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 : 16 [SSC CGL Prelim 1999]
A train is travelling at the rate of 45km/hr. How many seconds it will take to cover a distance of $4/5$ km ?
a) 120 sec.
b) 90 sec.
c) 64 sec.
d) 36 sec.
Answer »Answer: (b)
Using Rule 1,Distance = Speed × TimeSpeed = $\text"Distance"/\text"Time"$ , Time = $\text"Distance"/\text"Speed"$1 m/s = $18/5$ km/h, 1 km/h = $5/18$ m/s
Time taken = $\text"Distance"/ \text"time"$
= ${4/5}/45$ hour
= ${4 × 60 × 60}/{5 × 45}$ sec. = 64 seconds
Question : 17 [SSC SAS 2010]
A and B travel the same distance at speed of 9 km/hr and 10 km/ hr respectively. If A takes 36 minutes more than B, the distance travelled by each is
a) 66 km
b) 60 km
c) 54 km
d) 48 km
Answer »Answer: (b)
Let the distance between A and B be x km, then
$x/9 - x/10 = 36/60 = 3/5$
$x/90 = 3/5$
$x = 3/5 × 90$ = 54 km.
Using Rule 9,
Here, $S_1 = 9, t_1 = x, S_2 = 10, t_2 = x - 36/60$
$S_1t_ 1 = S_2t_ 2$
$9 × x = 10(x - 36/60)$
9x = 10x - 6 = 6
Distance travelled = 9 × 6 = 54 km
Question : 18 [SSC CGL Tier-I 2013]
A man rides at the rate of 18 km/ hr, but stops for 6 mins. to change horses at the end of every 7th km. The time that he will take to cover a distance of 90 km is
a) 6 hrs. 24 min.
b) 6 hrs. 18 min.
c) 6 hrs. 12 min.
d) 6 hrs.
Answer »Answer: (b)
90 km = 12 × 7km + 6 km.
To cover 7 km total time taken = $7/18$ hours + 6 min. = $88/3$ min.
So, (12 × 7 km) would be covered in $(12 × 88/3)$ min.
and remaining 6km is $6/18$ hrs or 20 min.
Total time = $1056/3$ + 20
= $1116/{3 × 60}$ hours = 6$1/5$ hours
= 6 hours 12 minutes.
Question : 19 [SSC CGL Prelim 2004]
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
Question : 20 [SSC CGL Tier-II 2014]
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
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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