Model 1 Basic Time & Distance using formula Section-Wise Topic Notes With Detailed Explanation And Example Questions
MOST IMPORTANT quantitative aptitude - 5 EXERCISES
The following question based on time & distance topic of quantitative aptitude
(a) ${ab}/200$ hours
(b) $b/a$ hours
(c) $b/{5a}$ hours
(d) ${200b}/a$ hours
The correct answers to the above question in:
Answer: (b)
Using Rule 1,
Man's speed = $\text"Distance"/ \text"Time"$
= $a/b$ kmph = ${1000a}/b$ m/hour
Time taken in walking 200 metre
= $200/{{1000a}/b} =b/{5a}$ hours
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Read more basic problems formulas Based Quantitative Aptitude Questions and Answers
Question : 1
A man crosses a road 250 metres wide in 75 seconds. His speed in km/hr is :
a) 15
b) 12.5
c) 12
d) 10
Answer »Answer: (b)
Using Rule 1,
Speed = $\text"Distance"/\text"Time" = 250/75$
= $10/3$ m/sec = $10/3 × 18/5$ km/hr
[Since, 1 m/s = $18/5$ km/hr]
= 2 × 6 km/hr. = 12 km/hr.
Question : 2
The speed 3$1/3$ m/sec when expressed in km/hour becomes
a) 12
b) 10
c) 9
d) 8
Answer »Answer: (a)
Since, 1 m/sec = $18/5$ kmph
$10/3$ m/sec
= $18/5 × 10/3$ = 12 kmph
Question : 3
An aeroplane covers a certain distance at a speed of 240 km hour in 5 hours. To cover the same distance in 1$2/3$ hours, it must travel at a speed of :
a) 720 km./hr.
b) 600 km./hr.
c) 360 km./hr.
d) 300 km./hr.
Answer »Answer: (a)
Using Rule 1,
Let the required speed is x km/ hr
Then, $240 × 5 = 5/3 × x$
x = 720 km/hr.
Question : 4
A bullock cart has to cover a distance of 120 km. in 15 hours. If it covers half of the journey in $3/5$th time, the speed to cover the remaining distance in the time left has to be
a) 15 km/hr
b) 10 km/hr
c) 6.67 km/hr
d) 6.4 km/hr
Answer »Answer: (c)
Using Rule 1,
Remaining time
= $2/5 × 15 = 6$ hours
Required speed
= $60/6$ = 10 kmph
Question : 5
A man travelled a distance of 80 km in 7 hrs partly on foot at the rate of 8 km per hour and partly on bicycle at 16km per hour. The distance travelled on the foot is
a) 44 km
b) 36 km
c) 48 km
d) 32 km
Answer »Answer: (d)
Journey on foot=x km
Journey on cycle = (80 –x)km
$x/8 + {80 - x}/16 = 7$
${2x + 80 - x}/16 = 7$
x + 80 = 16 × 7 = 112
x= 112 - 80 = 32 km.
Using Rule 13,Let a man take 't' hours to travel 'x' km. If he travels some distance on foot with the speed u km/h and remaining distance by cycle with the speed v km/h,then time taken to travel on foot.Time = ${(vt - x)}/{(v - u)}$Distance travelled on foot = Time × u
Here, x = 80, t = 7, u = 8, v = 16
Time = $({vt - x}/{v - u})$
=$({16 × 7 - 80}/{16 - 8})$
=$({112 - 80}/8) = 32/8$ = 4 hrs
Distance travelled
= 4 × 8 = 32 kms
Question : 6
A man travelled a certain distance by train at the rate of 25 kmph. and walked back at the rate of 4 kmph. If the whole journey took 5 hours 48 minutes, the distance was
a) 15 km
b) 20 km
c) 30 km
d) 25 km
Answer »Answer: (c)
Let the distance be x km.
Total time = 5 hours 48 minutes
= $5 + 48/60 = (5 + 4/5)$ hours
= $29/5$ hours
$x/25 + x/4 = 29/5$
${4x + 25x}/100 = 29/5$
5 × 29x = 29 × 100
$x = {29 × 100}/{5 × 29}$ = 20 km.
Using Rule 5,If a bus travels from A to B with the speed x km/h and returns from B to A with the speed y km/h,then the average speed will be $({2xy}/{x + y})$
Here, x = 25, y = 4
Average speed = ${2xy}/{x + y}$
= ${2 × 25 × 4}/{25 + 4} = 200/29$
Total Distance = $200/29 × 5{4}/5$
= $200/29 × 29/5$ = 40 km
Required distance = 20 km
GET time & distance PRACTICE TEST 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
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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