Question

A man travels from his house to his office at 5 km/h and reaches his office 20 minutes late. If his speed had been 7.5 km/h, he would have reached his office 12 minutes early. Find the distance from his house to his office.

Answer 1

Let the distance between house and office be 'd' km .

$First \:speed = 5 \:km/hr$

$Time (t_{1}) = \frac{distance}{speed} \\= \frac{d}{5}\: ---(1)$

$Second /:speed = 7.5 \:km/hr$

$Time (t_{2}) = \frac{d}{7.5} \: km/hr \:---(2)$

/* According to the problem given*/

$Difference \:in \: time = 20 \:mi - ( -12 ) \:mi \\= 20 + 12 \\= 32 \: mi \\= \frac{32}{60} \: hours \\= \frac{8}{15}$

$\implies \frac{d}{5} - \frac{d}{7.5} = \frac{8}{15}$

$\implies \frac{d}{5} - \frac{2d}{15} = \frac{8}{15}$

$\implies \frac{ 3d - 2d}{15} = \frac{8}{15}$

$\implies \frac{ d}{15} = \frac{8}{15}$

$\implies d = \frac{8\times 15}{15}$

$\implies d = 8 \: km$

Therefore.,

$\red { Distance \: from \: house \:to \: office }\green {= 8 \:km }$

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