Speed, Time and Distance
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Walking at 3 of his usual speed, 4
cover the same distance, (in hours) isa man is 1 1 hours late. His usual time to 2
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Time and speed are inversely proportional.
∴ 4 of usual time –usual time 3 = 3 2 ⇒ 1 × usual time = 3 3 2 ∴ Usual time = 3 × 3 = 9 2 2 = 4 1 hours 2
Second Method :Here, A = 3, B = 4, t = 3 2 Usual time = A × time Diff. of A and B = 3 × 3 (4 − 3) 2 = 4 1 hrs. 2 Correct Option: A
Time and speed are inversely proportional.
∴ 4 of usual time –usual time 3 = 3 2 ⇒ 1 × usual time = 3 3 2 ∴ Usual time = 3 × 3 = 9 2 2 = 4 1 hours 2
Second Method :Here, A = 3, B = 4, t = 3 2 Usual time = A × time Diff. of A and B = 3 × 3 (4 − 3) 2 = 4 1 hrs. 2
- Two trains start at the same time from A and B and proceed toward each other at the speed of 75 km/hr and 50 km/hr respectively. When both meet at a point in between, one train was found to have travelled 175 km more than the other. Find the distance between A and B.
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View Hint View Answer Discuss in Forum
Let the trains meet after t hours.
Distance = Speed × Time
According to the question,
75t – 50t = 175
⇒ 25t = 175⇒ t = 175 = 7 hours 25
∴ Distance between A and B
= 75t + 50t = 125t
= 125 × 7 = 875 km.
Second Method :
Here, a = 75, b = 50, d = 175Required distance = 
a + b 
× d a − b = 75 + 50 × 175 75 − 50 = 125 × 175 25
= 125 × 7 = 875 kmCorrect Option: A
Let the trains meet after t hours.
Distance = Speed × Time
According to the question,
75t – 50t = 175
⇒ 25t = 175⇒ t = 175 = 7 hours 25
∴ Distance between A and B
= 75t + 50t = 125t
= 125 × 7 = 875 km.
Second Method :
Here, a = 75, b = 50, d = 175Required distance = a + b × d a − b = 75 + 50 × 175 75 − 50 = 125 × 175 25
= 125 × 7 = 875 km
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late. The usual time taken by him to cover that distance isWalking 6 th of his usual speed, a man is 12 minutes 7
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View Hint View Answer Discuss in Forum
Time and speed are inversely proportional.
∴ Usual time × 7 – usual time = 12 minutes 6 ⇒ Usual time × 1 = 12 minutes 6
∴ Usual time = 72 minutes
= 1 hour 12 minutes
Second Method :t = 12 = 1 hrs. 60 5 Usual time = A × time Diff. of A and B = 6 × 1 = 1 1 hrs. (7 − 6) 5 5
= 1 hrs. 12 minutesCorrect Option: B
Time and speed are inversely proportional.
∴ Usual time × 7 – usual time = 12 minutes 6 ⇒ Usual time × 1 = 12 minutes 6
∴ Usual time = 72 minutes
= 1 hour 12 minutes
Second Method :t = 12 = 1 hrs. 60 5 Usual time = A × time Diff. of A and B = 6 × 1 = 1 1 hrs. (7 − 6) 5 5
= 1 hrs. 12 minutes
- A car travels from P to Q at a constant speed. If its speed were increased by 10 km/h, it would have been taken one hour lesser to cover the distance. It would have taken further 45 minutes lesser if the speed was further increased by 10 km/h. The distance between the two cities is
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Fixed distance = x km and certain speed = y kmph (let).
Case I,x = x – 1 y + 10 y ⇒ x + 1 = x --- (i) y + 10 y
Case II,x = x − 1 − 3 y + 20 y 4 = x − 4 + 3 y 4 ⇒ x + 7 = x --- (ii) y + 20 4 y
From equations (i) and (ii),x + 1 = x + 7 y + 10 y + 20 4 ⇒ x − x = 7 − 1 y + 10 y + 20 4 ⇒ x y + 20 − y − 10 (y + 10)(y + 20) = 7 − 4 = 3 4 4
⇒ 3 (y + 10) (y + 20) = 40 x⇒ 3 (y + 10) (y + 20) = x ---(iii) 40
From equation (i),3 (y + 10) (y + 20) + 1 40(y + 10) = 3 (y + 10) (y + 20) 40y
⇒ 3 (y +20) + 40= 3 (y + 10) (y + 20) y
⇒ 3y2+ 60y + 40 y = 3(y2 + 30y+ 200)
⇒ 3y2 + 100y = 3y2 + 90y + 600
⇒ 10y = 600 ⇒ y = 60
Again from equation (i),x + 1 = x y + 10 y ⇒ x + 1 = x 60 + 10 60 ⇒ x + 1 = x 70 60 ⇒ x + 70 = x 70 60
⇒ 6x + 420 = 7x
⇒ 7x – 6x = 420
⇒ x = 420 km.Correct Option: B
Fixed distance = x km and certain speed = y kmph (let).
Case I,x = x – 1 y + 10 y ⇒ x + 1 = x --- (i) y + 10 y
Case II,x = x − 1 − 3 y + 20 y 4 = x − 4 + 3 y 4 ⇒ x + 7 = x --- (ii) y + 20 4 y
From equations (i) and (ii),x + 1 = x + 7 y + 10 y + 20 4 ⇒ x − x = 7 − 1 y + 10 y + 20 4 ⇒ x y + 20 − y − 10 (y + 10)(y + 20) = 7 − 4 = 3 4 4
⇒ 3 (y + 10) (y + 20) = 40 x⇒ 3 (y + 10) (y + 20) = x ---(iii) 40
From equation (i),3 (y + 10) (y + 20) + 1 40(y + 10) = 3 (y + 10) (y + 20) 40y
⇒ 3 (y +20) + 40= 3 (y + 10) (y + 20) y
⇒ 3y2+ 60y + 40 y = 3(y2 + 30y+ 200)
⇒ 3y2 + 100y = 3y2 + 90y + 600
⇒ 10y = 600 ⇒ y = 60
Again from equation (i),x + 1 = x y + 10 y ⇒ x + 1 = x 60 + 10 60 ⇒ x + 1 = x 70 60 ⇒ x + 70 = x 70 60
⇒ 6x + 420 = 7x
⇒ 7x – 6x = 420
⇒ x = 420 km.
- A car covers four successive 7 km distances at speeds of 10 km/hour, 20 km/hour, 30 km/
hour and 60 km/hour respectively. Its average speed over this distance is
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Total distance = 7 × 4 = 28 km.
Total time
=7 + 7 + 7 + 7 hours 10 20 30 60 = 42 + 21 + 14 + 7 hours 60 = 84 hours = 7 hours 60 5 ∴ Average speed = Total distance Total time = 28 kmph 7/5 = 28 × 5 = 20 kmph 7 Correct Option: B
Total distance = 7 × 4 = 28 km.
Total time
=7 + 7 + 7 + 7 hours 10 20 30 60 = 42 + 21 + 14 + 7 hours 60 = 84 hours = 7 hours 60 5 ∴ Average speed = Total distance Total time = 28 kmph 7/5 = 28 × 5 = 20 kmph 7
