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27 Jun 2021, 22:21
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sahilvermani
We need "MUST BE" between, not could be between. In the case I mentioned above, s/v - m/p will not be between them. Even if in some cases it will be between them, it doesn't matter. We need those in which it will ALWAYS be between them.
This is how we deal with "must be true" questions. If we can find one case in which it is not true, that is enough to say that the statement does not hold.
Also, check this: https://www.gmatclub.com/forum/veritas-prep-resource-links-no-longer-available-399979.html#/2017/0 ... -question/
08 Aug 2021, 22:53
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imhimanshu
i would try with simple values through brute force which will take us faster to the options
let m=s=1 and p=8 and v=4
then the condition is met
in 2) 1/32 is recieved when we plug into the equation which is not between 1/8 and 1/4
in 3) 1/4 -1/8 =1/8 which is not in between so we safely eleminate
in 1) however we get 2/12 = 1/6 which is between 1/8 and 1/4 therefore we take this value
Therefore IMO B
17 Aug 2021, 20:57
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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.
\(\frac{m}{p}<\frac{s}{v} \to \frac{v}{p}<\frac{s}{m}\)
- manipulate to determine by adding p/p to left side and m/m to right side
\(\frac{v+p}{p}<\frac{s+m}{m} \to \frac{m}{p}<\frac{s+m}{v+p}\)
\(\frac{m}{p}<\frac{s}{v} \to \frac{m}{s}<\frac{p}{v}\)
- manipulate to determine by adding p/p to left side and m/m to right side
\(\frac{m+s}{s}<\frac{p+v}{v} \to \frac{m+s}{p+v}<\frac{s}{v}\)
I is always true.
II.
we know; \(\frac{m}{p}<\frac{s}{v} \) multiply both sides by \(\frac{m}{p} \)
=\((\frac{m}{p})^2<\frac{ms}{pv}\)
we know; \(\frac{m}{p}<\frac{s}{v} \) multiply both sides by \(\frac{s}{v} \)
=\(\frac{ms}{pv}<(\frac{s}{p})^2\)
combined
\((\frac{m}{p})^2<\frac{ms}{pv}<(\frac{s}{p})^2\)
squareroot
\(\frac{m}{p}<\sqrt{\frac{ms}{pv}}<\frac{s}{p}\)
it's not always true that
\(\sqrt{\frac{ms}{pv}} = \frac{ms}{pv}\)
hence II is not always true.
III.
let's start with the stem;
\(\frac{m}{p} <\frac{s}{v}\)
= \(0<\frac{s}{v} - \frac{m}{p} \)
= \(\frac{m}{p}-\frac{s}{v} <0 \)
= \(\frac{m}{p}-\frac{s}{v} < \frac{s}{v} - \frac{m}{p} \)
= \(\frac{2m}{p}-\frac{s}{v} < \frac{s}{v} \)
now lets evaluate III
if its between then that means
\(\frac{m}{p} < \frac{s}{v} - \frac{m}{p} <\frac{s}{v}\)
= add \(\frac{m}{p}\) and subtract \(\frac{s}{v}\)
= \(\frac{2m}{p}-\frac{s}{v} < \frac{m}{p} \)
therefore III is not inbetween. its less than.
so we know that its not always the case where
\(\frac{2m}{p}-\frac{s}{v} < \frac{m}{p} \)
based on our analysis of the stem it can be = \(\frac{m}{p} < \frac{2m}{p}-\frac{s}{v} < \frac{s}{v} \)
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.
\(\frac{m}{p}<\frac{s}{v} \to \frac{v}{p}<\frac{s}{m}\)
- manipulate to determine by adding p/p to left side and m/m to right side
\(\frac{v+p}{p}<\frac{s+m}{m} \to \frac{m}{p}<\frac{s+m}{v+p}\)
\(\frac{m}{p}<\frac{s}{v} \to \frac{m}{s}<\frac{p}{v}\)
- manipulate to determine by adding p/p to left side and m/m to right side
\(\frac{m+s}{s}<\frac{p+v}{v} \to \frac{m+s}{p+v}<\frac{s}{v}\)
I is always true.
II.
we know; \(\frac{m}{p}<\frac{s}{v} \) multiply both sides by \(\frac{m}{p} \)
=\((\frac{m}{p})^2<\frac{ms}{pv}\)
we know; \(\frac{m}{p}<\frac{s}{v} \) multiply both sides by \(\frac{s}{v} \)
=\(\frac{ms}{pv}<(\frac{s}{p})^2\)
combined
\((\frac{m}{p})^2<\frac{ms}{pv}<(\frac{s}{p})^2\)
squareroot
\(\frac{m}{p}<\sqrt{\frac{ms}{pv}}<\frac{s}{p}\)
it's not always true that
\(\sqrt{\frac{ms}{pv}} = \frac{ms}{pv}\)
hence II is not always true.
III.
let's start with the stem;
\(\frac{m}{p} <\frac{s}{v}\)
= \(0<\frac{s}{v} - \frac{m}{p} \)
= \(\frac{m}{p}-\frac{s}{v} <0 \)
= \(\frac{m}{p}-\frac{s}{v} < \frac{s}{v} - \frac{m}{p} \)
= \(\frac{2m}{p}-\frac{s}{v} < \frac{s}{v} \)
now lets evaluate III
if its between then that means
\(\frac{m}{p} < \frac{s}{v} - \frac{m}{p} <\frac{s}{v}\)
= add \(\frac{m}{p}\) and subtract \(\frac{s}{v}\)
= \(\frac{2m}{p}-\frac{s}{v} < \frac{m}{p} \)
therefore III is not inbetween. its less than.
so we know that its not always the case where
\(\frac{2m}{p}-\frac{s}{v} < \frac{m}{p} \)
based on our analysis of the stem it can be = \(\frac{m}{p} < \frac{2m}{p}-\frac{s}{v} < \frac{s}{v} \)
