Time and Distance Part 9

Problem 1: From a point on a circular track 5km long A, B, C  started running in the same direction at the same time will speed of 2\frac{1}{2}km/hr, 3km/hr,2km/hr then on the starting point all three will meet again?

Solution :

L.C.M of time A,B,C

T = \frac{5\times 2}{5},\frac{5}{3},\frac{5}{2}

T=2,\frac{5}{3},\frac{5}{2}

\frac{10}{1}=10hrs

Problem 2: A man covered a certain distance at some speed, if he had move 3km/h faster, he would have taken 40min less. If he moved 2km/h slower, he would have taken 40min more. Find the distance in km?

Solution:

+3           -40

-2            +40

Cross multiply +3 * +40 = 120

-2 * -40 = 80

120:80  = 3:2

1———> 5

3——–>?(\frac{5\times 3}{1})=15

2———->?(\frac{5\times 2}{1})=10

Distance=\frac{S_{1}\times S_{2}}{S_{1}\sim S_{2}}\times \Delta t

\frac{15\times 10}{5}\times \frac{80}{60}=40km

Problem 3: “A” leaves Mumbai at 6 AM and reaches Bangalore at 10 AM. “B” leaves Bangalore at 8 AM and reaches Mumbai at 11:30 AM. At what time do they cross each other?

Solution:

The time gap between 6 AM-10 AM=4 hours

The time gap between 8 AM and 11:30 AM=3.5 hours

A                       B

T            4                      3.5

S            3.5                     4

Distance = Speed * Time = 4*3.5 = 14

In two hours A travel 7 km (2*3.5=7km)

Time=\frac{Distance}{Speed}=\frac{7}{7.5}\times 60=56min

At 8:56 both trains will meet.

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