Question

In a circular race track of length 100 m, three persons A, B and C start together.
A and B start in the same direction at speeds of 10 m/s and 8 m/s respectively.
While C runs in the opposite at 15 m/s. When will all the three meet for the first time on the after the start?
a) After 4 s b) After 50 s c) After 100 s d) After 200 s

Answer 1

Let’s calculate the distance travelled by them after 4 s

A travels 10 × 4 = 40 m in 4 s

B travels 8 × 4 = 32 m in 4 s

C travels 15 × 4 = 60 m in 4 s

Let’s calculate the distance travelled by them after 50 s

A travels 10×50 = 500 m in 50 s

B travels 8×50 = 400 m in 50 s

C travels 15×50 = 750 m in 50 s

Let’s calculate the distance travelled by them after 100 s

A travels 10×100 = 1000 m in 100 s

B travels 8×100 = 800 m in 100 s

C travels 15×100 = 1500 m in 100 s

Let’s calculate the distance travelled by them after 200 s

A travels 10×200 = 2000 m in 200 s

B travels 8×200 = 1600 m in 200 s

C travels 15×200 = 3000 m in 200 s

Observe all the results approximately after 100 s A, B, C will meet for the first time on after start.

Answer 2

Answer:

C) 100 Sec

Step-by-step explanation:

A takes 10 sec to finish one complete round,

$A = \frac{100m}{10m/s} = 10sec$

Similarly, B takes 12.5 sec to finish one complete round,

$B = \frac{100m}{8m/s} = \frac{25}{2} sec = 12.5 sec$

Similarly, C takes 20/3 sec to finish one complete round

$C = \frac{100m}{15m/s} = \frac{20}{3} sec$

For them to meet, the time will be in LCM of above three times:

i.e. $LCM(\frac{10}{1} , \frac{25}{2}, \frac{20}{3} )$

$= \frac{LCM(10,20,25)}{HCF(1,2,3)}$

$= \frac{100}{1}$

$= 100 sec$

For the first time all of them meet at 100sec for second time they meet at 200sec after the start.