Trains A and B start traveling towards each other at the same time

Question

Trains A and B start traveling towards each other at the same time from stations X and Y, respectively, each moving at its own constant speed. Train A reaches station Y in 10 minutes, while train B takes 9 minutes to reach station X after meeting train A. What is the total time, in minutes, taken by train B to travel from station Y to station X?

A. 6
B. 8
C. 10
D. 12
E. 15

A. 6 
 B. 8 
 C. 10 
 D. 12 
 E. 15

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Purnank · Accepted Answer

use the meeting time formula here.
Lets assume x is the time that both took to meet.
Formula is,  x = \sqrt{Time taken by A to Reach Y after meet * Time taken by B to reach X after meet} 
Given - Train A reaches station Y in 10 minutes
Therefore, Time taken by A to Reach Y after meet = Total time - Meet time = 10 - x
and B takes 9 minutes to reach station X after meeting train A.
Therefore,  x = \sqrt{(10-x) * 9} 
 x^2 = (10-x)9 
solving,
 x^2 +9x - 90 = 0 
Get the value for x. Here x you will get 6 min.
Now the total time, in minutes, taken by train B to travel from station Y to station X = x + 9 = 6+9 = 15 min.
  Answer is E.

Derivation for the formula mentioned-  
 I will draw schematic for this

(X) A>-------------------meeting point p----------------------<B (Y) 
 let ap be the distance covered by A during meet similarly bp is distance covered by B during the meet. 
 i will keep variable x as same for time taken by them to meet. 
 Now as we know Distance = Speed*Time 
 ap = A*x and bp = B*x (Here A and B are speeds for train A and B) 
 Both trains after meet moving towards their respective destinations 
 So now A will cover the distance distance bx and B will cover distance ax and  lets take time they took to cover those distances after meeting at p are m and n respectively for both A and B  
Therefore,
ap = A*x = B*n and bp = B*x = A*m

now from here we get  \frac{A}{B} = \frac{n}{x} = \frac{x}{m} 
from here just take these two  \frac{n}{x} = \frac{x}{m} 
solve,  x^2 = m*n  
which means  x = \sqrt{mn} = \sqrt{Time taken by A to Reach Y after meet * Time taken by B to reach X after meet}

Hope this helps

Edoua · Answer

IMO E

X- - - - ————- ——Y
 D= x miles
A. B

Va= x/10. Vb= x/(t+9) where t is the time taken to meet each other

GAP= (Va+Vb)* Time taken to meet each other

—-> x=  *t
1= ((1/10)+1/(t+9))*t
10t+90 =t(t+19)

t^2 +9t-90 =0
t=6 or t=-15 only the positive value is taken

Thus the  time to travel from Y to X is t+9 = 6+9=15

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adewale223 · Answer

Pick LCM of 9 and 10 as distance between X and Y

D = 90

Speed of A is 90/10

A = 9

Let time it will take for them to meet be t

t = 9B/A

t = B

Therefore 9B/A = 90/A+ B

B = 90/9 + B

B^2 + 9B - 90 = 0

(B - 6)(B + 15) = 0

B = 6

So time it will take train B to travel from Y to X

9B/A + 9 = 15

Answer E

Adewale Fasipe, GMAT quant instructor ( YT Channel Quant Clinic) , Lagos Nigeria.

Ayeka · Answer

Let, total distance=d 
Speed of Train A= S1 & speed of train B= S2 
Time taken by train A= 10 mins 
Time taken by train B= 9 mins after meeting A

For train A, d=S1*10 mins 
For train B, d=S2*(t+9)

distance= (S1+S2)*t 
d=(d/10+d/(t+9))*t 
Solving above, we get t=6 or -15 (negative time, not possible.... So 6 mins)

Hence, total time taken by train B= (t+9) = 6+9 = 15 mins

Answer: E

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monster1810 · Answer

Let x be the distance between station X and Y, and t be the time both trains meet.
We have speed of train A: vA = x/10; and speed of train B: vB = x/(t+9)
X -------------> Z (trains meet) ------------> Y
Because it is the same distance from station X to the place both trains meet (Z), we have: 
t*vA=9*vB
=> t*(x/10)=9*(x/(t+9))
=> t/10=9/(t+9)
=> t^2+9t-90=0
=> t=6 (eliminate -15)
=> total time for train B = 6+9 = 15min => E

Krunaal · Answer

Speed = \frac{Distance}{Time}

For A:   Sa = \frac{D}{10}

For B:   Sb = \frac{D}{(t+9)}  .................................(where t = time for A and B to meet)

And we know when A and B meet,

t = \frac{D}{(Sa + Sb)}

t = \frac{10t + 90 }{ t + 19}

t^2 + 9t - 90 = 0 
t = 6 or t = -15
t = 6 (as time can't be negative)

Total Time taken by B was t + 9 => 6 + 9 =  15

Answer E.

pierjoejoe · Answer

here we have to apply two different types of solutions we saw in
https://gmatclub.com/forum/distance-spe ... 87481.html

hope you like my solution

first we know that both train travel the same distance with a constant velocity and a certain time:

(table 1)

train	D	S	T
A	D	V1	t1 = 10
B	D	V2	t2 = x+9

D represents the total distance between X and Y
t1 = 10
t2 = x +9
where x is the time when they meet

we also know that they meet at a certain time x and a certain distance that depends for each train by their speed:

(table 2)

train	D	S	T
A	D1	V1	x
B	D2	V2	x

where x is the time they meet, and therefore is the same for both trains. D1 is the distance to the meeting point of A, D2 is the distance to the meeting point of B

from the first table:
V1 = D/10
V2 = D/(x+9)

from the second table
D = D1 + D2 = V1*x + V2*x

now substitute V1 and V2 to the last equation:
D = V1*x + V2*x = x(D/10) + x(D/(x+9)) = D

x(D/10) + x(D/(x+9)) = D
simplify:
x(1/10) + x(1/(x+9)) = 1
x(1/10 + 1/(x+9)) = 1
x * ((x+19) / ((x+9)*10) = 1
x * (x+19) = 10x+90

x^2 + 19x -10x -90 = 0
x^2 +9x -90 = 0
x = -15 or x = 6, only the latter is an acceptable solution for time

thus the time they meet is at 6 minutes, meaning that (from the first table) B travels for:
t2 = x +9 = 6+9 = 15 minutes

madlol77777 · Answer

Could you please elaborate more on the " meeting time formula"?

Purnank · Answer

I have edited original post please refer that.

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Trains A and B start traveling towards each other at the same time

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