Question

A person walks from his house at a speed of $4$ km/hr and reaches his schools $5$ minutes late. If his speed has been $5$ km/hr, he would have reached $10$ minutes earlier. The distance of the school from his house is

• A. $5$ km
• B. $6$ km
• C. $7$ km
• D. $8$ km

Answer 1

Answer: (A)

$5$ km

Let the correct time to reach the school be ‘t’ hrs.

Then,the time taken by the man when he wakes at a speed of $4km$ will be ‘t’ hrs+5 mins,

which is nothing but $(t+\frac{5}{60})$hrs.

And,the time taken by the man when he walks at 5km/hr will similarly be $(t-\frac{10}{60})$hours.

As the distance in both the cases is same,

$4(t+\frac{5}{60}$)hrs=$5(t-\frac{10}{60})$hrs

$4t+\frac{20}{60}$=$5t-\frac{50}{60}$

Therefore$t=\frac{70}{60}$=$\frac{7}{6}$hrs

Actual distance=$4(t+\frac{5}{60})$

= $4(\frac{70}{60}+\frac{5}{60})$

=$4\times \frac{75}{60}$=$5kms.$