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If a, b, and c are real numbers, is a/b > b/c ?

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hemanthp
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If a, b, and c are real numbers, is a/b > b/c ? [#permalink] New post   Updated on: 21 May 2017, 05:56
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If a, b, and c are real numbers, is a/b > b/c ?

(1) a > c
(2) ac > b^2
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E

Originally posted by hemanthp on 10 Oct 2010, 03:02.
Last edited by Bunuel on 21 May 2017, 05:56, edited 2 times in total.
Renamed the topic and edited the question.
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Re: If a, b, and c are real numbers, is a/b > b/c ? [#permalink] New post   10 Oct 2010, 03:29
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hemanthp wrote:
If a, b, and c are real numbers, is a/b > b/c ?

(1) a > c
(2) ac > b^2

(A) Statement (1) ALONE is sufficient, but statement (2) is not sufficient.
(B) Statement (2) ALONE is sufficient, but statement (1) 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.

Again not convinced with the answer.
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Is a/b > b/c ?
\(\frac{a}{b} \gt \frac{b}{c}\)
\(\frac{a}{b} - \frac{b}{c} \gt 0\)
\(\frac{ac-b^2}{bc} \gt 0\)

(1) Clearly inssufficient, as it doesnt say anything about b. Consider the simple case where ac=b and where ac is not equal to b, in the former case the answer is always no and in the latter it could be yes or no.

(2) ac>b^2 ... Only sufficient if know about the sign of bc, which is unknown. Insufficient

(1+2) ac>b^2 so numerator is >0. But a>c tells us nothing about denominator. Insufficient, could be true or false.

Answer is (e)
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Re: If a, b, and c are real numbers, is a/b > b/c ? [#permalink] New post   11 Oct 2010, 01:58
ac > b^2 .... i have a doubt with the B option.

i can write the above one as ac/b > b
====>a/b > b/c

so i can say this is stright to to ans ....

Please let me know if i am wrong
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Re: If a, b, and c are real numbers, is a/b > b/c ? [#permalink] New post   11 Oct 2010, 02:20
vitamingmat wrote:
ac > b^2 .... i have a doubt with the B option.

i can write the above one as ac/b > b
====>a/b > b/c

so i can say this is stright to to ans ....

Please let me know if i am wrong


In an inequality, when you divide both sides, the sign of the inequality remains the same if the number you are dividing by is positive otherwise it flips. So you are assuming b>0, which is why you are going wrong.

Eg. 4>2 hence 4/2> 2/1 OR 2>1
4>2 hence 4/-2<2/-2 OR -2<-1
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Re: If a, b, and c are real numbers, is a/b > b/c ? [#permalink] New post   11 Oct 2010, 03:17
Thanks shrouded1 .. I did not consider it ...thxs for the reply
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Is a/b > b/c? [#permalink] New post   13 Jan 2017, 09:04
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Is a/b > b/c?

(1) a > c
(2) ac > b^2
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Re: Is a/b > b/c? [#permalink] New post   13 Jan 2017, 10:19
Bunuel wrote:
Is a/b > b/c?

(1) a > c
(2) ac > b^2



Is a/b-b/c>0
(ac-b^2) / bc > 0 ??

(1) no info of b ...thus insuff

(2) ac > b^2
or ac-b^2>0 thus we know numerator is >0 but we don't have info of denominator( bc)
let a=5 , c=3 && b=2 then YES
but if a=5 , c=3 && b= -2 then NO

thus insuff

Combining also we have no info on value b
thus insuff

Ans E
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Re: If a, b, and c are real numbers, is a/b > b/c ? [#permalink] New post   21 May 2017, 05:57
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hemanthp wrote:
If a, b, and c are real numbers, is a/b > b/c ?

(1) a > c
(2) ac > b^2


The highlighted part above is redundant. All numbers on the GMAT are real by default.
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Re: If a, b, and c are real numbers, is a/b > b/c ? [#permalink] New post   28 Nov 2023, 11:17
Hello from the GMAT Club BumpBot!

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Re: If a, b, and c are real numbers, is a/b > b/c ? [#permalink]
28 Nov 2023, 11:17
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