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Ashish Sharma

· started a discussion

· 1 Months ago

trick (Question: 113301.0 )

plz send short trick

Question:

A train leaves the station 1 hours before the scheduled time. The driver decreases its speed by 4 km/h. At the next station 120 km away, the train reached on time then the original speed of the train will be:

Options:
A)  21 km/h
B) 24 km/h
C) 25 km/h
D) 23 km/h
Solution:
Ans: (b)

Let the train takes x hours in second case.

The speed = \(\cfrac{120}{x}=\cfrac{120}{x-1}-4\)

or\(\cfrac{120}{x-1}-\cfrac{120}{x}=4\)

or \(\cfrac{120(x-x+1)}{(x-1)x}\ \ =\ \ 4\) 

or 120 = 4x2 - 4x = 0

or 4x2 - 4x - 120 = 0

or x2 - x - 30 = 0

or (x-6) (x+5) = 0 or

x = - 5, 6

\(\therefore\) x = 6

Therefore, the train takes 6 hours in second case i.e. (6 - 1) = 5 hours in original case.

\(\therefore\) original speed = \(\cfrac{120}{5}\) = 24 km/hr.

Knowledge Expert

· commented

· 1 Months ago

Dear student,
there is no short trick.
Keep learning,
Team TR

Rambalaji

· commented

· 1 Months ago

Go with options
Take b 24kmph as original speed, u will get 5 hrs time for 120 kms , since it had departed 1 hr early and reached on time the new time would be 6 hrs, for 6 hrs the speed would be 120/6 =20 kmph which is less than 24 by 4 kmph , which satisfies the question , so it is the right one

latika

· commented

· 1 Months ago

Yet to be approved!
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