Question

A student walks from his house at $2\frac{1}{2}$ km an hour and reaches his school 6 minutes late. The next day he increases his speed by 1 kilometer an hour and reaches 6 minutes early. How far is the school from the house ?

• A. $2.5$ km
• B. $3$ km
• C. $1.75$ km
• D. $1$ km

Answer 1

Answer: (C)

$1.75$ km

Let the distance between the home & the school of the boy = x km.

In the first case, the speed is $2\frac{1}{2}$km/hr = $\frac{5}{2}$km/hr.

$\therefore$ Time taken = $x÷\frac{5}{2}$hr.

=$\frac{2x}{5}$hr.

In the second case his speed= 1 + $\frac{3}{2}$km/hr. =$\frac{7}{2}$km/hr.

$\therefore$ Time taken=$x÷\frac{7}{2}$hr.=$\frac{2x}{7}$hr.

$\therefore$ The difference in time = ($\frac{2x}{5}-\frac{2x}{7}$)hr.=$\frac{4x}{35}$hr.=$\frac{4x}{35}\times 60$min.=$\frac{48x}{7}\times$min.

Now , in the first case he is $6$ min late = $-6$ min. and

In the second case, he is $6$ min early = $+6$ min.

$\therefore$ The difference in time$=[6-(-6\left)\right]$min.$=12$ min.

$\therefore$ $\frac{48x}{7}$min$.=12$ min.

Or $4x=7$

Or $x=$$\frac{7}{4}$$=\mathrm{1.75.}$

$\therefore$ The distance between the home & the school of the boy = $1.75$ km.