Model 6 Change in speed Vs Change with time travel Practice Questions Answers Test with Solutions & More Shortcuts
trains PRACTICE TEST [6 - EXERCISES]
Model 1 Train Vs Train in same direction
Model 2 Train Vs Train in opposite direction
Model 3 Train Vs Bridge/Platform
Model 4 Train Vs Pole/Signal Post/Man
Model 5 Train Vs Both platform and a man/a pole
Model 6 Change in speed Vs Change with time travel
Question : 11 [SSC CPO S.I.2003]
A student rides on bicycle at 8 km/hour and reaches his school 2.5 minutes late. The next day he increases his speed to 10 km/ hour and reaches school 5 minutes early. How far is the school from his house ?
a) 8 km
b) 5 km
c) 10 km
d) $5/8$ km
Answer »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
Question : 12 [SSC MTS 2013]
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 : 13 [SSC CGL Tier-1 2011]
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 : 14 [SSC CGL Prelim 1999]
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 : 15 [SSC CGL Prelim 2000]
If a man reduces his speed to 2/ 3, he takes 1 hour more in walking a certain distance. The time (in hours) to cover the distance with his normal speed is :
a) 1
b) 3
c) 1.5
d) 2
Answer »Answer: (d)
Since man walks at $2/3$ of usual speed, time taken wil be $3/2$ of usual time.
$3/2$ of usual time
= usual time + 1 hour.
$(3/2 –1)$ of usual time = 1
usual time = 2 hours.
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Top 489+ Time And Distance MCQ Problems on Train For BANK »
Model 6 Change in speed Vs Change with time travel Shortcuts »
Click to Read...Model 6 Change in speed Vs Change with time travel Online Quiz
Click to Start..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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