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 : 11 [SSC CPO S.I.2005]
A boy runs 20 km in 2.5 hours. How long will he take to run 32 km at double the previous speed ?
a) 5 hours
b) 4$1/2$ hours
c) 2$1/2$ hours
d) 2 hours
Answer »Answer: (d)
Using Rule 1,
The boy covers 20 km in 2.5 hours.
Speed = $20/{2.5}$ = 8 km/hr.
New speed = 16 km/hr
Time = $32/16$ = 2 hours.
Question : 12 [SSC CGL Prelim 2003]
A car travelling at a speed of 40 km/hour can complete a journey in 9 hours. How long will it take to travel the same distance at 60 km/hour ?
a) 4$1/2$ hours
b) 4 hours
c) 3 hours
d) 6 hours
Answer »Answer: (d)
Total distance covered
= Speed × Time
= 40 × 9 = 360 km.
The required time at 60 kmph
= $360/60 = 6$ hours.
Using Rule 9,Speed(s) ∝ $1/{time (t)}$ ⇒ s ∝ $1/t$$s_1t_1 = s_2t_2$(Provided distance is constant)
Here, $S_1 = 40, t_1 = 9, S_2 = 60, t_2$ = ?
$S_1t_1 = S_2t_2$
40 × 9 = 60 × $t_2$
$t_2 = {4 × 9}/6$ = 6 hours
Question : 13 [SSC MTS 2011]
Walking at the rate of 4 km an hour, a man covers a certain distance in 3 hours 45 minutes. If he covers the same distance on cycle, cycling at the rate of 16·5 km/hour, the time taken by him is
a) 45.55 minutes
b) 55.44 minutes
c) 54.55 minutes
d) 55.45 minutes
Answer »Answer: (b)
Using Rule 1,
Distance covered on foot
= 4 × 3$3/4$ km. = 15 km.
Time taken on cycle
= $\text"Distance"/\text"Speed" = 15/{16.5}$ hour
= ${15 × 60}/{16.5}$ minutes
= 54.55 minutes
Question : 14 [SSC CGL Tier-I 2013]
A man covers $2/15$ of the total journey by train, $9/20$ by bus and the remaining 10 km on foot. His total journey (in km) is
a) 12.8
b) 16.4
c) 24
d) 15.6
Answer »Answer: (b)
Let the total journey be of x km, then
${2x}/15 + {9x}/20 + 10 = x$
$x - {2x}/15 - {9x}/20$ = 10
${60x - 8x - 27x}/60$ = 10
${25x}/60$ = 10
$x = {60 × 10}/25$ = 24 km
Question : 15 [SSC CGL Prelim 2007]
A person started his journey in the morning. At 11 a.m. he covered $3/8$ of the journey and on the same day at 4.30 p.m. he covered $5/6$ of the journey. He started his journey at
a) 6.30 a.m.
b) 7.00 a.m.
c) 3.30 a.m.
d) 6.00 a.m.
Answer »Answer: (a)
Difference of time
= 4.30 p.m - 11.a.m.
= $5{1}/2$ hours $11/2$ hours
Distance covered in $11/2$ hrs
= $5/6 - 3/8 = {20 - 9}/24 = 11/24$ part
Since, $11/24$ part of the journey is covered in $11/2$ hours
$3/8$ part of the journey is covered in
= $11/2 × 24/11 × 3/8$
= $9/2$ hours = 4$1/2$ hours.
Clearly the person started at 6.30 a.m.
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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