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September 9, 2023
Speed Distance and Time is a part of our study since beginning but students generally fail to focus on this topic . This page will provide you with the most simple approach to Solve Speed Distance and Time questions quickly along with the help of some Examples
Before solving the question, we need to know about the Formulas of Speed Time and Distance.
Question 1 : Yamini runs from one side of a road to the other to save herself from a dog who was chasing her . The road was 100meters across. In order to save herself from the dog Yamini takes 25 seconds to cross the road. What is the speed of Yamini ?
Options:
A. 2 meters/second
B. 3.5 meters/second
C. 5.1 meters/second
D. 4 meters/second
Solution: S = \frac{d}{t} \
S = \frac{100.0}{25.0} \
⇒ 4 meters/second
Correct Answer: D
Question 2 : Two person X and a Y are working together in for a task and stay together in a nearby apartment. X takes 30 minutes and Y takes 20 minutes to walk from apartment to office. If one day X started at 10:00 AM and the Y at 10:05 AM from the apartment to office, when will they meet?
A. 10: 40 AM
B. 10: 15 AM
C. 10:00 AM
D. 10:05 am
Solution: Ratio of old man speed to young man speed = 3:2
The distance covered by the old man in 5 min = 10units
The 10 unit is covered with relative speed= \frac{10}{(3-2)} \ =10 min
so, they will meet at 10:15 am.
Correct Answer: B
Question 3 : A freight train takes 5 hours to travel a certain distance. If it reduces its speed by 10 km/h, it would take 6 hours to cover the same distance. Find the original speed of the freight train and the distance it traveled.
Options
A. 12Km/h, 60 km
B. 23 Km/h, 1 km
C.45Km/h, 30 km
D.18km/h, 1 km
Solution: Let the original speed of the freight train be S km/h, and let the distance it traveled
be D kilometers.
According to the problem, time taken to travel the distance at the original speed is 5 hours, and time taken to travel the same distance at a reduced speed (S – 10) km/h is 6 hours.
We can set up the equations: D / S = 5 D / (S – 10) = 6
Now, we can solve for S and D. From the first equation, we get: S = D / 5 Substitute the value of S into the second equation: D / (D / 5 – 10) = 6 D / (D – 50) = 6 D = 6D – 300 5D = 300 D = 60 km
Now, substitute the value of D back into the first equation to find S: S = D / 5 S = 60 km / 5 S = 12 km/h
Therefore, the original speed of the freight train is 12 km/h, and it traveled a distance of 60 kilometers.
Correct Answer: A
Question 4 Walking \frac{6}{7^{th}} \ of his usual speed, a man is 12 minutes late. What is the usual time taken by him to cover the distance?
A. 2 hrs
B. 1 hr 30 min
C. 1 hr 12 min
D. 39 min
Solution: New speed of man = \frac{6}{7} \ of usual speed
As we know speed & time are inversely proportional
Hence, new time= \frac{7}{6} \ of usual time
Hence \frac{7}{6} \ of usual time- usual time= 12 minutes⇒ \frac{1}{6} \ of usual time= 12 minutes
Therefore, usual time = 12×6 = 72 minutes
It means 1 hr 12 minutes
Correct Answer: C
Question 5: Walking barefoot shanti takes 6 hours for walking to a certain place and coming back. She would have taken 2 hours less by riding both ways. What will be the time required by her to walk both ways?
A. 6 hrs
B. 1 hr
C. 3 hrs
D. 8 hrs
Solution: Time taken by the kid in walking to a certain place & riding back is 6 hrs
Time walk+ Time ride = 6
2 hrs of the time is reduced if he rides both the ways.
Time ride+ Time ride = 4
2× ride= 4
ride = \frac{4}{2} \
ride= 2
Now by putting the value of ride, we get
time walk + 2= 6
timewalk = 6-2 =4
Here we are finding the time taken by the kid if he walk both the ways
4+4=8
Therefore, he will take 8 hrs to walk both the ways.
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