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

Question : 16 [SSC CGL Prelim 2007]

A student walks from his house at a speed of 2$1/2$ km per hour and reaches his school 6 minutes late. The next day he increases his speed by 1 km per hour and reaches 6 minutes before school time. How far is the school from his house ?

a) $7/4$ km

b) $9/4$ km

c) $11/4$ km

d) $5/4$ km

Answer: (a)

Let the required distance be x km.

Difference of time

= 6 + 6 = 12 minutes = $1/5$ hr.

According to the question,

$x/{5/2} - x/{7/2} = 1/5$

${2x}/5 - {2x}/7 = 1/5$

${14x - 10x}/35 = 1/5$

${4x}/35 = 1/5 ⇒ x = 35/20 = 7/4$ km.

Using Rule 10,

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

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

= ${5/2 × 7/2 × (6 + 6)}/{7/2 - 5/2}$

= $35/4 × 12/60 = 7/4$ km

Question : 17 [SSC CPO S.I.2009]

A boy is late by 9 minutes if he walks to school at a speed of 4 km/hour. If he walks at the rate of 5 km/hour, he arrives 9 minutes early. The distance to his school is

a) 5 km

b) 4 km

c) 6 km

d) 9 km

Answer: (c)

Let the required distance be x km.

According to the question,

$x/4 - x/5 = 18/60$

${5x - 4x}/20 = 3/10$

$x = 3/10 × 20$ = 6 km

Using Rule 10,

Here, $S_1 = 4, t_1 = 9, S_2 = 5, t_2$ = 9

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

= ${(4 × 5)(9 + 9)}/{5 - 4}$

= $20 × 18/60 = $6 km

Question : 18 [SSC CGL Tier-I 2014]

A train travelling at a speed of 55 km/hr travels from place X to place Y in 4 hours. If its speed is increased by 5 km/hr., then the time of journey is reduced by

a) 35 minutes

b) 20 minutes

c) 30 minutes

d) 25 minutes

Answer: (b)

Using Rule 1

Distance between stations X and Y

= Speed × Time = 55 × 4 = 220 km.

New speed = 55 + 5 = 60 kmph

Required time = $220/60$

= $11/3$ hours

= 3 hours 40 minutes.

Required answer

= 4 hours - 3 hours 40 minutes

= 20 minutes

IMPORTANT quantitative aptitude EXERCISES

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