SSC Selection Post 2022 PYQ — Question 51

SSC Selection Post - Higher Secondary Level 2022 - 11 Feb 2022 - Shift 1 • Quantitative Aptitude • Q51

If 2x - y = 2 and xy = 32,\frac{3}{2},  then what is the value of 12(x3−y38)?\frac{1}{2}(x^3 - \frac{y^3}{8}) ?

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Correct answer: B

Explanation

The correct answer is option 2 i.e. 138\frac{13}{8}.

⇒ 2x - y = 2

⇒ y = 2x - 2....(1)

⇒ xy = 32\frac{3}{2} ....(2)

Put the value of y in equation 2

⇒ x(2x - 2) = 32\frac{3}{2}

⇒ 2x2 - 2x = 32\frac{3}{2}

⇒ 4x2 - 4x - 3 = 0

⇒ 4x2 - 6x + 2 x - 3 = 0

⇒ 2x(2x - 3) + 1(2x - 3) = 0

⇒ (2x + 1)(2x - 3) = 0

⇒ x = 3/2

Then, y = 3 - 2 = 1

⇒ 12(x3−y38)\frac{1}{2}(x^3 - \frac{y^3}{8})  = 12((32)3−(1)38)\frac{1}{2}(({3\over2})^3 - \frac{(1)^3}{8})

⇒ 12(278−18)\frac{1}{2}({27\over8} - \frac{1}{8})

⇒ 12(268)\frac{1}{2}({26\over8})  = 138\frac{13}{8}