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) 15 km
(b) 20 km
(c) 30 km
(d) 25 km
The correct answers to the above question in:
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
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Read more basic problems formulas Based Quantitative Aptitude Questions and Answers
Question : 1
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 : 2
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 : 3
A man walks 'a' km in 'b' hours. The time taken to walk 200 metres is
a) ${ab}/200$ hours
b) $b/a$ hours
c) $b/{5a}$ hours
d) ${200b}/a$ hours
Answer »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
Question : 4
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 : 5
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 : 6
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
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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