QUESTIONS ON TIME, DISTANCE & SPEED (PART-II)

QUERY 11

A man starts A to B at 12 pm and another man starts B to A at 2 pm. They meet each other at 4:05 pm and both of them reach their destination at the same time, what is that time?

A) 6 pm
B) 7 pm
C) 6:45 pm
D) 6:15 pm

Ashwani Singh
Let speed of A and B respectively is s1 and s2 and they meet after x minutes.

Obviously the distance travelled by A in x minute and distance travelled by B in (2 pm to 4:05 m) = 2 hrs 5 minute i.e.125 minutes is equal; so s1*x = s2*125 ——–(1)

Similarly distance travelled by B in x minutes is equal to distance travelled by A in (12 pm to 4:05 pm) = 4 hrs 5 minutes i.e. 245 minutes; so s2*x = s1*245 ——–(2)

Dividing (1) by (2)
(s1*s1)/(s2*s2) = 125/245
=> (s1/s2)² = 25/49
=> (s1/s2)² = (5/7)²
=> s1/s2 = 5/7

Now letting s1 = 5 m/minute and s2 = 7m/minute’ total distance between A and B = 245*5 + 125*7 = 2100 m

Time taken to reach the destination for both A and B is the same; so find time taken of any, that will be the answer. Let’s find that of A

Total time taken by A =2100/5 = 420 i.e. 7 hours; therefore time at which A reach the destination = 12 pm+7 hours = 7  pm (option ‘B’)

QUERY 12

Two persons A and B are at two places P and Q, respectively. A walks at x km/h and B is 2 km/h faster than A, starting simultaneously from where they stand. If they walk towards each other, they meet in 72 minutes. If they walk in the same direction, the faster overtakes the slower in 6 hrs. What are their respective speeds in km/h?

A) 4 km/h and 6 km/h
B) 3 km/h and 5 km/h
C) 5 km/h and 7 km/h
D) 6 km/h and 8 km/h

MAHA GUPTA
CASE-I
As both of them are walking in opposite directions their relative speed = (x+2) + x = (2x+2) km/h
Time taken to meet = 72 minutes = 6/5 hours

Letting distance between them as d km; d = (6/5)(2x+2)              —-DISTANCE = TIME*SPEED

CASE-II
Both of them meet in 6 hours walking in the same direction
Therefore, distance covered by the faster = d + distance covered by the slower

So, 6(x+2) = d +6x
=> d = 12

Now putting this value of d in the equation of case-I; x = 4

Therefore their respective speeds = 4 km/h and 4+2 = 6 km/h (option ‘A’)

QUERY 13

Two cars start simultaneously from place A and B 100 km apart, towards each other. A bird sitting on one car starts at the same time towards the other car, and as soon as it reaches the second car, it flies back to the first car and it continues in this manner flying backwards and forwards from one car to the other until the cars meet. Both cars travel at a speed of 50 km/h and the bird flies at 100 km/h. Total distance covered by the bird will be?

A) 100 km
B) 200 km
C) 50 km
D) 80 km

MAHA GUPTA
Here we need to know the relative speed of the two cars. If two objects are moving in opposite directions (either towards each other or away from each other) there relative speed is equal to the sum of their speeds.

So relative speed of the cars 50+50 = 100 km/h
Obviously the distance of 100 km has to be covered by the cars at 100 km/h
Therefore time taken by the cars to cover it = DISTANCE/SPEED = 100/100 = 1 hour

It also means the sparrow flies for 1 hour
Now, speed of the sparrow = 100 km/h
Hence the distance flown by the sparrow until both the cars meet = SPEED*TIME = 100*1 = 100 km (option ‘A’)

QUERY 14

A man leaves point ‘P’ and reaches the point ‘Q’ in 5 hrs. Another man leaves the point ‘Q’ 2 hrs later and reaches the point ‘P’ in 7 hrs. Find the time in which the first man meets the second man?

A) 3 hours 30 minutes
B) 3 hours 45 minutes
C) 4 hours
D) 4 hours 15 minutes

MAHA GUPTA
The second man starts his journey after 2 hours of the first man, means the first man has travelled some distance already

Now let the total distance be 1 km
The speed of the first man = DISTANCE/TIME = ¹⁄₅ km/h
Distance travelled by him in 2 hours = SPEED*TIME = (¹⁄₅)*2 = ²⁄₅ km
Remaining distance = 1 – ²⁄₅ = ³⁄₅ km

Here we need to know the relative speed of both the men
Now speed of the second men = DISTANCE/TIME = 1/7 km/h
Therefore the relative speed = ¹⁄₅ + ¹⁄₇ = ¹²⁄₃₅ km/h

So time after which both of them will meet = DISTANCE/SPEED
[Here the distance will be the remaining distance as both of them together have to travel this distance only; and the speed will be the relative speed.]

= (³⁄₅)/(¹²⁄₃₅) = ⁷⁄₄ hours = 1.75 hours = 1 hour 45 minutes

But the first man has travelled for 2 hours already
So, the time in which the first man meets the second man = 2hr + 1hr 45 minutes = 3 hours 45 minutes (option ‘B’)

QUERY 15

Arun, Barun and Kiranmala start from the same place and travel in the same direction at speeds of 30, 40 and 60 km/h respectively. Barun starts two hours after Arun. If Barun and Kiranmala overtake Arun at the same instance, how many hours after Arun did Kiranmala start?

A) 1 hour
B) 3 hours
C) 6 hours
D) 4 hours

SHIV KISHORE
Let Arun walks for ‘t’ hours

Distance covered by all obviously is the same
We know that DISTANCE = TIME*SPEED

According to the question Barun starts two hours after Arun
Therefore, 30t = 40(t-2); where ‘t’ is time taken
=> t = 8 hours

Thus, the total distance = 30*8 = 240 km
Hence time taken by Kiranmala to cover the distance = 240/60 = 4 hours

So, The number of hours Kiranmala started after Arun = 8 – 4 = 4 hours (option ‘D’)

QUERY 16

Two buses start from a bus terminal with a speed of 20 km/hr at an interval of 10 minutes. What is the speed of a man coming from the opposite direction towards the bus terminal if he meets the buses at interval of 8 minutes?

A) 3 km/h
B) 4 km/h
C) 5 km/h
D) 7 km/h 

MAHA GUPTA
Time interval between two buses = 10 minutes
Time interval between the two consecutive buses as observed by the man = 8 minutes

Obviously the distance covered by the man in 8 minutes will be equal to the distance covered by bus in (10 – 8) i.e. 2 minutes

Now let’s calculate the distance covered by bus in 2 minutes as the speed of the bus too is given

Distance covered by the bus in 60 minutes = 20 km
Therefore the distance covered by the bus in 2 minute = (20/60)2 = 2/3 km

But the man in 8 minutes travels this distance of 2/3 km
Therefore in 60 minutes distance traveled by him = [(2/3)/8]60 = 5 km/h (option ‘C’)

QUERY 17

A, B, C start from the same place and travel in the same direction with speed 40, 48, 60 km/h respectively. B starts 2 hours after A. If B and C overtake A at the same instant, how many hours after A did C start?

A) 1 hour
B) 3 hours
C) 6 hours
D) 4 hours

MAHA GUPTA
Let A walks for ‘t’ hours

Distance covered by all obviously is the same
We know that DISTANCE = TIME*SPEED

According to the question B starts two hours after A
Therefore, 40t = 48(t-2); where ‘t’ is time taken
=> t = 12 hours

Thus, the total distance = 40*12 = 480 km
Hence time taken by C to cover the distance = 480/60 = 8 hours

So, The number of hours C started after A = 12 – 8 = 4 hours (option ‘D’)

QUERY 18

When I travel at 60 m/s I reach office 2 hr early. And when I travel at 30 m/s I reach office on time. What should be my speed so that I reach office one hour early?

A) 35 m/s
B) 38 m/s
C) 40 m/s
D) 45 m/s

MAHA GUPTA
Let the distance to be covered = x m/s; and
the speed of the man at which he reaches office 1 hour early = y m/s

Therefore the time that the man takes if he walks at 30 m/s (it’s normal speed here as he reaches office in time at this speed)= Distance/speed = x/30
And the time taken at 60 m/s = x/60
Time taken at the speed when he reaches office 1 hour early = x/y

So according to the question:
x/30 – x/60 = 2 => x = 120 ——-(i)
x/30 – x/y = 1 ——–(ii)

putting x = 120 in (ii)
120/30 – 120/y = 1
=> y = 40 m/s (option ‘C’)

QUERY 19

Climbing up on an going down escalator I take 60 minutes to reach to the top from bottom. Climbing down on the same going down escalator I take 30 minutes to reach bottom from the top. How many minutes I will take to reach the top from bottom if escalator is still? My own speed upward and downward is same.

A) 36 minutes
B) 42 minutes
C) 45 minutes
D) 40 minutes

RITIK DALAL
Let k = distance traveled
And speed of myself = x
And speed of the escalator = y

We know the TIME = DISTANCE/SPEED
Now relative speed going upward = x – y
And relative speed while climbing downward = x + y

Therefore, k/(x – y) = 60 —- (i)
And k/(x + y) = 30 —- (ii)
Solving (i) and (ii) x = 3y and k = 120y
If escalator is still then y = 0

Hence time taken by me if escalator is still = DISTANCE/SPEED = k/x = 120y/3y = 40 (option ‘D’)

QUERY 20

Walking at 5/6 of usual speed a man is 10 minutes late. Find out the usual time to cover the entire journey.

A) 43 minutes
B) 42 minutes
C) 52 minutes
D) 50 minutes

MAHA GUPTA
Let the usual time = x minutes
If he walks 5/6 of his usual speed means the time will be taken 6/5 of the usual time.

Therefore 6/5 of the usual time = usual time + 10 minutes

=> 6x/5 = x + 10
=> x = 50 minutes (option ‘D’)