Vallabhm001 wrote:
A bus left point X for point Y. Two hours later a car left point X for Y and arrived at Y at the same time as the bus. If the car and the bus left simultaneously from the opposite ends X and Y towards each other, they would meet 1.33 hours after the start. How much time did it take the bus to travel from X to Y?
(A) 2 h
(B) 4 h
(C) 6 h
(D) 8 h
(E) 10 h
Algebraic method
Let the speed of the bus and car be b and c.
Since the car starts after 2 hrs and covers the lead of bus in remaining time, the distance b-c covers in 2hrs is 2*c => \(\frac{2c}{b-c}=2...........2c=2b-2c......b=2c\)
That is , the speed of car is twice the speed of bus. c:b=2:1.
If the car and the bus left simultaneously from the opposite ends X and Y towards each other, they would meet 1.33 hours after the start.
\(\frac{d}{c+b}=1.33......\frac{d}{c+2c}=1.33.......\frac{d}{c}=3*1.33=4\)
Also, b:c=2:1 means bus will cover 1/3 and bus 2/3 of total distance => 1/3 in 1.33 h means 1 in 3*1.33=3.99~4.
Choices
A) 2 hr
This would mean that the car reaches in 2-2 or 0 hr. NOT possible
B) 4 hr
This would mean the bus takes 4-2 or 2 hr.
Combined distance both will travel in 1 hr = 1/2+1/4=3/4.
Or
3/4 in 1 hr OR total distance in 4/3 or 1.33 hrMatches with our statement..ANSWER
B