Question
Download Solution PDFA 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:
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFDetailed 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)
Last updated on Jul 12, 2025
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