Question: A student goes to school at the rate of $ \frac{5}{2}km/h $ and reaches 6 min late. If he travels at the speed of $ 3km/h, $ he reaches 10 min earlier. The distance of the school is
Options:
A) 46 km
B) 20 km
C) 10 km
D) 4 km
Show Answer
Answer:
Correct Answer: D
Solution:
- [d] Let the distance to school be D km and time taken is t.
Case I When he goes at a rate of $ \frac{5}{2}km/h. $
He reaches 6 min late.
$ \therefore $ $ \frac{D}{\frac{5}{2}}=t+\frac{6}{60} $
$ \Rightarrow $ $ \frac{2D}{5}=t+\frac{1}{10} $
$ \Rightarrow $ $ t=\frac{2D}{5}-\frac{1}{10} $
Case II When he goes at a rate of 3 km/h.
He reaches 10 min earlier
$ \therefore $ $ \frac{D}{3}=t-\frac{10}{60} $
$ \Rightarrow $ $ \frac{D}{3}=t-\frac{1}{6} $
On putting value of t from Eq. (i).
$ \frac{D}{3}=\frac{2D}{5}-\frac{1}{10}-\frac{1}{6} $
$ \Rightarrow $ $ \frac{2D}{5}-\frac{D}{3}=\frac{1}{10}+\frac{1}{6} $
$ \Rightarrow $ $ \frac{6D-5D}{15}=\frac{3+5}{30} $
$ \Rightarrow $ $ D=\frac{8\times 15}{30}=4km $
BETA
