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22 Feb 2020, 09:41
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A
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
45%
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Question Stats:
63% (01:39) correct
37%
(01:56)
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If \(y ≠ 0\), what is the value of \(\frac{6^{(x-y)^2}}{6^{(x+3)(x-3)}}\)?
(1) \(y = 2x\).
(2) \(y = 2\).
(1) \(y = 2x\).
(2) \(y = 2\).
22 Feb 2020, 13:08
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If \(y ≠ 0\), what is the value of \(\frac{6^{(x-y)^2}}{6^{(x+3)(x-3)}}\)?
\(\frac{6^{(x-y)^2}}{6^{(x+3)(x-3)}}=6^{(x-y)^2-(x+3)(x-3)}=6^{x^2-2xy+y^2-x^2+9}=6^{y^2-2xy+9}\)
(1) \(y = 2x\). The expression above becomes: \(6^{y^2-2xy+9}=6^{4x^2-4x^2+9}=6^9\). Sufficient.
(2) \(y = 2\). The expression above becomes: \(6^{y^2-2xy+9}=6^{4-4x+9}\). No info about x. Not sufficient.
Answer: A.
\(\frac{6^{(x-y)^2}}{6^{(x+3)(x-3)}}=6^{(x-y)^2-(x+3)(x-3)}=6^{x^2-2xy+y^2-x^2+9}=6^{y^2-2xy+9}\)
(1) \(y = 2x\). The expression above becomes: \(6^{y^2-2xy+9}=6^{4x^2-4x^2+9}=6^9\). Sufficient.
(2) \(y = 2\). The expression above becomes: \(6^{y^2-2xy+9}=6^{4-4x+9}\). No info about x. Not sufficient.
Answer: A.
24 Feb 2020, 03:04
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Asad
If \(y ≠ 0\), what is the value of \(\frac{6^{(x-y)^2}}{6^{(x+3)(x-3)}}\)?
(1) \(y = 2x\).
(2) \(y = 2\).
(1) \(y = 2x\).
(2) \(y = 2\).
Solution
Step 1: Analyse Question Stem
- • \(y≠0\)
• We need to find the value of \(\frac{6^{(x-y)^2}}{6^{(x+3)(x-3)}}\)
- o Or, \(\frac{6^{(x-y)^2}}{6^{(x+3)(x-3)}} = \frac{6^{(x^2+y^2–2xy)}}{6^{(x^2 – 9)}}= \frac{6^{x^2} *6^{y^2}*6^9}{6^{x^2}*6^{2xy}}= \frac{6^{y^2} *6^9}{6^{2xy}}\)
Step 2: Analyse Statements Independently (And eliminate options) – AD/BCE
Statement 1: \(y = 2x\)
• \(\frac{6^{y^2}*6^9}{6^{2xy}} = \frac{6^{4*x^2}*6^9}{6^{4x^2}} = 6^9\).
Hence, statement 1 is sufficient and we can eliminate answer options B, C and E
Statement 2: \(y = 2\).
• Here, \(\frac{6^{y^2}*6^9}{6^{2xy} }= \frac{6^{2^2}*6^9}{6^{4x}}\)
• However, we don’t know the value of x.
Hence, statement 2 is NOT sufficient.
Thus, the correct answer is Option A.
