Question

Two bicycle riders A and B make a 36 km trip and
reach the destination at the same time. While cyclist
A travelled non-stop at an average speed of
18 km/h, cyclist B travelled with a lunch break of
30 min. The average speed of B for the time interval
he was in motion is
(1) 18 km/h
(2) 24 km/h
(3) 72/5 km/h
(4) 12 km/h​

Answer 1

Answer:

(2). 24km/h

Explanation:

d travelled by A=36 km

d travelled by B=36 km

v of A=18 km/h

so,

time taken by A=36/18

=2 h

therefore time taken by B=2-1/2

=3/2 h

so,

v of B =36×2/3

=24 km/h

Answer 2

Answer:

The average speed of B for the time interval  is 24 km/h.

(2) is correct option.

Explanation:

Given that,

Distance = 36 km

Time = 30 min = 1800 sec

Speed of cyclist A = 18 km/h

We need to calculate the time

Using formula of average speed

$v = \dfrac{d}{t}$

$t=\dfrac{d}{v}$

Put the value into the formula

$t=\dfrac{36}{18}$

$t=2\ hr$

We need to calculate the average speed of B for the time interval

The time taken is,

$t'=t-30$

$t'=2- 0.5$

$t'=1.5\ hr$

Using formula of speed again

$v=\dfrac{d}{t}$

$v=\dfrac{36}{1.5}$

$v=24\ km/h$

Hence, The average speed of B for the time interval  is 24 km/h.