If c ≠ 0 and (a*b)/c

Question

If c ≠ 0 and  \frac{(a*b)}{c}<0 , is  \frac{a}{c}<0  ?

(1) a < 0

(2) b < 0

QuantMadeEasy · Answer

Step 1: Understanding the question stem

If (a∗b)c<0, following are the cases:
Case 1: a>0, b>0, c<0
Case 2: a>0, b<0, c>0
Case 3: a<0, b>0, c>0
Case 4: a<0, b<0, c<0

Step 2: Analysing the statements
(1) a < 0
Case 3 or Case 4
Insufficient

(2) b < 0
Case 2 or Case 4
In both the cases a/c > 0
Sufficient

B is correct

LipiSinha · Answer

I solved this using numbers.

S1) a<0

If a = -2 and b = -3, then c should be <0. Let us assume c to be -1. In this case \(\frac{a}{c }\)>0
If a = -2 and b = 3, then c should be >0. Let c be 1. Here, \(\frac{a}{c }\)<0

Insufficient.

S2) b<0

If b = -2 and a = -3, c should be <0. Let c be -1. Here, \(\frac{a}{c} \)>0
If b = -2 and a = 3, c should be >0. Let c be 1. In this case too, \(\frac{a}{c }\)>0

Sufficient.

The answer will be B.

TarunKumar1234 · Answer

If c ≠ 0 and \frac{(a*b)}{c}<0 , is \frac{a}{c}<0 ? Stat1: a < 0 then, for the condition, \frac{(a*b)}{c}<0 ; \frac{b}{c} >0, so b =+ve, c =+ve or, b =-ve, c =-ve \frac{a}{c}<0 = +ve, if a <0 and c <0 & \frac{a}{c}<0 = -ve, if a <0 and c >0. Not sufficient Stat2: b < 0 then, for the condition, \frac{(a*b)}{c}<0 ; \frac{a}{c} =+ve always. Sufficient So, I think B.

SaquibHGMATWhiz · Answer

Solution: Pre Analysis: We are given \frac{(a*b)}{c}<0 such that c ≠ 0 We are asked if \frac{a}{c}<0 or not \frac{a}{c}<0 is possible if one and only one of either a or c is negative Statement 1: a<0 Let us take 2 cases that satisfy a<0 and \frac{(a*b)}{c}<0 Case 1: a<0,b<0 and c<0 . In this case, \frac{a}{c}>0 Case 2: a<0,b>0 and c>0 . In this case, \frac{a}{c}<0 Thus, statement 1 alone is not sufficient and we can eliminate options A and D Statement 2: b < 0 If b<0 , then for \frac{(a*b)}{c}<0 to be true a and c either has to be both positive or both negative And in both cases, \frac{a}{c}>0 Thus, statement 2 alone is sufficient Hence the right answer is Option B

If c ≠ 0 and (a*b)/c < 0, is a/c < 0

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