Model 6 Change in speed Vs Change with time travel Practice Questions Answers Test with Solutions & More Shortcuts

Question : 6 [SSC CHSL 2012]

A train covers a distance of 10 km in 12 minutes. If its speed is decreased by 5 km/hr, the time taken by it to cover the same distance will be :

a) 13 minutes 20 sec

b) 13 minutes

c) 11 minutes 20 sec

d) 10 minutes

Answer: (a)

Using Rule 1,

Speed of train = $\text"Distance"/\text"Time"$

= $10/{12/60}$ kmph

= ${10 × 60}/12 = 50$ kmph

New speed = 45 kmph

∴ Required time = $10/45$ hour

= $2/9 × 60$ minutes

= $40/3$ minutes or 13 minutes 20 seconds

Question : 7 [SSC CGL Tier-1 2011]

A student goes to school at the rate of 2$1/2$ km/h and reaches 6 minutes late. If he travels at the speed of 3 km/h. he is 10 minutes early. The distance (in km) between the school and his house is

a) 4

b) 3

c) 1

d) 5

Answer: (a)

Let the required distance be x km.

$x/{5/2} - x/3 = 16/60$

${2x}/5 - x/3 = 4/15$

${6x - 5x}/15 = 4/15 ⇒ x = 4$ km.

Using Rule 10,

Here, $S_1 = 2{1}/2 , t_1 = 6, S_2 = 3, t_2$ = 10

Distance = ${(S_1 × S_2)(t_1 + t_2)}/{S_2 - S_1}$

= ${5/2 × 3(6 + 10)}/{3 - 5/2}$

= $15 × 16/60$ km = 4 km

Question : 8 [SSC CGL Tier-1 2011]

Walking at 5 km/hr a student reaches his school from his house 15 minutes early and walking at 3 km/hr he is late by 9 minutes. What is the distance between his school and his house ?

a) 8 km

b) 3 km

c) 2 km

d) 5 km

Answer: (b)

Let the required distance be x km.

Then, $x/3 - x/5 = 24/60$

${5x - 3x}/15 = 2/5$

${2x}/3 = 2$ ⇒ 2x = 2 × 3

x = 3 km

Using Rule 10,

Here, $S_1 = 3, t_1 = 9, S_2 = 5, t_2$ = 15

Distance = ${(S_1 × S_2)(t_1 + t_2)}/{S_2 - S_1}$

= ${(3 × 5)(9 + 15)}/{5 - 3}$

= $15/2 × 24/60$ = 3 km

Question : 9 [SSC CGL Tier-II 2012]

A train covers a distance between station A and station B in 45 minutes. If the speed of the train is reduced by 5 km/hr, then the same distance is covered in 48 minutes. The distance between station A and B is

a) 64 km

b) 80 km

c) 55 km

d) 60 km

Answer: (d)

Using Rule 1,
Distance = Speed × Time
Speed = $\text"Distance"/\text"Time"$ , Time = $\text"Distance"/\text"Speed"$
1 m/s = $18/5$ km/h, 1 km/h = $5/18$ m/s

Let the distance between stations be x km, then speed of train

= $x/{45/60} = {4x}/3$ kmph

$x/{{4x}/3 - 5} = 48/60$

${3x}/{4x - 15} = 4/5$

16x - 60 = 15x

x = 60 km

Question : 10 [SSC CGL Tier-I 2014]

If a train runs at 70 km/hour, it reaches its destination late by 12 minutes. But if it runs at 80 km/ hour, it is late by 3 minutes. The correct time to cover the journey is

a) 2 hours

b) 1 hour

c) 59 minutes

d) 58 minutes

Answer: (b)

Distance of journey = x km

Difference of time

= 12 - 3 = 9 minutes

= $9/60$ hour = $3/20$ hour

$x/70 - x/80 = 3/20$

$x/7 - x/8 = 3/2$

${8x - 7x}/56 = 3/2$

$x/56 = 3/2$

$x = 3/2 × 56$ = 84 km

∴ Required correct time

= $84/70$ hours - 12 minutes

= $(84/70 × 60 - 12)$ minutes

= 72 - 12 = 60 minutes

= 1 hour

IMPORTANT quantitative aptitude EXERCISES

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