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# If m, p, s and v are positive, and m/p < s/v, which of the following

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Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
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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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30 Jan 2019, 01:30
Manukaran wrote:
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.

Thanks VeritasKarishma. Wouldn't this logic also hold true for "could be true" questions.

The given statement may not be true for the particular values we consider, but how do we know that the given statement will not be true for any values?

Absolutely! The logic will hold for could be true.

Must be true for all - Usually, the values you will try would be true. You might be hard pressed to find a value for which it doesn't hold. Here lies the challenge.

Could be true for some value of x - Usually, the values you will try would not be true. You will need to find a value for which it will hold. Doing that is often a bit easier by putting in 0, 1 etc. But yes, there could be a challenge here too.
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If m, p, s and v are positive, and m/p < s/v, which of the following  [#permalink]

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03 Mar 2019, 12:48
Let's analyze statement 1.

Let me first ask you this. What would happen IF we had the fraction ([m+am]/[p+ap])?
We could factor 1+a and get m/p.

Now let ma=s.
Now let's compare the fraction ma/pa to s/v. The numerators are equal. V has to be less than pa. If two positive fractions have the same numerator the fraction
with the lesser denominator is greater. So pa>v.
Now let's compare ([m+am]/[p+ap]) to ([m+s]/[p+v]). ([m+s]/[p+v]) must be greater since the numerators are equal but p+ap is greater than p+v.

We can use very similar reasoning to show that ([m+s]/[p+v]) is less than s/v. So it must lie between m/p and s/v.

Statement 2 can easily be shown to be false using the numbers .9 and .1

Statement 3 can be shown to be false using the numbers .9 and .8
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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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13 Apr 2019, 19:32
all the number are positive, we multiple all the inequalities to find the answer
this can be done is 1 minutes.
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Joined: 09 Jun 2019
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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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25 Jul 2019, 01:11
what if m,p,s,v are 1,4,4,1? then it becomes equal.
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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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03 Dec 2019, 13:31
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.

Hi, thanks for the explanation. I had a doubt, especially for the 1st option. Can't we simple subtract $$\frac{m}{p}$$ from $$\frac{m+s}{p+v}$$ which basically gives a positive result. Similarly we can check it for the other side too and the result comes out to be negative there.
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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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05 Dec 2019, 15:16
jlgdr wrote:
This is one of those very few problems were picking numbers is almost your only weapon

Let's say that m/p<s/v,

Let's say m=4, p=2, s=3 and v=1

So let's begin with statements

I. 4+3/3=7/3=2.3 is it between 2<x<3?

Yes, so this one is true

2. Clearly not true
3. Clearly not true

Therefore only I is our best answer choice

Hope this helps
Cheers
J

@BANUEL - Can we plug in and come to a conclusion? However, it gets very messy when we have to try fractions, decimals etc (when is is not given that they are integers ). In this question only one option satisfies when we plug in integers 1 2 3 4, but is this the right approach?

thx
Re: If m, p, s and v are positive, and m/p < s/v, which of the following   [#permalink] 05 Dec 2019, 15:16

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