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A dog after traveling 50 km meets a swami who counsels him to go slower. He then proceeds at $$\frac{3}{4}$$ of his former speed and arrives at his destination 35 min late. Had the meeting occurred 24 km further the dog would have reached its destination 25 min late. The speed of dog is:

Answer & Solution
Correct Answer: Option A
He proceeds at $$\frac{3}{4}$$ S where S is his usual speed means $$\frac{1}{4}$$ decrease in the speed which will lead to $$\frac{1}{3}$$ increase in the time. Now the main difference comes in those 24km and the change in difference of time = 35 - 25 m = 10 m.

⇒ $$\frac{1}{3}$$ × T = 10 where T is the time required to cover the distance of (74 - 50) = 24 km.
T = 30 min = 0.5 hours.

Speed of the dog = $$\frac{{24}}{{0.5}}$$  = 48 kmph.
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4 Comments
Ashraful Islam
Ashraful Islam 7 years ago
Acq, let before meet speed = 4x
After meet = 3x
24/4x - 24/3x = 35-25 = 10

1/12x = 10/24
x = 1/5 km/min
4x = 4*60/5 = 48 km/hr
Pulak Fakir
Pulak Fakir 7 years ago
24/(3x/4)-24/x=10/60
On solving. X=48kmph
MD Rakib
MD Rakib 8 years ago
SPEED;4:3
TIME=3:4[GAP 1]
BUT WE NEED TO GAP 10
T=30:60
S=74-50/30/60=48KM/HR...(ANS)
Pavanmanesh Mylavarapu
Pavanmanesh Mylavarapu 10 years ago
can u elaborate? 1/3 increase in time?