Subject: General Aptitude · Chapter: Time and Distance · Exam: · Marks: · Difficulty:
A man travels for 5 hours 15 minutes. If he covers the first half of the journey at 60km/h and rest at 45km/h. Find the total distance travelled by him. [SSC-CHSL (10 + 2) Exam, 2015]
(a)$1028 \frac{6}{7}$ km
(b)189 km
(c)378 km
(d)270 km
Answer
Answer (as printed): D
Explanation
Let the distance covered be $2 x \mathrm{km}$. Time $=\frac{\text { Distance }}{\text { Speed }}$ Time taken to covers the first half and second half of the journey in $t_{1}$ and $t_{1}$ hours $$\begin{aligned} & \Rightarrow \frac{a}{60}=t_{1} \\ & \Rightarrow \frac{a}{45}=t_{2} \end{aligned}$$Adding (i) and (ii) we get $$\begin{aligned} & \frac{a}{60}+\frac{a}{45}=t_{1}+t_{2} \\ & \frac{a}{60}+\frac{a}{45}=5 \frac{15}{60}=5 \frac{1}{4} \\ & \Rightarrow \frac{3 a+4 a}{180}=\frac{21}{4} \\ & \Rightarrow 7 a=\frac{21}{4} \times 180 \\ & \Rightarrow a=\frac{21 \times 180}{4 \times 7}=135 \mathrm{km} \end{aligned}$$ ∴ Length of total journey. $=2 \times 135=270 \mathrm{km}$.