A ship is travelling between two port cities with a speed

Question

A ship is travelling between two port cities with a speed such that it will reach its destination on time. After covering 25% of the distance, the ship faces an emergency and returns to its starting point with a different speed v. It then resumes travelling towards its destination without any further delay with the same speed v. By what percentage is v more than the original speed, if the ship reaches its destination in time?

A. 33.33
B. 50
C. 66.67
D. 75
E. 87.5

taken from CL GMAT

A. 33.33
B. 50
C. 66.67
D. 75
E. 87.5

taken from CL GMAT

chetan2u · Answer

After 25% of distance or D/4 is covered, time to cover remaining 3D/4 at original speed, say x, and time to cover D + D/4 at a speed of v is same.

Thus,  \frac{\frac{3D}{4}}{x}=\frac{\frac{5D}{4}}{v}

3v=5x………v=\frac{5x}{3}

Increase =  v-x=\frac{5x}{3}-x=\frac{2x}{3}

Increase as a % of initial speed =  \frac{\frac{2x}{3}}{x}*100=66.67

C

Bunuel · Answer

Can you please clarify what is " L GMAT" and provide the link and/or screenshot of the question ? Thank you.

VarshithaRS · Answer

May I please know, how are we considering that , 
 \frac{\frac{3D}{4}}{x}

It is not mentioned in the question explicitly that, ship would have taken the same time as travelling in the velocity 'v', if the ship had started from the 25% of the distance which it had alredy covered?
@ chetan2u  , can you please explain me, how have you made this assumption.

Thank you

zlishz · Answer

Let the total distance between the 2 ports be  4D .
Now,  25% of 4D = 1D 
Let the speed of the ship at the first half of the journey be  x  when the ship travelled distance D and then returned because of an emergency. 
If the emergency wouldn't have occurred, the rest of the distance would have been covered at speed  x only. 
Thus, 3D distance at speed x. 
Time, in this case = 3D/x

Given that the speed after restarting the journey was  v 
Since, the ship will reach at the same time as before, it means that total distance 4D + D (while going back to the port) = 5D will be covered at speed v
Time, in this case = 5D/v

Both time are equal, so,
3D/x = 5D/v
v/x = 5/3

%= (v-x/x)*100
=(5-3/3)*100
=200/3 =  66.67%, option C

A ship is travelling between two port cities with a speed

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A ship is travelling between two port cities with a speed

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