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

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New post 31 Jan 2019, 15:41
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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


The key word here is MUST.
So, if we can show that a certain statement is NOT TRUE we can eliminate some answer choices.

GIVEN: m/p < s/v

So, let's see what happens when m = 1, p = 3, s = 1 and v = 2
We get the inequality: 1/3 < 1/2, which works.

Now test the 3 statements:

I) (m+s)/(p+v) = (1+1)/(3+2) = 2/5
Since it IS the case that 1/3 < 2/5 < 1/2, statement I COULD be true

II) ms/pv = (1)(1)/(3)(2) = 1/6
Here, 1/6 < 1/3 < 1/2
In other words, 1/6 is NOT between 1/3 and 1/6
So, statement II need not be true.
ELIMINATE II

III) s/v - m/p = 1/2 - 1/3 = 1/6
Here, 1/6 < 1/3 < 1/2
In other words, 1/6 is NOT between 1/3 and 1/6
So, statement III need not be true.
ELIMINATE III

At this point, only answer choices A and B remain.
We COULD try testing more values in the hopes that statement I may not be true.
Or we can try to convince ourselves that statement I IS true.
Let's try the latter.

I) (m+s)/(p+v)
Is it the case that (m+s)/(p+v) is BETWEEN m/p and s/v?
Let's first see whether (m+s)/(p+v) < s/v
Since v is POSITIVE, we can safely multiply both sides by v to get: v(m+s)/(p+v) < s
Since (p+v) is POSITIVE, we can safely multiply both sides by (p+v) to get: v(m+s) < s(p+v)
Expand to get: vm + sv < sp + sv
Subtract sv to get: vm < sp
Divide both sides by p to get: vm/p < s
Divide both sides by v to get: m/p < s/v
So, our (m+s)/(p+v) < s/v turns into the inequality m/p < s/v, which is GIVEN information.
Since we know that the inequality m/p < s/v is true, it must also be the case that the inequality (m+s)/(p+v) < s/v is also true

Using the same technique to show that m/p < (m+s)/(p+v) [I'll leave it to you to do that :-)]

Since we can show that m/p < (m+s)/(p+v) < s/v, we can conclude that statement I is true.

Answers: B

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

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New post 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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New post 03 Dec 2019, 13:31
VeritasKarishma 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


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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New post 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
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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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