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SSC Selection Post 2022 PYQ — Question 58

SSC Selection Post - Higher Secondary Level - 14 March 2022 - Shift 2 • Quantitative Aptitude • Q58

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P and Q travel a distance of 120 km such that the speed of P is faster than Q. The sum of their speeds is 75 km/h, and together they took a total time of 6 hours and 40 minutes. Then the speed of Q (in km/h) is:

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Correct answer: C

Explanation

The correct answer is option 3 i.e. 30.

Let Q's speed = x

P's speed = 75 - x

Time = 6 hr 40 min = 203\frac{20}{3}

According to the question,

⇒ 120x+12075−x=203\frac{120}{x} + \frac{120}{75 - x} = \frac{20}{3}

⇒ 360x+36075−x=20\frac{360}{x} + \frac{360}{75 - x} = 20

⇒ 360(75−x) + 360x = 20x(75−x)

⇒​​​​​​​ 27000 − 360x + 360x = 1500x − 20x2

⇒​​​​​​​ 27000 = 1500x − 20x2

⇒​​​​​​​ x2 − 75x + 1350 = 0

⇒​​​​​​​ x2 - 45x - 30x + 1350 = 0

⇒​​​​​​​ x(x - 45) - 30(x - 45) = 0

⇒​​​​​​​ (x - 45)(x - 30) = 0

⇒​​​​​​​ x = 45, 30

The speed of Q = 30 km/hr