A bus left point X for point Y. Two hours later a car left point X for

Question

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

(A) 2 h
(B) 4 h
(C) 6 h
(D) 8 h
(E) 10 h

rsrighosh · Answer

I am getting 8 hrs instead of 4. Can anybody please help me where i am making mistake?

I used relative speed. Let speed of Bus = b mph and speed of car = c mph X ------------------- Y Distance = d Two hours later a car left point X for Y and arrived at Y at the same time as the bus. This can be written as \frac{d }{ (c-b)} = 2 => d = 2(c-b) ---> eq 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. This can be written as \frac{d }{ (c+b)} = 1.33 => d = 1.33(c+b) ---> eq 2 So.. eq 1 = eq 2 2(c-b) = 1.33(c+b) => 2c - 2b = 1.33c+ 1.33b => (2-1.33)c = (2+1.33)b => 0.67c = 3.33b => c = 5b Time taken by bus to cover distance d \frac{d}{b} = \frac{2(c-b)}{b} =\frac{2(5b-b)}{b} = \frac{2(4b)}{b} = 8 Please help

prateekchugh · Answer

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

I would suggest in this type of question, use the answer options to solve the question fast

Let the Distance = d, Speed of bus= b, speed of car =c

Write Word problem into equation:

Two hours later a car left point X for Y and arrived at Y at the same time as the bus-->  \frac{d}{b} = \frac{d}{c} + 2 .....equation 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}{(b+c)} = 1.33 .....equation 2

We have two equations: Now let take the option D: 6 hrs --> and let take d=100 miles--->  \frac{100}{b}=6 -->  b=\frac{50}{3}  miles/hr
From equation, we need to find that is  \frac{d}{c}  =4 or c= 25 miles/hr

enter the values in equation 2:  \frac{100}{(50/3 +c)}  =1.33--> \frac{100}{1.33}= \frac{50}{3 }+c  --> c= 58.5 miles/hr---> but we need c= 25 miles/hr. Therefore we need some less time

Lets take option B: 4 hrs-->and let take d=100 miles--->  \frac{100}{b} =4--> b=25 miles/hr
From equation, we need to find that is  \frac{d}{c } =2 or c= 50 miles/hr

enter the values in equation 2:  \frac{100}{(25 +c)}  =1.33--> \frac{100}{1.33} = 25 +c --> c= 50.18 miles/hr---> BINGO

Option B is Correct.

chetan2u · Answer

You are wrong in the highlighted portion.
The distance that is covered by relative speed c-b is the lead taken by the bus, 2*c.

chetan2u · Answer

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 hr 
Matches with our statement..ANSWER

B

rihsat · Answer

Let,

Distance from X to Y = d
Speed of Bus = b
Speed of Car = c
Time taken by bus = t =?

as speed = distance/time
b=d/t
c=d/(t-2)

1.33 hours = 4/3 hours

now as they meet after the same time so

distance covered by bus in 4/3 hours + distance covered by car in 4/3 hours = total distance

=> d/t*4/3+d/(t-2)*4/3=d
=> 4/3{1/t+1/(t-2)}=1
=> (t-2+t)/{t(t-2)}=3/4
=>2t-2/t^2-2t=3/4
=>3t^2-6t=8t-8
=>3t^2-14t+8=0
=>(t-4) (3t-2)=0
so, t=4 or t=2/3

4 is given in option b.

Muzamil94 · Answer

Let total distance be 1 units, speed of bus = b, car = c and t be the time taken by bus and car, then by question .

b = 1/t and c = 1/(t - 2).....(1)

Also (b + c)* 1.33 = 1......(2)

Substitute values of 1 in 2, we get

3t^2 - 14t + 8 = 0
=> t = 4 and t = 2/3. t = 4 is the required answer.

Posted from my mobile device

effatara · Answer

Let the time taken by the Bus and the Car to cover XY be 't' hrs and (t-2) hrs respectively and their meeting point be M. Since we know (or can easily calculate) the times the Bus and the Car take to cover MX and MY respectively there is a handy formula we can use:

(Time taken by Bus to cover MX)*(Time taken by Car to cover MY) = Square of time they take to meet at M (4/3 hrs)
(t-4/3)(t-2-4/3)=(16/9)...> 9t^2 - 42t +24...> 3t^2 - 14t +24...> 3t^2 -12t - 2t +8 =0...> 3t(t-4) -2(t-4)...> (t-4)(3t-2)=0
So either t=4 or 2/3 but it cannot be the latter because that yields a negative value for the speed of the car. T/4, t=4.

ANS:B

ashishpathak · Answer

We can back solve this question after scanning the answer choices
if the bus took 4 hours the car would have taken 2 hours. let the distance be 8 km(4 X2) with this distance the speed of the bus is 2 km /hr and the car is 4 km /hr.when they are closing in the combined speed is 6 km /hr. the distance is 8 km time will be 8/6=1.33.
Will appreciate any feedback thanks

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