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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The following question based on trains topic of quantitative aptitude
(a) 40 km/hour
(b) 45 km/hour
(c) 60 km/hour
(d) 50 km/hour
The correct answers to the above question in:
Answer: (a)
Let the initial speed of the car be x kmph and the distance be y km.
Then, y = $9/2$ x ...(i)
and, y = 4 (x + 5) ...(ii)
${9x}/2 = 4(x + 5)$
9x = 8x + 40
x = 40 kmph
Discuss Form
Read more changing speed with time travel Based Quantitative Aptitude Questions and Answers
Question : 1
If a train runs at 40 km/hour, it reaches its destination late by 11 minutes. But if it runs at 50 km/ hour, it is late by 5 minutes only. The correct time (in minutes) for the train to complete the journey is
a) 15
b) 19
c) 21
d) 13
Answer »Answer: (b)
If the distance be x km,
then $x/40 - x/50 = 6/60$
$x/4 - x/5 = 1$
x = 20 km.
Required time
= $(20/40)$ hour - 11 minutes
= $(1/2 × 60 - 11)$ minutes
= 19 minutes
Question : 2
When a person cycled at 10 km per hour he arrived at his office 6 minutes late. He arrived 6 minutes early, when he increased his speed by 2 km per hour. The distance of his office from the starting place is
a) 7 km
b) 12 km
c) 16 km
d) 6 km
Answer »Answer: (b)
Let the distance be x km.
$x/10 - x/12 = 12/60$
${6x - 5x}/60 = 1/5$
$x = 1/5 × 60$ = 12 km.
Using Rule 10,
Here, $S_1 = 10, t_1 = 6, S_2 = 12, t_2$ = 6
Distance = ${(S_1 × S_2)(t_1 + t_2)}/{S_2 - S_1}$
= ${(10 × 12)(6 + 6)}/{12 - 10}$
= ${120 × 12}/2$
= 60 × $12/60$ km = 12km
Question : 3
If a boy walks from his house to school at the rate of 4 km per hour, he reaches the school 10 minutes earlier than the scheduled time. However, if he walks at the rate of 3 km per hour, he reaches 10 minutes late. Find the distance of his school from his house.
a) 4 km
b) 6 km
c) 4.5 km
d) 5 km
Answer »Answer: (a)
Let the distance of school be x km, then
$x/3 - x/4 = 20/60$
$x/12 = 1/3 ⇒ x =12/3 = 4$ km
Using Rule 10,
Here, $S_1 = 3, t_1 = 10, S_2 = 4, t_2$ = 10
Distance = ${(S_1 × S_2)(t_1 + t_2)}/{S_2 - S_1}$
= ${(3 × 4)(10 + 10)}/{4 - 3}$
= $12 × 20/60$ = 4 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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