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Re: If m, p, s and v are positive, and m/p < s/v, which of the following [#permalink]

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22 Sep 2013, 07:31

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imhimanshu wrote:

If m, p, s and v are positive, and \(\frac{m}{p} <\frac{s}{v}\), which of the following must be between \(\frac{m}{p}\) and \(\frac{s}{v}\)

I. \(\frac{m+s}{p+v}\) II. \(\frac{ms}{pv}\) III. \(\frac{s}{v} - \frac{m}{p}\)

A. None B. I only C. II only D. III only E. I and II both

Given that \(\frac{m}{p}<\frac{s}{v} \to \frac{m}{s}<\frac{p}{v}\)Adding 1 on both sides we have \(\frac{m+s}{s}<\frac{p+v}{v} \to \frac{m+s}{p+v}<\frac{s}{v}\)

Again,\(\frac{m}{p}<\frac{s}{v} \to \frac{v}{p}<\frac{s}{m}\) Adding 1 on both sides, we have \(\frac{v+p}{p}<\frac{s+m}{m} \to \frac{m}{p}<\frac{s+m}{v+p}\) . Thus, I is always true. We just have to check for Option II now.

Assming it to be true, we should have\(\frac{ms}{pv}<\frac{s}{v} \to \frac{m}{p}<1\) Which is not always true. Thus the answer is B.
_________________

Re: If m, p, s and v are positive, and m/p < s/v, which of the following [#permalink]

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23 Sep 2013, 00:37

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igotthis wrote:

How can I solve this using numbers?

You can take numbers such 1,2,3 and 4 as m,p,s and v find the value between m/p and s/v try to enter the numbers in the answer choices. only option B will satisfy .

If m, p, s and v are positive, and \(\frac{m}{p} <\frac{s}{v}\), which of the following must be between \(\frac{m}{p}\) and \(\frac{s}{v}\)

I. \(\frac{m+s}{p+v}\) II. \(\frac{ms}{pv}\) III. \(\frac{s}{v} - \frac{m}{p}\)

A. None B. I only C. II only D. III only E. I and II both

Responding to a pm:

You can work on this question using some number line and averaging concepts. Let's look at statement II and III first since they are very easy.

We know \(\frac{m}{p} <\frac{s}{v}\)

On the number line: .............0....................m/p ........................s/v (since m, p, s and v are all positive so m/p and s/v are to the right of 0)

II. \(\frac{ms}{pv}\) Think of the case when m/p and s/v are both less than 1. When you multiply them, they will become even smaller. Say .2*.3 = .06. So the product may not lie between them.

III. \(\frac{s}{v} - \frac{m}{p}\) Think of a case such as this: .............0..............................m/p .......s/v \(\frac{s}{v} - \frac{m}{p}\) will be much smaller than both m/p and s/v and will lie somewhere here: .............0.......Here...........................m/p .......s/v So it needn't be between them.

Now only issue is (I). You can check some numbers for it including fractions and non fractions. Or try to understand it using number line. Think of 4 numbers as N1, N2, D1, D2 for ease and given fractions as N1/D1 and N2/D2.

\(\frac{m+s}{p+v} = \frac{m+s/2}{p+v/2}\) = (Avg of N1 and N2)/(Avg of D1 and D2)

Now numerator of avg will lie between N1 and N2 and denominator of avg will lie between D1 and D2. So Avg N/Avg D will lie between N1/D1 and N2/D2. Try to think this through.

If N1/D1 < N2/D2, it could be because N1 < N2 and D1 = D2. So AvgN will lie between N1 and N2 and AvgD = D1 = D2. It could also be because N1 < N2 and D1 > D2. AvgN will be larger than N1 but smaller than N2. AvgD will be smaller than D1 but greater than D2 so AvgN/AvgD will be greater than N1/D1 but smaller than N2/D2. It could also be because N1 << N2 and D1 < D2 i.e. N1 is much smaller than N2 as compared to D1 to D2. It could be because N1=N2 but D1>D2. Again, AvgD will lie between D1 and D2 and AvgN = N1 = N2. It could also be because N1 > N2 but D1 >> D2. Take some numbers to understand why this makes sense.
_________________

Re: If m, p, s and v are positive, and m/p < s/v, which of the following [#permalink]

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19 Nov 2013, 00:08

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rahultripathi2005 wrote:

igotthis wrote:

How can I solve this using numbers?

You can take numbers such 1,2,3 and 4 as m,p,s and v find the value between m/p and s/v try to enter the numbers in the answer choices. only option B will satisfy .

Thanks Rahul

Hi,

I make the plug in as follows but it does not work: 1/2 < 3/4

1. (1+3)/(2+4) = 4/6 = 2/3 so in between 2. (1*3)/(2*4) = 3/8 less than 1/2 3. (3-1)/(4-2) = 1 > 3/4

So this problem cannot solve by using plugin?? Please help to correct me if I went wrong.

I make the plug in as follows but it does not work: 1/2 < 3/4

1. (1+3)/(2+4) = 4/6 = 2/3 so in between 2. (1*3)/(2*4) = 3/8 less than 1/2 3. (3-1)/(4-2) = 1 > 3/4

So this problem cannot solve by using plugin?? Please help to correct me if I went wrong.

Thanks!

Statement 3 is (s/v) - (m/p) When you take values as 1, 2, 3 and 4, it becomes (3/4) - (1/2) = 1/4 (This is not between 1/2 and 3/4 and hence you know that statement 3 may not hold always)

Plugging in numbers is not the best strategy for 'must be true' questions. You know that statement 1 holds for these particular values of m , p, s and v (1, 2, 3 and 4) but how do you know that it will be true for every set of valid values of m, p, s and v? You cannot try every set. You can certainly ignore statements II and III since you have already got values for which they are not satisfied. But you must focus more on statement I and try to figure out using logic whether it must always hold.
_________________

which of the following 'must be' between m/p and s/v?

I didn't understand the question. What does the question mean here by 'must be'?

"Must be" means for any (possible) values of m, p, s and v. So, \(\frac{m}{p}< (option) <\frac{s}{v}\), must hold true fo any positive values of m, p, s and v.

Re: If m, p, s and v are positive, and m/p < s/v, which of the following [#permalink]

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17 May 2014, 09:52

I couldn't get my head around this one, so I used numbers:

\(\frac{m}{p}\) = \(\frac{1}{2}\) and \(\frac{s}{v}\) = \(\frac{3}{4}\)

This fulfils the condition that \(\frac{m}{p}\) \(<\) \(\frac{s}{v}\)

Now,

\(\frac{m+s}{p+v}\) is \(\frac{1+3}{2+4}\) \(=\)\(\frac{4}{6}\) \(=\)\(\frac{2}{3}\), so sufficient as it is between \(\frac{1}{2}\) and \(\frac{3}{4}\).

\(\frac{ms}{pv}\) is \(\frac{1*3}{2*4} = \frac{3}{8}\), which is less than \(\frac{1}{2}\), so not sufficient

\(\frac{s}{v} - \frac{m}{p}\) is \(\frac{3}{4}-\frac{1}{2} = \frac{1}{4}\), which is less than \(\frac{1}{2}\), so not sufficient

Answer is B (1 only).

Is this approach suitable for such questions? Both in terms of accuracy and time: is this valid solution valid for all fractions and is there a faster way to solve this?

Re: If m, p, s and v are positive, and m/p < s/v, which of the following [#permalink]

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23 Apr 2015, 14:25

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mau5 wrote:

imhimanshu wrote:

If m, p, s and v are positive, and \(\frac{m}{p} <\frac{s}{v}\), which of the following must be between \(\frac{m}{p}\) and \(\frac{s}{v}\)

I. \(\frac{m+s}{p+v}\) II. \(\frac{ms}{pv}\) III. \(\frac{s}{v} - \frac{m}{p}\)

A. None B. I only C. II only D. III only E. I and II both

Given that \(\frac{m}{p}<\frac{s}{v} \to \frac{m}{s}<\frac{p}{v}\)Adding 1 on both sides we have \(\frac{m+s}{s}<\frac{p+v}{v} \to \frac{m+s}{p+v}<\frac{s}{v}\)

Again,\(\frac{m}{p}<\frac{s}{v} \to \frac{v}{p}<\frac{s}{m}\) Adding 1 on both sides, we have \(\frac{v+p}{p}<\frac{s+m}{m} \to \frac{m}{p}<\frac{s+m}{v+p}\) . Thus, I is always true. We just have to check for Option II now.

Assming it to be true, we should have\(\frac{ms}{pv}<\frac{s}{v} \to \frac{m}{p}<1\) Which is not always true. Thus the answer is B.

You have the best approach in this thread, but I would never have thought of your solution to #1. I think you made it more complicated than it needed to be.

This is actually a very easy problem once you spend a few moments understanding what you need to do.

You're told the following: 1) m, p, s and v are positive 2) \(\frac{m}{p} <\frac{s}{v}\)

So, you need to know whether each option is both less than \(\frac{s}{v}\) and greater than \(\frac{m}{p}\). That's it. Once that clicks, it's very easy.

I. \(\frac{m+s}{p+v}\)

So part A:

Is \(\frac{m+s}{p+v} < \frac{s}{v}\) ? Is \(mv+sv < sp + sv\) ? Is \(mv < sp\) ? Hmmm, that looks familar. In fact, it's the second thing they gave us: \(\frac{m}{p} <\frac{s}{v}\). So yes!

Part B:

Is \(\frac{m+s}{p+v} > \frac{m}{p}\) ? Is \(mp+sp > mp + mv\) ? Is \(sp > mv\) ? Hmmm, that looks familar too. In fact, it's also the second thing they gave us: \(\frac{m}{p} <\frac{s}{v}\). So yes!

Then you perform the same (quicker) process for II.

Re: If m, p, s and v are positive, and m/p < s/v, which of the following [#permalink]

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21 Jun 2015, 10:13

Hi guys,

I got this problem on an exam pack 1 CAT and failed to solve it correctly due to a technical error (DRIVING ME CRAZY). Otherwise I picked numbers and was 99% on the right path. One question I have for you is:

Do you think testing one pari of numbers is enough? I picked 2 pairs: - m=1, p=3, s=3 and v=4 - m=3, p=1, s=4 and v=1

These 2 options are more conservative as I wanted to test these equations for fractions and integers. Do you think this was overkill (question stem says we're talking about positive numbers and not integers)? As far as I can see, all of you tested just one pair of numbers, which of course almost certainly decreases the time needed to solve the problem.

Thanks!
_________________

Thank you very much for reading this post till the end! Kudos?

I got this problem on an exam pack 1 CAT and failed to solve it correctly due to a technical error (DRIVING ME CRAZY). Otherwise I picked numbers and was 99% on the right path. One question I have for you is:

Do you think testing one pari of numbers is enough? I picked 2 pairs: - m=1, p=3, s=3 and v=4 - m=3, p=1, s=4 and v=1

These 2 options are more conservative as I wanted to test these equations for fractions and integers. Do you think this was overkill (question stem says we're talking about positive numbers and not integers)? As far as I can see, all of you tested just one pair of numbers, which of course almost certainly decreases the time needed to solve the problem.

Thanks!

Testing numbers for "must be true" doesn't always work. One set of values could help you establish that statements II and III do not hold. But how can you PROVE that statement I will always hold just because it holds for one set of values? You do need to establish algebraically or logically that it will hold.

Then I. = 0.3, which is between the two values Then II. = 0.05, which is not between the two values Then III. = -0.05, which is not between the two values
_________________

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Then I. = 0.3, which is between the two values Then II. = 0.05, which is not between the two values Then III. = -0.05, which is not between the two values

Not necessarily. A sub-600 question can be categorized as a 650-700 level question if you end up taking more than usual time for solving it. Some people might not see the value of plugging in different numbers for this question and spend time on using algebra for the same. As you can see out of 443 attempts, only 50% people were able to answer this correctly. Alternately, this question might be a difficult question for people who are not gfood with inequalities in general.

So, the current categorization is somewhat accurate.

Re: If m, p, s and v are positive, and m/p < s/v, which of the following [#permalink]

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11 Oct 2015, 19:45

VeritasPrepKarishma wrote:

Cee0612 wrote:

I make the plug in as follows but it does not work: 1/2 < 3/4

1. (1+3)/(2+4) = 4/6 = 2/3 so in between 2. (1*3)/(2*4) = 3/8 less than 1/2 3. (3-1)/(4-2) = 1 > 3/4

So this problem cannot solve by using plugin?? Please help to correct me if I went wrong.

Thanks!

Statement 3 is (s/v) - (m/p) When you take values as 1, 2, 3 and 4, it becomes (3/4) - (1/2) = 1/4 (This is not between 1/2 and 3/4 and hence you know that statement 3 may not hold always)

Plugging in numbers is not the best strategy for 'must be true' questions. You know that statement 1 holds for these particular values of m , p, s and v (1, 2, 3 and 4) but how do you know that it will be true for every set of valid values of m, p, s and v? You cannot try every set. You can certainly ignore statements II and III since you have already got values for which they are not satisfied. But you must focus more on statement I and try to figure out using logic whether it must always hold.

I tried 2/4 < 3/4 (I avoided typical 1/2 and matched the denominators)

I.- With the numbers I chose, this gives me 5/8. I multiply my initial numbers by 2/2 and I get:

4/8 < 5/8 < 6/8

By having the numbers nicely and tightly between them, Isn't it a good indication that it must be true?

They didn't mention, however, that m,p,s and v must be integers. If they had, then this approach must certainly be sufficient, since you can divide everything by 8 and end up with 4<5<6. Am I right?

And by the way, once you get I.- to hold true, you end up between B and E and you can forget about testing III.-. Once you find II.- does not hold true, you know It's B

Re: If m, p, s and v are positive, and m/p < s/v, which of the following [#permalink]

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28 Aug 2016, 07:08

imhimanshu wrote:

If m, p, s and v are positive, and \(\frac{m}{p} <\frac{s}{v}\), which of the following must be between \(\frac{m}{p}\) and \(\frac{s}{v}\)

I. \(\frac{m+s}{p+v}\) II. \(\frac{ms}{pv}\) III. \(\frac{s}{v} - \frac{m}{p}\)

A. None B. I only C. II only D. III only E. I and II both

This question is quite easy: III. is clearly out, because this will be smaller than m/p. II. is clearly out. Test this with m/p=0 or s/v=1 I. Struggled a minute here, but then I got this idea:

If we pick a number for m/p and a number for s/v: For example: m/p = 1/3 and s/v = 3/4

Now, I we try to get these numbers to the same denominator we get: 4/12 < 9/12 If these numbers are added ab (4+9)/(12+12), we always get a number that is higher than m/p but lower than s/v. This is how I solved this

This works for every number pair. We add the same denominator, but an enumerator which is bigger than the original one. Therefore this has to be bigger than m/p but smaller than s/v!

Re: If m, p, s and v are positive, and m/p < s/v, which of the following [#permalink]

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27 Sep 2017, 04:07

Nowhere it says m,p,s and v can't be same numbers. Therefore I plugged in numbers to disprove the statements. Shall they not mention that the numbers are distict.