Two buses P and Q start from cities A and B towards each other. They m

Question

Two buses P and Q start from cities A and B towards each other. They meet at a distance of 22 km from City B, cross each other to reach city B and A and turn back. During the second time they meet at a distance of 10 km from City A. Find the distance between cities A and B.

A. 24 km
B. 56 km
C. 60 km
D. 72 km
E. 84 km

A. 24 km 
B. 56 km 
C. 60 km 
D. 72 km 
E. 84 km

chetan2u · Accepted Answer

let's concentrate on Bus Q that starts from city B..  
when they meet first time  they both combined have traveled entire distance say x, out of which Q travels 22.. 
when they meet second time,  a total of 2 time the entire distance over and above the initial has been covered, so 3x..

If Q covers 22 in x, it will cover 3*22 = 66 in 3x..
But they meet 10 kms from A, so Q has traveled x+10...
x+10=66...x=56

B

KhushiWGupta · Answer

First meeting:  22 km from B  
 Second meeting:  10 km from A   
Key idea:
 👉  Distance covered by buses between meetings is the same ratio 
So we compare  cross distances :
Distance from A at 1st meet + distance from A at 2nd meet
 = Distance from B at 1st meet + distance from B at 2nd meet
So:
(D−22)+10=22+(D−10)(D - 22) + 10 = 22 + (D - 10)(D−22)+10=22+(D−10)
Simplify:
D−12=D+12❌D - 12 = D + 12 ❌D−12=D+12❌
Instead, use shortcut:
👉  Difference between meeting points = difference between end distances

saswata4s · Answer

Can you please explain this part a little more?

chetan2u · Answer

Q starts from B and P starts from A and meet 22 kms from B
so when they meet each other, they would have covered a combine of x and B covers 22..

Now when they meet second time..
Q reaches A from B and turns back, so covers x and some distance
In the same time, P covers x distance from A to B and turns back..
Now when they meet they have again covered the distance x..
so TOTAL 3x..
Q covers 22 when they cover x distance, so another 22 when they have covered another x that is x+x
another 22, when they have travelled another x or 3x.. so total 22+22+22=66

saswata4s · Answer

Thanks. Its clear now..

afa13 · Answer

I have figured out the algebraic approach:

Let us say that train P leaves A at speed a and covers distance x.
Q leaves B at the same time P leaves A at at constant velocity b and meet at 22 from B (let us call this point R.
We can get these equations: x/a=22/b => a/b=x/22.
After P leaves R it will cover 22 and the distance from B to miles away from A. Thus, it will cover 22+(x+22)-10=x+34
After Q leaves R it will cover x then reach A then the 10 miles to meet with P. Therefore, it will cover x+10.
we can set up these equations: (x+34)/p=(x+10)/q => x+34=(p/q)(x+10) => x+34=(x/22)(x+10) leading to the following quadratic equation:
x^2-12x-(22)(34)=0
(x-34)(x+22)=0
since distance is greater than zero then x=34 and total distance is 34+22=56.

Is there a shorter way?

zflodeen · Answer

Seems to me that chetan2u has the shortest method. You have to think about it differently, but I like how quick you can get an answer. I solved a little longer method.

We know that d=r*t. We can assign Rp as rate for P and Rq as rate for Q. We also know that meet at first at one time, call it T1, and then meet again at another time, call it T2.

At the first meeting point 22=Rq*T1 and d-22=Rp*T1. 
I solved each for T1 and then substituted for T1 to get 22/Rq = (d-22)/Rp....I

At the second meeting point we have d+10=Rq*T2 and 2d-10=Rp8T2
Here I solved for T2 and substituted for T2 to get (d+10)/Rq = (2d-10)/Rp......II

Since Rq and Rp are constant, their ratio also has to be constant, so I solved I and II for Rp/Rq to get the following:

Rp/Rq = (d-22)/22 using I
Rp/Rq = (2d-10)/(d+10) using II

I first thought to make these two equal and solve, but the equation looked ugly and I already spent too much time, so I noticed that the first equation is divisible by 22, so the second equation must also be divisible by a multiple of 22. I then asked what value of d in the answer choices will make (d+10) a multiple of 22 and 56 was the only option (choice B)

With all that, chetan2u's method is much quicker and works out the same.

karanladla · Answer

2x-10/x+10=d-22/22. Ratio of velocities will be const.

jaykayes · Answer

What does "x" represent here?

Bunuel · Answer

Here x represents the total distance between City A and City B. So the equation should be written as:

(2x - 10)/(x + 10) = (x - 22)/22

KhushiWGupta · Answer

Two buses meet  22 km from B  first time.
 After turning back, they meet  10 km from A .
Let total distance =  D 
At 1st meet:
 
 From A = D−22D - 22D−22 
 From B = 222222  
At 2nd meet:
 
 From A = 101010 
 From B = D−10D - 10D−10  
Distance ratio stays same, so:
D−2222=D−1010\frac{D - 22}{22} = \frac{D - 10}{10}22D−22=10D−10
Solve:
10(D−22)=22(D−10)10(D - 22) = 22(D - 10)10(D−22)=22(D−10) 10D−220=22D−22010D - 220 = 22D - 22010D−220=22D−220 12D=440⇒D=6012D = 440 \Rightarrow D = 6012D=440⇒D=60

pappal · Answer

let the speeds of the two busses be p and q; and the distance between A & B is d.
for the first meet if Q travels 22 miles so P travels (d-22) miles, since time is constant
so (d-22)/p = 22/q or q=22p/(d-22)------i
now for the second meet P travels 22+d-10=d+12
and Q travels d-22+10=d-12 
since time is constant so (d+12)/p=(d-12)/q-----ii
from i and ii
we can simplify d=56
B

Two buses P and Q start from cities A and B towards each other. They m

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