SSC Selection Post 2020 PYQ — Question 58

SSC Selection Post - Matriculation Level 2020 - 6 Nov 2020 - Shift 1 • Quantitative Aptitude • Q58

If 60% of (x – y) = 45% (x + y) and y = k% of x, then 49% of k is equal to:

Show answer & explanation

Correct answer: A

Explanation

The correct answer is option 1 i.e. 7.

60% of (x – y) = 45% (x + y)

⇒ 4(x - y) = 3(x + y)

⇒ 4x - 4y = 3x + 3y

⇒ x = 7y

⇒ y = x/7

⇒ y = k% of x

⇒ x/7 = k% of x

⇒ k = 14.28

⇒ 49% of k = 49100×14.28{49\over100} \times14.28 = 6.99 ≈ 7