Speed, Time and Distance
- 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.
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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 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.
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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
- 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 ?
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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
- 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.
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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.
- 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.
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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.
