# Exercise : 2

^{1}⁄

_{2}hours. What is the distance covered?

(b) 480 km

(c) 420 km

(d) Cannot be determined

(e) None of these

**Ans.b**

Here ^{D}⁄_{7.5} - ^{D}⁄_{8} = 4

(where D is the distance in km)

⇒ 0.5 D = 4 × 8 × 7.5

⇒ D = 2 × 4 × 8 × 7.5 = 480 km

(b) 650 metres

(c) 350 metres

(d) Data inadequate

(e) None of these

**Ans.c**

When a train crosses a platform, it crosses a distance equal to the sum of the length of the platform and that of the train. But when a train crosses a signal pole, it crosses the distance equal to its length only.

Here, time taken by the train to cross a signal pole = 18 seconds

Hence, speed of the train = ^{300}⁄_{18} m/sec

The train takes 21 = (39 – 18) seconds extra in order to cross the platform.

Hence, length of platform = \(\frac{21 \times 300}{18} = 350 m\)

(b) 49 km/hr

(c) 72 km/hr

(d) 70 km/hr

(e) None of these

**Ans.c**

Speed of train = \(\frac{260 + 120}{19} \times \frac{18}{5} = 72 \; km/hr\)

(b) 320 metres

(c) 260 metres

(d) 230 metres

(e) None of these

**Ans.d**

Relative speed = 120 + 80 kmph = 200 × ^{5}⁄_{18} m/sec

t = \(\frac{Distance}{Speed} = \frac{\left(270 + x \right) \times 9}{500}\)

or 270 + x = \(\frac{9 \times 500}{9}x\) = 500 – 270 = 230 m

(b) 22nd

(c) 23rd

(d) 24th

(e) None of these

**Ans.b**

In 2 minutes, he ascends = 1 metre

∴ 10 metres, he ascends in 20 minutes.

∴ He reaches the top in 21st minute.

^{1}⁄

_{4}h. He can walk both ways in 7

^{3}⁄

_{4}h. How long it would take to ride both ways ?

(b) 4

^{1}⁄

_{2}hours

(c) 4

^{3}⁄

_{4}hours

(d) 6 hours

(e) None of these

**Ans.c**

We know that, the relation in time taken with two different modes of transport is

t_{walk both} + t_{ride both} = 2 (t_{walk} + t_{ride})

^{31}⁄_{4} + t_{ride both} = 2 × ^{25}⁄_{4}

⇒ t_{ride both} = ^{25}⁄_{2} - ^{31}⁄_{4} = ^{19}⁄_{4} = 4^{3}⁄_{4} hours

(b) 17.45 s

(c) 35.75 s

(d) 41.45 s

(e) None of these

**Ans.d**

Let the distance between each pole be x m.

Then, the distance up to 12th pole = 11 xm

Speed = ^{11x}⁄_{24} m/s

Time taken to covers the total distance of 19x

= \(\frac{19x \times 24}{11x} = 41.45 \; s\)

(b) 50 min.

(c) 45 min.

(d) 55 min.

(e) None of these

**Ans.b**

Rest time = Number of rest × Time for each rest

= 4 × 5 = 20 minutes

Total time to cover 5km

= (^{5}⁄_{10} × 60)minutes + 20 minutes = 50 minutes

(b) 30 km/h

(c) 5 km/h

(d) 20 km/h

(e) None of these

**Ans.d**

Let the average speed be x km/h.

and Total distance = y km. Then,

^{0.2}⁄_{10}y + ^{0.6}⁄_{30}y + ^{0.2}⁄_{20}y = ^{y}⁄_{x}

⇒ x = ^{1}⁄_{0.05} = 20 km/h

(b) 15 min

(c) 10 min

(d) 5 min

(e) None of these

**Ans.c**

After 5 minutes (before meeting), the top runner covers 2 rounds i.e., 400 m and the last runner covers 1 round

i.e., 200 m.

∴ Top runner covers 800 m race in 10 minutes.