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Re: If abc = b3 , which of the following must be true? [#permalink]

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20 Feb 2011, 07:41

5

This post received KUDOS

abc = b^3 implies abc - b^3 = 0 or b*(ac-b^2)=0 which implies either b=0 or ac=b^2 so, either of them or both of them can be true, but none of them must be true. Answer A

Hi Bunuel, Request you to post your reasoning. Here is how I approached this one- abc = b^3 abc-b^3 =0 b(ac-b^2)=0 Hence either b=0 or ac=b^2

However b=0 is not a solution. Could you please shed some light on how to approach must be true or could be true questions. Thanks H

If abc = b^3 , which of the following must be true? I. ac = b^2 II. b = 0 III. ac = 1

A. None B. I only C. II only D. I and III E. II and III

\(abc = b^3\) --> \(b(ac-b^2)=0\) --> EITHER \(b=0\) OR \(ac=b^2\), which means that NONE of the option MUST be true.

For example if \(b=0\) then \(ac\) can equal to any number (not necessarily to 0 or 1), so I and III are not always true, and if \(ac=b^2\) then \(b\) can also equal to any number (not necessarily to 0), so II is not always true.

OA is definitely A. Folks may be little bit confused at option *i and iii* (3) tells us that ac=1. So what? Is b also 1 or equal to 0? Can't figure it out. Move further. If some part of multiplication is 0, there is no need to ans. That question. It doesnt satisfy the condition: abc=b^3.

NOTE:WHEN YOU ARE TACKLING "MUST BE TYPE" QUESTION, always try to find out " entire set" instead of "sub set" cz subset answers "what should be" rather than "what must be". If you are an accountant, you may consider it equivalent to " Bad debt reserve".

This tells us that either b = 0 OR \(ac = b^2\). At least one of these 2 'must be true'.

So can I say b must be equal to 0? No! It is possible but it is also possible that instead, \(ac = b^2\). So can I say ac must be equal to \(b^2\)? No! It is possible but it is also possible that instead, b = 0. So can I say that b = 0 AND \(ac = b^2\)? No! It is possible but it is also possible that only one of them is true. So can I say that b = 0 OR \(ac = b^2\)? Yes, I can! One of them must be true.

If one of the options were 'I or II', that would have been true.
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Re: If abc = b^3 , which of the following must be true? [#permalink]

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08 Oct 2014, 21:47

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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Re: If abc = b^3 , which of the following must be true? [#permalink]

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10 Oct 2015, 19:57

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
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Why is it wrong to divide both sides by b at the start?

Another way to understand this:

When you divide by b, you are assuming that b is not 0. Is it a fair assumption? No. You are losing on a possible solution of the equation.

How can I make abc = b^3? I can do that by either making b = 0 or making ac = b^2.

But if you divide by b, you get only one solution: ac = b^2 and that means I must be true. Actually, the answer is none because there is another possibility and that is b = 0.
_________________

Re: If abc = b^3 , which of the following must be true? [#permalink]

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14 Mar 2016, 23:51

nikhilsrl wrote:

If abc = b^3 , which of the following must be true?

I. ac = b^2 II. b = 0 III. ac = 1

A. None B. I only C. II only D. I and III E. II and III

This is from Kaplan CAT. I am not quite sure how they arrived at the answer.

HERE the question is asking us which statement must be true Hence we can discard all of them as we can make each one of them untrue by taking the other leftover equations true. Hence A
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Re: If abc = b^3 , which of the following must be true? [#permalink]

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02 Dec 2016, 14:26

Top Contributor

STEMBusiness wrote:

Why is it wrong to divide both sides by b at the start?

Here's an illustrative example: (5)(0) = (6)(0) TRUE Now divide both sides by 0 to get: 5 = 6 NOT TRUE

Likewise, if xy = xz, we can't divide both sides by x and conclude that y = z, because it's possible that x = 0 So, while it's true that (0)(y) = (0)(z), we can't then conclude that y = z
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Re: If abc = b^3 , which of the following must be true? [#permalink]

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02 Dec 2016, 14:34

Top Contributor

1

This post was BOOKMARKED

nikhilsrl wrote:

If abc = b³, which of the following must be true?

I. ac = b² II. b = 0 III. ac = 1

A. None B. I only C. II only D. I and III E. II and III

Let's answer this question by Process Of Elimination.

I. ac = b² Must this be true? No. Consider the case where a = 1, b = 0 and c = 1 This set of values satisfies the given equation (abc = b³), HOWEVER it is not the case that ac = b² Plug in the values to get (1)(1) = 0² - NOT TRUE

II. b = 0 Must this be true? No. Consider the case where a = 1, b = 1 and c = 1 This set of values satisfies the given equation (abc = b³), HOWEVER it is not the case that b = 0

NOTE: At this point, we need not examine statement III, because none of the answer choices state that only statement III must be true.

However, let's examine statement III for "fun"

III. ac = 1 Must this be true? No. Consider the case where a = 0, b = 0 and c = 0 This set of values satisfies the given equation (abc = b³), HOWEVER it is not the case that ac = 1

Re: If abc = b^3 , which of the following must be true? [#permalink]

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10 Dec 2016, 04:57

I considered three different integers:

When '1' is considered, Statement 1 & 3 are correct & when '3' is considered, statement 1 alone holds. Finally, considered a=-3,b=3 & c=3. None of the statements hold true.

Given that only 49% of the folks got this simple looking question right, understood that there is a trap.

gmatclubot

Re: If abc = b^3 , which of the following must be true?
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10 Dec 2016, 04:57

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