SOLU lets first find out when they both will meet
as we all know both cyclist are coming towards each other they will meet soon so total distance must be divided by sum of their speed.
total distance will be 400 meters circumference of circle
so 400/10+5= 400/15
=80/3 in every 80/3 seconds they will meet
so in 40 min they will meet
40*60(convert in seconds)
we will divide total time with they took time to meet once so to identify how many times they met.
40*60*3/80
=90
so OA IS D

400m/15 combined mps=26 2/3 seconds until first crossing
40 min*60 sec=2400 seconds
2400/(26 2/3)=90 crossings
D

Find combined rate per minute. (R + R). Find time required to cross paths one time (D/R).
Finally, divide total time by time per crossing. Answer equals the # of times they meet.

Convert seconds to minutes

If time units are different, you can convert X seconds to one minute. Multiply the rate's number of seconds by N to get to 60 seconds. Multiply the numerator by N, too.

Rate of Boy 1:
\((\frac{10m}{1sec}*\frac{60}{60})=(\frac{600m}{60secs})=\frac{600m}{1min}\)

Rate of Boy 2:
\((
\frac{5m}{1sec})=(\frac{300m}{60secs})=\frac{300m}{1min}\)

Add rates

When travelers move in opposite directions, whether towards or away from one another, add rates (speeds). Combined rate:

\(\frac{600m}{1min} + \frac{300m}{1min}=\frac{900m}{1min}\)

Time it takes for them to cross paths once?

\(R*T = D\), so \(T=\frac{D}{R}\)

\(T=\frac{400m}{900\frac{meters}{minute}}=\frac{400}{900}mins=\frac{4}{9}min\)

for them to meet once

Number of times they cross/meet?

\(\frac{TotalMinutes}{MinutesPerOneMeet}=\) # of times they meet

\(\frac{40}{\frac{4}{9}}=40*\frac{9}{4}=90\) times that they meet

Answer D
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We can let t = the time they will meet for the first time:

10t + 5t = 400

15t = 400

t = 400/15 = 80/3 seconds

Since 40 minutes = 40 x 60 = 2,400 seconds, they will meet 2,400/(80/3) = (2,400 x 3)/80 = 30 x 3 = 90 times.

Answer: D
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total time taken by two cyclists ; 10+5 ; 15 mps
400/15 ; 80/3 seconds
in 40 mins total circles ; 40 *60 ; 2400 sec
2400/80 *3 ; 90 times
IMO D

Given: Two boys start cycling around a circular track in opposite directions at constant speeds.

Asked: if the circumference of the track is 400 meters, the faster boy cycles at 10 meters per second and the slower boy cycles at 5 meters per second, how many times would they have crossed each other after cycling for 40 minutes?

40 mins = 40 *60 = 2400 seconds

Time taken to meet each other = 400/15 = 26 10/15 = 26 2/3 seconds

Number of times they met each other in 40 mins = 2400/(400/15) = 24*15/4 = 6*15 = 90 times

IMO D
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Speed of A= 10m/sec
Speed of B= 5m/sec

Distance=400m (One revolution)
For A
speed=distance/time
10m/sec=400/time
Time=40sec
Total Time Given 40mins(2400secs)
No. of revolutions made by A in 40mins(2400sec)
If it takes A 40secs for one revolution so in 40mins that is 2400secs A will make 2400secs/40sec= 60 revolutions

For B
Speed=distance/time
5m/sec=400/time
time=80secs
Total Time Given 40mins(2400)
No. of revolutions made by B in 40mins(2400secs)
If it takes B 80secs for one revolution then in 2400secs it will make 2400/80=30 revolutions.

Adding no. of revolution of A and B= 60+30=90 revolutions in total.

Please check and let me know If I am wrong, I tried on a few problems and I got the answer through this method.
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total time they cycle for is 40*60=2400 secs
to find: 2400/how long they take to meet the first time

Relative speed= 15 (since they are traveling in opp directions)
total distance= 400
Time they meet = 400/15 = 80/3

answer = 2400*/80 = 90