Model 6 Change in speed Vs Change with time travel Section-Wise Topic Notes With Detailed Explanation And Example Questions
MOST IMPORTANT quantitative aptitude - 6 EXERCISES
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Top 489+ Time And Distance MCQ Problems on Train For BANK »
The following question based on trains topic of quantitative aptitude
(a) 8 km
(b) 5 km
(c) 10 km
(d) $5/8$ km
The correct answers to the above question in:
Answer: (b)
Let x km. be the required distance.
Difference in time
= 2.5 + 5 = 7.5 minutes
= $7.5/60$ hrs. = $1/8$ hrs.
Now, $x/8 - x/10 = 1/8$
= ${5x - 4x}/40 = 1/8$
$x = 40/8$ = 5 km.
Using Rule 10,If a man travels at the speed of $s_1$, he reaches his destination $t_1$ late while he reaches $t_2$ before when he travels at $s_2$ speed, then the distance between the two places is D = ${(S_1 × S_2)(t_1 + t_2)}/{S_2 - S_1}$
Here, $S_1 = 8, t_1 = 2.5, S_2 = 10, t_2 = 5$
Distance = ${(S_1 × S_2)(t_1 + t_2)}/{S_2 - S_1}$
= ${(8 × 10)(2.5 + 5)}/{10 - 8}$
= $40 × {7.5}/60$ = 5 km
Discuss Form
Read more changing speed with time travel Based Quantitative Aptitude Questions and Answers
Question : 1
Walking at a speed of 5 km/hr, a man reaches his office 6 minutes late. Walking at 6 km/hr, he reaches there 2 minutes early. The distance of his office is
a) 4 km
b) 3.5 km
c) 2 km
d) 3 km
Answer »Answer: (a)
Let the distance of the office be x km, then
$x/5 - x/6 = 8/60$
${6x - 5x}/30 = 2/15$
x = 2 × 2 = 4 km
Using Rule 10,
Here, $S_1 = 5, t_1 = 6, S_2 = 6, t_2$ = 2
Distance = ${(S_1 × S_2)(t_1 + t_2)}/{S_2 - S_1}$
= ${(6 × 5)(6 + 2)}/{6 - 5}$
= $30 × 8/60$ = 4 km
Question : 2
Shri X goes to his office by scooter at a speed of 30km/h and reaches 6 minutes earlier. If he goes at a speed of 24 km/h, he reaches 5 minutes late. The distance of his office is
a) 21 km
b) 22 km
c) 24 km
d) 20 km
Answer »Answer: (b)
Let the distance of office be x km.
$x/24 - x/30 = 11/60$
${5x - 4x}/120 = 11/60$
$x/120 = 11/60$
$x = 11/60$ × 120 = 22 km.
Using Rule 10,
Here, $S_1 = 24, t_1 = 5, S_2 = 30, t_2$ = 6
Distance = ${(S_1 × S_2)(t_1 + t_2)}/{S_2 - S_1}$
= ${24 × 30(5 + 6)}/{30 - 24}$
= ${720 × 11}/{6 × 60}$ = 22 km
Question : 3
If a man walks 20 km at 5 km/ hr, he will be late by 40 minutes. If he walks at 8 km/hr, how early from the fixed time will he reach?
a) 25 minutes
b) 50 minutes
c) 1$1/2$ hours
d) 15 minutes
Answer »Answer: (b)
Time taken to cover 20 km at the speed of 5km/hr
= 4 hours.
Fixed time = 4 hours - 40 minutes
= 3 hour 20 minutes
Time taken to cover 20 km at the speed of 8 km/hr
= $20/8$ = 2 hours 30 minutes
Required time
= 3 hours 20 minutes - 2 hours 30 minutes
= 50 minutes
Question : 4
If a train runs at 70 km/hour, it reaches its destination late by 12 minutes. But if it runs at 80 km/ hour, it is late by 3 minutes. The correct time to cover the journey is
a) 2 hours
b) 1 hour
c) 59 minutes
d) 58 minutes
Answer »Answer: (b)
Distance of journey = x km
Difference of time
= 12 - 3 = 9 minutes
= $9/60$ hour = $3/20$ hour
$x/70 - x/80 = 3/20$
$x/7 - x/8 = 3/2$
${8x - 7x}/56 = 3/2$
$x/56 = 3/2$
$x = 3/2 × 56$ = 84 km
∴ Required correct time
= $84/70$ hours - 12 minutes
= $(84/70 × 60 - 12)$ minutes
= 72 - 12 = 60 minutes
= 1 hour
Question : 5
A train covers a distance between station A and station B in 45 minutes. If the speed of the train is reduced by 5 km/hr, then the same distance is covered in 48 minutes. The distance between station A and B is
a) 64 km
b) 80 km
c) 55 km
d) 60 km
Answer »Answer: (d)
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
Let the distance between stations be x km, then speed of train
= $x/{45/60} = {4x}/3$ kmph
$x/{{4x}/3 - 5} = 48/60$
${3x}/{4x - 15} = 4/5$
16x - 60 = 15x
x = 60 km
Question : 6
Walking at 5 km/hr a student reaches his school from his house 15 minutes early and walking at 3 km/hr he is late by 9 minutes. What is the distance between his school and his house ?
a) 8 km
b) 3 km
c) 2 km
d) 5 km
Answer »Answer: (b)
Let the required distance be x km.
Then, $x/3 - x/5 = 24/60$
${5x - 3x}/15 = 2/5$
${2x}/3 = 2$ ⇒ 2x = 2 × 3
x = 3 km
Using Rule 10,
Here, $S_1 = 3, t_1 = 9, S_2 = 5, t_2$ = 15
Distance = ${(S_1 × S_2)(t_1 + t_2)}/{S_2 - S_1}$
= ${(3 × 5)(9 + 15)}/{5 - 3}$
= $15/2 × 24/60$ = 3 km
trains Shortcuts and Techniques with Examples
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Model 1 Train Vs Train in same direction
Defination & Shortcuts … -
Model 2 Train Vs Train in opposite direction
Defination & Shortcuts … -
Model 3 Train Vs Bridge/Platform
Defination & Shortcuts … -
Model 4 Train Vs Pole/Signal Post/Man
Defination & Shortcuts … -
Model 5 Train Vs Both platform and a man/a pole
Defination & Shortcuts … -
Model 6 Change in speed Vs Change with time travel
Defination & Shortcuts …
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