another math question!?

@150m/min boy1 will take 8min(1200/150) to complete 1.2km path

@80m/min boy2 will take 15min(1200/80) to complete 1.2km path

For reaching starting point together u need to find LCM of 8 and 15

which is 120min

which is 2hrs

I hope you got the explaination

The faster boy does a lap in

(1200m)/(150m/min) = 8 min,

slower boy in

(1200m)/(80m/min) = 15 min.

LCM of 8 and 15 =

LCM of 2*2*2 and 3*5 =

2*2*2*3*5 = 120 minutes = 2 hours

.

Assuming a closed loop for a path, of distance 1.2 km, boy A will complete the loop in time:

T-a = 1.2 km /(150 m/min) = (1.2/.15) minutes = 8 minutes.

Likewise, boy B will complete the loop in time:

T-b = 1.2 km/(80 m/min) = (1.2/0.08) minutes = 15 minutes.

So the question boils down to: Will these two ever be at the end of a complete loop at the same time?

This would mean:

N-a * 8 = N-b * 15

or

N-a = (15/8)*N-b

where both N-a and N-b must be integers.

Now this is clear: when N-b = 8, N-a will be 15.

That means the time will be:

time = 8*15 = 8*(hour/4) = 2 hours

So, they will meet in 2 hours.

answer: after t = 0 hrs they will both be simultaneously at the starting point.

Let t be the number of hours.

150(60t) - 80(60t) = 1200

Solve for t,

t = 0.2857 hours