Speed, Time and Distance
 A railway engine is proceeding towards A at uniform speed of 30 km/hr. While the engine is 20 kms away from A an insect starting from A flies again and again between A and the engine relentlessly. The speed of insect is 42 km per hr. Find the distance covered by the insect till the engine reaches A.

 25 km.
 32 km.
 30 km.
 28 km.

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Relative speed of insect
= 30 + 42 = 72 km per hr.
Distance between railway engine and insect = 20 km.
Engine and insect will meet for the first time after= 20 hr. 72
Distance covered in this period= 20 × 42 = 35 km 72 3 The insect will cover 35 km in returning to A. 3
The distance covered by engine in this period= 20 × 30 = 25 km 72 3
Since, the insect when reaches A,the engine will cover 25 to A. 3
∴ Remaining distance between A and engine= 20 − 25 + 25 3 3 Again, engine and insect will meet after 10 = 5 hr. 3 × 72 108
The distance covered by the insect in this period= 5 × 42 = 35 km 108 18 tand again the insect will cover 35 km in returning. 18 ∴ Total distance covered by the insect = 70 + 70 +..... 3 18 = 70 1 + 1 +.......∞ 3 6
It is a Geometric Progression to infinity with common ratio 1/6.= 70 1 3 1 − (1/6) ∴ S_{∞} = a 1 − r = 70 × 1 3 (5/6) = 70 × 6 = 28 km 3 5 Correct Option: D
Relative speed of insect
= 30 + 42 = 72 km per hr.
Distance between railway engine and insect = 20 km.
Engine and insect will meet for the first time after= 20 hr. 72
Distance covered in this period= 20 × 42 = 35 km 72 3 The insect will cover 35 km in returning to A. 3
The distance covered by engine in this period= 20 × 30 = 25 km 72 3
Since, the insect when reaches A,the engine will cover 25 to A. 3
∴ Remaining distance between A and engine= 20 − 25 + 25 3 3 Again, engine and insect will meet after 10 = 5 hr. 3 × 72 108
The distance covered by the insect in this period= 5 × 42 = 35 km 108 18 tand again the insect will cover 35 km in returning. 18 ∴ Total distance covered by the insect = 70 + 70 +..... 3 18 = 70 1 + 1 +.......∞ 3 6
It is a Geometric Progression to infinity with common ratio 1/6.= 70 1 3 1 − (1/6) ∴ S_{∞} = a 1 − r = 70 × 1 3 (5/6) = 70 × 6 = 28 km 3 5
 A man covered a distance of 3990 km partly by air, partly by sea and remaining by land. The time spent in air, on sea and on land is in the ratio 1 : 16 : 2 and the ratio of average speed is 20 : 1 : 3 respectively. If total average speed is 42 km per hr, find the distance covered by sea.

 1720 km.
 1620 km.
 1520 km.
 1820 km.

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Total distance travelled = 3990 km
Distance = Time × Speed
Ratio of time spent = 1 : 16 : 2
Ratio of speed = 20 : 1 : 3
∴ Ratio of time × speed
= 20 × 1 : 16 × 1 : 2 × 3
= 20 : 16 : 6
Sum of the ratios
= 20 + 16 + 6 = 42
∴ Distance covered by sea= 3990 × 16 = 1520 kms 42 Correct Option: C
Total distance travelled = 3990 km
Distance = Time × Speed
Ratio of time spent = 1 : 16 : 2
Ratio of speed = 20 : 1 : 3
∴ Ratio of time × speed
= 20 × 1 : 16 × 1 : 2 × 3
= 20 : 16 : 6
Sum of the ratios
= 20 + 16 + 6 = 42
∴ Distance covered by sea= 3990 × 16 = 1520 kms 42
 A motorist and a cyclist start from A to B at the same time. AB is 18 km. The speed of motorist is 15 m per hr. more than the cyclist. After covering half the distance, the motorist rests for 30 minutes and thereafter his speed is reduced by 20%. If the motorist reaches the destination B, 15 minutes earlier than that of the cyclist, then find the speed of the cyclist.

 16 kmph
 12 kmph
 14 kmph
 15 kmph

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Let the speed of the cyclist be x km per hr.
Speed of the motorist= (x + 15) km per hr.
Time taken by the motorist to cover half of the distance= 18 = 9 hrs. 2 × (x + 15) x + 15
After covering 9 kms, the speed of motorist gets reduced by 20%∴ New speed = (x + 15) × 80 100 = 4(x + 15) km per hr. 5
Time taken by the motorist to cover the remaining half distance= 9 × 5 = 45 hrs. 4 (x + 15) 4 (x + 15)
Total time taken by the motorist= 9 + 1 + 45 hrs. x + 15 2 4 (x + 15) Total time taken by the cyclist = 18 hrs. x Motorist reaches 15 minutes, i.e., 1 hr. earlier. 4 ∴ 18 − 9 − 1 − 45 = 1 x x + 15 2 4 (x + 15) 4 ⇒ 18 × 4 (x + 15) − 36x − 2x (x + 15) − 45x = 1 4x (x + 15) 4
⇒ 72x + 1080 – 36x – 2x^{2} – 30x – 45x = x^{2} + 15
⇒ 3x^{2} + 54x – 1080 = 0
⇒ x^{2} + 18x – 360 = 0
⇒ x^{2} + 30x – 12x – 360 = 0
⇒ x (x + 30) – 12 (x + 30) = 0
⇒ (x + 30) (x – 12) = 0
⇒ x = – 30, 12
The speed cannot be negative.
∴ The speed of the cyclist = 12 km per hr.Correct Option: B
Let the speed of the cyclist be x km per hr.
Speed of the motorist= (x + 15) km per hr.
Time taken by the motorist to cover half of the distance= 18 = 9 hrs. 2 × (x + 15) x + 15
After covering 9 kms, the speed of motorist gets reduced by 20%∴ New speed = (x + 15) × 80 100 = 4(x + 15) km per hr. 5
Time taken by the motorist to cover the remaining half distance= 9 × 5 = 45 hrs. 4 (x + 15) 4 (x + 15)
Total time taken by the motorist= 9 + 1 + 45 hrs. x + 15 2 4 (x + 15) Total time taken by the cyclist = 18 hrs. x Motorist reaches 15 minutes, i.e., 1 hr. earlier. 4 ∴ 18 − 9 − 1 − 45 = 1 x x + 15 2 4 (x + 15) 4 ⇒ 18 × 4 (x + 15) − 36x − 2x (x + 15) − 45x = 1 4x (x + 15) 4
⇒ 72x + 1080 – 36x – 2x^{2} – 30x – 45x = x^{2} + 15
⇒ 3x^{2} + 54x – 1080 = 0
⇒ x^{2} + 18x – 360 = 0
⇒ x^{2} + 30x – 12x – 360 = 0
⇒ x (x + 30) – 12 (x + 30) = 0
⇒ (x + 30) (x – 12) = 0
⇒ x = – 30, 12
The speed cannot be negative.
∴ The speed of the cyclist = 12 km per hr.
 A boatman takes his boat in a river against the stream from a place A to a place B where AB is 21 km and again returns to A. Thus he takes 10 hours in all. The time taken by him downstream in going 7 km is equal to the time taken by him against stream in going 3 km. Find the speed of river.

 2 kmph
 2.5 kmph
 3 kmph
 3.5 kmph

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Let the speed of boat and river be x km per hr. and y km per hr. respectively.
Then, The speed of boatman downstream = (x + y) km per hr.
and the speed of boatman upstream = (x – y) km per hr.
Time taken by boatman in going 21 km downstream= 21 hours x + y
Time taken by boatman in going 21 km upstream= 21 hrs. x − y
According to the question,= 21 = 21 = 10 ...(i) x + y x − y Now, time taken for 7 kms downstream = 7 hrs. x + y and time taken for 3 kms upstream = 3 hrs. x − y
According to the question7 − 3 = 0 ...(ii) x + y x − y
By (ii) × 7 + (i)49 − 21 = 21 = 21 = 10 x + y x − y x + y x − y ⇒ 70 = 10 x + y
⇒ x + y = 7 ...(iii)
Putting x + y = 7 in equation (ii)
we have7 − 3 = 0 7 x − y ⇒ 1 − 3 = 0 x − y
⇒ x – y = 3 ...(iv)
On adding (iii) and (iv), we have
2x = 10
⇒ x = 5
∴ y = 7 – x = 7 – 5 = 2
∴ Speed of river = 2 km per hr.Correct Option: A
Let the speed of boat and river be x km per hr. and y km per hr. respectively.
Then, The speed of boatman downstream = (x + y) km per hr.
and the speed of boatman upstream = (x – y) km per hr.
Time taken by boatman in going 21 km downstream= 21 hours x + y
Time taken by boatman in going 21 km upstream= 21 hrs. x − y
According to the question,= 21 = 21 = 10 ...(i) x + y x − y Now, time taken for 7 kms downstream = 7 hrs. x + y and time taken for 3 kms upstream = 3 hrs. x − y
According to the question7 − 3 = 0 ...(ii) x + y x − y
By (ii) × 7 + (i)49 − 21 = 21 = 21 = 10 x + y x − y x + y x − y ⇒ 70 = 10 x + y
⇒ x + y = 7 ...(iii)
Putting x + y = 7 in equation (ii)
we have7 − 3 = 0 7 x − y ⇒ 1 − 3 = 0 x − y
⇒ x – y = 3 ...(iv)
On adding (iii) and (iv), we have
2x = 10
⇒ x = 5
∴ y = 7 – x = 7 – 5 = 2
∴ Speed of river = 2 km per hr.
 A person can row a boat 32 km upstream and 60 km downstream in 9 hours. Also, he can row 40 km upstream and 84 km downstream in 12 hours. Find the rate of the current.

 3 kmph
 2.5 kmph
 1.5 kmph
 2 kmph

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Let the upstream speed be x km per hr. and downstream speed be y km per hr.
Then, we can write,32 + 60 = 9 x y and, 40 + 84 = 12 x y Let 1 = m and 1 = n x y
The above two equations can now be written as
32 m + 60 n = 9 ...(i)
and, 40 m + 84 n = 12 ...(ii)
7 × (i) – 5 × (ii) gives 24 m = 3or m = 1 or x = 8 km per hr. 8
4 × (ii) – 5 × (i) gives 36 n = 3or n = 1 or y = 12 km per hr. 12
Rate of current= y − x = 12 − 8 = 2 km. per hr. 2 2 Correct Option: D
Let the upstream speed be x km per hr. and downstream speed be y km per hr.
Then, we can write,32 + 60 = 9 x y and, 40 + 84 = 12 x y Let 1 = m and 1 = n x y
The above two equations can now be written as
32 m + 60 n = 9 ...(i)
and, 40 m + 84 n = 12 ...(ii)
7 × (i) – 5 × (ii) gives 24 m = 3or m = 1 or x = 8 km per hr. 8
4 × (ii) – 5 × (i) gives 36 n = 3or n = 1 or y = 12 km per hr. 12
Rate of current= y − x = 12 − 8 = 2 km. per hr. 2 2