Question

14. A sparrow flies to see a friend at a speed of 4 km/h. His friend is
out, so the sparrow immediately returns home at a speed of
5 km/h. The complete journey took 54 minutes. How far away
does his friend live?​

Answer 1

hope it's helps you

answer in the attachment

Answer 2

$Let \: the \: distance = d \: km$

$Speed \: of \: the \: sparrow \: to \: visiting$

$his \: friend v_{1} = 4 \: km/h$

$Time \: taken \: to \: reach \: his \: friend$

$t_{1} = \frac{d}{v_{1}}$

$= \frac{d}{4} \: --(1)$

$Speed \: of \: the \: sparrow \: returning$

$from \: his \: friend = 5 \: km/h$

$Time \: taken \: to \: reach \: his \: house$

$t_{2} = \frac{d}{v_{2}}$

$= \frac{d}{5} \: --(2)$

$Total \: time \: of \: his \: journey = 54 \: mi$

$\implies t_{1} + t_{2} = \frac{54}{60} \: hours$

$\implies \frac{d}{4} + \frac{d}{5} = \frac{54}{60} \: hours$

$\implies \frac{5d+4d}{20} = \frac{54}{60} \: hours$

$\implies \frac{9d}{20} = \frac{54}{60} \: hours$

$\implies d= \frac{54}{60} \times \frac{20}{9}$

$\implies d = 2 \: km$

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