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3 brothers A, B, and C started simultaneously from a common point to go to a fair, which is 60 miles away from that point. A and B started on a bike that moves at a speed of 40 miles/hour, while C started walking at a constant speed of 10 miles/hour. B dropped A at a point X somewhere along the route and returned back on the bike to pick up C. B picked up C at point Y, and reached the fair at the same time A reached the fair. If B had traveled for 2 hours by the time he picked up C, find the distance that A had to walk?
A. 5
B. 10
C. 15
D. 20
E. 25
The OA is B.
In two hours C has already traveled 2*10=20 miles.
B drops A at X and comes to pick C means B has traveled X (forward)+(X-20) (backward to pick C)= X+(X-20). However, C in total has covered 2*40=80 miles going back and forth which leads to below equation.
X+(X-20)=80 -> 2X=80 -> X=50 the point @ which A is dropped by C. So, distance left for A to cover is 60-50=10 miles. Option B.
Has anyone another strategic approach to solve this PS question? Regards!
A. 5
B. 10
C. 15
D. 20
E. 25
The OA is B.
In two hours C has already traveled 2*10=20 miles.
B drops A at X and comes to pick C means B has traveled X (forward)+(X-20) (backward to pick C)= X+(X-20). However, C in total has covered 2*40=80 miles going back and forth which leads to below equation.
X+(X-20)=80 -> 2X=80 -> X=50 the point @ which A is dropped by C. So, distance left for A to cover is 60-50=10 miles. Option B.
Has anyone another strategic approach to solve this PS question? Regards!
