For integers a, b, and c, if ab = bc

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BTGmoderatorDC: Moderator; Posts: 7187; Joined: Thu Sep 07, 2017 4:43 pmFollowed by: 23 members

by BTGmoderatorDC » Mon Feb 12, 2018 8:14 pm

For integers a, b, and c, if ab = bc, then which of the following must also be true?

A. a = c
B. a^2*b=b*c^2
C. a/c = 1
D. abc > bc
E. a + b + c = 0

Can some experts show me how to solve this?

OA B

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EconomistGMATTutor: GMAT Instructor; Posts: 555; Joined: Wed Oct 04, 2017 4:18 pmFollowed by: 12 members; Thanked: 180 times

by EconomistGMATTutor » Tue Feb 13, 2018 2:33 am

lheiannie07 wrote:For integers a, b, and c, if ab = bc, then which of the following must also be true?

A. a = c
B. a^2*b=b*c^2
C. a/c = 1
D. abc > bc
E. a + b + c = 0

Can some experts show me how to solve this?

OA B

Hi lheiannie07.

Let's take a look at your question.

First, the question is which of the options listed MUST be true. Hence, it has to be true always.

If we choose a=1, b=0 and c=2, then ab=bc=0. Now

(A) a=c is FALSE.
(B) a^2*b=b*c^2=0 is true, but it has to be true always.
(C) a/c = 1 is FALSE, because a/c=1/2.
(D) abc>bc is FALSE, because abc=0=bc.
(E) a+b+c=0 is FALSE, because a+b+c=1+0+2=3.

Hence, the only option that was true is (B). It implies that the correct option is B.

But, if we want to show that it is always true we have to do the following: let be a, b and c general integers, hence $$\text{if}\ ab=bc\ \text{then}\ \ a\cdot\left(ab\right)=a\cdot\left(bc\right)$$ $$\Leftrightarrow\ a^2b=\left(ab\right)\cdot c$$ $$\Leftrightarrow\ a^2b=\left(bc\right)\cdot c$$ $$\Leftrightarrow\ a^2b=bc^2.$$ I hope it helps you.

I'm available if you'd like a follow up.

Regards.

GMAT Prep From The Economist
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GMATGuruNY: GMAT Instructor; Posts: 15539; Joined: Tue May 25, 2010 12:04 pmFollowed by: 1906 members; Location: New York, NY; Thanked: 13060 times; GMAT Score:790

by GMATGuruNY » Tue Feb 13, 2018 5:21 am

lheiannie07 wrote:For integers a, b, and c, if ab = bc, then which of the following must also be true?

A. a = c
B. a^2*b=b*c^2
C. a/c = 1
D. abc > bc
E. a + b + c = 0

ab = bc
ab - bc = 0
b(a-c) = 0.

Implication:
Either b=0 or a-c=0.
Given this constraint, prove that four of the answer choices DON'T have to be true.

Case 1: b=0, a=2, c=1
In this case, A, C, D and E are not true.
Eliminate A, C, D and E.

The correct answer is B.

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by Scott@TargetTestPrep » Wed Feb 14, 2018 10:10 am

lheiannie07 wrote:For integers a, b, and c, if ab = bc, then which of the following must also be true?

A. a = c
B. a^2*b=b*c^2
C. a/c = 1
D. abc > bc
E. a + b + c = 0

If ab = bc, then:

ab - bc = 0

b(a - c) = 0

b = 0 or a = c

Now, let's analyze the answer choices.

A. a = c

We see that this might not be true since b can be 0.

B. a^2*b = b*c^2

If b = 0, we have a^2*b = b*c^2 since both sides will be 0.

If a = c, then a^2 = c^2 and certainly we will have a^2*b = c^2*b.

Answer: B

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