A student goes to school at the rate of \[2\frac{1}{2}\,\,km/h\] and reaches 6 min late. If he travels at the speed of 3 km/h, he is 10 min early. The distance (in km) between the school and his house is |
A) 5
B) 4
C) 3
D) 1
Correct Answer: B
Solution :
Let the required distance be x km. |
Difference between time \[=\frac{16}{60}\] |
Then, \[\frac{x}{5}-\frac{x}{3}=\frac{16}{60}\]\[\Rightarrow \]\[\frac{2x}{5}-\frac{x}{3}=\frac{4}{15}\] |
\[\Rightarrow \] \[\frac{6x-5x}{15}=\frac{4}{15}\] |
\[x=4\,km\] |
Alternate Method |
Here, \[{{t}_{1}}=6\,\,\min ,\]\[{{t}_{2}}=10\,\,\min ,\] |
\[{{S}_{1}}=2\frac{1}{2}=\frac{5}{2}\,\,km/h\] and \[{{S}_{2}}=3\,km/h\] |
Distance \[=\,\,\frac{({{t}_{1}}+{{t}_{2}}){{S}_{1}}{{S}_{2}}}{({{S}_{2}}-{{S}_{1}})\times 60}=\frac{(6+10)\times \frac{5}{2}\times 3}{\left( 3-\frac{5}{2} \right)\times 60}\] |
\[=\,\,\frac{\frac{16\times 15}{2}}{\frac{1}{2}\times 60}=\frac{16\times 15}{60}=4\,km\] |
You need to login to perform this action.
You will be redirected in
3 sec