A student walks from his house at speed of $2\frac{1}{2}$ 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
| Day | Speed | What happened | |---|---|---| | First | $2\dfrac{1}{2}=\dfrac{5}{2}$ km/h | 6 min late | | Second | $\dfrac{5}{2}+1=\dfrac{7}{2}$ km/h | 6 min early | The two journeys therefore differ by $6+6=12\text{ min}=\dfrac{12}{60}=\dfrac{1}{5}$ hour. Let the distance be $d$ km. Using $\text{time}=\dfrac{\text{distance}}{\text{speed}}$, the slower time minus the faster time is that difference: $\dfrac{d}{\frac{5}{2}}-\dfrac{d}{\frac{7}{2}}=\dfrac{1}{5}$ $\Rightarrow \dfrac{2d}{5}-\dfrac{2d}{7}=\dfrac{1}{5}$ $\Rightarrow \dfrac{14d-10d}{35}=\dfrac{1}{5}$ $\Rightarrow \dfrac{4d}{35}=\dfrac{1}{5}$ $\Rightarrow d=\dfrac{35}{20}=\dfrac{7}{4}$ km