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705-805 (Hard)|   Arithmetic|   Exponents|      
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EgmatQuantExpert
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If a, b and x are integers such Updated on: 31 May 2017, 05:26
Originally posted by EgmatQuantExpert on 24 May 2017, 11:45.
Last edited by EgmatQuantExpert on 31 May 2017, 05:26, edited 1 time in total.
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D
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!

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75% (hard)

Question Stats:

55% (02:14) correct 45% (02:17) wrong based on 471 sessions

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Q.

If a, b and x are integers such that \(a^6=b^3=\frac{|x|}{x}\), what is the value of a - b

(1) \(a^3b^7 > 0\)

(2) \(a + b > 0\)


Answer Choices



A. Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient to answer the question asked.
B. Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient to answer the question asked.
C. BOTH statements (1) and (2) TOGETHER are sufficient to answer the question asked, but NEITHER statement ALONE is sufficient to answer the question asked.
D. EACH statement ALONE is sufficient to answer the question asked.
E. Statements (1) and (2) TOGETHER are NOT sufficient to answer the question asked, and additional data specific to the problem are needed.


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Official Answer
D
Pure algebraic questions like this one are no longer part of the GMAT Data Sufficiency syllabus.
For relevant Data Sufficiency questions, please check the Data Sufficiency forum:
Data Sufficiency Forum
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If a, b and x are integers such Updated on: 31 May 2017, 05:43
Originally posted by EgmatQuantExpert on 24 May 2017, 11:47.
Last edited by EgmatQuantExpert on 31 May 2017, 05:43, edited 1 time in total.
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The official solution has been posted. Looking forward to a healthy discussion..:)
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If a, b and x are integers such 24 May 2017, 12:00
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IMO D
\(a^6 = b^3 = \frac{|x|}{x}\)

\(\frac{|x|}{x}\) = 1 or -1 depending on the sign of x,but since this fraction is equal to \(a^6\),it has to be equal to 1.

Therefore x=some postive,b =1 and a =+1 or -1 => a-b=0 or a-b=-2 depending on the value of a.

statement 1:

\(a^3b^7\)>0,

since b=1, \(a^3\)>0, not possible for this statment.

therefore a>0 => a=1.

a-b=0


statement 2:

a+b >0 ,

if a=-1,then a+b=0. Therefore a=1.

a-b=0.

Both the statement satisfies ,so D.
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If a, b and x are integers such 31 May 2017, 05:42
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Official Solution



Steps 1 & 2: Understand Question and Draw Inferences

Given: Integers a,b,x

    • \(a^6=b^3=\frac{|x|}{x}\)
To Find: Value of a - b ?

\(\frac{|x|}{x}\) can take two possible values:

    • If \(x > o, |x| = x, \frac{|x|}{x}=\frac{x}{x}=1\)

    • If \(x < o, |x| = -x, \frac{|x|}{x}=−\frac{x}{x}=−1\)

    • \(a^6=\frac{|x|}{x}\)

    • As \(a^6\) is always positive,\(a^6 = 1\), i.e. \(a = 1\) or \(-1\)

    • So, we can reject the value of \(\frac{|x|}{x}=−1\)

    • \(b^3=\frac{|x|}{x}=1\)

    • b = 1

Possible values of \(a – b\)

    • If \(a = 1\) and \(b = 1, a – b = 0\)

    • If \(a = -1\) and \(b = 1, a- b = -2\)

So, we need to find the unique value of a to find the value of \(a – b\).

Step 3: Analyze Statement 1 independently

(1) \(a^3*b^7 > 0\)

    • Rewriting \(a^3b^7\) as \(ab(a^2b^6)\)

    • Therefore, \(ab(a^2b^6)>0\)

    • We know that \(a^2b^6\) is always \(> 0\) (even power of any number is always positive)

      • So, for \(ab(a^2b^6)> 0\)

      • \(ab > 0\)

    • This tells us that a and b have same signs.

    • Since \(b > 0\), therefore a will also be greater than 0, so the value of \(a = 1\).

      • \(a – b = 1 -1 = 0\)

Sufficient to answer

Step 4: Analyze Statement 2 independently

(2) \(a + b > 0\)

    • If \(a = 1\) and \(b = 1\), \(a + b = 2 > 0\)

    • If \(a = -1\) and \(b = 1\), \(a + b = 0\), is not greater than zero

      • Hence, we have a unique answer, where \(a =1\) and \(b = 1\)

    • Thus \(a – b = 1 – 1 = 0\).

Sufficient to answer.

Answer: D


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If a, b and x are integers such 24 Aug 2017, 21:43
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Since \(a^6\) >=0
\(\frac{|x|}{x}\) will be positive, and |x| >=0 so x must be positive
=> |x| = x (also x can't be 0) and the fraction can be reduced to 1.

\(b^3\)=1 => b = 1
\(a^6\) = 1 => a =1, or -1

After analyzing, we can see the true question here is whether a is 1 or -1.

1- \(a^3\) \(b^7\) >0
or \(a^3\) x 1 >0
so a>0
=> a=1
SUFF

2-a+b>0
or a+1 > 0
a can only be -1 or 1, so a =1
SUFF

In sum, answer is D.
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If a, b and x are integers such 26 Aug 2017, 11:24
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From the question, without looking at the statements, we can infer that the a=b=1 due to |x|/x. It cannot be 0 as 0/0 is undefined. It can't be -1 too due to b^3. Is this question legit?


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If a, b and x are integers such 27 Aug 2017, 02:42
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antztheman
From the question, without looking at the statements, we can infer that the a=b=1 due to |x|/x. It cannot be 0 as 0/0 is undefined. It can't be -1 too due to b^3. Is this question legit?


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b must be 1 but a can be 1 ass well as -1. For both values, 1 and -1, a^6 is still 1.

Possible values of \(a – b\)

    • If \(a = 1\) and \(b = 1, a – b = 0\)

    • If \(a = -1\) and \(b = 1, a- b = -2\)
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If a, b and x are integers such 27 Aug 2017, 15:57
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I thought a^6 = |x|/x => as a6 is positive x has to be positive: => a = 1?
x>0 => be is +1 and x<0 means b= -1

Is my reasoning wrong?

Can someone help me..
Thanks,
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If a, b and x are integers such 18 Mar 2019, 11:13
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I thought the absolute value operator will always give a positive value. Can someone please clarify?
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If a, b and x are integers such 18 Mar 2019, 11:14
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I thought the absolute value operator will always give a positive value. Can someone please clarify?
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If a, b and x are integers such 12 Nov 2019, 06:10
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EgmatQuantExpert
If a, b and x are integers such that \(a^6=b^3=\frac{|x|}{x}\), what is the value of a - b?

(1) \(a^3b^7 > 0\)
(2) \(a + b > 0\)
Show more

\(a^6=b^3=\frac{|x|}{x}\)
\(x>0:\frac{|x|}{x}=\frac{x}{x}=1\)
\(x<0:\frac{|x|}{x}=\frac{-x}{x}=-1\)
\(a^6≥0…a^6=1…|a|=1…a=(1,-1)\)
\(a^6=1…a^6=b^3…b^3=1…b=1\)

case 1: \(a=1:a-b=1-1=0\)
case 2: \(a=-1:a-b=-1-1=-2\)

(1) \(a^3b^7 > 0\) sufic
\((a,b)>0…a>0…a=1…a-b=1-1=0\)

(2) \(a + b > 0\) sufic
\(a=-1:a+b=-1+1=0<0…invalid\)
\(a=1:a+b=1+1=2>0…valid\)
\(a-b=1-1=0\)

Ans (D)
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If a, b and x are integers such 11 Jan 2026, 08:46
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