How to Go From a 48 to 51 in GMAT Quant

By Karishma, Veritas Prep

Another method of saving time on simple questions – use data given in one statement to examine the other!

Now you might think we have lost it! After all, you know very well that in Data Sufficiency questions of GMAT, you must examine each statement independently. You CANNOT use data from one to analyze the other – absolutely correct. So you should ignore the other statement completely while examining one – hmm, not necessarily!

Sometimes, one statement could give us ideas about the next one such that we could save time while examining it. Needless to say, we need to be very careful but it certainly is a useful strategy. Also, it could help us verify that our calculations are correct. Here is why…

When we say DS question, think of a puzzle. The question stem gives you the statement of a puzzle ending with something like “What is the value of x?” or “Is x 7?” etc. You have to answer the question asked in the puzzle. Think of the two statements that come with the question as clues to the puzzle. So the puzzlemaster gives you the first clue (statement 1) and asks you: can you answer the question now? If you are able to, your answer is either (A) or (D).

Then he tells you to ignore the first clue and gives you another clue (statement 2). Again he asks you: can you answer the question now? Again, you may or may not able to. If you are able to, your answer will be (B) or (D) depending on how you fared in statement 1. If you are unable to answer the question, he tells you to consider both statements together and then try to answer. If you are able to, your answer is (C).

The point to note here is that both clues lead you to answer the same puzzle. Say if the puzzle is: What is x? If clue 1 tells you that x is 6, clue 2 cannot tell you that x is 9. They both must lead you to the same value of x. Clue 1 could tell you that x is either 6 or 8 and clue 2 could tell you that x is either 8 or 9. In this case, when we use both clues together, we find that x must be 8 to satisfy both. Hence the statements never contradict each other. This means, if we get possible values of x from statement 1, we know that statement 2 will also give us at least one of those values.

This is how one statement could give us a starting point for the next one. Now that you understand the “why”, let’s go on to “how”, using a question.

Question: If K is a positive integer less than 10 and N = 4,321 + K, what is the value of K?

Statement 1: N is divisible by 3
Statement 2: N is divisible by 7

Solution:

Given: \(N = 4321 + K\)

\(1 \leq K \leq 10\)

So N could range from 4322 (when \(K = 1\)) to 4331 (when \(K = 10\)). To find the value of K, we need to find the unique value of N.

Statement 1 tells us that N is divisible by 3.

4321 is not divisible by 3 since the sum of its digits is 4+3+2+1 = 10. It is 1 more than a multiple of 3. So the next multiple of 3 will be 4323. Hence N could be 4323. But there are some other multiples of 3 which could be the value of N. After 4323, 4326 and 4329 could also be the values of N since they are multiples of 3 too. We know this because if A is a multiple of 3, A+3, A+6, A+9, A-3, A-6 etc are also multiples of 3. So since 4323 is a multiple of 3, 4326 and 4329 will also be multiples of 3. We did not get a unique value for N so statement 1 alone is not sufficient.

Now let’s go on to statement 2. This tells us that N must be a multiple of 7. In 10 consecutive numbers, there will be either one multiple of 7 or two multiples of 7. If there is only one multiple of 7 in the range 4322 to 4331, statement 2 alone will be sufficient to give us the value of N. If there are two multiples of 7 in this range, then statement 2 alone will not be sufficient.

Recall that from statement 1, we already know that N will take one of three values: 4323, 4326 or 4329.

Let’s check for 4326 because it is in the middle. If 4326 is divisible by 7, there will be no other multiple of 7 in the range 4322 to 4331 because the closest multiples of 7 to 4326 will be 4326 – 7 and 4326 + 7. When we divide 4326 by 7, we find that it is divisible. This means that statement 2 gives us a single value of N. Hence statement 2 alone is sufficient.

Hypothetically, what if we had found that 4326 is not divisible by 7? Then we would have known that either 4323 or 4329 must be a multiple of 7. In both cases, statement 2 would have given us 2 multiples of 7 because both 4330 (7 more than 4323) and 4322 (7 less than 4329) are in the possible range. Then we would have known that the answer will be (C) i.e. we will need both statements to answer the question since the possible values from the two statements will have only one overlap in either case.

Note that what we gleaned from statement 1 helped us quickly examine statement 2 and get to the answer right away. But this is an advanced technique and you should use it only if you understand it very well. Else, it is best to stick to completely ignoring one statement while working on the other.

This question is discussed HERE.

Other Articles On This topic:

Improving from 30-35 to 40-44
Improving from 44 to 50

Other discussions dedicated to this issue:

How about taking number of students at A as 400 and at B as 200 no. of boys will be 240 and 80 respectively hence the difference after transfer will be 220-100=110
How about taking number of students at A as 100 and at B as 50 no. of boys will be 60 and 20 respectively hence the difference after transfer will be 40-40=0

it's actually tough to get ratio as 5:3 after transfer if it doesn't set well there are chances of panic during the exam.

I have seen with my own experience and from expert's views on various forums that one can get a score of Q50 while still missing 10-15 questions. However, with regards to Q51 I have not seen such more or less conclusive opinion. I do remember reading a post on BTG (I think by a chicago based tutor) that one can't hope to score Q51 while answering more than 5 questions wrong.
Do the experts agree yet on the following?
"How many question one can't afford to get wrong while still hoping to score Q51?"
Thanks!

By Karishma, Veritas Prep

Today, we bring another tip for you to help get that dream score of 51 – if you must write down the data given, write down all of it! Let us explain.

If you think that you will need to jot down the data given in the question and then solve it on your scratch pad (instead of in your mind), you must jot down every single detail. It is easy to overlook small things which are difficult to express algebraically such as ‘x is an integer’. These details are often critical and could make all the difference between an ‘unsolvable’ question and a ‘solvable within 2 minutes’ one. Once you start solving the question on your scratch pad, you will not refer back to the original question again and again and hence might forget these details. Have them along with the rest of the data. Read every word of the question carefully, and ensure that it is consolidated on your scratch pad. For example, look at this question:

A set of five positive integers has an arithmetic mean of 150. A particular number among the five exceeds another by 100. The rest of the three numbers lie between these two numbers and are equal. How many different values can the largest number among the five take?

It is a difficult question because it incorporates statistics as well as max-min – both tricky topics. On top of it, people often overlook the ‘are equal’ part of the question here. The reason for that is that they are actively looking for implications of the sentences and the moment they read “The rest three numbers lie between these two numbers”, they go back to the previous sentence which tells us “A particular number among the five exceeds another by 100”. They then make a note of the fact that 100 is the range of the five positive integers. In all this excitement, they miss the three critical words “and are equal”. Ensure that when you go to the sentence above, you pick the next sentence from the point where you left it. Another thing to note here is that all numbers are positive integers. This information will be critical to us.

Let’s demonstrate how you will solve this question after incorporating all the information given.

Question: A set of five positive integers has an arithmetic mean of 150. A particular number among the five exceeds another by 100. The rest of the three numbers lie between these two numbers and are equal. How many different values can the largest number among the five take?

(A) 18
(B) 19
(C) 21
(D) 42
(E) 59

Solution:

Let’s assume that the 5 natural numbers in increasing order are: a, b, b, b, a+100

We are given that \(a < b < a+100\).

Also, we are given that a and b are positive integers. This information is critical – we will see later why.

The average of the 5 numbers is \(\frac{(a+b+b+b+a+100)}{5} = 150\)

\((a+b+b+b+a+100) = 5*150\)

\(2a+3b = 650\)

We need to find the number of distinct values that a can take because a+100 will also take the same number of distinct values.

Now there are two methods to proceed. Let’s discuss both of them.

Method 1: Pure Algebra – Write b in terms of a and plug it in the inequality

\(b = \frac{(650 – 2a)}{3}\)

\(a < \frac{(650 – 2a)}{3} < a+100\)

\(3a < 650 – 2a < 3a + 300\)

Now split it into two inequalities: \(3a < 650 – 2a\) and \(650 – 2a < 3a + 300\)

Inequality 1: \(3a < 650 – 2a\)

\(5a < 650\)

\(a < 130\)

Inequality 2: \(650 – 2a < 3a + 300\)

\(5a > 350\)

\(a > 70\)

So we get that \(70 < a < 130\). Since a is an integer, can we say that a can take all values from 71 to 129? No. What we are forgetting is that b is also an integer. We know that

\(b = \frac{(650 – 2a)}{3}\)

For which values will be get b as an integer? Note that 650 is not divisible by 3. You need to add 1 to it or subtract 2 out of it to make it divisible by 3. So a should be of the form 3x+1.

\(b = \frac{(650 – 2*(3x+1))}{3} = \frac{(648 – 6x)}{3} = 216 – 2x\)

Here, for any positive integer x, b will be an integer.

From 71 to 129, we have the following numbers which are of the form 3x+1:

73, 76, 79, 82, 85, … 127

This is an Arithmetic Progression. How many terms are there here?

Last term = First term + (n – 1)*Common Difference

\(127 = 73 + (n – 1)*3\)

\(n = 19\)

a will take 19 distinct values so the last term i.e. (a+100) will also take 19 distinct values.

Method 2: Using Transition Points

Note that \(a < b < a+100\)

Since a < b, let’s find the point where a = b, i.e. the transition point

\(2a + 3a = 650\)

\(a = 130 = b\)

But b must be greater than a. If we increase b by 1, we need to decrease a by 3 to keep the average same. But decreasing a by 3 decreases the largest number i.e. a+100 by 3 too; so we need to increase b by another 1.
We get \(a = 127\) and \(b = 132\). This give us the numbers as 127, 132, 132, 132, 227. Here the average is 150

Since \(b < a+100\), let’s find the point where \(b = a+100\)

\(2a + 3(a+100) = 650\)

\(a = 70\), \(b = 170\)

But b must be less than a+100. If we decrease b by 1, we need to increase a by 3 to keep the average same. But increasing a by 3 increases the largest number, i.e. a+100 by 3 too, so we need to decrease b by another 1.

We get \(a = 73\) and \(b = 168\). This gives us the numbers as 73, 168, 168, 168, 173. Here the average is 150

Values of a will be: 73, 76, 79, ….127 (Difference of 3 to make b an integer)

This is an Arithmetic Progression.

Last term = First term + (n – 1)*Common difference

\(127 = 73 + (n – 1)*3\)

\(n = 19\)

a will take 19 distinct values so the last term i.e. (a+100) will also take 19 distinct values.

Answer (B) This question is discussed HERE.

By Karishma, Veritas Prep

Today’s post the next part in our “How to Go From 48 to 51 in Quant” series. Again, we will learn a technique that can be employed by the test-taker at an advanced stage of preparation – requiring one to understand the situations in which one can use this simplifying technique.

We all love to use the plug-in method on GMAT Quant questions. We have an equation given, and if the answer choices are the possible values of x, we just plug in these values to find the one that satisfies the equation.

But what if the answer choices are all complicated values? What if it seems that five times the calculation (in the worst case) will be far more time consuming than actually solving the given equation? Then one is torn between using the favorite plug-in method and using algebra. Let’s take an example to review the methods we can use to solve the question and learn how to simplify the plug-in process by approximating the five available options:

Question: If |4x−4|=|2x+30|, which of the following could be a value of x?

(A) –35/3
(B) −21/2
(C) −13/3
(D) 11/5
(E) 47/5

Solution: This question is an ideal candidate for the “plug-in” method. Here, you have the absolute value equation with the potential values of x given in the answer choices. The problem is that the values of x given are fractional. Of course, if we do plan to solve the equation rather than “plug-in”, we can still solve it using our holistic approach rather than pure algebra. Let’s take a look at that now, and later we will discuss the trick to making the answer choices easier for us to plug in.

Method 1:

\(|4x – 4| = |2x + 30|\)

\(4 * |x – 1| = 2 * |x + 15|\)

\(2 * |x – 1| = |x + 15|\)

This is how we rephrase the equation in our words: twice the distance of x from 1 should be equal to the distance of x from -15.

——————(-15) —————————————————(0)——(1)——————

There are two ways to find the value of x:

Case 1: x could be between -15 and 1 such that the distance between them is split in the ratio 2:1.

or

Case 2: x could be to the right of 1 such that the distance between x and -15 is twice the distance between x and 1.

Let’s examine both of these cases in further detail:

Case 1: The distance from -15 to 1 is of 16 units – this can be split into 3 sections of 16/3 units each. So, the distance of x from 1 should be 16/3, which would make the distance of x from -15 two times 16/3, i.e. 32/3.

So, x should be at a point 16/3 away from 1 toward the left.

\(x = 1 – \frac{16}{3} = -\frac{13}{3}\)

This is one of our answer choices and, hence, the correct answer. Normally, we would just move on to the next question at this point, but had we not found -13/3 in the answer options, we would have moved on to Case 2:

Case 2: The distance between -15 and 1 is 16 units. x should be an additional 16 units to the right of 1, so the distance between x and 1 is 16 and the distance between x and -15 is two times 16, i.e. 32. This means that x should be 16 units to the right of 1, i.e. x = 17. If you would not have found -13/3 in the answer choices, then you would have found 17.

Now let’s move on to see how we can make the plug-in method work for us in this case by examining each answer choice we are given:

Method 2:

\(|4x – 4| = |2x + 30|\)

\(2 * |x – 1| = |x + 15|\)

(A) -35/3

It is difficult to solve for \(x = -\frac{35}{3}\) to see if both sides match. Instead, let’s solve for the closest integer, -12.

\(2 * |-12 – 1| = |-12 + 15|\)

On the left-hand side, you will get 26, but on the right-hand side, you will get 3.

These values are far away from each other, so x cannot be -35/3. As the value of x approaches the point where the equation holds – i.e. where the two sides are equal to each other – the gap between the value of the two sides keeps reducing. With such a huge gap between the value of the two sides in this case, it is unlikely that a small adjustment of -35/3 from -12 will bring the two sides to be equal.

(B) -21/2

For this answer choice, let’s solve for the nearest integer, x = -10.

\(2 * |-10 – 1| = |-10 + 15|\)

On the left-hand side, you will get 22; on the right-hand side, you will get 5.

Once again, these values are far away from each other and, hence, x will not be -21/2.

(C) -13/3

For this answer choice, let’s solve for x = -4.

\(2 * |-4 -1| = |-4 + 15|\)

On the left-hand side, you will get 10; on the right-hand side, you will get 11.

Here, there is a possibility that x can equal -13/3, as the two sides are so close to one another – plug in the actual value of -13/3 and you will see that the left-hand side of the equation does, in fact, equal the right-hand side. Therefore, C is the correct answer. This question is discussed HERE.

Basically, we approximated the answer choices we were given and shortlisted the one that gave us very close values. We checked for that and found that it is the answer.

We can also solve this question using pure algebra (taking positive and negative signs of absolute values) but in my opinion, the holistic Method 1 is almost always better than that. Out of the two methods discussed above, you can pick the one you like better, but note that Method 2 does have limited applications – only if you are given the actual values of x, can you use it. Method 1 is far more generic for absolute value questions.

By Karishma, Veritas Prep

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

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

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

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

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

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

Let’s take an official example:

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

Statement 1: She has $282 worth of packages.

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

Solution:

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

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

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

So, the test-taker will start working on finding the first solution. We are told:

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

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

Statement 1: She has $282 worth of packages

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

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

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

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

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

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

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

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

Hi Bunuel,

Thanks for this! Very very helpful.

In the 7th part - could you please explain this - 'This is where the 51 level scorer stops because they never lose sight of the big picture. The “lightbulb” switches on, and now he or she knows that there will be only one set of values that can satisfy this equation. Why? Because y will be less than 17 in the first set of values that satisfies this equation. So if we want to get the next set that satisfies, we will need to subtract y by 17 (and add 13 to x), which will make y negative.' ??

Additionally, could you please help me with my approach? Whether I could use this across questions?

Statement 1: 17x + 13y = 282

-> Here, without actually calculating the values for x and y, we can establish that this equation must have only one solution since:
Statement 2 provides us with x=2y => x-2y = 0 [ Without using this statement, we know that this statement must be true ] and since, (a/b) =/= (c/d) => eq 1 will only have one such solution.

And so, the answer is A. Please let me know !

What does this mean? I don't follow.

Dear VeritasPrepKarishma,

My question is about highlighted portion. How can we be sure about these extreme points in fractions with variables like this one?

\(\frac{y}{x+y}\)

Since both x and y are positive, then the denominator is greater than numerator. So, \(\frac{y}{x+y}<1\).

Next, if x were equal to y, then we'd have y/(2y) = 1/2 but x < y, so denominator is less than 2y, which makes the fraction greater than 1/2.

Hope it's clear.

I am assuming your question is about this:

Now since y is greater than x, y/(x + y) will be more than 1/2 but definitely less than 1 (x and y are positive numbers).


Had both x and y been equal, say x = 3, y = 3, then y /(x+y) = 3/6 (= exactly 1/2)

But in our question, x and y are positive numbers such that x < y. Say x = 2 and y = 3
x + y = 5 (in this sum, y has contributed more than x)
y/(x + y) = 3/5 (will be more than 1/2 since y's contribution is more than half of the sum)
But since x is a positive number, x+y will be greater than y so y/(x+y) will be less than 1.

P.S.- Have already written the post so might as well put it up though Bunuel has already explained it.

Hello Bunnuel,


Can you please explain that part ? I do not understand what you mean. Why? Because y will be less than 17 in the first set of values that satisfies this equation. So if we want to get the next set that satisfies, we will need to subtract y by 17 (and add 13 to x), which will make y negative.


Thank you very much for your work

I dont get
Why? Because y will be less than 17 in the first set of values that satisfies this equation. So if we want to get the next set that satisfies, we will need to subtract y by 17 (and add 13 to x), which will make y negative.

what does it mean that y will be less than 17 in the first set of values ..
how did i conclude this

thanks

Pmista
As x increases, y is decreasing (because they both add up to make 282)
So as we keep increasing x to 4/5/6/7... y will keep getting smaller and smaller. Since when x = 3, y = 17.something, as x increases, y will be less than 17.

Say we get x = 5, y = 12 as the first solution (just assuming).
The next solution will be x = 5+13 and y = 12 - 17 (a negative number)
Hence, only one solution is possible.

VeritasKarishma

isn't possible that there are no integral solutions, or should I take for a fact the statement (1) that she has bought corn and rice for 282 usd

You are given that she has bought corn and rice packets for $282. Obviously, you cannot break a packet. Hence there has to be at least one intgeral solution.