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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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01 Oct 2017, 04:19
Picking nos is the best approach.
m,p,s & v are +ve.
m/p<s/v.
Let m/p=1/2=0.5 ie m=1, p=2
and s/v=4/5=0.8 ie s=4 and v=5
0.5<0.8
Plugin the values in the answer choices.
I. (m+s)/(p+v)=(1+4)/(2+5)=5/7=0.71 this is between 0.5 and 0.8
II. ms/pv=4/10=0.4 not between 0.5 and 0.8
III. s/v−m/p=4/5-1/2=0.3 again not between 0.5 and 0.8.
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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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08 Dec 2017, 10:28
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

Let’s analyze each statement using specific values for the variables.

We can let m = 2, s = 3, p = 4, and v = 5. Thus:

m/p = 2/4 = 0.5 and s/v = 3/5 = 0.6.

Notice that m/p = 0.5 is less than s/v = 0.6. Now let’s analyze each statement.

I. (m+s)/(p+v)

(2 + 3)/(4 + 5) = 5/9 = 0.555… is between m/p = 0.5 and s/v = 0.6.

II. (ms)/(pv)

(2 x 3)/(4 x 5) = 6/20 = 0.3 is NOT between 0.5 and 0.6.

III. s/v - m/p

3/5 - 2/4 = 0.6 - 0.5 = 0.1 is NOT between 0.5 and 0.6.

From the above, we see that only statement I is true. However, this was illustrated by using one set of numbers (m = 2, s = 3, p = 4, and v = 5). It’s possible that it could be false when we use another set of values for m, s, p, and m.

However, we can prove that (m+s)/(p+v) is between m/p and s/v; that is, we can prove that m/p < (m+s)/(p+v) < s/v regardless of the values we use for m, s, p, and m, as long as the values are positive.

Notice that m/p < (m+s)/(p+v) < s/v means m/p < (m+s)/(p+v) and (m+s)/(p+v) < s/v. Also, keep in mind that we are given that m/p < s/v, which is equivalent to mv < ps.

Let’s prove that m/p < (m+s)/(p+v):

m/p < (m+s)/(p+v) ?

m(p+v) < p(m + s) ?

mp + mv < mp + ps?

mv < ps ? (YES)

Since mv < ps is true, m/p < (m+s)/(p+v) is true. Finally, let’s prove that (m+s)/(p+v) < s/v:

(m+s)/(p+v) < s/v ?

v(m+s) < s(p+v)?

mv + sv < sp + sv ?

mv < ps ? (YES)

Again, since mv < ps is true, (m+s)/(p+v) < s/v is true. Thus we have shown that m/p < (m+s)/(p+v) < s/v is always true.

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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 Dec 2017, 08:29
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 IS HARD
the point tested here is that we need to change from ratio from to factor form
m/p<s/v
multiple two sides with s.v
mv<sp

from this point you can solve the problem.
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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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07 Jan 2018, 05:02
One very easy way to make I correct is as below
Any number can be expressed in form of A/1 (lets say --3/2 it can be written as 1.5/1)
so basically if you see option 1 is asking us A/1 < (A+B)/(2) < B/1

The average is always in btw A & B

II & III can simply be rejected by taking values 1,2,3,4.
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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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18 Apr 2018, 23:36
1
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

I. if $$\frac{m+s}{p+v}$$ is to be more than $$\frac{m}{p}$$,

==> $$\frac{m}{p}$$<$$\frac{m+s}{p+v}$$
==> pm+mv<pm+ps and since all numbers are positive,
==> $$\frac{m}{p}$$<$$\frac{s}{v}$$ which is true as per question stem.

if $$\frac{m+s}{p+v}$$ is to be less than $$\frac{s}{v}$$,
==> $$\frac{s}{v}$$>$$\frac{m+s}{p+v}$$
==> sp+sv>mv+sv
==> $$\frac{m}{p}$$<$$\frac{s}{v}$$ which is true as per question stem.

II. if $$\frac{ms}{pv}$$ is to be more than $$\frac{m}{p}$$,

==> $$\frac{s}{v}$$ must be greater than 1, which is not conclusive as per stem of question.

III. if $$\frac{s}{v} - \frac{m}{p}$$ is to be more than $$\frac{m}{p}$$,

==> $$\frac{s}{v} must be greater than 2 X [m]\frac{m}{p}$$, which is not conclusive as per stem of question.

So only (I) can be derived with 100% confidence.
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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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11 May 2018, 11:36
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.

Must or Could be True Questions to practice: http://gmatclub.com/forum/search.php?se ... tag_id=193

Hope it helps.

Do you have a more elegant solution to this problem Bunuel? Most of the answers here are either choosing numbers or doing complicated calculations that I would find difficult to try during the test.
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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 May 2018, 15:44
DarkBlizzard 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

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! :wink:

is III clearly out? 1 < 100, but 100-1 is in between the 2. am i misreading something?

I picked 1/8 < 2/3 to plug in, and III does work here. with 1/2 < 3/4 however it does not.
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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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14 May 2018, 04:57
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

As the question asks for Must, we can take test with only one set of value -

s = 5, m = 4, p,v = 1

Only Statement II Satisfies. Hence, Option B.
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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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14 May 2018, 06:16
rahul16singh28 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

As the question asks for Must, we can take test with only one set of value -

s = 5, m = 4, p,v = 1

Only Statement II Satisfies. Hence, Option B.

Yeah I understand. I guess what I'm saying is that picking the right numbers is essential on this one, as if the wrong numbers are picked, III works too. Just realized that there isn't a I and III option though, so it were down to I and III one could pick new numbers
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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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15 May 2018, 11:17
bp2013 wrote:
rahul16singh28 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

As the question asks for Must, we can take test with only one set of value -

s = 5, m = 4, p,v = 1

Only Statement II Satisfies. Hence, Option B.

Yeah I understand. I guess what I'm saying is that picking the right numbers is essential on this one, as if the wrong numbers are picked, III works too. Just realized that there isn't a I and III option though, so it were down to I and III one could pick new numbers

Which is why when picking numbers, it's always best to choose at least 2 set. I picked: 1, 2, 3, 9 for easy math and they made B true and everything else false. If you have time, best to verify with a second set of numbers or algebraically.
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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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28 Sep 2018, 21:09
DarkBlizzard 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

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! :wink:

How is the third part getting satisfied. 9/12- 4/12= 5/12 this falls between m/p and s/v
This third part is giving me quite a headache, it's getting satisfied for some pairs and some not. I guess, like karishma said, we should avoid plugging in numbers and proceed with logic.
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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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28 Sep 2018, 22:10
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.

Thank you. Very clear explanation for (I)
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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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17 Oct 2018, 11:53
Amanrohra wrote:
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.

Dear Amanrohra,

I'm happy to respond here,
the question doesn't explicitly mentions that m,p,s and v are distinct, but the question gives us a condition:
$$\frac{m}{p}< \frac{s}{v}$$
which can not be true with map,s and v being the same numbers.
Hence they need to be distinct, at least in a manner to fulfil the condition.
for e.g. $$\frac{2}{5} < \frac{5}{2}$$
I hope this helps.
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Re: If m, p, s and v are positive, and m/p < s/v, which of the following &nbs [#permalink] 17 Oct 2018, 11:53

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