# Exercise : 3

(b)

^{1332}⁄

_{67}km

(c) 40 km

(d) 28 km

(e) None of these

**Ans.a**

Let the length of the journey = x km.

∴ Journey rides by horse cart = x(1 - ^{1}⁄_{2} - ^{1}⁄_{3})

= ^{1}⁄_{6}x km

Then, total time taken to complete journey = ^{31}⁄_{5}hr

⇒ t_{1} + t_{2} + t_{3} = ^{31}⁄_{5}

⇒ ^{x}⁄_{2} × ^{1}⁄_{4} + ^{x}⁄_{3} × ^{1}⁄_{12} + ^{x}⁄_{(6 × 9)} = ^{31}⁄_{5}

⇒ x = ^{31}⁄_{5} × ^{216}⁄_{37} = 36.2 km ≈ 36 km

(b) 11 : 30 AM

(c) 1 : 30 PM

(d) 12 : 30 PM

(e) None of these

**Ans.d**

Let after t hours they meet then,

3t + 4t = 17.5 ⇒ t = 25

∴ Time = 10 am + 2.5 h = 12 : 30 am

(b) 40 km/h

(c) 45 km/h

(d) 36.5 km/h

(e) None of these

**Ans.a**

Let original speed = S km/h

Here, distance to be covered is constant

∴ S × 8 = (S + 5)(^{20}⁄_{3})

⇒ 8S - ^{20}⁄_{3}S = ^{100}⁄_{3} ⇒ S = ^{100}⁄_{4} = 25 km/h

(b) 7 : 56 am

(c) 8 : 36 am

(d) 8 : 56 am

(e) None of these

**Ans.b**

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

the speed of A is ^{x}⁄_{4} km/h and that of B is ^{2x}⁄_{7} km/h.

Relative speeds of the trains

= (^{x}⁄_{4} + ^{2x}⁄_{7}) = ^{15x}⁄_{28} km/h

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

= x - ^{x}⁄_{2} = ^{x}⁄_{2} 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.

_{1}and C

_{2}travel to a place at a speed of 30 and 45 km/h respectively. If car C

_{2}takes 2

^{1}⁄

_{2}hours less time than C

_{1}for the journey, the distance of the place is

(b) 400 km

(c) 350 km

(d) 225 km

(e) None of these

**Ans.d**

Let C_{1} takes t hrs. Then,

∴ Distance is same.

∴ 30t = 45(t - ^{5}⁄_{2})

⇒ t = ^{15}⁄_{2} hrs.

∴ Distance = 30 × ^{15}⁄_{2} = 225 km

(b) 4.5 km

(c) 5

^{1}⁄

_{4}km/h

(d) Cannot be determined

(e) None of these

**Ans.a**

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

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

[Here, –ve sign indicates before the schedule time]⇒ d = 5 km

(b) 40 kmph

(c) 30 kmph

(d) 42 kmph

(e) None of these

**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.

(b) 18

(c) 10

(d) 16

(e) None of these

**Ans.a**

Due to stoppages, it covers 20 km less .

Time taken to cover 20 km = ^{20}⁄_{80}h = ^{1}⁄_{4}h

= ^{1}⁄_{4} × 60 min = 15 min

^{1}⁄

_{2}hr

(b) 1 hr

(c)

^{3}⁄

_{4}hr

(d) 2 hrs

(e) None of these

**Ans.b**

If new speed is ^{a}⁄_{b} of original speed, then usual time × (^{b}⁄_{a} - 1) = change in time

∴ usual time × (^{4}⁄_{3} - 1) = ^{1}⁄_{3}

⇒ usual time = ^{1}⁄_{3} × 3 = 1 hr

(b) 2

^{1}⁄

_{2}hrs

(c) 1 hr 50 min

(d) 2 hrs 15 min

(e) None of these

**Ans.b**

usual time × (^{4}⁄_{5} - 1) = ^{-30}⁄_{60}

⇒ usual time = ^{1}⁄_{2} × 5 = 2^{1}⁄_{2} hrs