Model 1 Basic Time & Distance using formula Practice Questions Answers Test with Solutions & More Shortcuts
time & distance PRACTICE TEST [5 - EXERCISES]
Model 1 Basic Time & Distance using formula
Model 2 Vehicles in x/y of its usual speed
Model 3 Problems on average speed
Model 4 Time & Distance with Ratios
Model 5 Problems with Races
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 »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 »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 »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 »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 »Answer: (d)
30.6 kmph = $(30.6 × 5/18)$ m/sec.
= 8.5 m/sec
IMPORTANT quantitative aptitude EXERCISES
Model 1 Basic Time & Distance using formula Shortcuts »
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Click to Start..time & distance Shortcuts and Techniques with Examples
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Model 1 Basic Time & Distance using formula
Defination & Shortcuts … -
Model 2 Vehicles in x/y of its usual speed
Defination & Shortcuts … -
Model 3 Problems on average speed
Defination & Shortcuts … -
Model 4 Time & Distance with Ratios
Defination & Shortcuts … -
Model 5 Problems with Races
Defination & Shortcuts …
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