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
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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?

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.

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.
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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
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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.