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) 6.30 a.m.
(b) 7.00 a.m.
(c) 3.30 a.m.
(d) 6.00 a.m.
The correct answers to the above question in:
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.
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Read more basic problems formulas Based Quantitative Aptitude Questions and Answers
Question : 1
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 : 2
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 : 3
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 : 4
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 : 5
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 : 6
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
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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