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A student walks from his house at speed of 2122\frac{1}{2} km/hour and reaches his school late by 6 min. Next day, he increases his speed by 1 km/hour and reaches 6 min before school time. How far is the school from his house?

Solution

✅ Correct Option: 2
DaySpeedWhat happened
First212=522\dfrac{1}{2}=\dfrac{5}{2} km/h6 min late
Second52+1=72\dfrac{5}{2}+1=\dfrac{7}{2} km/h6 min early

The two journeys therefore differ by

6+6=12 min=1260=156+6=12\text{ min}=\dfrac{12}{60}=\dfrac{1}{5} hour.


Let the distance be dd km. Using time=distancespeed\text{time}=\dfrac{\text{distance}}{\text{speed}}, the slower time minus the faster time is that difference:

d52−d72=15\dfrac{d}{\frac{5}{2}}-\dfrac{d}{\frac{7}{2}}=\dfrac{1}{5}

⇒2d5−2d7=15\Rightarrow \dfrac{2d}{5}-\dfrac{2d}{7}=\dfrac{1}{5}

⇒14d−10d35=15\Rightarrow \dfrac{14d-10d}{35}=\dfrac{1}{5}

⇒4d35=15\Rightarrow \dfrac{4d}{35}=\dfrac{1}{5}

⇒d=3520=74\Rightarrow d=\dfrac{35}{20}=\dfrac{7}{4} km

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