In the figure above, car A and car B travel around a circular park

Circumference = 2*(pi)*(20) = 40(pi) miles.

Distance travelled by Car A in 20 minutes => 20 miles.

Remaining distance = (40*3.14) - 20 = 105.6 miles.

Relative speed = (60+40) = 100 mph.

Time taken to meet = 105.6/100 = 1.056 hours = 1.056*60 = 63 (approx).

Car A had travelled for 20 mins initially, so total time = 63 + 20 = 83 mins. Ans (C).

r=20 miles --> total track length = \(2\pi*r= 40\pi\)
After 20 mins, car A traveled (distance)= \(\frac{20}{60}*60=20\) (miles) --> Distance btw A and B when B leaves = \(40\pi-20=20*(2\pi-1)\)
Relative speed of both cars: \(60+40=100\) kph
A and B meet after (time) = \(\frac{{20*(2\pi-1)}}{100}=\frac{(2\pi-1)}{50}\) (hrs) or in minutes = \(\frac{{(2*3.14-1)*60}}{50}\approx{63}\) --> After car A start, they meet in: \(20+63=83\) (mins) --> Answer C

800score Official Solution:

To visualize this question, think about a circle as just a line segment joined at its two ends. Let's first cut the circle at the starting point and make it in to a straight line. Notice that cars A and B are at opposite ends of the line traveling towards each other.

A [ ______________________ ] B

We start by solving for the circumference of the park:
C = 2πr.

Plugging 20 into the equation, we get:
C = 2 × 3.14 × 20 = 125.6 miles.

To calculate how long it takes the cars to meet, use the variable T as the time, in hours, that it takes for the cars to meet after car A starts.

When the cars meet, car A will have traveled 60T miles (distance = rate × time) and car B will have traveled 40 × (T – (1/3)) miles (since it left 20 minutes later, it will have been traveling 1/3 of an hour less). Furthermore, the distance both cars will travel combined is 125.6 miles. So we have the equation:

Total Distance = Distance A travels + Distance B travels
125.6 = 60T + 40(T – (1/3))
125.6 = 100T – (40/3).

Then we have approximately: 139 = 100T. So, T = 1.39 hours.This is equivalent to 1.39 × 60 = 83 minutes.

The correct answer is choice (C).

Similar questions to practice:
in-the-figure-above-car-a-and-car-b-simultaneously-begin-traveling-ar-201859.html (DS)
in-the-figure-above-car-a-and-car-b-simultaneously-begin-traveling-ar-190712.html (PS)

Hi Bunuel

can I subtract the distance that A traveled in 20 minutes ealier than B which is 20 miles. then combine both speed rates and get how many minutes required A & B to meet based on 105.6 miles only ?
I hope you understand my language.

sorry plus 20 minutes

total track distance 2*20*pi ; 40 * pi
in 20 mins A would have covered distance of 20 miles
net distance b/w A & B = 40*pi -20 ; 105.6 miles
relative speed = 100 mph
time they meet 1.056 hr or say 63 mins
total time taken when they meet from start 63+20
83 mins
option C

Length of the racetrack = 2*π*20 = 40π
Since Car B starts after 20 minutes of A, the distance covered by A before B = 60*20/60 = 20m

Now, the total distance covered by both the cars = 40π - 20 m
Since both the cars are moving in opposite directions, the time taken to meet after B starts = (40π - 20)*60/100 = 63.36 minutes.

Thus, time taken to meet after A started = 20 + 63.3 = 83.3 minutes.
The correct option is C.

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