A student walks from his house at 4 km per hour and reaches his school late by 5 minutes. If his speed has been 5 km per hour then he would have reached 10 .minutes early. The distance of the school from his house is

\frac{5}{3}

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5 km

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6 km

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4 km

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Solution

The correct option is A 5 km
$L e t t h e d i s t a n c e f r o m h o m e t o s c h o o l b e d k m . T h e n t i m e t a k e n a t t h e s p e e d o f 4 k m / h r i s t_{1} = \frac{d}{4} h r . A n d t i m e t a k e n a t t h e s p e e d o f 5 k m / h r i s t_{2} = \frac{d}{5} h r . ∴ t_{1} - t_{2} = (\frac{d}{4} - \frac{d}{5}) h r . = \frac{d}{20} h r . - - - - (1) I n t h e f i r s t c a s e h e i s l a t e b y 5 m i n . a n d i n t h e s e c o n d c a s e h e i s e a r l i e r b y 10 m i n . ∴ D i f f e r e n c e t_{1} - t_{2} i s [10 - (- 5)] m i n . \Rightarrow t_{1} - t_{2} = 15 m i n . \Rightarrow t_{1} - t_{2} = \frac{1}{4} h r . - - - - (2) f r o m (1) a n d (2) w e g e t \frac{d}{20} = \frac{1}{4} \Rightarrow d = 5 k m (A n s)$

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