Target Exam

CUET

Subject

-- Applied Mathematics - Section B2

Chapter

Numbers, Quantification and Numerical Applications

Question:

A man travelled 91 km downstream in a river using a speedboat and then returned back. It took him 20 hours altogether. If the speed of his boat is 10 km/hr in still water, then rate of flow of water in the river is :

Options:

3 km/hr

9 km/hr

2.27 km/hr

7.4 km/hr

Correct Answer:

3 km/hr

Explanation:

The correct answer is Option (1) → 3 km/hr

Let speed of boat = 10 km/h

speed of river = x km/h

Downstream speed = 10 + x km/h

Upstream speed = 10 - x km/h

$Time=\frac{Distance}{Speed}$

Time downstream = $\frac{91}{10+x}$

Time upstream = $\frac{91}{10-x}$

Total time is $\frac{91}{10+x}+\frac{91}{10-x}=20$

$91(10-x)+91(10+x)=20(10-x)(10+x)$

$910-91x+910+91x=20(100-x^2)$

$91(20)=20(100-x^2)$

$91=100-x^2$

$x^2=9$

$x=3$