# Exercise : 3

1. A man walks half of the journey at 4 km/h by cycle does one third of journey at 12 km/h and rides the remainder journey in a horse cart at 9 km/h, thus completing the whole journey in 6 hours and 12 minutes. The length of the journey is
(a) 36 km
(b) 133267 km
(c) 40 km
(d) 28 km
(e) None of these

#### View Ans & Explanation

Ans.a

Let the length of the journey = x km.

∴ Journey rides by horse cart = x(1 - 12 - 13)

= 16x km

Then, total time taken to complete journey = 315hr

⇒ t1 + t2 + t3 = 315

x2 × 14 + x3 × 112 + x(6 × 9) = 315

⇒ x = 315 × 21637 = 36.2 km ≈ 36 km

2. R and S start walking each other at 10 AM at the speeds of 3 km/h and 4 km/h respectively. They were initially 17.5 km apart. At what time do they meet?
(a) 2 : 30 PM
(b) 11 : 30 AM
(c) 1 : 30 PM
(d) 12 : 30 PM
(e) None of these

#### View Ans & Explanation

Ans.d

Let after t hours they meet then,

3t + 4t = 17.5 ⇒ t = 25

∴ Time = 10 am + 2.5 h = 12 : 30 am

3. A train does a journey without stoppage in 8 hours, if it had travelled 5 km/h faster, it would have done the journey in 6 hours 40 minutes. Find its original speed.
(a) 25 km/h
(b) 40 km/h
(c) 45 km/h
(d) 36.5 km/h
(e) None of these

#### View Ans & Explanation

Ans.a

Let original speed = S km/h

Here, distance to be covered is constant

∴ S × 8 = (S + 5)(203)

⇒ 8S - 203S = 1003 ⇒ S = 1004 = 25 km/h

4. A train leaves station X at 5 a.m. and reaches station Y at 9 a.m. Another train leaves station Y at 7 a.m. and reaches station X at 10 : 30 a.m. At what time do the two trains cross each other ?
(a) 7 : 36 am
(b) 7 : 56 am
(c) 8 : 36 am
(d) 8 : 56 am
(e) None of these

#### View Ans & Explanation

Ans.b

Let the distance between X and Y be x km. Then,

the speed of A is x4 km/h and that of B is 2x7 km/h.

Relative speeds of the trains

= (x4 + 2x7) = 15x28 km/h

Therefore the distance between the trains at 7 a.m.

= x - x2 = x2 km

Hence, time taken to cross each other

= $\frac{\frac{x}{2}}{\frac{15x}{28}} = \frac{x}{2} \times \frac{28}{15x} = \frac{14}{15} \times 60 = 56 \; min.$

Thus, both of them meet at 7 : 56 a.m.

5. Cars C1 and C2 travel to a place at a speed of 30 and 45 km/h respectively. If car C2 takes 212 hours less time than C1 for the journey, the distance of the place is
(a) 300 km
(b) 400 km
(c) 350 km
(d) 225 km
(e) None of these

#### View Ans & Explanation

Ans.d

Let C1 takes t hrs. Then,

∴ Distance is same.

∴ 30t = 45(t - 52)

⇒ t = 152 hrs.

∴ Distance = 30 × 152 = 225 km

6. If I walk at 4 km/h, I miss the bus by 10 minutes. If I walk at 5 km/h, I reach 5 minutes before the arrival of the bus. How far I walk to reach the bus stand ?
(a) 5 km
(b) 4.5 km
(c) 514 km/h
(d) Cannot be determined
(e) None of these

#### View Ans & Explanation

Ans.a

d = product of speed $\frac{difference \; of \; time}{difference \; of \; speed}$

d = $\frac{4 \times 5}{60}\left[\frac{10 - \left(-5 \right)}{5 - 4} \right]$

⇒ d = 5 km

7. A goods train leaves a station at a certain time and at a fixed speed. After 6 hours, an express train leaves the same station and moves in the same direction at a uniform speed of 90 kmph. This train catches up the goods train in 4 hours. Find the speed of the goods train.
(a) 36 kmph
(b) 40 kmph
(c) 30 kmph
(d) 42 kmph
(e) None of these

#### View Ans & Explanation

Ans.a

Let the speed of the goods train be x kmph.

Distance covered by goods train in 10 hours

= Distance covered by express train in 4 hours.

∴ 10x = 4 × 90 or x = 36.

So, speed of goods train = 36 kmph.

8. Without stoppages, a train travels certain distance with an average speed of 80 km/h, and with stoppages, it covers the same distance with an average speed of 60 km/h. How many minutes per hour the train stops ?
(a) 15
(b) 18
(c) 10
(d) 16
(e) None of these

#### View Ans & Explanation

Ans.a

Due to stoppages, it covers 20 km less .

Time taken to cover 20 km = 2080h = 14h

= 14 × 60 min = 15 min

9. If a man walks to his office at 3/4 of his usual rate, he reaches office 1/3 of an hour later than usual. What is his usual time to reach office.
(a) 12 hr
(b) 1 hr
(c) 34 hr
(d) 2 hrs
(e) None of these

#### View Ans & Explanation

Ans.b

If new speed is ab of original speed, then usual time × (ba - 1) = change in time

∴ usual time × (43 - 1) = 13

⇒ usual time = 13 × 3 = 1 hr

10. If a man walks to his office at 5/4 of his usual rate, he reaches office 30 minutes early than usual. What is his usual time to reach office.
(a) 2 hrs
(b) 212 hrs
(c) 1 hr 50 min
(d) 2 hrs 15 min
(e) None of these

#### View Ans & Explanation

Ans.b

usual time × (45 - 1) = -3060

⇒ usual time = 12 × 5 = 212 hrs