Speed, Time and Distance
 Two places A and B are 80 km apart from each other on a highway. A car starts from A and another from B at the same time. If they move in the same direction, they meet each other in 8 hours. If they move in opposite directions towards each other, they meet in 1 hour 20 minutes. Determine the speed of the faster car.

 20 kmph
 25 kmph
 35 kmph
 30 kmph

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Case : I
When the cars are moving in the same direction.
Let A and B be two places and C be the place of meeting.
Let the speed of car starting from A be x kmph, and that of car starting from B be y kmph.
Relative speed = (x – y) kmph
According to the question.
(x – y) × 8 = 80
⇒ x – y = 10 ...(i)
Case : II
When the cars are moving in the opposite directions and they meet at point C.
Relative speed = (x + y) kmph
Time taken = 1 hour 20 minutes= 1 + 1 = 4 hours 3 3 ∴ (x + y) × 4 = 80 3 ⇒ x + y = 80 × 3 4
⇒ x + y = 60 ...(ii)
Adding equations (i) and (ii),
2x = 70
⇒ x = 35
From equation (ii),
x + y = 60
⇒ 35 + y = 60
⇒ y = 60 – 35 = 25
∴ Speed of the faster car = 35 kmphCorrect Option: C
Case : I
When the cars are moving in the same direction.
Let A and B be two places and C be the place of meeting.
Let the speed of car starting from A be x kmph, and that of car starting from B be y kmph.
Relative speed = (x – y) kmph
According to the question.
(x – y) × 8 = 80
⇒ x – y = 10 ...(i)
Case : II
When the cars are moving in the opposite directions and they meet at point C.
Relative speed = (x + y) kmph
Time taken = 1 hour 20 minutes= 1 + 1 = 4 hours 3 3 ∴ (x + y) × 4 = 80 3 ⇒ x + y = 80 × 3 4
⇒ x + y = 60 ...(ii)
Adding equations (i) and (ii),
2x = 70
⇒ x = 35
From equation (ii),
x + y = 60
⇒ 35 + y = 60
⇒ y = 60 – 35 = 25
∴ Speed of the faster car = 35 kmph
 In a 200 metre race, A beats B by 20 metres; while in a 100 metres race, B beats C by 5
metres. Assuming that the speed of A, B and C remain the same in various races, by how many metres will A beat C in one kilometre race ?

 140 metre
 145 metre
 135 metre
 125 metre

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According to the question,
when A covers the distance of 200 metres, B covers only 200– 20 = 180 metres
Again, in 100 metre race, B beats C by 5 metres.
Hence, if B runs 100 metres, C runs 100–5 = 95 metres
∵ If B runs 100 m, C runs = 95 m
∴ If B runs 180 m, C runs= 95 × 180 = 171 m 100
∴ A : B : C = 200 : 180 : 171
Hence, A will beat C by
= 200–171 = 29 m in 200 m race.
i.e., 29 × 5 = 145 m in 1 km race.Correct Option: C
According to the question,
when A covers the distance of 200 metres, B covers only 200– 20 = 180 metres
Again, in 100 metre race, B beats C by 5 metres.
Hence, if B runs 100 metres, C runs 100–5 = 95 metres
∵ If B runs 100 m, C runs = 95 m
∴ If B runs 180 m, C runs= 95 × 180 = 171 m 100
∴ A : B : C = 200 : 180 : 171
Hence, A will beat C by
= 200–171 = 29 m in 200 m race.
i.e., 29 × 5 = 145 m in 1 km race.
 A steamer goes downstream from one port to another in 4 hours. It covers the same distance upstream in 5 hours. If the speed of the stream be 2 km/hr, find the distance between the two ports.

 60 km.
 45 km.
 80 km.
 65 km.

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Let the speed of steamer in still water = x kmph
∴ Rate downstream = (x + 2) kmph
Rate upstream = (x – 2) kmph
Obviously, distance covered downstream and upstream are equal
⇒ 4 (x + 2) = 5 (x – 2)
⇒ 4x + 8 = 5x – 10
⇒ 5x – 4x = 10 + 8 ⇒ x = 18
∴ Rate downstream
= 18 + 2 = 20 kmph
Therefore, the required distance
= Speed downstream × Time
= 20 × 4 = 80 km.Correct Option: C
Let the speed of steamer in still water = x kmph
∴ Rate downstream = (x + 2) kmph
Rate upstream = (x – 2) kmph
Obviously, distance covered downstream and upstream are equal
⇒ 4 (x + 2) = 5 (x – 2)
⇒ 4x + 8 = 5x – 10
⇒ 5x – 4x = 10 + 8 ⇒ x = 18
∴ Rate downstream
= 18 + 2 = 20 kmph
Therefore, the required distance
= Speed downstream × Time
= 20 × 4 = 80 km.
 A person travels a certain distance on a bicycle at a certain speed. Had he moved 3 km/hour faster, he would have taken 40 minutes less. Had he moved 2km/hour slower, he would have taken 40 minutes more. Find the distance.

 45 km.
 40 km.
 50 km.
 55 km.

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Let the original speed of the person be x km/hr. and the distance be y km.
Case : Iy − y = 40 minutes x x + 3 or, 40 hr. 60 or, y − y = 40 = 2 x x + 3 60 3 or, y 1 − 1 = 1 x (x + 3) 3 or, y x + 3 − x = 2 x (x + 3) 3 or, 3y = 2 x (x + 3) 3
or, 2 x (x + 3) = 9y ...(i)
Case : II= y − y = 40 x − 2 x 60 or, y 1 + 1 + 2 x − 2 x 3 or, y x − x + 2 = 2 x (x − 2) 3 or, 2y = 2 x (x − 2) 3
or, x (x – 2) = 3y ... (ii)
On dividing equation (i) by (ii) we have,2x(x + 3) = 9y x (x − 2) 3y or, 2(x + 3) = 3 (x − 2)
or, 2x + 6 = 3x – 6
or, 3x – 2x = 6 + 6 = 12
or, x = 12 km/hr.
∴ Original speed of the person
= 12 km/hr.
Putting the value of x in equation (ii)
12 (12 – 2) = 3y
or, 3y = 12 × 10or, y = 12 × 10 = 40 3
∴ The required distance = 40km.Correct Option: B
Let the original speed of the person be x km/hr. and the distance be y km.
Case : Iy − y = 40 minutes x x + 3 or, 40 hr. 60 or, y − y = 40 = 2 x x + 3 60 3 or, y 1 − 1 = 1 x (x + 3) 3 or, y x + 3 − x = 2 x (x + 3) 3 or, 3y = 2 x (x + 3) 3
or, 2 x (x + 3) = 9y ...(i)
Case : II= y − y = 40 x − 2 x 60 or, y 1 + 1 + 2 x − 2 x 3 or, y x − x + 2 = 2 x (x − 2) 3 or, 2y = 2 x (x − 2) 3
or, x (x – 2) = 3y ... (ii)
On dividing equation (i) by (ii) we have,2x(x + 3) = 9y x (x − 2) 3y or, 2(x + 3) = 3 (x − 2)
or, 2x + 6 = 3x – 6
or, 3x – 2x = 6 + 6 = 12
or, x = 12 km/hr.
∴ Original speed of the person
= 12 km/hr.
Putting the value of x in equation (ii)
12 (12 – 2) = 3y
or, 3y = 12 × 10or, y = 12 × 10 = 40 3
∴ The required distance = 40km.
 Distance between two stations X and Y is 220 km. Trains P and Q leave station X at 8 a.m. and 9.51 a.m. respectively at the speed of 25 kmph and 20 kmph respectively for journey towards Y. A train R leaves station Y at 11.30 a.m. at a speed of a 30 kmph. for journey towards X. When will P be at equal distance from Q and R ?

 12:48 pm.
 12:30 pm.
 12:45 pm.
 11:48 pm.

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Distance covered by P till 11.30 a.m.
= (11.30 a.m. – 8 a.m.) × 25km= 3 1 × 25 = 87.5 km. 2
Distance covered by Q till 11.30 a.m.
= (11.30 – 9.51 am) × 20= 1 39 hrs. × 20 = 33 km 60
So, at 11.30 a.m. the three trains will be at positions shown below :
P gains 5 km every hour over Q.
Relative speed of P w.r.t. R = 20 + 30 = 50 km per hr
Let P be at equal distance from Q and R after t hours.
∴ (87.5 – 33) + 5t
= 132.5 –55t
or, 54.5 + 5t = 132.5 – 55t
or, 60 t = 78or, t = 78 hrs. 60
= 1 hr 18 minutes
11.30 am + 1 hr. 18 min.
= 12.48 pm
At 12.48 pm, P would have covered a distance
= (12.48 pm – 8 am) × 25
= 120 km
Therefore, P will be at equal distance from Q and R at 12.48 pmCorrect Option: A
Distance covered by P till 11.30 a.m.
= (11.30 a.m. – 8 a.m.) × 25km= 3 1 × 25 = 87.5 km. 2
Distance covered by Q till 11.30 a.m.
= (11.30 – 9.51 am) × 20= 1 39 hrs. × 20 = 33 km 60
So, at 11.30 a.m. the three trains will be at positions shown below :
P gains 5 km every hour over Q.
Relative speed of P w.r.t. R = 20 + 30 = 50 km per hr
Let P be at equal distance from Q and R after t hours.
∴ (87.5 – 33) + 5t
= 132.5 –55t
or, 54.5 + 5t = 132.5 – 55t
or, 60 t = 78or, t = 78 hrs. 60
= 1 hr 18 minutes
11.30 am + 1 hr. 18 min.
= 12.48 pm
At 12.48 pm, P would have covered a distance
= (12.48 pm – 8 am) × 25
= 120 km
Therefore, P will be at equal distance from Q and R at 12.48 pm