Two cars, each moving at their respective constant speeds, leave simul

Question

Two cars, each moving at their respective constant speeds, leave simultaneously from points A and B towards each other. After they pass each other, car from A reaches B in 25 minutes, while car from B reaches A in 36 minutes. How many minutes it took the faster car to cover the whole distance between the two points ?

A. 44 minutes
B. 45 minutes
C. 50minutes
D. 55 minutes
E. 66 minutes

Bunuel · Accepted Answer

Official Solution:

Two cars, each moving at their respective constant speeds, depart simultaneously from points A and B towards each other. After passing each other, the car from point A reaches point B in 25 minutes, while the car from point B reaches point A in 36 minutes. How many minutes did it take for the faster car to cover the entire distance between the two points?

A. 44 minutes
B. 45 minutes
C. 50 minutes
D. 55 minutes
E. 60 minutes

Since the cars departed simultaneously and after meeting, the car from point A reached point B in less time than the car from point B took to reach point A, it's evident that the car from point A is faster.

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Assuming they met after  t  minutes, then the car from A spent 25 minutes covering the same distance the car from B covered in  t  minutes. Since they travel at constant speeds, the ratio of their speeds would be equal to the reciprocal of the ratio of their times: the ratio of speeds of A to that of B =  \frac{t}{25} . For instance, if one car took  t  minutes to cover 10 kilometers, and another took 25 minutes to cover the same distance, then the ratio of the speed of the first car to that of the second car would be  \frac{(\frac{10}{t})}{(\frac{10}{25})}=\frac{25}{t} .

Similarly, since the car from A spent  t  minutes covering the same distance the car from B covered in 36 minutes, we'd have the ratio of speeds of A to that of B =  \frac{36}{t} .

Equating these two gives:  \frac{t}{25} = \frac{36}{t} , which yields  t^2 = 25 * 36 , and finally,  t = 5 * 6 = 30  minutes.

Therefore, the faster car, the one from point A, took  t + 25 = 30 + 25 = 55  minutes to cover the entire distance between the two points.

Answer: D

SaquibHGMATWhiz · Answer

This is a very interesting question that can become tricky to solve using the conventional approach.

I want to give an approach using the usage of ratio.

Let's try to visualize the scenario.

Car 1 ..........X minutes ...........| .......25 minutes ..........

A_______________________ M___________________B

Car 2 ....... 36 minutes ...........| ......... X minutes .........

Let's understand the scenario, 
 Car 1 and car 2 travel for X minutes to meet at M. 
 So, car 1 travels AM distance and car 2 travels MB distance 
Car 1 then takes 25 minutes to travel from M to B and car 2 takes 36 minutes to travel from M to A

Since they are traveling at a uniform speed so the time taken to travel the same distance will be in proportion.
  \frac{The Time Taken By Car 1 To TravelAM}{The Time Taken By Car 2 To TravelAM } 
=\frac{The Time Taken By Car 1 To Travel MB}{The Time Taken By Car 2 To TravelMB }  
 Or,  \frac{X}{36} = \frac{25}{X}   
 Or,  x^2 = 36 	imes 25  
Or,  x = 30

We can see the faster car is 1. Hence the time taken = 25 + 30 = 55 minutes.

Answer: D

Abulikabuli · Answer

Logic is of converging problems….
Let they cross each other after t minutes, total distance be D and let x be the distance travelled by CAR1 in 25 mins.
So car1 travels D-x distance in t mins to meet car2.
Speed of car1 is (D-x)/t or x/25
So (D-x)/t = x/25

In convergence sums, car2 did a pretty similar thing…it travelled (D-x) in 36 mins and X in t mins.
So (D-x)/36 = x/t

Find value of X from each equation and you can equate to get,
 T/36 = 25/t
 T^2 = 25x36
 T= 30….
Add 25 mins for faster car’s time… Ans. 55 mins (option D)

Vordhosbn · Answer

Assume they meet at time T.

From the problem we can tell Car A (the one traveling from point A) is the faster one.

The key to this problem is realizing that the amount that each has travelled upon meeting each other can be written as an equation related to the other car. (i.e, the amount that A has traveled can be written based on B, and the amount that B has travelled can be written as an equation based on A).

In other words, what the problem is really saying is that the amount A covered upon meeting B, is equivalent to B * 3/5 -> *Since the amount that B has left is equal to the amount A has travelled* and we know B covers that part in in 3/5 of an hour.
Similarly, the amount that B has travelled is equal to the amount that A has left and we know that A travels it at its constant speed in 5/12 of an hour.

Thus:
A * t = B * 3/5
B * t = A * 5/12
(where T is the amount that took them to meet each other)

Separate variable T, and we get:

(1) t = B/A * 3/5
(2) t = A/B * 5/12

-> B/A * 3/5 = A/B * 5/12
-> A^2/B^2 = 36/25
-> A/B = 6/5

Plugging this ratio into (1) we get: t = 1/2 hr, which is 30 minutes.
So they met after traveling 30 minutes. A finished after 30 + 25 = 55 minutes. B finished after 30 + 36 minutes= 66 minutes.

Again, the faster car finished in 55 minutes.

minecraftsalzburg · Answer

There is no calculation necessary. It is evident that the car traveling from A to B is faster. That also means that it has covered more distance than the car traveling from B to A until meeting (faster car covers more distance in same time). Therefore solutions A, B, C cannot be true because all of them would imply that the distance already covered by the car traveling from A to B is shorter than the distance, covered by the car traveling from B to A (because the time it took the car to travel that distance would be shorter than the time it will take the car to travel the remaining distance after meeting).

This leaves us with D and E. But again we know that the car traveling from A to B is faster. Therefore it has to be faster than 36+25=61 minutes. Anwer E (66) would imply that the car traveling from A to B took more time traveling the route from A to the meeting point than the time it will take the car traveling from B to A (slower one) to get from the meeting point to A.

Therefore the solution is D.

Two cars, each moving at their respective constant speeds, leave simul

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Two cars, each moving at their respective constant speeds, leave simul

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