Subject: General Aptitude · Chapter: Time and Distance · Exam: 2007 · Marks: · Difficulty:
A student walks from his house at a speed of $2 \frac{1}{2}$ km per hour and reaches his school 6 minutes late. The next day he increases his speed by 1 km per hour and reaches 6 minutes before school time. How far is the school from his house?
(a)$1 \frac{1}{4} \mathrm{km}$
(b)$1 \frac{3}{4} \mathrm{km}$
(c)$2 \frac{1}{4} \mathrm{km}$
(d)$2 \frac{3}{4} \mathrm{km}$
Answer
Answer (as printed): B
Explanation
Let the distance be $x \mathrm{km}$. $$\begin{aligned} & \text { Difference in timings }=12 \mathrm{min}=\frac{12}{60} \mathrm{hr}=\frac{1}{5} \mathrm{hr} . \\ & \therefore \quad \frac{x}{\left(\frac{5}{2}\right)}-\frac{x}{\left(\frac{7}{2}\right)}=\frac{1}{5} \Leftrightarrow \frac{2 x}{5}-\frac{2 x}{7}=\frac{1}{5} \\ & \quad \Leftrightarrow 14 x-10 x=7 \Leftrightarrow x=1 \frac{3}{4} \mathrm{km} . \end{aligned}$$