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/vSo, 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