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

Question : 11 [SSC CPO S.I.2003]

A student rides on bicycle at 8 km/hour and reaches his school 2.5 minutes late. The next day he increases his speed to 10 km/ hour and reaches school 5 minutes early. How far is the school from his house ?

a) 8 km

b) 5 km

c) 10 km

d) $5/8$ km

Answer: (b)

Let x km. be the required distance.

Difference in time

= 2.5 + 5 = 7.5 minutes

= $7.5/60$ hrs. = $1/8$ hrs.

Now, $x/8 - x/10 = 1/8$

= ${5x - 4x}/40 = 1/8$

$x = 40/8$ = 5 km.

Using Rule 10,
If a man travels at the speed of $s_1$, he reaches his destination $t_1$ late while he reaches $t_2$ before when he travels at $s_2$ speed, then the distance between the two places is D = ${(S_1 × S_2)(t_1 + t_2)}/{S_2 - S_1}$

Here, $S_1 = 8, t_1 = 2.5, S_2 = 10, t_2 = 5$

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

= ${(8 × 10)(2.5 + 5)}/{10 - 8}$

= $40 × {7.5}/60$ = 5 km

Question : 12 [SSC MTS 2013]

Walking at a speed of 5 km/hr, a man reaches his office 6 minutes late. Walking at 6 km/hr, he reaches there 2 minutes early. The distance of his office is

a) 4 km

b) 3.5 km

c) 2 km

d) 3 km

Answer: (a)

Let the distance of the office be x km, then

$x/5 - x/6 = 8/60$

${6x - 5x}/30 = 2/15$

x = 2 × 2 = 4 km

Using Rule 10,

Here, $S_1 = 5, t_1 = 6, S_2 = 6, t_2$ = 2

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

= ${(6 × 5)(6 + 2)}/{6 - 5}$

= $30 × 8/60$ = 4 km

Question : 13 [SSC CGL Tier-1 2011]

Shri X goes to his office by scooter at a speed of 30km/h and reaches 6 minutes earlier. If he goes at a speed of 24 km/h, he reaches 5 minutes late. The distance of his office is

a) 21 km

b) 22 km

c) 24 km

d) 20 km

Answer: (b)

Let the distance of office be x km.

$x/24 - x/30 = 11/60$

${5x - 4x}/120 = 11/60$

$x/120 = 11/60$

$x = 11/60$ × 120 = 22 km.

Using Rule 10,

Here, $S_1 = 24, t_1 = 5, S_2 = 30, t_2$ = 6

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

= ${24 × 30(5 + 6)}/{30 - 24}$

= ${720 × 11}/{6 × 60}$ = 22 km

Question : 14 [SSC CGL Prelim 1999]

If a man walks 20 km at 5 km/ hr, he will be late by 40 minutes. If he walks at 8 km/hr, how early from the fixed time will he reach?

a) 25 minutes

b) 50 minutes

c) 1$1/2$ hours

d) 15 minutes

Answer: (b)

Time taken to cover 20 km at the speed of 5km/hr

= 4 hours.

Fixed time = 4 hours - 40 minutes

= 3 hour 20 minutes

Time taken to cover 20 km at the speed of 8 km/hr

= $20/8$ = 2 hours 30 minutes

Required time

= 3 hours 20 minutes - 2 hours 30 minutes

= 50 minutes

Question : 15 [SSC CGL Prelim 2000]

If a man reduces his speed to 2/ 3, he takes 1 hour more in walking a certain distance. The time (in hours) to cover the distance with his normal speed is :

a) 1

b) 3

c) 1.5

d) 2

Answer: (d)

Since man walks at $2/3$ of usual speed, time taken wil be $3/2$ of usual time.

$3/2$ of usual time

= usual time + 1 hour.

$(3/2 –1)$ of usual time = 1

usual time = 2 hours.

IMPORTANT quantitative aptitude EXERCISES

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