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

Question : 1 [SSC CPO S.I.2009]

A car can cover a certain distance in 4$1/2$ hours. If the speed is increased by 5 km/hour, it would take $1/2$ hour less to cover the same distance. Find the slower speed of the car.

a) 40 km/hour

b) 45 km/hour

c) 60 km/hour

d) 50 km/hour

Answer: (a)

Let the initial speed of the car be x kmph and the distance be y km.

Then, y = $9/2$ x ...(i)

and, y = 4 (x + 5) ...(ii)

${9x}/2 = 4(x + 5)$

9x = 8x + 40

x = 40 kmph

Question : 2 [SSC FCI Assistant Grade-III 2012]

If a train runs at 40 km/hour, it reaches its destination late by 11 minutes. But if it runs at 50 km/ hour, it is late by 5 minutes only. The correct time (in minutes) for the train to complete the journey is

a) 15

b) 19

c) 21

d) 13

Answer: (b)

If the distance be x km,

then $x/40 - x/50 = 6/60$

$x/4 - x/5 = 1$

x = 20 km.

Required time

= $(20/40)$ hour - 11 minutes

= $(1/2 × 60 - 11)$ minutes

= 19 minutes

Question : 3 [SSC MTS 2011]

When a person cycled at 10 km per hour he arrived at his office 6 minutes late. He arrived 6 minutes early, when he increased his speed by 2 km per hour. The distance of his office from the starting place is

a) 7 km

b) 12 km

c) 16 km

d) 6 km

Answer: (b)

Let the distance be x km.

$x/10 - x/12 = 12/60$

${6x - 5x}/60 = 1/5$

$x = 1/5 × 60$ = 12 km.

Using Rule 10,

Here, $S_1 = 10, t_1 = 6, S_2 = 12, t_2$ = 6

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

= ${(10 × 12)(6 + 6)}/{12 - 10}$

= ${120 × 12}/2$

= 60 × $12/60$ km = 12km

Question : 4 [SSC CGL Tier-II 2013]

If a boy walks from his house to school at the rate of 4 km per hour, he reaches the school 10 minutes earlier than the scheduled time. However, if he walks at the rate of 3 km per hour, he reaches 10 minutes late. Find the distance of his school from his house.

a) 4 km

b) 6 km

c) 4.5 km

d) 5 km

Answer: (a)

Let the distance of school be x km, then

$x/3 - x/4 = 20/60$

$x/12 = 1/3 ⇒ x =12/3 = 4$ km

Using Rule 10,

Here, $S_1 = 3, t_1 = 10, S_2 = 4, t_2$ = 10

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

= ${(3 × 4)(10 + 10)}/{4 - 3}$

= $12 × 20/60$ = 4 km

Question : 5 [SSC CGL Prelim 2003]

A man covered a certain distance at some speed. Had he moved 3 km per hour faster, he would have taken 40 minutes less. If he had moved 2 km per hour slower, he would have taken 40 minutes more. The distance (in km) is :

a) 35

b) 36$2/3$

c) 40

d) 20

Answer: (c)

Let the distance be x km and initial speed be y kmph.

According to question,

$x/y - x/{y + 3} = 40/60$ ...(i)

and, $x/{y - 2} - x/y = 40/60$ ...(ii)

From equations (i) and (ii),

$x/y - x/{y + 3} = x/{y - 2} - x/y$

$1/y - 1/{y + 3} = 1/{y - 2} - 1/y$

${y + 3 -y}/{y(y + 3)} = {y - y + 2}/{y(y - 2)}$

3 (y - 2) = 2 (y + 3)

3y - 6 = 2y + 6

y = 12

From equation (i),

$x/12 - x/15 =40/60$

= ${5x - 4x}/60 = 2/3$

$x = 2/3 × 60$ = 40

Distance = 40 km.

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