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DS 4

by gmater29 » Fri Oct 23, 2009 4:48 am
If vmt ≠ 0, is v^2m^3t^-4 > 0?
(1) m>v^2
(2) m>t^-4

A. Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.
B. Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient.
C. BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.
D. EACH statement ALONE is sufficient.
E. Statements (1) and (2) TOGETHER are NOT sufficient.

OA D

Can you tell me how ?
Last edited by gmater29 on Fri Oct 23, 2009 5:31 am, edited 1 time in total.
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Re: DS 4

by uttam.albela » Fri Oct 23, 2009 4:59 am
gmater29 wrote:If vmt ≠ 0, is v2m3t-4 > 0?
(1) m>v2
(2) m>t-4

A. Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.
B. Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient.
C. BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.
D. EACH statement ALONE is sufficient.
E. Statements (1) and (2) TOGETHER are NOT sufficient.

OA D


Can you tell me how ?
Hello,

let us first simplify the wording of this question.

vmt ≠ 0, it means that none of v, m or t is zero.

Question is whether (square of v) * (cube of m) / (double square of t) is greater than zero ?

square of v = always positive
double square of t = always positive
so cube of m will decide whether the whole expression is negative or positive.
or we can say m will decide. if m is +ve, m3 will be +ve and if m is -ve, m3 will be -ve.

So the question is actually whether m is negative or +ve?

option a ) m > v2 implies m is greater than zero as v2 is always +ve. So a gives us the answer.

now option b) m > t-4
m> 1/t4
t4 is always +ve
so 1/t4 is always +ve
so m is +ve according to option b
so b also is sufficient alone.

Thus each of a and b are sufficient alone.

D correct answer.
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by gmater29 » Fri Oct 23, 2009 5:32 am
Apologies again,, Its all ^ typo..
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by uttam.albela » Fri Oct 23, 2009 5:45 am
gmater29 wrote:Apologies again,, Its all ^ typo..

while solving it has been considered the same.
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