A) 46 km
B) 20 km
C) 10 km
D) 4 km
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). |
\[\frac{D}{3}=\frac{2D}{5}-\frac{1}{10}-\frac{1}{6}\] |
\[\Rightarrow \] \[\frac{2D}{5}-\frac{D}{3}=\frac{1}{10}+\frac{1}{6}\] |
\[\Rightarrow \] \[\frac{6D-5D}{15}=\frac{3+5}{30}\]\[\Rightarrow \]\[D=\frac{8\times 15}{30}=4km\] |
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