# QUESTIONS ON TIME, DISTANCE & SPEED (PART-I)

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## QUESTIONS ON TIME, DISTANCE & SPEED (PART-I)

#### QUERY 1

**Supposing that telegraph poles on a railroad are 50 m apart, how many will be passed by a train in 4 hours if the speed of the train is 45 km an hour.**

A) 3601

B) 3600

C) 360

D) 361

**JAYANTCHARAN CHARAN**

The distance travelled by the train = 45 × 4 = 180 km = 180000 m

Now the number of poles passed by that train in 4 hours = (180000/50) + 1 = 3600 + 1 = 3601 (1 is added because one pole was also at the starting point)

So option ‘A’ is correct.

QUERY 2

**A, B, C walk 1 km in 5 minutes, 8 minutes and 10 minutes respectively. C starts walking from a point, at a certain time; B starts from the same point 1 minute later and A starts from the same point 2 minutes later than C; then A meets B and C at times?**

A) 2 minutes, 3 minutes

B) 4/3 minutes, 3 minutes

C) 5/3 minutes, 2 minutes

D) 1 minute, 2 minutes

**MAHA GUPTA**

Speed of A, B, C is 1/5, 1/8, 1/10 in km/min respectively

B walks 1 km in 8 minutes, means he covers 1/8 km in 1 minute

As A starts after 1 minute than B does

So the initial distance between A & B= 1/8 km

Here we need to know their relative speed; and we know if two bodies are moving in the same direction the relative speed is the difference of their actual speed

So the relative speed of A & B = 1/5 – 1/8 = 3/40 km/min

But we have to find the time taken by A to meet B; obviously A has to make up the distance that B covers in 1 minute i.e. 1/8 km

We know Time = Distance/Speed

So the time taken by A to meet B = (1/8)*(40/3) = 5/3 minutes

If we see, the answer options 5/3 is only in ‘C’; hence it’s the correct answer. (option ‘C’)

Otherwise you can calculate the time taken by A to meet C like this; see how

Distance between A & C = 1/10 + 1/10 = 2/10 km

Their relative speed= 1/5 – 1/10 = 1/10 km/min

Therefore the time= (2/10)*(10/1) = 2 minutes

So the answer 5/3 minutes; 2 minutes (option ‘C’)

QUERY 3

**A man covered a certain distance at some speed. Had he moved 3 Km per hour faster, he would have taken 40 minutes less. If he had moved 2 Km per hour slower, he would have taken 40 minutes more. The distance in Km is?**

A) 30

B) 40

C) 50

D) 60

**MAHA GUPTA**

Let the distance to be covered = x km; and

the normal spreed of the man = y kmph

Therefore the time that the man would have taken if he walked at normal speed = Distance/speed = x/y hours

If there is an increase in speed by 3 kmph the speed = y+3 kmph; and the time to be taken this way = x/(y+3) hours

If there is a decrease in speed by 2 kmph the speed = y – 2 kmph; and the time to be taken this way = x/(y – 2) hours

So according to the question:

x/y – x/(y+3) = 40/60 => 2y(y + 3) = 9x ——–(i) ……………..(40 minutes = 40/60 hours)

x/(y-2) – x/y = 40/60 => y(y – 2) = 3x ——–(ii)

Now dividing (i) by (ii)

[2y(y + 3)]/[(y(y – 2)] = 9x/3x

=> y = 12

Now by putting this in equation (ii)

x = 40 (option ‘B’)

QUERY 4

**A person walks a distance in 114 days, when he rests 9 hours a day. How long will he take to walk twice the distance if he walks twice as fast and rests twice as long each day as before?**

A) 57 days

B) 228 days

C) 285 days

D) 324 days

**HITESH ARORA
**Speed of the person and the distance covered by him are doubled. Obviously it will cancel out the effect of one another, so time he spends walking only will matter.

Now time spent walking per day in the first condition = 24 – 9 = 15 hours

And time spent walking per day in the second condition = 24 – 9*2 = 6 hours

Total time spent on walking in the first condition = 114*15 = 1,710 hours

But now he walks only 6 hours a day

Hence time to be taken now = 1710/6 = 285 days (option ‘C’)

QUERY 5

**A man who is seen 100 meters in the fog is walking @2m/s and a motorcycle crosses him in the same direction of the man it sees in 30s. what is the speed of the motorcycle?**

A) 16 m/s

B) 16/3 m/s

C) 5 m/s

D) 6 m/s

**MAHA GUPTA**

The motorcycles crosses the man in 30 seconds; means we need to find the distance traveled by that man in 30 seconds.

In one second the distance traveled by the man = 2 meters

Therefore in 30 seconds the distance covered by him = 2*30 = 60 meters

As the motorcycles sees the man 100 meters away, therefore the distance covered by it in 30 seconds = 100 meters + the distance covered by the man in 30 seconds i.e. 60 meters = 160 meters

Speed = Distance/Time

So the speed of the motorcycle = 160/30 = 16/3 m/s (option ‘B’)

QUERY 6

**A man travel 360 km in 4 hrs partly by air and partly by train. If he had travelled all the way by air, he would have saved 4/5 of the time he was in train and would have arrived at his destination 2 hrs early. Find the distance he travelled by air and train.**

A) 270 km and 90 km

B) 280 km and 100 km

C) 280 km and 80 km

D) 270 km and 80 km

**Rohan Kashyap
**If the man had travelled all the way by air he would have reached 2 hours early; means he would have covered 360 km in 2 hr by air

So speed in air = 360/2 = 180 km/h

But this 2 hrs is 4/5 of the time of the journey he would have covered by train

Therefore time to be taken by the train for whole journey = 2*(5/4) = 2.5 hrs.

Thus, time for he actually went by air = 4 – 2.5 = 1.5 hours

Therefore the distance that he covered by air = 180*(1.5) = 270 km, and the distance covered by him in the train = 360 – 270 = 90 km (option ‘A’)

QUERY 7

**Two cars moving with speed v1 and v2 towards a crossing along two roads. If their distances from the crossing be 40 meter and 50 meter at an instance of time then they do not collide if their speed are such that?**

A) v1 : v2 = 5 : 4

B) v1 : v2 = 25 : 16

C) v1 : v2 = 16 : 25

D) v1 : v2 = 4 : 5

**MAHA GUPTA
**They will collide only if the following pair of equations is satisfied

Time*v1 = 40 (i)

Time*v2 = 50 (ii)

Dividing (i) by (ii)

v1/v2 = 4/5

Therefore, the ratio of their speeds when they do not collide

v1/v2 ≠ 4 : 5 (option ‘D’)

#### QUERY 8

**If a person leaving his home for the bus stop at 4 km/h, notices that the bus has gone 7 minutes before already. And if he leaves his home at 6 km/h, he reaches the bus stop early by 8 minutes. Find the distance between his house and the bus stop.**

A) 1 km

B) 1.5 km

C) 2 km

D) 3 km

**MAHA GUPTA
**Let the required distance = x km.

Difference in time taken at two speeds = 7 + 8 = 15 minutes

= ¼ hours

We know TIME = DISTANCE/SPEED

Now therefore, time taken to reach the bus stop in first case = x/4

And time taken to reach the bus stop in the second case = x/6

Hence, x/4 – x/6 = ¼

=> x = 3 (option ‘D’)

#### QUERY 9

**A policeman on his jeep chases a thief on his bike. Jeep runs at a speed of 2160 m/5 sec and bike at 360 m/sec. If thief is initially 12 km ahead then how far policeman will travel to catch the thief?**

A) 60 km

B) 65 km

C) 72 km

D) 50 km

**MAHA GUPTA
**Speed of the policeman per second = 2160 m/5s = 432 m/s

Let the distance travelled by the thief till he is caught by the policeman = x meters

Therefore total distance travelled by the policeman till the thief is caught = (12000 + x) meters

We know, TIME = DISTANCE/SPEED

Hence time taken by the policeman = (12000 + x)/432 seconds

And time taken by the thief = x/360 seconds

Obviously time taken by both must be equal

So, (12000 + x)/432 = x/360

Solving, x = 60000 meters = 60 km

Therefore total distance travelled by the policeman till the thief is caught = 12 + 60 = 72 km (option ‘C’)

QUERY 10

**While going in a car, a person sees a milestone that has a 2 digit number written on it. After travelling 1 hour, he sees the same digits… but in reverse order. After travelling another 1 hour, he sees the same digit but with 0 between them. Find the speed of the car.**

A) 55 km/h

B) 60 km/h

C) 45 km/h

D) 40 km/h

**MAHA GUPTA**

Letting unit place digit as y and tens digit as x, find the numbers at each milestone

The number at the first milestone wherefrom he starts = 10x + y

As the digits reverse at the second milestone, the number at it = 10y + x

At the third milestone the digits are same but 0 is in the middle, means digit that was at tens place earlier comes at hundreds place; so the number = 100x + 0 + y = 100x + y

Its easy to understand that the distance covered in the first hour = Number at the second milestone – Number at the first milestone, i.e. (10y + x) – (10x + y)

And distance covered in the second hour = Number at the third milestone – Number at the second milestone, i.e, (100x + y) – (10y + x)

Obviously the speed is uniform therefore

Distance covered in first hour = Distance covered in second hour

(10y + x) – (10x + y) = (100x + y) – (10y + x)

Solving, y = 6x

As x and y both are digits, so value of each can’t be more than 9; and being such x = 1 only

Therefore y = 6

Hence the number at the first milestone = 10x + y = 16

Number at the second milestone = 10y + x = 61

The number at the third milestone = 100x + y = 106

Therefore the distance covered in the first and second hour each = 61 – 16 = 45

But this distance is covered in 1 hour; so the speed of the car = 45 km/h (option ‘C’)

SHORT

Since first 2 values are two digit numbers, the speed can’t be greater than 100.

Thus difference between 2nd and 3rd value is also less than 100.

It means value at the third milestone is in 100’s

So x = 1

Now checking by hit and trial method y = 6

So the numbers at all the three milestones respectively = 16, 61, 106

Hence the speed of the car = 61 – 16 = 45 km/h (option ‘C’)