A train after traveling for 50 km meets with an accident

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BTGmoderatorLU: Moderator; Posts: 2505; Joined: Sun Oct 15, 2017 1:50 pm; Followed by:6 members

A train after traveling for 50 km meets with an accident

by BTGmoderatorLU » Fri Jul 06, 2018 2:01 pm

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A train after traveling for 50km meets with an accident and then proceeds at 3/4 of its former speed and arrives at its destination 35 minutes late. Had the accident occurred 24 km farther, it would have reached the destination only 25 minutes late. What is the speed of the train?

A. 45
B. 33
C. 48
D. 55
E. 61

The OA is C.

Please, can someone assist me with this PS question? I appreciate your help. Thanks!

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Source: Beat The GMAT — Problem Solving | Fri Jul 06, 2018 2:01 pm

swerve: Legendary Member; Posts: 2499; Joined: Sun Oct 29, 2017 2:04 pm; Followed by:6 members

by swerve » Sat Jul 07, 2018 10:09 am

Let the two endpoints between which this train is running be A and B. Also suppose that accident occurs at C which is at a distance of 50km from A. thus we have

A---------50km----------C----------------------------B 1)

Time taken to travel the distance after the accident i.e. CB is 35 minutes more than the usual time.

Now, if the accident were to occur at 24 km farther, then the train would have reached only 25 minutes late. i.e.

A--------50 km----------C---------24--------D---------------B 2)

If we look at the situation depicted in 1) and 2) we will see that 10-minute difference in time is occurring because of this 24 Km. which is traveled by its usual speed (=x) in the later case. Thus we have,
$$\frac{24\cdot4}{3x}-\frac{24}{x}=\frac{10}{60}$$
$$\frac{32}{x}-\frac{24}{x}=\frac{1}{6}$$
Solving this we get x = 48 km/h. Regards!

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Scott@TargetTestPrep: GMAT Instructor; Posts: 8086; Joined: Sat Apr 25, 2015 10:56 am; Location: Los Angeles, CA; Thanked: 43 times; Followed by:29 members

by Scott@TargetTestPrep » Sun Jul 22, 2018 5:32 pm

BTGmoderatorLU wrote:A train after traveling for 50km meets with an accident and then proceeds at 3/4 of its former speed and arrives at its destination 35 minutes late. Had the accident occurred 24 km farther, it would have reached the destination only 25 minutes late. What is the speed of the train?

A. 45
B. 33
C. 48
D. 55
E. 61

We can let the speed of the train be r kmph and the distance to its destination be d km. We can create the following equations for the time:

50/r + (d - 50)/(3r/4) = d/r + 35/60

and

74/r + (d - 74)/(3r/4) = d/r + 25/60

Multiplying both equation by 60r, we have:

3000 + 80(d - 50) = 60d + 35r

and

4440 + 80(d - 74) = 60d + 25r

Subtracting these two equations, we have:

-1440 + 1920 = 10r

480 = 10r

48 = r

Answer: C

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