Question

An aeroplane travels 1000 km with tailwind in same time that it travels

600 km against a headwind. If the speed of aeroplane is 500 km/h in still

air, what is the speed of wind?
​

Answer 1

Answer:

Speed of the wind = 125 km/hr

Step-by-step explanation:

Given:

• The aeroplane travels 1000 km with tailwind in the same time it travels 600 km against a headwind.
• Speed of the aeroplane = 500 km/hr

To Find:

• Speed of the wind

Solution:

Let speed of the wind be x km/hr

Let the time taken to travel in both cases given by y hours.

Now,

Speed with tailwind = (500 + x) km/hr

Speed against headwind = (500 - x) km/hr

We know that,

Time = Distance/Speed

In the first case time taken is given by,

$\sf Time =\dfrac{1000}{500+x}---(1)$

In the second case given,

$\sf Time =\dfrac{600}{500-x}---(2)$

But by given time taken in case 1 and case 2 is same.

Hence,

$\sf \dfrac{1000}{500+x} =\dfrac{600}{500-x}$

$\implies \sf \dfrac{10}{500+x} =\dfrac{6}{500-x}$

Cross multiplying,

10 (500 - x) = 6 (500 + x)

5000 - 10x = 3000 + 6x

5000 - 3000 = 16 x

2000 = 16x

x = 2000/16

x = 125 km/hr

Hence the speed of the wind is 125 km/hr.

Answer 2

Answer:

$\huge \bf \: required \: answer$

Distance of aeroplane with tailwind = 1000 km

Distance of aeroplane with headwind at same time = 60 km

Speed of aeroplane = 500 km/h

Speed of wind be x

Now,

$\sf \: speed \: with \: tailwind \: = 500 + x$

$\sf \: speed \: agains \: headwind \: = 500 - x$

We know that

${ \huge {\bf {\underline{time \: = \dfrac{distance}{speed} }}}}$

First case

$\sf \: time \: = \dfrac{1000}{500+x}$

Second case

$\sf \: time \: = \dfrac{600}{500 - x}$

Time taken in case 1 and case 2 is same.

$\sf\dfrac{1000}{500 + x } = \dfrac{600}{500 - x}$

Cross multiplication

10 (500 - x) = 6 (500 -x)

5000 - 10 x = 3000 - 6x

5000 - 3000 = 10x + 6x

5000 - 3000 = 16x

2000 = 16x

x = 2000 ÷ 16

x = 125 km/hr

$\bf \: speed \: of \: wind \: = 125 \: km \: per \: hour$