BANK & INSURANCE (SPEED TIME AND DISTANCE) PART 1

Total Questions: 45

31. Tom, Charlie and Alan decide to run around a circular track. Three of them run in same direction and they started running at same time. Tom is the quickest and when Tom finishes a lap, Alan is as much behind Charlie as Charlie is behind Tom. When tom completes 3 laps, Alan is at the exact same position on the circular track as Charlie was when tom finishes 1 lap. Find the ratio of speed of Tom, Charlie and Alan.

Correct Answer: (d) 5:3:1 
Solution:

Let track length be equal to t
When Tom completes a lap, let us assume Charlie has run a distance of (t - d).
At this time, Alan should have run a distance of (t - 2d).
Distance for Alan after 3 lap = 3 × (t - 2d) = 3t - 6d.
After 3 laps, Alan is in the same position as Charlie was at the end of one lap.
So he could have travelled a distance of either t - d.
3t - 6d = t - d
=> 2t = 5d
=> d = 0.4t
the ratio of their speeds is 5 : 3 : 1.

32. A man X is driving to work. In order to reach office for meeting he drives faster and the speed is increased by 5 km/h and he reached office 30 minutes early. Assuming that due to traffic his speed is decreased by 3 km/h he will be late for the meeting by 30 minutes. Calculate the speed and the distance to the office?

Correct Answer: (a) 15 km/h and 30 km
Solution:

Let the distance be = x
Speed be = y
Time be = z
So according to question
x/y - x/(y + 5) = x/(y - 3) - x/y
1/y - 1/(y + 5) = 1/(y - 3) - 1/y
(y+5) - y / [y(y+5)] = (y - y+3) / [(y)(y - 3)]
5/[y(y+5)] = 3/[y(y - 3)]
2y = 30
y = 15km/h
Putting the value in the equation:
(y+5)x(z-30/60) = (y-3)x(z+30/60)
(y+5)x(2z-1) = (y-3)x(2z+1)
20x(2z-1) = 12x(2z+1)
10z-5 = 6z+3
4z = 8
z = 2hours
z = 120 minutes
Distance = 15×120/60 = 30 km.

33. The taxis plying in Delhi have the following fare structure: fixed charge of Rs. 15 for the first four kms, Rs.7 for every km in excess of 4 km and up to 15 km, and Rs.12 for every km in excess of 15 km. Autos on the other hand charge Rs. 5 per km. Abhijeet takes a taxi from the New Delhi railway station to his home. On the way, at a distance of 18 km from the railway station, he gets down from the taxi and takes an auto to reach his home. He spends a total of Rs.148 to reach his home from railway station. How far is his home from the railway station?

Correct Answer: (a) 22 km 
Solution:Let the distance between home and railway station be x km. He uses taxi for 18 km, he must have paid = 15 + 7×11 + 3×12 = Rs.128.
Therefore, he must have paid the rest of 148-128 = Rs.20 to the auto driver, thus he must have travelled 20/5 = 4 km by auto.
The distance between railway station and home = 18 + 4 = 22 km

34. A train covered a certain distance at a uniform speed. If the train could have been 20 km/hr faster, it would have taken 2 hours less than the scheduled time to cover the same distance and if the train was slower by 20 km/hr, it would have taken 4 hours more than the scheduled time to cover the same distance. The distance covered by the train is?

Correct Answer: (c) 480 km
Solution:

Let the scheduled distance travelled = d,
Let time taken = t, uniform speed = s
(s+20)(t-2) = st – (1)
(s - 20)(t+4) = st – (2)
From (1), 20t - 2s = 40
From (2), -20t + 4s = 80
On solving the equations, we get t = 8 hrs and s = 60km/hr
Distance travelled = 60 × 8 = 480 km

35. An ant is crawling up a pillar to reach the top. The top of the pillar is 1800 cm from its position. After every minute of crawling it halts for half a minute. In every halt it slides down by 30 cm from its positions. In how many minutes can the ant reach the top of the pillar if it can crawl 150 cm per minute?

Correct Answer: (d) 21 min 48 sec
Solution:

Number of trials = 1800/(150 - 30) = 1800/120 = 15
Let us take 14 trails of sliding up.
For every trail of these 14 trails its effective upward movement is (150 - 30) = 120 cm
The time taken for this = (14 × 1 + 14 × 1/2)
= 21 minutes.
Total distance = (14 × 120) = 1680 cm
Remaining = 1800 – 1680 = 120 cm

Remaining 120 cm it can reach in (60 × 120)/150
= 48 seconds
Total time = 21 min 48 sec.

36. Two trains one from Hyderabad to Cochin and another from Cochin to Hyderabad start simultaneously. After they meet, Trains reach their destinations after 4 hrs and 9 hrs respectively. Find the ratio of the speeds?

Correct Answer: (b) 3:2 
Solution:

A : B = S₁/S₂ = √(T₂/T₁) = √(9/4) = 3 : 2

37. Two cars start from a place with a speed of 40 kmph at an interval of 10 minutes. What is the speed of a man coming from the opposite direction towards the place if he meets the cars at an interval of 8 minutes?

Correct Answer: (a) 10 kmph 
Solution:

Distance covered in 10 minutes at 40 kmph = distance covered in 8 minutes at (40+x) kmph.
40×10/60 = 8/60 × (40+x), 20 × 5 = 80 + 2x
x = 10 kmph

38. Waking 3/4 of his normal speed, Ravi was 18 minutes late in reaching his office. The usual time took to cover the distance between his home and office was:

Correct Answer: (d) 54 minutes 
Solution:3/4 of speed = 4/3 of original time
4/3 of original time = original time + 18 minutes
1/3rd of original time = 18 minutes
Thus, original time = 18×3 = 54 minutes.

39. Mr. Ravi completes a certain journey by a car. If he covered 40% of the distance at the speed of 20kmph, 50% of the distance at 25 kmph and the remaining of the distance at 10 kmph, then what will be the speed?

Correct Answer: (b) 20 kmph 
Solution:Assume Total distance = 100 km.
So speed = 100/[(40/20)+(50/25)+(10/10)]
speed = 100/[(2)+(2)+(1)]
= 100/5 = 20 kmph.

40. Ajay walked at 10 kmph for certain part of the journey and then he took an auto for the remaining part of the journey travelling at 30 kmph. If he took 10 hours for the entire journey, what part of journey did he travelled by auto if the average speed of the entire journey be 18 kmph?

Correct Answer: (d) 120 km 
Solution:Total distance = 18×10 = 180
Journey traveled by auto = x hours
30x + (10-x) 10 = 180
30x + 100 - 10x = 180
20x = 80, x = 4 hours
Distance travelled by auto = 4 × 30 = 120 km