Time and Distance problems
- 1. Akbar and Birbal setout at the same time to walk towards each other respectively from Agra and Banaras 144 km apart. Akbar walks at the constant speed of 8 km/h, while Birbal walks 4 km in the first hour, 5 km in the second hour, 6 km in the third hour and so on. Then the Akbar and Birbal will meet?
Options- A. In 6 hr.
- B. In 8 hr.
- C. Midway between Agra and Banaras
- D. 80 km away from Banaras Discuss
Correct Answer: Midway between Agra and Banaras
Explanation:
Distance travelled by them in first hour = 12 km
Distance travelled by them in second hour = 13 km
Distance travelled by them in third hour = 14 km and so on
Thus in 9 hours they will cover exactly 144 km and in 9 h each will cover half the total distance.
( 8 x 9 = 72 and 4 + 5 + 6 + 7 + 8 + 9 + 10 + 11 + 12 = 72)
- 2. Amarnath express left Amritsar for Gorakhpur. Two hours later Gorakhnath express left from Amritsar to Gorakhour. Both trains reached Gorakhpur simultaneously. If Amarnath express had started from Amritsr and Gorakhnath express had started from Gorakhpur at the same time and travelled towards each other they would meet in 1 h 20 min. Find the time taken by Amarnath express to travel from Amritsar to Gorakhpur(in hour)?
Options- A. 2
- B. 4
- C. 5
- D. 6 Discuss
Correct Answer: 4
Explanation:
Let Amarnath express takes H hours, then Gorakhnath express takes (H - 2) hours.
? 1/H + 1/(H - 2) = 60/80
H = 4 hrs.
- 3. A man goes to the fair in Funcity with his son and faithful dog. Unfortunately man misses his son which he realises 20 minutes later. The son comes back towards his home at the speed of 20 m/min and man follows him at 40 m/min. The dog runs to the son (child) and comes back to the man (father) to show him the direction of his son. It keeps moving to and fro at 60 m/min between son and father, till the man meets the son. what is the distance travelled by the dog in the direction of the son?
Options- A. 800 m
- B. 1675 m
- C. 848 m
- D. 1000 m Discuss
Correct Answer: 1000 m
Explanation:
In 20 minutes the difference between man and his son = 20 x 20 = 400 m
Distance travelled by dog when he goes towards son = (400/40) x 60 = 600 m and time required is 10 minutes
In 10 minutes the remaining difference between man and son = 400 - (20 x 10) = 200 m
Note: Relative speed of dog with child is 40 km/h and the same with man is 100 km/h.
Time taken by dog to meet the man = 200/100 = 2 min
In 2 minute the remaining distance between child and man = 200 - ( 2 x 20) = 160 m
Now, the time taken by dog to meet the child again = 160/40 = 4 min
In 4 minutes he covers 4 x 60 = 240 m distance while going towards the son.
In 4 minute the remaining distance between man and child = 160 - (4 x 20) = 80 m
Time required by dog to meet man once again = 80/100 = 0.8 min
In 0.8 min remaining distance between man and child = 80 - (0.8 x 20) = 64 min
Now, time taken by dog to meet the child again = (64/40) x (8/5) min
? Distance travelled by dog = (8/5) x 60 = 96 m
Thus, we can observe that every next time dog just go 2/5th of previous distance to meet the child in the direction of child.
So, we can calculate the total distance covered by dog in the direction of child with the help of GP formula.
Here, first term (a) = 600 and common ratio (r) = 2/5
Sum of the infinite GP = a/(1 - r)
= 600/(1 - 2/5) = (600 x 5)/3 = 1000 m
- 4. A hunter fired two shots from the branch of a tree at an interval of 76 seconds. A tiger separating too fast hears the two shots at an interval of 83 seconds. If the velocity of the sound is 1195.2 km/h, then find the speed of tiger?
Options- A. 112.8 km/h
- B. 100.8 km/h
- C. 80.16 km/h
- D. none of these Discuss
Correct Answer: 100.8 km/h
Explanation:
In case of increasing gap between two objects.
Speed of sound / speed of tiger = Time utilised / Difference in time
? 1195.2 / x = 83 / 7
x = 100.8 km/h
- 5. A soldier fired two bullets at an interval of 335 seconds moving at a uniform speed v 1. A terrorist who was running ahead of the soldier in the same direction, hears the two shots at an interval of 330 seconds? If the speed of sound is 1188 km/h, then who is the faster and by how much?
Options- A. Soldier, 22 km/h
- B. Terrorist, 25 km/h
- C. Soldier, 18 km/h
- D. Terroist, 20 km/h Discuss
Correct Answer: Soldier, 18 km/h
Explanation:
Speed of wind (sound) / Relative speed of soldier and terrorist = Time utilised / Difference in time
1188/x = 330/5
x = 18 km/h and solder is faster.
- 6. In reaching the Purnagiri a man took half as long again to climb the second third as he did to climb the first third and a quarter as long again for the last third as for the second third. He took altogether 5 hr 50 minutes. Find the time he spent on the first third of the journey?
Options- A. 72 min
- B. 80 min
- C. 81 min
- D. 88 min Discuss
Correct Answer: 80 min
Explanation:
Let the time taken in first third part of the journey be x minutes, then the time required in second third part of the journey is 3x/2 and in the last third part of the journey time required is 15x/8.
Therefore, x + (3x/2) + (15x/8) = 350min
? x = 80 min
- 7. A man travels the first part of his journey at 20 km/h and the next at 70 km/h, covering the entire journey at an average speed of 50 km/h. What is the ratio of the distance that he covered at 20 km/h to that he covered at 70 km/h?
Options- A. 4 : 21
- B. 3 : 22
- C. 1 : 4
- D. 3 : 5 Discuss
Correct Answer: 4 : 21
Explanation:
x/20 + y/70 = (x + y)/50
? (70x + 20y)/1400 = (x + y)/50
? 42x = 8y
? x/y = 4/21
- 8. A man decides to travel 80 km in 8 h party by foot and party on a bicycle. If his speed on foot is 8 km/h and on bicycle 16 km/h, what distance would he travel on foot?
Options- A. 20 km
- B. 30 km
- C. 48 km
- D. 60 km Discuss
Correct Answer: 48 km
Explanation:
Here, S1 = 8 km/h
and S2 = 16 km/h
? d1 =s1 x t1 = 8t1 ...(i)
and d2 = s2 x t2 = 16t2 ...(ii)
We know that,
t1 + t2 = 8 ...(iii)
and
d1 + d2 = 80 (given) ...(iv)
From Eqs.(i) and (ii) put the value of d1 and d2 in Eq. (iv), we get
d1 + d2 = 80
8t1 + 16t2 = 80
? 8t1 + 8t2 + 8t2 = 80
8(t1 + t2) + 8t2 = 80
From Eq (iii),
8 x 8 + 8t2 = 80
8t2 = 80 - 64 = 16
? t2 = 16/8 = 2h
? t1 = 8 - 2 = 6 h
? Distance travelled by foot = d1 = 8 x 6 = 48 km.
- 9. A car driver covers a distance between two cities at a speed of 60 km/h and on the return his speed is 40 km/h. He goes again from the 1st to the 2nd city at twice the original speed and returns at half the original return speed. Find his average speed for entire journey?
Options- A. 55 km/h
- B. 50 km/h
- C. 48 km/h
- D. 40 km/h Discuss
Correct Answer: 40 km/h
Explanation:
Required average speed = 4/(1/60 + 1/40 + 1/120 +1/20)
= (4 x 120)/12 = 40 km/h
- 10. The distance between two station 'X' and 'Y' is 450 km. A train L starts at 6:00 pm from X and moves towards Y at an average speed of 60 km/h. Another train M starts from Y at 5:20 pm
and moves towards X at an average speed of 80 km/h How far from X will the two trains meet and at what time?
Options- A. 170 km, 8 : 50 pm
- B. 150 km, 7 : 50 pm
- C. 170 km, 6 : 50 pm
- D. 150 km, 9 : 50 pm Discuss
Correct Answer: 170 km, 8 : 50 pm
Explanation:
Let two trains meet at a km from 'X'
[Time taken by M to cover (450 - a)km] - [Time taken by the L to cover a km ] = 40/60
? (450 - a)/80 - a/60 = 40/60
? (450 - a)/80 = 40/60 + a/60
? (450 - a)/8 - (a + 40) / 6 = 0
? 3(450 - a) - 4(a + 40) = 0
? 7a = 1190
? a = 1190 / 7 = 170
Time taken by L to cover 170 km = 170/60 h = 2 h 50 min
So, the two trains will meet 2 h 50 min after 6:00 pm. it means that the two trains will meet at 8 : 50 pm .
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