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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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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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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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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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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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I thought the absolute value operator will always give a positive value. Can someone please clarify?
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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 that \(a^6=b^3=\frac{|x|}{x}\), what is the value of a - b?

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

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

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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