Two points, N and Q (not shown), lie to the right of point M on line ℓ

Hello pclawong,

It is not possible for the line to be Q-----M------------N. Here's why.

Stem says that N and Q are to the the right of M. Two cases are possible

Case 1: M-----------Q---------------N

Case 2: M-----------N---------------Q

Lets see which of the above two cases holds true

S1 says - Twice the length of MN is 3 times the length of MQ. Algebraically this means

\(2MN = 3MQ\)

\(MN = \frac{3}{2}* MQ\)

Since \(\frac{3}{2}\) is a number greater than 1, we get

\(MN > MQ\)

This clearly means the Line HAS to look like this M-------Q-----------N

Do let me know in case I was unable to give an answer.

Ans is C , as we do not know the exact positions of N and Q , in the question it is given that both lie to the right of M however we do not know whether N lie to the right of Q or Q lie to the right of N , so without knowing the exact position of N and Q we cannot ans the question. Therefore both statements are necessary for the conclusion.

From statement 1 we know that "Twice the length of MN is 3 times the length of MQ" -> 2MN = 3MQ it means that MN>MQ -> hence the position of Q is between M and N.
ans:A

I solved this by picking numbers (I know it may look like a overkill on DS but it helped me to know the positions of N and Q )

S1:

I see 2 and 3 in the statement and pick a Number 6 for MN.
2(MN) = 3(MQ)
2(6) = 3(MQ) => MQ = 4
MQ=4 and MN=6 => QN = 2
QN/MQ = 2/4 = 1/2

S1 Suff.

S2:

Insuff. since we don't know the exact location of Q

Answer : A

Statement 1:
\(2(MN)=3(MQ)\)
\(\frac{2}{3}=\frac{MQ}{MN}\)

Let \(MQ=2d\) and \(MN=3d\).
The result is the following number line:
M<--2d-->Q<--d-->N

Thus:
\(\frac{QN}{MQ}=\frac{d}{2d}=\frac{1}{2}\)
SUFFICIENT.

Statement 2:
No information about any lengths.
INSUFFICIENT.

Hi Genoa2000,

DS questions are interesting because they're built to 'test' you on a variety of skills (far more than just your 'math' skills), including organization, accuracy, attention-to-detail, thoroughness and the ability to prove that your answer is correct. DS questions also have no 'safety net' - meaning that if you make a little mistake, then you will convince yourself that one of the wrong answers is correct.

If you choose to select an answer after just reading the information in Fact 2 (and not do any work to PROVE what the correct answer is), then one of two outcomes is likely... You might understand the information perfectly and have the correct answer.... OR.... you have made some type of silly/little mistake and you are about to let some easy points slip away. Doing work "in your head" is the WORST way to approach a GMAT question - so the more often you take that approach, the more likely you lose some 'gettable' points on Test Day.

GMAT assassins aren't born, they're made,
Rich

Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and independent equations ensures a solution.

Two points, N and Q (not shown), lie to the right of point M on line ℓ. What is the ratio of the length of QN to the length of MQ?

(1) Twice the length of MN is 3 times the length of MQ.
(2) Point Q is between points M and N.



------|---------------------------------- line L

M

When you modify the original condition and the question, the que is ratio of QN:MQ, which is for 1) ratio and for 2) number(equation). In a case like this, it is most likely that ratio is the answer.

In 1),

------|--------|--------------------------|----- line L

M Q N

If MQ=2d, 2MN=3*2d, MN=3d and QN=d. Therefore, the que, QN:MQ=d:2d=1:2 is derived, which is unique and sufficient.
Therefore, the answer is A.


-> Once we modify the original condition and the question according to the variable approach method 1, we can solve approximately 30% of DS questions.

Point Q is between points M and N does NOT mean that QM = QN. If it were so, the answer would be D, not A.

Video solution from Quant Reasoning:
Subscribe for more: https://www.youtube.com/QuantReasoning? ... irmation=1

Hi eshan333,

While you ARE correct that we don't know the exact positions of N and Q, the question does NOT ask us for them (so you have to be careful about when you choose to stop working). The prompt asks for the RATIO of two lengths, NOT the exact measure of either of them. With a bit of 'playing around' and TESTing VALUES, you might find that you change your answer.

GMAT assassins aren't born, they're made,
Rich

Can you please explain how you go from "2MN = 3MQ" to "2MQ + 2QN = 3MQ"? Thank you.

how can we know Q is between MN,
Can it be like Q-----M------------N ? then MN still longer than QM.

Hello guys ,
a prompt question does "Point Q is between points M and N" mean that QM=QN , or is Q just at a random place between M and N ?

Solution:

Question Stem Analysis:

We need to determine the value of QN/MQ. We can let MQ = x, QN = y. If Q is between M and N, then MN = x + y. If N is between M and Q, then MN = x - y. In any case, we need to determine the value of y/x.

Statement One Alone:

With statement one, we see that N is further to the right of M than Q is. Therefore, Q is between M and N. In that case, MN = x + y and MQ = x, and we have:

2(x + y) = 3x

2x + 2y = 3x

2y = x

y/x = 1/2

Statement one alone is sufficient.

Statement Two Alone:

Knowing only that Q is between M and N is not sufficient to answer the question. For example, if Q is exactly the midpoint of M and N, then the ratio QN/MQ = 1. However, if Q is closer to M than it is to N, then the ratio is greater than 1.

Answer: A

Answer: Option A

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ScottTargetTestPrep
For the math

"2y = x

y/x = 1/2"

To confirm my understanding, I have a few questions:

Question 1: When you divide by x there is a 1 left on the right hand-side of the equal sign, correct? so it is 2y=x --> divide by x --> 2y/x=1 --> divide by 2= --> 1/2=y/x
Question 2: Why can you just divide by 2 when it is grouped with y/x on the left-hand side? In other words, for 2y/x doesn't the division of x apply to both the 2 and the y?

Thank you for your help.

For your first question, we do get a 1 on the right hand side after we divide each side by x. We can obtain y/x = 1/2 from 2y = x in two steps, like the way you showed; or we can simply divide each side of 2y = x by 2x to obtain y/x = 1/2 in a single step.

For your second question, the division of x does not apply to both 2 and y. You must be thinking of the case where we divide a summation by some number, for instance, (2 + y)/x. In this case, the division indeed applies to both 2 and y, so we can rewrite this expression as 2/x + y/x. However, 2y/x is equal to (2/x) * y or 2 * (y/x), but not (2/x) * (y/x) (which you can verify simply by observing that the product of 2/x and y/x is not equal to 2y/x).