e. the estimate loses approximately 1 per number. Since there are 20 such numbers, the estimate is 1*20 = 20 less than actual.

Overall, the estimate will be approximately 20 – 1 = 19 less than actual.

Minimum value of E – S = -19

Step 2: Getting Maximum Value of E – S

Now let’s try to make E as large as possible. For that, we need to do two things:

1. When we round up the decimals (even tenths digit), the difference between actual and estimate should be very high. Say the numbers are something similar to 3.000001 (tenths digit is the smallest even digit) and they will be rounded up to 4 i.e. the estimate gains 1 per number. Since there are 10 even tenths digit numbers, the estimate will be approximately 1*10 = 10 more than actual.

2. When we round down the decimals (odd tenths digit), the difference between actual and estimate should be very little. Say the numbers are something similar to 3.1 (tenths digit is the smallest odd digit). They will be rounded down to 3 i.e. the estimate loses approximately 0.1 per number. Since there are 20 such numbers, the estimate is approximately 0.1*20 = 2 less than actual.

Overall, the estimate will be approximately 10 – 2 = 8 more than actual.

Maximum value of E – S = 8.

The minimum value of E – S is -19 and the maximum value of E – S is 8.

So E – S can take the values -16 and 6 but cannot take the value 10.

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 series!

Converting Non-Terminating Repeating Decimals to Fractions

Quarter Wit, Quarter WisdomLast week we discussed the properties of terminating decimals. We also discussed that non-terminating but repeating decimals are rational numbers.

For GMAT, we must know how to convert these non-terminating repeating decimals into rational numbers. We know how to do vice versa i.e. given a rational number, we can divide the numerator by the denominator to find its decimal equivalent.

 

For example:

1/3 = 0.333333333… (infinite number of 3s)

But given 0.555555555…, how will you convert it to its exact fraction equivalent?

Take a look at this file: Non-terminating Decimals

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!

Terminating Decimals in Data Sufficiency on the GMAT

Quarter Wit, Quarter Wisdom

Last week, we discussed the basics of terminating decimals. Let me review the important points here:

–  To figure out whether the fraction is terminating, bring it down to its lowest form.

–  Focus on the denominator – if it is of the form 2^a * 5^b, the fraction is terminating, else it is not.

 

Keeping this in mind, let’s look at a couple of DS questions on terminating decimals.

Question 1: If a, b, c, d and e are integers and m = 2^a*3^b and n = 2^c*3^d*5^e, is m/n a terminating decimal?

Statement 1: a > c
Statement 2: b > d

Solution:

Given: a, b, c, d and e are integers

Question: Is m/n a terminating decimal?

Or Is (2^a*3^b)/(2^c*3^d*5^e)?

We know that powers of 2 and 5 in the denominator are acceptable for the decimal to be terminating. If there is a power of 3 in the denominator after reducing the fraction, then the decimal in non- terminating. So our question is basically whether the power of 3 in the denominator gets canceled by the power of 3 in the numerator. If b is greater than (or equal to) d, after reducing the fraction to lowest terms, it will have no 3 in the denominator which will make it a terminating decimal. If b is less than d, even after reducing the fraction to its lowest terms, it will have some powers of 3 in the dominator which will make it a non-terminating decimal.

Question: Is b >= d?

Statement 1: a > c

This statement doesn’t tell us anything about the relation between b and d. Hence this statement alone is not sufficient.
Statement 2: b > d

This statement tells us that b is greater than d. This means that after we reduce the fraction to its lowest form, there will be no 3 in the denominator and it will be of the form 2^c * 5^e only. Hence it will be a terminating decimal. This statement alone is sufficient.
Answer (B)

Now onto another DS question.

Question 2: If 0 < x < 1, is it possible to write x as a terminating decimal?

Statement 1: 24x is an integer.

Statement 2: 28x is an integer.

Solution:

Given: 0 < x < 1

Question: Is x a  terminating decimal?

Again, x will be a terminating decimal if it is of the form m/(2^a * 5^b)

Statement 1: 24x is an integer.

24x = 2^3 * 3 * x = m (an integer)

x = m/(2^3 * 3)

Is x a terminating decimal? We don’t know. If m has 3 as a factor, x will be a terminating decimal. Else it will not be. This statement alone is not sufficient.

Statement 2: 28x is an integer.

28x = 2^2 * 7 * x = n (an integer)

x = n/(2^2 * 7)

Is x a terminating decimal? We don’t know. If n has 7 as a factor, x will be a terminating decimal. Else it will not be. This statement alone is not sufficient.

Taking both together,

m/24 = n/28

m/n = 6/7

Since m and n are integers, m will be a multiple of 6 (and thereby of 3 too) and n will be a multiple of 7. So x will be a terminating decimal.

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!

How to Identify Terminating Decimals on the GMAT

Quarter Wit, Quarter WisdomWe know the basics of decimals and rational numbers.

–   Decimals can be rational or irrational.

– Decimals which terminate and those which are non-terminating but repeating are rational. They can be written in the form a/b.

–  Decimals which are non-terminating and non-repeating are irrational such as ?2, ?3 etc.

The problem comes when we get a question based on these basics. That’s when we realize that our basics are not as strong as we assumed them to be. For example, look at this question:

Question: Which of the following fractions has a decimal equivalent that is a terminating decimal?

(A) 10/189
(B) 15/196
(C) 16/225
(D) 25/144
(E) 39/128

If your first thought is that we will simply divide the numerator by the denominator in each case and figure out which terminates and which doesn’t, you must realize that that is a very time consuming process. There has to be another logical approach to this problem. Well, here it is:

A fraction in its lowest term can be expressed as a terminating decimal if and only if the denominator has powers of only 2 and/or 5. Let’s try to understand the logic behind it.

Say, a and b are two integers.

a/b = a * 1/b

For a/b to be terminating, 1/b must be a terminating decimal. What happens when you start dividing 1 by b? You add a decimal point and start adding 0s. You will get 1 followed by as many 0s as you require in the numerator. 10/100/1000/10000 etc have only two prime divisors: 2 and 5. If the denominator has 2s or 5s or both, we will be able to terminate the decimal by choosing the required multiple of 10. If there are any other primes, we will never be able to divide a multiple of 10 completely and hence the decimal will not terminate. It is obvious, isn’t it?

1/3 = .333333333333333333…

1/7 = .142857142857142857…

1/11 = .09090909090909090…

Now the question we posed above is quite simple. Let’s look at it again.

Question 1: Which of the following fractions has a decimal equivalent that is a terminating decimal?

(A) 10/189
(B) 15/196
(C) 16/225
(D) 25/144
(E) 39/128

Only option (E) has a denominator of the form 2^a*5^b.

128 = 2^7

Therefore, 39/128 will terminate. All the other denominators have other prime numbers as well and hence will not terminate.

Using the same concepts, let’s look at another question.

Question 2: If 1/(2^11 * 5^17) is expressed as a terminating decimal, how many non-zero digits will the decimal have?

(A) 1
(B) 2
(C) 4
(D) 6
(E) 11

Solution:

First realize that 2^11 * 5^17 = 2^11 * 5^11 * 5^6 = 10^11 * 5^6
So 1/(10^11 * 5^6)  is just 0.00…001/5^6.

Now let’s try to figure out the answer intuitively:

What do you get when you divide .01 by 5? You get .002. You write 0s till you get 10 and then you get a non-zero digit.

What do you get when you divide .01 by 125 (which is 5^3)? You get .00008.

Do you notice something? The non zero term is 8 = 2^3

The reason is this: You have 1 followed by as many 0s as you require in the dividend. 125 = 5^3 so you will need 2^3 i.e. you will need 10^3 as the dividend and then 125 will be able to divide it completely (i.e. the decimal will terminate).

Now, using the same logic, what will be the non zero digits if you are dividing .00001 by 625?
625 = 5^4. You will need 2^4 = 16 to get 10^4 and that will end the terminating decimal. So you will have two non 0 digits: 16

What will you get when you divide .000…0001 by 5^6? Your non zero digits will be 2^6 = 64 i.e. you will have 2 non-zero digits.

Another way to look at the problem is this:

1/(10^11 * 5^6)  = 2^6/(10^17) (multiply and divide by 2^6)

= 64/(10^17)

Since the denominator is a power of 10, it will just move the decimal 17 places to the left. The non-zero digits will remain 64 only i.e. 2 digits.

Answer (B)

We will look at some DS questions on terminating and non terminating decimals 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!

Can You Find the Correct Answer to This Tricky GMAT Question?

Quarter Wit, Quarter WisdomThis is hard to confess publicly but I must because it is a prime example of how GMAT takes advantage of our weaknesses – A couple of days back, I answered a 650 level question of weighted averages incorrectly. Those of you who have been following my blog would understand that it was an unpleasant surprise – to say the least. I know my weighted averages quite well, thank you! For this comedown, I blame the treachery of GMAT because it knows how to get you when you become too complacent. The takeaway here is – no matter how easy and conventional the question seems, you MUST read it carefully.

Let me share that particular question with you. I will also share two solutions which give you two different answers. It is an exercise for you to figure out which one is the correct solution (that is, if one of them is the correct solution). Needless to say, the error in the solution(s) is conceptual and very easy to see (not some sly calculation mistake). It’s just that in your haste, it’s very easy to miss this important point. I hope to see some comments with some good explanations.

Question: The price of each hair clip is ¢ 40 and the price of each hair band is ¢ 60. Rashi selects a total of 10 clips and bands from the store, and the average (arithmetic mean) price of the 10 items is ¢ 56. How many bands must Rashi put back so that the average price of the items that she keeps is ¢ 52?

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

Solution 1:

Price of each clip (Pc) = 40

Price of each band (Pb) = 60

Average price of each item (Pavg) = 56

Wc/Wb = (Pb – Pavg)/(Pavg – Pc) = (60 – 56)/(56 – 40) = 1/4 (our weighted average formula)

Since the total number of items is 10, number of clips = 1*2 = 2 and number of bands = 4*2 = 8

If the average price is changed to 52,

Wc/Wb = (Pb – Pavg)/(Pavg – Pc) = (60 – 52)/(52 – 40) = 2/3

Now the ratio has changed to 2:3. This gives us number of clips as 4 and number of bands as 6.

Since previously she had 8 bands and now she has 6 bands, she must have put back 2 bands.

Answer (B)

Solution 2:

Say the number of hair clips is C and the number of hair bands is 10 – C.

(40C + 60(10 – C))/10 = 56 (Using the formula: Average = Sum/Number of items)

On solving, you get C = 2

Number of clips is 2 and number of bands is (C – 2) = 8.

Now, let’s consider the scenario when she puts back some bands, say x.

(2*40 + (8 – x)*60)/(10 – x) = 52

On solving, you get x = 5

So she puts back 5 bands so that the average price is 52.

Answer (E)

Obviously, there is only one correct answer. It’s your job to figure out whether it is (B) or (E) or some third option. Also what’s wrong with one or both of these solutions?

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!

Elementary, My Dear Watson!

Quarter Wit, Quarter WisdomWhile eagerly awaiting the kick off of season 3 of BBC’s Sherlock, let’s put our time to good use. Though we have already spent a lot of it speculating over what really happened to Sherlock (HOW did he come back?!), perhaps we can take a leaf out of his book and learn to notice little things in whatever is leftover. There is a good reason to do that – there are little clues in some questions that the test maker unwittingly leaves to bring clarity to the question. If we understand those clues, a seemingly mysterious problem could be easily unraveled. Let us show you with an example.

Question: Peter and Jacob are at the northwest corner of a field, which is a rectangle 300 ft long and 160 ft wide. Peter walks in a straight line directly to the southeast corner of the field. If Jacob walks 180 ft down the west side of the field and then walks in a straight line directly to the southeast corner of the field, what is the difference in the distance traveled by the two?

(A) 20
(B) 40
(C) 80
(D) 120
(E) 140

Solution: The first thing we do in these “direction” questions is draw the diagram. But there is a problem here: how do we decide the orientation of the rectangle? It could be either of these two.

A few things help us decide this. There are two definitions of length:
1. Length is the longest side of the rectangle.
2. Width is from side to side and length is whatever width isn’t (i.e. the side from up to down in a rectangle) (this definition is less embraced than the first one)

If the side from up to down is the longest side, then there is no conflict.
Keeping this in mind, when drawing the figure, given that length is the longer of the two, one could make the rectangle on the left and there will be no conflict. But the question maker may not want to take for granted that you know this.

So he/she leaves a clue – the question mentions that ‘Jacob walks 180 ft down the west side of the field’. There needs to be at least 180 ft on the west side of the field for him to travel that much. So the orientation on the left makes sense. This is something the question maker would have put to try to give you a hint of the orientation. Now that we know what our diagram should look like, we can proceed to solve this question.

If you just remember some of your pythagorean triplets, this question can be solved in moments (and that’s why we suggest you to remember them!) If not, it would involve some calculations.
QR = 160, RS = 300
So QR:RS = 8:15
Remember 8-15-17 pythagorean triplet? (the third triplet after 3-4-5 and 5-12-13)
Since the two sides are in the ratio 8:15, the hypotenuse must be 17. The common multiplier is 20 so QS  should be 17*20 = 340
Therefore, Peter traveled 340 feet.

TP = 120, PS = 160
TP:PS = 3:4
Does it remind you of 3-4-5 triplet?
120 is 3*40 and 160 is 4*40 so TS will be 5*40 = 200
So Jacob traveled a total distance of 180 + 200 = 380 feet.

Difference between the distance traveled = 380 – 340 = 40 feet

Note: The following triplets come in handy: (3, 4, 5) (5, 12, 13) (8, 15, 17) (7, 24, 25) (20, 21, 29) and (9, 40, 41)

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!

Hard Quadratic Equations

Quarter Wit, Quarter WisdomWhen faced with an unusual quadratic equation, some people waste a lot of time while trying to ‘split the middle term’. The common refrain is ‘I am just not good at it.’ Actually it has little to do with intuition and a lot to do with understanding how numbers work. If I am looking at a quadratic equation and am unable to find the required factors, I will go back to check my quadratic to see if it is correct rather than try to use the esoteric quadratic formula.

To solve a quadratic, you need to find the two factors that add up to give the middle term and multiply to give the product of the constant term and co-efficient of x^2. What you may have problems with is ‘splitting the middle term’. Even though solving a quadratic is a basic skill one must possess to crack the GMAT, we have seen people struggle with it especially if the coefficient of x^2 is something other than 1. Let’s discuss how we can split the middle term quickly in such cases.

Question 1: Solve for x: 5x^2 – 34x + 24 = 0

To factorize, we need to find two numbers a and b such that:
a + b = -34
a*b = 5*24

Step 1: Prime factorize the product.
a*b = 5*24 = 2*2*2*3*5

Step 2: Check the signs and decide what you need. Here the sum of a and b is negative (-34) while product is positive. This means a and b both are negative. Both will add to give a negative number and multiply to give a positive number (more on this at the end*). It also means that both a and b are smaller than 34 (since both are negative, their absolute values will be added to give 34).

Step 3: Try to split the prime numbers into two groups such that their sum is 34. Try the most obvious group first i.e.
2*2*2 and 3*5 -> 8 and 15

If a = 8 and b  = 15, we get a + b = 23
But the sum needs to be 34, i.e. a number greater than 23.

Before we discuss the next step, let’s talk about how adding numbers works:
Let’s say the prime factorization we have is 2*2*5*5.

We split it into 2 groups -> 2*5 and 2*5 (10 and 10). The sum of 10 and 10 is 20.
We split it in another way -> 2*2 and 5*5 (4 and 25). Their sum is 29. The sum increased.
We split it in yet another way -> 2 and 2*5*5 (2 and 50). Their sum is 52. The sum increased further.

Notice that further apart the numbers are, the greater is their sum. We get the least sum (20) when the numbers are equal. If we need a higher sum, we increase the distance between the numbers.

Going back to the original question, the prime factorization is 2*2*2*3*5 and we split it as 2*2*2 and 3*5. This gave us a sum of 8 + 15 = 23. We need 34 so we need to get the numbers farther from each other but not too far either. Let’s say, we pick a 2 from 8 and give it to 15. We get two numbers 4 and 30. They are farther apart and their sum is 34. So the numbers we are looking for are -4 and -30 (to get -34 as sum).

Now the quadratic is simply: 5*(x –  4/5)*(x – 30/5) = 0 (a shortcut to the usual ‘5x^2 – 4x – 30x + 24 = 0 and proceed’ method)

x = 4/5 or 6

Question 2: Solve for x: 8x^2 – 47x – 63 = 0

To factorize, we need to find two numbers a and b such that:

a + b = -47
a*b = 8*(-63) = – 2*2*2*3*3*7

The product of a and b is negative which means one of a and b is negative. The sum of a and b is also negative therefore, the number with higher absolute value is negative and the other is positive. When these two numbers will be added, the difference of their absolute values will be the sum and the sign of the sum will be negative. So at least one of a and b is greater than 47. The greater one will be negative and the smaller one will be positive.

Let’s try to split the factors now. To start, we split the primes into two easy groups: 2*2*2 and 3*3*7 to get 8 and 63 but -63 + 8 = -55. We need the sum to have lower absolute value than 55 so we need to get the numbers closer together.

Take off a 3 from 63 and give it to 8 to get 2*2*2*3 and 3*7. Now a and b are -24 and 21; the numbers are too close.

Instead, take off 7 from 63 and give it to 8 to get 2*2*2*7 and 3*3. Now a and b are -56 and 9.

-56 + 9 = -47 -> the required sum.

Now the quadratic is 8*(x + 9/8)*(x – 56/8) = 0

x = -9/8 or 7

With a little bit of practice, the hardest questions can be quickly solved.

*How do you decide the sign of a and b:

Product is positive – This means both a and b have the same sign. If sum is negative, both a and b are negative; if sum is positive, they both are positive.

Product is negative – This means a and b have opposite signs; one is negative, the other is positive. If sum is positive, the number which is positive has a higher absolute value. If sum is negative, the number which is negative has a higher absolute value.

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 – Cumulative Graphs

Quarter Wit, Quarter WisdomComing back to Integrated Reasoning question types, let’s discuss a cumulative graph today. They are usually a little trickier than your usual line/pie/bar graphs since you have to focus on not the data points but ‘the change’ from one data point to another.  Every subsequent data point will be either above or at the same level as the previous data point.

Let’s try to understand what a cumulative graph is before we look at questions on one.

Here, the x axis gives the hourly wage and the y axis gives us the cumulative number of people. This means that the point giving the number of people with hourly wage of $30 actually gives the number of people whose hourly wage is UP TO $30. The point giving the number of people with hourly wage of $40 gives the number of people whose hourly wage is up to $40 and hence includes the number of people whose hourly wage is up to $30. That is the reason the graph will always be flat or will have a positive slope.

If we want to focus on the people whose hourly wage varies from $30 to $40, we need to look at the slope of the graph in between these two points. The difference between these two points gives us the number of people with hourly wage in the range of $30 – $40. This implies that if the slope is steep, many people lie in this range.

Set 1: 200 people were surveyed to find out their hourly wage. 100 people had college degrees while other 100 were those who had not completed high school. The following graph gives the cumulative number of people and their hourly wages.

Question 1: True/False: As per the given graph, the average hourly wages of people who did not complete high school is higher than the average hourly wages of people who have college degrees.

Question 2: The median wage of the people with college degrees lies between ___________ while the median wage of people who did not complete high school lies between ___________ (select two options, one for each blank)

(A)   $10–$20 per hr

(B)   $20–$30 per hr

(C)   $30–$40 per hr

(D)   $40–$50 per hr

(E)    $50–$60 per hr

Question 3: Approximately what percentage of ‘people with college degrees’ have hourly wage in the range $50 – $60?

(A) 25%

(B) 35%

(C) 45%

Solutions:

Solution 1: False

Just because the graph of ‘people who did not complete high school’ lies above the graph of ‘people who have college degrees’, it doesn’t mean that average hourly wage of people who did not complete high school is higher. Since the y axis gives the cumulative number of people, having a higher graph early on implies that many people have lower salaries. For example, about 45 people who did not complete high school have hourly wage up to $10. On the other hand, only 18 people with college degrees have hourly wage up to $10. Looking at the data, we can say that the average hourly wage of people with college degree will be much higher than the average hourly wage of people who did not complete high school.

Solution 2: The median wage will be the average of the wage of the 50th person and 51st person. The thick black line shows the range in which these two lie.

Let’s first look at people who have college degrees. The 50th and 51st people will have wages lying in the range $50 – $60. Answer (E)

What about people who did not complete high school? The 50th person and the 51st person lie in the range $10 – $20. Answer (A)

Solution 3: Number of people with hourly wage up to $50 is 45. Number of people with hourly wage up to $60 is 80. Hence number of people whose hourly wage lies in the range $50 – $60 is about 80 – 45 = 35. Since 100 people were surveyed, the required percentage is about 35%. Answer (B)

Ensure you understand these graphs 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!

‘Which’ vs ‘That’ Debate

Quarter Wit, Quarter WisdomI know I promised that I will bring you some tricky Integrated Reasoning questions this week, but I am really irked by the ‘which’ vs ‘that’ debate and would like to put it to rest once and for all. Hence, in this post I would like to talk about restrictive and non-restrictive clauses, about ‘which’ and ‘that’, about when to use a comma and some other such things.

First of all, it is NOT necessary that ‘which’ has to be preceded by a comma.  Just because you see a ‘which’ clause without commas, it does not mean the option is wrong.

To understand the uses of ‘which’ and ‘that’, we need to understand defining and non-defining relative clauses.

What is a Relative Clause? It is the clause that begins with a relative pronoun (who, which, that, whom, whose)! We use relative clauses to clarify which person or thing we are talking about or to add extra information about a noun.

For example:

My father, who is 70, goes running every day.

My youngest son, whose work takes him all over the world, is coming home tomorrow.

My son who works for a consultancy is coming home tomorrow.

I’m going to wear the shirt that I bought in Paris.

The relative clauses have been underlined. Note that some are surrounded by commas and some are not.

The ones that are not surrounded by commas clarify which person or thing we are talking about. These are defining relative clauses.

The ones that are surrounded by commas provide extra information about a noun. They are called non-defining relative clauses.

Defining relative clauses: They define the noun. What do we mean by that? Let’s see.

Example: My son who works for a consultancy is coming home tomorrow.

‘My son who works for a consultancy’ implies that I probably have more than one son and one of them works for a consultancy. He is the one who is coming.

Example: I’m going to wear the shirt that I bought in Paris.

‘the shirt that I bought in Paris’ defines the shirt. I have many shirts but I am going to wear the one I bought in Paris.

  • Defining relative clauses can begin with ‘who’, ‘which’ or ‘that’. You use ‘who’ or ‘that’ for people and ‘which’ or ‘that’ for things.

For example: All the sentences given below are correct.

My son who works for a consultancy is coming home tomorrow.

I’m going to wear the shirt which I bought in Paris.

I’m going to wear the shirt that I bought in Paris.

  • Also, sometimes you can omit the relative pronoun of defining relative clauses. When the relative pronoun acts as the object of the relative clause, you can omit the relative pronoun.

Example: I’m going to wear the shirt I bought in Paris.– Correct

‘shirt’ here is the object of the verb ‘bought’. The relative clause is ‘I bought the shirt in Paris.’ The relative pronoun replaces ‘the shirt’ which is the object of this clause.

  • When the relative pronoun is the subject of the relative clause, you cannot omit it.

My son who works for a consultancy is coming home tomorrow. – Incorrect

‘who’ is the subject of the verb ‘works’. You cannot omit the relative pronoun here.

Non-defining Relative Clauses: They provide extra information about the noun. In these cases, we already know the person/thing we are talking about.

Example: My father, who is 70, goes running every day.

‘My father’ clearly talks about my father (who we assume is unique). ‘who is 70’ only gives us more information about my father.

Example: My youngest son, whose work takes him all over the world, is coming home tomorrow.

My youngest son’ already clarifies that we are talking about my youngest son. ‘whose work…’ only tells us more about him.

  • Non defining relative clauses can use most relative pronouns but they cannot use ‘that’. Also, you cannot omit the pronoun.

My father, that  is 70, goes running every day. – Incorrect. Cannot use ‘that’

My father,  is 70, goes running every day. – Incorrect. Cannot work without the pronoun

We hope this clarifies the use of ‘which’ and ‘that’ and also when to use commas.

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!

IR Questions: Multi Source Reasoning

Quarter Wit, Quarter WisdomNow that we have seen some basic Integrated Reasoning question types, let’s start working on tricky Integrated Reasoning questions. The first set we would like to discuss is from GMAT Prep Software’s practice questions. This question has elements of RC, CR, PS and DS, all combined in one!

It was one of the first IR questions I had come across and I thought to myself – GMAT just got more complicated! Not because the given information was hard to understand but because there was a lot of it which needed to be analyzed together to arrive at a conclusion – not unlike our real life situations.

Anyway, let’s take a close look at this set to get a feel of what official IR questions are like.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

From our discussion of the previous few weeks, it must be apparent that this is a multi source reasoning set. There are three tabs containing data about the techniques, artifacts and budget.

A quick read of the three tabs gives us the following general idea:

Techniques – Discusses techniques of dating different types of material

Artifacts – Discusses the collection of artifacts found and the accuracy of dating them

Budget – Discusses the cost of the techniques

In this way, we have got the lay of the land – so to speak. Now we can come back to the relevant information after reading the question.

Question 1: Which one of the following pieces of information would, on its own, provide the strongest evidence that the given artifact was actually produced on Kaxna?

(A)   A radiocarbon date of 1050 BC for a wooden bowl

(B)   IRMS analysis of a necklace made from animal bones and teeth.

(C)   A TL date for a fired-clay brick that places it definitively in the period of the Kaxna Kingdom

(D)   ICP-MS analysis of a metal tool that reveals element ratios unique to a mine on Kaxna

(E)    Determination that a stone statue was found near a quarry known to produce stone statues during the Kaxna Kingdom

Solution 1: The options give us various artifacts and ask us to find the one which has the strongest evidence of being produced on Kaxna. We would probably need the data given in the first two tabs – Techniques (to tell us which technique gives is what information about the artifact) and Artifacts (to tell us the accuracy of the techniques)

Let’s look at each option:

(A)   A radiocarbon date of 1050 BC for a wooden bowl

From tab 1, we know that radiocarbon dating gives us the approximate date of the plant’s death (and hence the approximate date of the creation of the object – within two years) but it doesn’t tell us the location of the creation of the object. We don’t know whether the bowl was made on the Kaxna islands.

(B)   IRMS analysis of a necklace made from animal bones and teeth.

IRMS analysis gives us clues about the animal’s diet and mineral content in water. Knowing these ratios for Kaxna island can give us an idea of whether the artifact was created on Kaxna. But this ratio of mineral content may not be unique to Kaxna and hence we cannot say whether the artifact was in fact created on Kaxna.

(C)   A TL date for a fired-clay brick that places it definitively in the period of the Kaxna Kingdom

TL dating may place the brick in the period of the Kaxna Kingdom but it doesn’t tell us the location where it was created. It may not have been created on the Kaxna island.

(D)   ICP-MS analysis of a metal tool that reveals element ratios unique to a mine on Kaxna.

ICP-MS analysis reveals ratios unique to a mine on Kaxna. This means the tool must have come from that mine on Kaxna. We can say with fair bit of certainty that the artifact came from Kaxna. Hence this seems to be the correct option. Let’s still take a look at option (E) too.

(E)    Determination that a stone statue was found near a quarry known to produce stone statues during the Kaxna Kingdom

Just because the stone statue was found near the quarry, it doesn’t mean that it was produced in the quarry. Hence we cannot say that the statue belongs to the Kaxna period.

Answer (D)

We need to understand the given data really well to be able to answer this question. Let’s look at another question which uses the third Tab now.

Question 2: For each of the following combinations of Kaxna artifacts, select Yes if, based on the information provided, the cost of all pertinent techniques described can be shown to be within the museum’s first-year Kaxna budget. Otherwise, select No.

Yes                 No

O                     O             2 bone implements and 5 fired-clay cups decorated with gold

O                     O             7 wooden statues and 20 metal implements

O                     O             15 wooden statues decorated with bone

Solution 2: This question is similar to the Two-Part Analysis questions we have seen before. You have to answer three parts correctly here to get the right answer.

To answer this one, we need the first and the third tabs – Techniques to match the artifacts with the technique and Budget to get the cost of the technique. Let’s find out.

Total Budget: Unlimited IRMS + $7000 (4TL + 15RC OR 40 ICP-MS)

Note that we have no idea about the costs of TL, RC, ICP-MS tests. We don’t know how their costs are related either.

1. 2 bone implements and 5 fired-clay cups decorated with gold

Bone implements need IRMS (unlimited available).

Fired clay objects need TL technique tests (5).

We know that $7000 is enough for 4 TL tests but we don’t know whether it is enough for 5 TL tests. E.g. each TL may cost $1500 or it may cost $150 – we don’t know. Hence, we cannot say that the cost is within the budget.

2. 7 wooden statues and 20 metal implements

Plant matter needs RC dating tests (7).

Metals need ICP-MS  tests (20).

We know that $7000 is enough for 40 ICP-MS tests so we know that 20 ICP-MS tests will cost less than or equal to $3500.

We also know that $7000 is enough for 4TL + 15RC tests which means $3500 is enough for 2TL + 7.5 RC tests. So we can be certain that $3500 is enough for 7 RC tests.

Therefore, the 7 RC tests and the 20 ICP-MC tests can be done within the $7000 budget.

3. 15 wooden statues decorated with bone

Plant matter needs RC dating tests (15)

Animal bones need IRMS tests which happen in-house.

We know that $7000 is enough for 4TL + 15RC tests which means it is certainly enough for 15RC tests.

Therefore, the 15 RC tests and 15 IRMS tests can be done within the $7000 budget.

So we will click on ‘No’ in the first row, ‘Yes’ in the second row and ‘Yes’ in the second row.

We hope you realize that IR questions may seem hard but when you get down to it, they are pretty much like the other question types with which we are already very familiar.

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!

Critical Reasoning: Some Common Mistakes

Quarter Wit, Quarter WisdomNow that we have seen some basic Integrated Reasoning question types, we will look at some tricky questions but not this week. This week, we would like to discuss a Critical Reasoning question. This question is simple and straight forward but still many people falter in it. The reasons for this are not hard to find. Let’s analyze this question in detail.

Question:

People often criticize their local government for not providing enough funds to the public libraries in their district. They complain of lacking infrastructure and out-of-date and worn out reading material. Surprisingly, the most frequent and vociferous complaints come from those who live in districts where the libraries are most well maintained and kept current.

All of the following, considered individually, help to explain the apparent paradox EXCEPT:

(A) People from districts of well maintained libraries are more likely to use the public libraries.
(B) People have no knowledge of the facilities and infrastructure provided by the other libraries in their district.
(C) Good facilities cause people’s expectations to rise leading them to demand even more.
(D) The people in districts with well maintained libraries are likely to complain when the library they use is not as well maintained as the other libraries in that district.
(E) Most complaints about libraries come from political activists, most of who live in districts with well maintained libraries.

Solution:

The first thing to note here is that it is an ‘explain the paradox’ question but with an ‘EXCEPT’. This means that four of the five options will explain the paradox and our answer will be the one which will NOT. Test takers are often used to looking for the option that does explain the paradox and hence select an option which does this well. They often forget that they were actually required to do the opposite.  This is the first reason why test takers answer this question incorrectly.

Let’s try to understand the argument now:

People blame their local government for not maintaining their public libraries. The surprising thing is that most of these people come from districts with most well maintained libraries. You would expect that people living in districts with better libraries will have less to complain about and would be happier! Hence, here is the paradox.

Let’s do some pre-thinking now. How can you explain this paradox?

Some reasons come to mind: People who visit well maintained libraries have even higher expectations. They visit the library more and expect more out of it. If they find out that their friends in the same district have a library with great facilities, they become unhappy with their own library etc.

Let’s look at the options now:

(A)   People from districts of well maintained libraries are more likely to use the public libraries.

People who visit a place more are more aware of its follies. They get used to the amenities and worry about facilities that are not available. Also, if a library is frequented by many people, its books, videos and infrastructure in general will wear out faster. Another thing that could explain the paradox is that if more people visit the library, the probability of some people complaining about it is higher. Hence this option definitely explains why these people may complain more.

(B) People living in districts with good libraries have no knowledge of the facilities and infrastructure provided by the other libraries in their district.

This does not help explain our paradox. It doesn’t matter whether they know about the facilities provided by other libraries. If they do know the status of other libraries in their district, their behavior might be different but if they do not know, it has no effect on their behavior. Knowing may make them complain less or even more.Not knowing doesn’t make them complain more. In fact, if they do know about the facilities provided by other libraries in their district, they might start complaining even more if they like the facilities of other libraries more. It doesn’t help explain why they complain a lot right now.

(C) Good facilities cause people’s expectations to rise leading them to demand even more.

This option tells us that people get used to amenities and start expecting even more. This explains the paradox.

(D) The people in districts with well maintained libraries are likely to complain when the library they use is not as well maintained as the other libraries in that district.

This helps explain the paradox too. If many people in their district are getting better facilities than them, they might complain about the not equally good or better facilities available at their own library.

(E) Most complaints about libraries come from political activists, most of who live in districts with well maintained libraries.

This option is the most frequently chosen incorrect answer. Test takers reason that it is out of scope since it doesn’t matter who lives where. Actually, this explains the paradox too. Think about it – the option doesn’t talk about doctors or hair dressers; it talks about political activists. Political activists are usually very vocal and critical of things around them. These are the people who complain about everything and will complain about their library too. They live in districts with well maintained libraries and visit them. Since they visit well maintained libraries, many complaints will come from districts with well maintained libraries.

Also note that most does not necessarily imply just a little more than 50%. It could imply 80%, 90% etc too. Some test takers feel that 51% complaints come from political activists and more than half of them (i.e. approx. 26% of total) live in districts with well maintained libraries. This accounts for only 26% complaints and hence doesn’t explain the paradox.  But ‘most’ could just as easily imply 90% and hence this option certainly helps explain the paradox.

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 – Two Part Analysis Questions

Quarter Wit, Quarter WisdomLet’s continue our series and look at another Integrated Reasoning question type today – two part analysis. As complicated as it sounds, it’s actually the simplest of the IR question types in my opinion. The reason for this is that it tests no new skills; it checks your ability to handle the same old PS and CR questions.

The only reason it is new is that it reduces the probability of guessing correctly and it puts more time pressure on you! Your probability of guessing correctly is 20% in PS/CR questions; it goes down to 4% in two part analysis because you have to guess correctly twice. As for time pressure, you get about 2 mins for every PS and about 1.5 mins for every CR question. For each part of two part analysis, you have only 1.25 mins.

Anyway, let’s look at a sample question to get familiar with this question type.

Question: A grocery store sells fruits in pre packed closed bags such that individual pieces of fruit are not sold. Mangoes are sold at the rate of $5 per bag (each bag contains two mangoes) and apples at the rate of $8 per bag (each bag contains five apples). During a particular day, the store started with some mangoes and apples and sold them all by the end of the day. The revenue at the end of the day from selling mangoes and apples that day was $128. Which of the following could represent the number of mangoes and the number of apples that were in the store at the beginning of that day?

Choose only one from each column:

 

 

Solution:

Note that it is a PS question. Only the format of the question is different. Also, the use of the word ‘could’ in the question stem suggests that there could be multiple solutions to this problem. Let’s take a closer look at how to solve it.

Say, number of bags of mangoes is ‘m’ and number of bags of apples is ‘p’.

Then 5m + 8p = $128 (total revenue)

Each bag of mangoes has 2 mangoes and each bag of apples has 5 apples.

So number of mangoes sold = 2m (to be selected in the first column)

Number of apples sold = 5p (to be selected in the second column)

We need to solve for this equation: 5m + 8p = 128

It is easy to see that one solution to this equation is m = 0, p = 16. The next solution will be m = 8, p = 11. Another will be m = 16, p = 6. Yet another will be m = 24, p = 1. If you are wondering how we are landing on one solution after another so effortlessly, you need to check out a previous post of QWQW – Integral Solutions to Equations in Two Variables.

So there are three solutions possible to our question: Which of the following could represent the number of mangoes and the number of apples that were in the store at the beginning of that day?

There are three different cases possible:

Case 1: Number of mangoes sold could be 16 (= 2m when m is 8). In that case number of apples sold will be 55 ( = 5p when m is 8, p is 11)

Case 2: Number of mangoes sold could be 32 (= 2m when m is 16). In that case number of apples sold will be 30 ( = 5p when m is 16, p is 6)

Case 3: Number of mangoes sold could be 48 ( = 2m when m is 24). In that case number of apples sold will be 5 (= 5p when m is 24, p is 1)

The case we select should be that of which both numbers are included in the options. Case 3 satisfies this condition. So we select 48 in the first column and 5 in the second column.

This is the only ‘exotic’ step of the two part analysis. The rest of the question is just like any other PS 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!

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 = 1: sqrt(3) = r : a/2.

Therefore, a = 2 * sqrt(3) * r

Side of the triangle = 2 * sqrt(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 (sqrt(3)/2)*a. The incenter is at a distance 2/3rd of the altitude so OD (radius) = (1/3)* (sqrt(3)/2)*a = a/2*sqrt(3)

Or Side of the triangle = 2*sqrt(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: sqrt(3)

Hence, a = 2r/sqrt(3)

Side of the hexagon = (2/sqrt(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!