A journey of 96 km takes one hour less by a fast train (A) than by a slow train (B). If the average speed of B is 16 km/h less than that of A, then the average speed (in km/h) of A is:

Question

Question

This question was previously asked in

SSC CGL Previous Paper 51 (Held On: 7 June 2019 Shift 3)

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Answer 1

Detailed solution:

Let the speed of train A be x km/h.

Speed of train B = (x – 16) km/h

According to the question

\(\Rightarrow \frac{{96}}{{x - 16}} - \frac{{96}}{x}{\rm{}} = {\rm{}}1\)

\(\Rightarrow \frac{{96{\rm{\;}} \times {\rm{\;}}16}}{{x\left( {x - 16} \right)}}{\rm{}} = {\rm{}}1\)

⇒ x² – 16x = 96 × 16

⇒ x² – 16x – 1536 = 0

⇒ (x – 48) (x + 32) = 0

⇒ x = 48 and x = -32 (not possible)

∴ Speed of train A = 48 km/h

Shortcut Trick Go through the options

Let the speed of train be 48 km/h.

And speed of train B = 48 – 16 = 32

⇒ (96/32) – (96/48) = 1

⇒ 3 – 2 = 1

⇒ 1 = 1 (Satisfied)