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

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just took a practice test in a Kaplan book and had a few [#permalink]  06 Sep 2004, 07:28
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just took a practice test in a Kaplan book and had a few issues with it, so I thought I would post it here...you all may have addressed it prior, if so I would appreciate a link or a re-do:

If ab<ac, which is greater, b or c?

1. a<0
2. c<0

I didn't agree with the answer they give...thanks for the input in advance...
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Manager
Joined: 27 Aug 2004
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A

As per the questions ab<ac means depending on the sign of a this inequality will change

if a is + ive then, divide the inequality with a on both sides, it becomes
b<c

if a is - ive then, divide the inequality with a on both sides, it becomes

b>c

1) says a is negative - sufficient
2) insufficient. we don't know the value of b.
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I used some quick numbers and I have A as answer too.

Regards,

Alex
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Quote:
If ab<ac, which is greater, b or c?

1. a<0
2. c<0

As per the questions ab<ac means depending on the sign of a this inequality will change

if a is + ive then, divide the inequality with a on both sides, it becomes
b<c

if a is - ive then, divide the inequality with a on both sides, it becomes

b>c

1) says a is negative - sufficient
2) insufficient. we don't know the value of b.

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?
Manager
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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.
Manager
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lovely wrote:
Quote:
If ab<ac, which is greater, b or c?

1. a<0
2. c<0

As per the questions ab<ac means depending on the sign of a this inequality will change

if a is + ive then, divide the inequality with a on both sides, it becomes
b<c

if a is - ive then, divide the inequality with a on both sides, it becomes

b>c

1) says a is negative - sufficient
2) insufficient. we don't know the value of b.

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?

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.
Joined: 31 Dec 1969
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DS qns from Kaplan & Princeton [#permalink]  07 Sep 2004, 02:31
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 ?
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newuserabc..... 0 is an even number
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Jim

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