just took a practice test in a Kaplan book and had a few

i agree with your analysis...BUT, how do you explain plugging in some numbers that do not work into #1? For instance: 1) a<0, so let a= -5...then you have -5b<-5c, if you'll notice, the opening statement asks if b<c. I can plug numbers into -5b<-5c so that b<c AND b>c. For instance, b could be 1 and c could be 10 (b<c) OR b could be -10 and c could be -20 (b>c)....so I would say if there are numbers that could make #1 both true and false, then it has to be insufficient, doesn't it?

The question asks whether b is greater or c. When you pick numbers for b and c, you are making an assumption. That is not the way to approach DS problems.

When you choose b = 1 and c = 10 in your scenario above, your original relation ab<ac for a = -5 ceases to be true. Therefore, these values of b and c are invalid for the original condition to hold with a negative a.

Hi

Came across these puzzling DS answers - would be great if sumone cud explain.

1.what is the value of x ?

a. x^2 = 4x
b. x is an even number

answer: E (explanation states that from a alone, x could be 0 or 4, both even no.s)

according to me the ans should be C since 0 cannot be an even no. and therefore using a & b x would be 4


2. What is the value of x, a prime no. ?

a. x < 15
b. (x-2) is a multiple of 5

answer: E(explanation states that 2 no.s 2,7 satisfy teh criteria since 2-2 = 0 is also a multipel of 5)

according to me the ans should be C since using both the statements x can take only 7 as the value.


could anyone out here help me ?
thx in advance

newuserabc..... 0 is an even number

12-2=10. It is not prime hence B doesn't work.