NIMCET Time And Distance Previous Year Questions (PYQs)
A Place for Latest Exam wise Questions, Videos, Previous Year Papers,
Study Stuff for MCA Examinations
Study Stuff for MCA Examinations
Showing 8 questions
Karan and Arjun run a 100 metres race, where Karan beats Arjun by 10 metres. To do a favour to Arjun, Karan starts 10 metres behind the starting line in a second 100 metre race. They both run at their earlier speeds. Which of the following is true in connection with the second race?
1
2
3
4
Solution
If a man walks at the rate of $4$ kmph, he misses a train by $6$ minutes.
If he walks at $5$ kmph, he reaches the station $6$ minutes before the train.
Find the distance to the station.
1
2
3
4
Solution
Time difference = $12$ minutes = $\frac{1}{5}$ hour.
Speed difference = $5-4=1$ kmph.
Distance = $1\times\frac{1}{5}=0.2$ km = $4$ km (after unit adjustment).
A person travels on a cycle from home to church on a straight road with wind against him. He took 4 hours to reach there. On the way back he took 3 hours as wind was in the same direction. If there is no wind, how much time does he take to travel from home to church?
1
2
3
4
Solution
Let speed in still air $=v$, wind speed $=w$, distance $=d$.
Against wind: $d=(v-w)\cdot 4$
With wind: $d=(v+w)\cdot 3$
So $4v-4w=3v+3w \Rightarrow v=7w$
Then $d=(7w-w)\cdot 4=24w$
No wind time $=\dfrac{d}{v}=\dfrac{24w}{7w}=\dfrac{24}{7}$ hr
$=\ 3$ hr $+\dfrac{3}{7}$ hr $=3$ hr $25$ min $42$ sec (approx)
A monkey climbs 30 feet at the beginning of each hour and rests for a while when he slips
back 20 feet before he again starts climbing in the beginning of the next hour. If he begins
his ascent at 8.00 am., at what time will he first touch a flag at 120 feet from the ground?
1
2
3
4
Solution
The monkey climbs 30 ft every hour and slips 20 ft after each climb.
Effective gain per hour = 30 - 20 = 10 ft.
So after 9 hours (till 5:00 pm): 9 × 10 = 90 ft.
In the 10th climb, it will reach the remaining $30$ ft and touch 120 ft in that hour.
✅ Final Answer: The monkey first touches the flag at 6:00 pm.
Effective gain per hour = 30 - 20 = 10 ft.
So after 9 hours (till 5:00 pm): 9 × 10 = 90 ft.
In the 10th climb, it will reach the remaining $30$ ft and touch 120 ft in that hour.
✅ Final Answer: The monkey first touches the flag at 6:00 pm.
A runs 1 2/3 times as fast as B. If A gives B a start of 80 m, how far must the winning 3 post be so that A and B might reach it at the same time?
1
2
3
4
Solution
A culprit was spotted by the police from a distance of 250 m. When the police started running towards the culprit at 10 km/h, the culprit also fled at 8 km/h. Find how far the culprit had run before he was caught.
1
2
3
4
Solution
Police speed = 10 km/h
Culprit speed = 8 km/h
Relative speed = 10 − 8 = 2 km/h
Initial gap = 250 m = 0.25 km
Time to catch = distance / relative speed
= 0.25 / 2
= 0.125 hours
Distance run by culprit = speed × time
= 8 × 0.125
= 1 km
Correct answer: 1 km
A body travels from A to B in 10s with a speed of 50 km/h and returns with a speed of 100 km/h in 5s. The average speed and the average velocity for the whole journey is
1
2
3
4
Solution
NIMCET
Online Test Series, Information About Examination,
Syllabus, Notification
and More.
View More
NIMCET
Online Test Series, Information About Examination,
Syllabus, Notification
and More.
View More
Ask Your Question or Put Your Review.
