# PE 3 - Time, Speed & Distance | Arithmetic - Time, Speed & Distance

**PE 3 - Time, Speed & Distance | Arithmetic - Time, Speed & Distance**

A candle of 10 cm long burns at the rate of 4 cm in 4 hours and another candle of 12 cm long burns at the rate of 10 cm in 5 hours. If both the candles start burning at the same time; after how long will they be of same length.

- (a)
1 hour

- (b)
1.5 hours

- (c)
2 hours

- (d)
None of these

Answer: Option C

**Explanation** :

Burning rate of 10 cm long candle = 4/4 = 1 cm/hr.

Burning rate of 12 cm long candle = 10/5 = 2 cms/hr.

Let the candles be of equal length after t hours.

∴ 10 – t × 1 = 12 – t × 2

⇒ t = 2 hours.

∴ It will takes 2 hours for the two candles to become equal in length.

Hence, option (c).

Workspace:

**PE 3 - Time, Speed & Distance | Arithmetic - Time, Speed & Distance**

Two persons start from the opposite ends of a 60 km straight track and run to and from between the two ends. The speed of first person is 10 m/s and the speed of other is 125/18 m/s. They continue their motion for 10 hours. How many times they pass each other?

- (a)
5

- (b)
4

- (c)
3

- (d)
None of these

Answer: Option A

**Explanation** :

The speed of two persons is 36 km/h and 25 km/h.

When 2 people start from opposite ends and to and fro between the two ends, time taken to meet for nth time

= $\frac{\left(2n-1\right)D}{a+b}$ [Here, ½< a/b < 2]

∴ $(2n-1)\times \frac{60}{36+25}$ ≤ 10

⇒ 2n – 1 ≤ 61/6

⇒ n ≤ 5.6

∴ Both of them will meet 5 times in 10 hours.

Hence, option (a).

Workspace:

**PE 3 - Time, Speed & Distance | Arithmetic - Time, Speed & Distance**

A bus leaves from Chennai to Bangalore and Bangalore to Chennai every hour, the first one at 6 a.m. The trip from one city to other takes 3½ hours, and all buses travel at the same speed. How many buses will you pass while going from Chennai to Bangalore if you start at 12 noon? (excluding the buses at Chennai or Bangalore)

- (a)
8

- (b)
6

- (c)
9

- (d)
7

Answer: Option D

**Explanation** :

The last bus to arrive in Chennai before we leave would’ve arrived at 11:30. This bus would’ve left Bangalore at 8 a.m. We will not be able to meet this bus but will meet other buses which start after this one.

∴ We will meet buses which left Bangalore at 9 a.m. and after that.

If we start at 12 noon, we will reach Bangalore at 3:30 p.m.

∴ We will mee all buses which left till 3 p.m.

⇒ We meet all buses starting from 9 a.m. till 3 p.m. i.e., 7 buses.

Hence, option (d).

Workspace:

**PE 3 - Time, Speed & Distance | Arithmetic - Time, Speed & Distance**

Two ants start simultaneously from two ant holes towards each other. The first ant covers 8% of the distance between the two ant holes in 3 hours, the second ant covered 7/120 of the distance in 2 hours 30 minutes. Find the speed (feet/h) of the second ant if the first ant travelled 800 feet to the meeting point.

- (a)
15 feet/hr

- (b)
25 feet/hr

- (c)
45 feet/hr

- (d)
35 feet/hr

Answer: Option D

**Explanation** :

Let the distance between the two ant holes be 600d feet.

∴ Speed of first ant = $\frac{600d\times \frac{8}{100}}{3}$ = 16d feet/hr

∴ Speed of second ant = $\frac{600d\times \frac{7}{120}}{2.5}$ = 14d feet/hr

Till meeting point first ant travels 800 ft and second ant travels 600d – 800 ft.

∴ $\frac{800}{600d-800}=\frac{16d}{14d}$

⇒ 700 = 600d – 800

⇒ d = 2.5

∴ Speed of second ant and = 14 × 2.5 = 35 feet/hr

Hence, option (d).

Workspace:

**PE 3 - Time, Speed & Distance | Arithmetic - Time, Speed & Distance**

A train after travelling 100 km meets with an accident and then proceeds with 2/3rd of its former speed and arrives at its destination 2 hours late. Had the accident occurred 50 kms further, it would have reached the destination 1 hour late. What is the total distance travelled by the train?

- (a)
400 kms

- (b)
250 kms

- (c)
200 kms

- (d)
150 kms

Answer: Option C

**Explanation** :

Accident occurs after travelling 100 kms. Let the remaining distance to be travelled is d kms.

Train gets late by 4 hours in travelling d kms at reduced speed.

Also, train will get late by 2 hours in travelling d – 50 kms at reduced speed.

⇒ Train will get late by 2 hours in travelling 50 kms at reduced speed.

Since, train gets late by same time while travelling 50 or d – 50 kms.

⇒ 50 = d – 50

⇒ d = 100 kms.

∴ The total distance to be travelled by train = 100 + 100 = 200 kms.

**Alternately,**

Accident occurs after travelling 100 kms. Let the remaining distance to be travelled is d kms.

Since speed become 2/3rd it would take 3/2 times the normal time.

Let the normal time taken to travel d kms is t. Since the train gets delayed by 2 hours,

∴ 3t/2 – t = 2

⇒ t = 4 hours.

∴ It takes 4 hours to travel d kms …(1)

Now, had the accident occurred after 50 kms, the distance yet to travel would have been d – 50 kms.

Train gets late by 1 hour during this journey.

∴ 3t’/2 – t’ = 1

⇒ t’ = 2 hours.

∴ It takes 2 hours to travel d – 50 kms …(2)

From (1) and (2) we see that train takes 2 hours to travel 50 kms.

∴ Normal speed of the train = 50/2 = 25 kmph

Now, train travels d kms in 4 hours

∴ d = 25 × 4 = 100 kms.

The total distance to be travelled by train = 100 + 100 = 200 kms.

Hence, option (c).

Workspace:

**PE 3 - Time, Speed & Distance | Arithmetic - Time, Speed & Distance**

A and B start from the same point in opposite directions round a circular field 1000 meters in circumference. A did not start till B has run 200 m. They pass each other when A has run 500 m. Who will come first at the starting point and what distance will they then be apart?

- (a)
A, 240 m

- (b)
A, 200 m

- (c)
B 300 m

- (d)
A, 160 m

Answer: Option B

**Explanation** :

Let speed of A = a and that of B = b.

When B has run 200 meters A and B will be 800 meters apart.

Now when they meet A has run 500 meters hence B would have run 800 – 500 = 300 meters.

∴ Ratio of speeds of A and B = 500 : 300 = 5 : 3

They meet at a distance of 500 meters from the starting point i.e., at a point exactly opposite the starting point.

Now for them to come at the starting point both of them have to travel 500 meters. Since A’s speed is greater than b’s speed A will cover 500 meters before B and in the same time B will cover only 300 meters.

∴ When A reaches starting point B will be still 200 meters away from the starting point.

Hence, option (b).

Workspace:

**PE 3 - Time, Speed & Distance | Arithmetic - Time, Speed & Distance**

A and B run a 1 km race. In the first heat A gives B a start of 100 meters and beats him by 40 seconds; In the second heat A gives B a start of 20 seconds and is beaten by 300 meters. In what time (in seconds) can each run 1 kilometer?

- (a)
A = 45, B = 90

- (b)
A = 30, B = 60

- (c)
A = 40, B = 80

- (d)
A = 50, B = 100

Answer: Option D

**Explanation** :

Let speeds of A and B be a and b respectively.

In the first heat A gives B a start of 100 meters and beats him by 40 seconds

∴ $\frac{1000}{a}$ + 40 = 900/b …(1)

In the second heat A gives B a start of 20 seconds and is beaten by 300 meters

∴ $\frac{1000}{a}$ + 20 = 700/b …(2)

Solving (1) and (2), we get,

a = 20 m/s and b = 10 m/s

∴ Time taken by them to complete 1 km is

A = 1000/20 = 50 seconds

B = 1000/10 = 100 seconds

Hence, option (d).

Workspace:

**PE 3 - Time, Speed & Distance | Arithmetic - Time, Speed & Distance**

Two people leave a junction simultaneously, one travelling North and the other travelling West. One of them runs at a speed of 2 km/h more than the other. If after two hours from start, they are 20 kms apart, the speed of the faster person is:

- (a)
14 kmph

- (b)
8 kmph

- (c)
16 kmph

- (d)
10 kmph

Answer: Option B

**Explanation** :

Let the speed of the faster person be x kmph, then speed of the slower person be (x - 2) km/h.

Distance travelled by them in 2 hours will be, 2x and 2(x - 2) kms.

∴ (2x)^{2} + (2(x - 2))^{2} = 20^{2}

⇒ 8x^{2} – 16x + 16 = 400

⇒ 8x^{2} – 16x - 384 = 0

⇒ x^{2} – 2x - 48 = 0

⇒ x = 8 or -6. (-6 will be rejected)

∴ Speed of the faster person is 8 kmph.

Hence, option (b).

Workspace:

**PE 3 - Time, Speed & Distance | Arithmetic - Time, Speed & Distance**

A boatman can row 36 km with the current in the time he would take to row 18 km against it but if the speed of current were 2 km/h more he would row five times as rapidly with the current as against it. What is his rate of rowing in still water?

- (a)
2 kmph

- (b)
9 kmph

- (c)
5 kmph

- (d)
6 kmph

Answer: Option D

**Explanation** :

Let speed of boatman in still water be b and speed of current = r.

Initially,

Speed of boat against the current = b – r

Speed of boat with the current = b + r

Since the boat covers twice the distance downstream as compared to upstream in the same time, it means downstream speed is twice that of upstream speed.

∴ b + r = 2(b - r)

⇒ b = 3r …(1)

Now, when speed of current is increased by 2 kmph, upstream speed is 5 times downstream speed.

∴ (b + r + 2) = 5(b – (r + 2))

⇒ 4r + 2 = 5(2r - 2)

⇒ r = 2 and b = 6 kmph.

Hence, option (d).

Workspace:

**PE 3 - Time, Speed & Distance | Arithmetic - Time, Speed & Distance**

A car driving in fog passed a man walking at 2 kmph in the same direction. He could see the car for 1 minutes and up to a distance of 300 m. What was the speed of the car?

- (a)
20 kmph

- (b)
18 kmph

- (c)
21 kmph

- (d)
None of these

Answer: Option A

**Explanation** :

Man could see the car for 3 minutes and up to a distance of 300 m.

∴ Relative speed of car w.r.t. man = 0.3/(1/60) = 18 kmph.

⇒ Relative speed = 18 = S_{car} – S_{man} = S_{car} – 2

⇒ S_{car} = 20 kmph.

Hence, option (a).

Workspace:

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