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Re: If m, p, s and v are positive, and m/p<s/v, which of the fol [#permalink]
22 Sep 2013, 07:31

8

This post received KUDOS

Expert's post

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 fol [#permalink]
23 Sep 2013, 00:37

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 .

Re: If m, p, s and v are positive, and m/p<s/v, which of the fol [#permalink]
24 Sep 2013, 01:53

8

This post received KUDOS

Expert's post

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

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 fol [#permalink]
19 Nov 2013, 00:08

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.

Re: If m, p, s and v are positive, and m/p<s/v, which of the fol [#permalink]
19 Nov 2013, 19:56

Expert's post

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.

Re: If m, p, s and v are positive, and m/p<s/v, which of the fol [#permalink]
01 Mar 2014, 06:16

Expert's post

siriusblack1106 wrote:

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 fol [#permalink]
05 Apr 2014, 09:45

Bunuel wrote:

siriusblack1106 wrote:

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: m, p, s, and v are positive and m/p < s/v [#permalink]
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: m, p, s, and v are positive and m/p < s/v [#permalink]
17 May 2014, 23:53

Expert's post

dhirajx wrote:

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?