# Exercise : 4

(b) 50 km

(c) 60 km

(d) 70 km

(e) None of these

**Ans.b**

Let the distance between the two stations be x km.

Then, ^{x}⁄_{50} - ^{10}⁄_{60} = ^{x}⁄_{30} - ^{50}⁄_{60}

⇒ ^{x}⁄_{50} - ^{1}⁄_{6} = ^{x}⁄_{30} - ^{5}⁄_{6}

or ^{x}⁄_{30} - ^{x}⁄_{50} = ^{2}⁄_{3} or x = 50 km

(b) 25 km/h and 60 km/h respectively

(c) 25 km/h and 50 km/h respectively

(d) 25 km/h and 70 km/h respectively

(e) None of these

**Ans.c**

Let the speed of the bus be x km / h.

then speed of the car = (x + 25) km / h

∴ \(\frac{500}{x} = \frac{500}{x + 25} + 10\)

⇒ x^{2} + 25x – 1250 = 0 ⇒ x = 25

Thus speed of the bus = 25 km/h

Speed of the car = 50 km/h

**Alternative:**

Difference in speeds 25 km / hr is in only option (c).

(b) 12 sec

(c) 15 sec

(d) 20 sec

(e) None of these

**Ans.c**

Length of train = 12 × 15 = 180 m.

Then, speed of train = ^{180}⁄_{18} = 10 m/s

Now, length of train = 10 × 15 = 150m

∴ Required time = ^{150}⁄_{10} = 15 sec.

^{1}⁄

_{4}min

(b) 3 min

(c) 4

^{1}⁄

_{4}min

(d) 4

^{1}⁄

_{2}min

(e) None of these

**Ans.a**

Time = \(\frac{225}{6 \times \frac{5}{18}}\) = 135 sec = 2^{1}⁄_{4} min.

(b) 58 kmph

(c) 52 kmph

(d) 50 kmph

(e) None of these

**Ans.a**

Relative speed = (^{280}⁄_{9}) m/sec = (^{280}⁄_{9} × ^{18}⁄_{5}) kmph

= 112 kmph.

∴ Speed of goods train = (112 – 50) kmph = 62 kmph.

(b) 4 : 3

(c) 6 : 7

(d) 9 : 16

(e) None of these

**Ans.b**

Let us name the trains as A and B. Then,

(A's speed) : (B's speed) = √b : √a = √16 : √9 = 4 : 3.

(b) 2.5 km/h

(c) 4 km/h

(d) 1.5 km/h

(e) None of these

**Ans.a**

Let actual speed of train S_{T} km/h.

Then, S_{T} - 6 = ^{75}⁄_{18} × ^{18}⁄_{5} = 15

⇒ S_{T} = 21 km/h

Now, let speed of second man = S_{m}

21 - S_{m} = ^{75}⁄_{15} × ^{18}⁄_{5} = 18

⇒ S_{m} = 3 km/h

(b) 18 sec

(c) 36 sec

(d) 72 sec

(e) None of these

**Ans.c**

Speed of train relative to jogger

= (45 – 9) km/h = 36 km/h

= (36 × ^{5}⁄_{18}) m/sec = 10 m/sec

Distance to be covered = (240 + 120) m = 360 m.

∴ Time taken = (^{360}⁄_{10}) sec = 36 sec.

(b) 23

^{2}⁄

_{9}m

(c) 27 m

(d) 27

^{7}⁄

_{9}m

(e) None of these

**Ans.d**

Relative speed = (40 – 20) km/h

= (20 × ^{5}⁄_{18}) m/sec = (^{50}⁄_{9}) m/sec

Length of faster train

= (^{50}⁄_{9} × 5) m = ^{250}⁄_{9}m = 27^{7}⁄_{9} m

(b) 32 m/s, 48 m/s

(c) 40 m/s, 44 m/s

(d) 38 m/s, 42 m/s

(e) None of these

**Ans.d**

Let speed of trains are S_{1} m/s and S_{2} m/s.

Then, s_{1} - s_{2} = \(\frac{130 + 110}{60} = 4\) ....(i)

and s_{1} + s_{2} = \(\frac{130 + 110}{3} = 80\) ....(ii)

on solving (i) and (ii), we get

S_{1} = 42 m/s , S_{2} = 38 m/s