Model 1 Basic Time & Distance using formula Practice Questions Answers Test with Solutions & More Shortcuts

Question : 6 [SSC CGL Tier-II 2014]

A car driver leaves Bangalore at 8.30 A.M. and expects to reach a place 300 km from Bangalore at 12.30 P.M. At 10.30 he finds that he has covered only 40% of the distance. By how much he has to increase the speed of the car in order to keep up his schedule?

a) 30 km/hr

b) 35 km/hr

c) 40 km/hr

d) 45 km/hr

Answer: (a)

Distance covered by car in2hours

=${300 × 40}/100$ = 120km

Remaining distance

= 300 - 120 = 180 km

Remaining time= 4 - 2= 2 hours

Required speed

=$180/2$= 90kmph

Original speed of car

=$120/2$= 60kmph

Required increase in speed

= 90 - 60 = 30 kmph

Question : 7 [SSC CGL Prelim 2004]

A boy goes to his school from his house at a speed of 3 km/hr and returns at a speed of 2 km/ hr. If he takes 5 hours in going and coming, the distance between his house and school is :

a) 6.5 km

b) 5.5 km

c) 5 km

d) 6 km

Answer: (d)

Let the required distance be x km. Then,

$x/3 + x/2$ = 5

${2x + 3x}/6$ = 5

5x = 6 × 5

x = ${6 × 5}/5$ = 6 km

Using Rule 5,

Here, x = 3, y = 2

Average Speed = ${2 × x × y}/{x + y}$

= ${2 × 3 × 2}/{3 + 2} = 12/5$ km/hr

Total distance = $12/5 × 5$ = 12km

Required distance = $12/2$ = 6 km

Question : 8 [SSC CGL Tier-I 2014]

You arrive at your school 5 minutes late if you walk with a speed of 4 km/h, but you arrive 10 minutes before the scheduled time if you walk with a speed of 5 km/h. The distance of your school from your house (in km) is

a) 2

b) 10

c) 5

d) 4

Answer: (b)

If the required distance be = x km, then

$x/4 - x/5 = {10 + 5}/60$

${5x - 4x}/20 = 1/4$

$x/20 = 1/4 ⇒ x= 1/4 × 20$ = 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, $S_1 = 4, t_1 = 5, S_2 = 5, t_2$= 10

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

=${(4 × 5)(5 + 10)}/{5 - 4}$

=20 × $15/60$ = 5 kms

Question : 9 [SSC CGL Tier-I 2014]

Sarita and Julie start walking from the same place in the opposite directions. If Julie walks at a speed of 2$1/2$ km/hr and Sarita at a speed of 2 km/hr, in how much time will they be 18 km apart ?

a) 4.8 hrs

b) 5.0 hrs

c) 4.5 hrs

d) 4.0 hrs

Answer: (d)

Using Rule 12,
If both objects run in opposite direction then, Relative speed = Sum of speeds.
If both objects run in the same direction then, Relative speed = Difference of Speeds.
Time taken in meeting = $\text"Distance between them"/\text"Relative Speed"$

Relative speed

=$(5/2 + 2)$ kmph = $9/2$ kmph

Time = $\text"Distance"/ \text"Relative speed"$

=$18/{9/2} = {18 × 2}/9$ = 4 hours

Question : 10 [SSC Constable (GD) 2013]

A speed of 30.6 km/.hr is the same as

a) 15.5 m/sec.

b) 12 m/sec.

c) l0 m/sec.

d) 8.5 m/sec.

Answer: (d)

30.6 kmph = $(30.6 × 5/18)$ m/sec.

= 8.5 m/sec

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