# PE 4 - Time, Speed & Distance | Arithmetic - Time, Speed & Distance

**PE 4 - Time, Speed & Distance | Arithmetic - Time, Speed & Distance**

The circumferences of front wheel and rear wheel of a wagon are 315 cm and 420 cm, respectively. What is the distance travelled by the wagon, if the front wheels have made 10 rotations more than the rear wheels?

- (a)
126 m

- (b)
105 m

- (c)
108 m

- (d)
95 m

Answer: Option A

**Explanation** :

If x is the number of rotations made by the smaller wheel, and y is the number of rotations made by the larger wheel in covering a certain distance, then

(x) × (315) = y × 420 …(1)

x = y + 10 …(2) (Given)

From (1) and (2),

⇒ (y + 10) (315) = y × 420 …(3)

∴ y = 30

So, distance covered by the wagon = 30 × 420 = 126 m.

Hence, option (a).

Workspace:

**PE 4 - Time, Speed & Distance | Arithmetic - Time, Speed & Distance**

There was a race between a hare and a tortoise. The length of the race was 2400 m. The speeds of hare and tortoise were 24 km/hr and 0.4 km/hr, respectively. But, along the way, the hare felt tired and slept for some time. After some time, he woke up and completed the race with the same speed as before. But, the race ended in a tie. For how long did the hare sleep?

- (a)
370 mins

- (b)
354 mins

- (c)
338 mins

- (d)
242 mins

- (e)
180 mins

Answer: Option B

**Explanation** :

Let the tortoise run for T seconds.

So, 0.4 × 5/18 × T = 2400

⇒ T = 21,600 seconds

Let the time spent by the hare running = t seconds

So, 24 × 5/18 × t = 2400

or, t = 360 seconds

So, time spent by the hare sleeping = (T - t) seconds = (21600 - 360) seconds = 21240 seconds = 354 mins.

Hence, option (b).

Workspace:

**PE 4 - Time, Speed & Distance | Arithmetic - Time, Speed & Distance**

A ship is 72 kms away from the shore. It springs a leak in the bottom which admits 3 ⅖ tons of water in 12 minutes. 60 tons of water would suffice to sink her, but the ship’s pump can throw out 12 tons of water in an hour. Find the minimum speed of the ship, so that she may reach the shore just before ship begins to sink.

- (a)
4 kmph

- (b)
4.5 kmph

- (c)
5 kmph

- (d)
None of these

Answer: Option D

**Explanation** :

3 ⅖ tons of water is admitted in 12 mins.

So, in 60 mins (17/5 × 60/12) = 17 tons of water is admitted i.e. 17 tons/hr.

So, net inlet of water = 17 – 12 = 5 tons/hr to sink

Hence, to admit 60 tons it has 60 ÷ 5 = 12 hours

The ship needs to cover a distance of 72 kms in 12 hours to avoid sinking.

Speed of ship = 72/12 = 6 km/hr

Hence, option (d).

Workspace:

**PE 4 - Time, Speed & Distance | Arithmetic - Time, Speed & Distance**

A hare pursued by a hound is 50 of her own leaps ahead of him. When the hare takes 4 leaps, the hound takes 3. In one leap the hare goes 1¾ meters and the hound 2¾ meters. In how many leaps will the hound overtake the hare?

- (a)
200 leaps

- (b)
205 leaps

- (c)
210 leaps

- (d)
None of these

Answer: Option C

**Explanation** :

50 leaps of Hare = 50 × 13/4 = 175/2 m

So, the hound should gain 175/2 m over hare.

Now, when the hound travels 3 × 2¾ m hare travels 4 × 1¾ m

∴ In 3 leaps of Hound, the hound gain =33/4 - 28/4 = 5/4 m

∴ Number of leaps required = (175∕2)/(5∕4) × 3 = 210 leaps

Hence, option (c).

Workspace:

**PE 4 - Time, Speed & Distance | Arithmetic - Time, Speed & Distance**

Two guns are fired from a fort after an interval of 15 mins. and a man running towards the fort hears the reports of the gun after an interval of 14 mins. Find the rate of progress, sound travelling at the rate of 300 meters per sec.

- (a)
7.2 kmph

- (b)
5.5 kmph

- (c)
4.5 kmph

- (d)
None of these

Answer: Option A

**Explanation** :

Let us consider the distance from the point where the person hears first shot till he hears the second shot.

This distance is covered by man in 15 mins while the same distance is covered by sound in (15 – 14 =) 1 min.

Distance traveled by sound in 1 min = 60 × 300 = 18000 meters.

So, man covers in 15 mins = 1800 m = 1.8 km

∴ In 1 hour distance covered by the man = 4 × 1.8 = 7.2 kms.

Hence, option (a).

Workspace:

**PE 4 - Time, Speed & Distance | Arithmetic - Time, Speed & Distance**

Two trains X and Y start from stations A and B towards B and A respectively. After passing each other, they take 4 hours 48 minutes and 3 hours 20 minutes to reach B and A respectively. If train X is moving at 45 km/h, the speed of train Y is:

- (a)
60 kmph

- (b)
54 kmph

- (c)
64 kmph

- (d)
37.5 kmph

Answer: Option B

**Explanation** :

Here $\frac{\mathrm{X}\text{'}\mathrm{s}\mathrm{speed}}{\mathrm{Y}\text{'}\mathrm{s}\mathrm{speed}}$ = $\sqrt{\frac{TimetakenbyYtoreachAaftermeeting}{TimetakenbyXtoreachBaftermeeting}}$

⇒ $\frac{45}{\mathrm{Y}\text{'}\mathrm{s}\mathrm{speed}}$ = $\sqrt{\frac{10/3}{24/5}}$ = $\frac{5}{4}$

⇒ Y's speed = 54 kmph

Hence, option (b).

Workspace:

**PE 4 - Time, Speed & Distance | Arithmetic - Time, Speed & Distance**

In a race of 600 m, A can beat B by 50 m and in a race of 500 m B can beat C by 60 m. By how many meters will A beat C in a race of 400 m?

- (a)
48 m

- (b)
56 m

- (c)
71 m

- (d)
58 m

Answer: Option A

**Explanation** :

Ratio of speeds of A : B = 600 : 550 = 12 : 11

Ratio of speeds of B : C = 500 : 440 = 25 : 22

Ratio of speeds of A : C = 12 × 25 : 11 × 22 = 150 : 121

∴ 150/121 = 300/D_{C}

⇒ D_{C} = 242

⇒ A beats C by (300 - 242) = 58 meters in a 300 meter race.

Hence, option (a).

Workspace:

**PE 4 - Time, Speed & Distance | Arithmetic - Time, Speed & Distance**

Excluding stoppages, the speed of a bus is 54 km/h and including stoppages, it is 45 km/h. For how many minutes does the bus stop per hour?

- (a)
9

- (b)
10

- (c)
12

- (d)
20

Answer: Option B

**Explanation** :

Due to stoppages it covers 9 km less

Now, time to cover 9 km = 9/54 × 60 = 10 minutes.

**Alternately,**

With average speed of 45 kmph, the bus can cover 45 kms in 1 hour.

The running time to cover 45 kms = 45/54 = 5/6 hours or 50 mins.

Hence, the bus stops for 10 mins in every 1 hour.

Hence, option (b).

Workspace:

## Feedback

Help us build a Free and Comprehensive Preparation portal for various competitive exams by providing us your valuable feedback about Apti4All and how it can be improved.