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Test: Speed, Time & Distance - 2 - CAT MCQ


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10 Questions MCQ Test - Test: Speed, Time & Distance - 2

Test: Speed, Time & Distance - 2 for CAT 2024 is part of CAT preparation. The Test: Speed, Time & Distance - 2 questions and answers have been prepared according to the CAT exam syllabus.The Test: Speed, Time & Distance - 2 MCQs are made for CAT 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test: Speed, Time & Distance - 2 below.
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Test: Speed, Time & Distance - 2 - Question 1

A Man travelled a distance of 61 km in 9 hours. He travelled partly on foot at 4 km/hr and partly on bicycle at 9 km/hr. What is the distance travelled on foot?

Detailed Solution for Test: Speed, Time & Distance - 2 - Question 1

Let the time in which he travelled on foot = x hour
Time for travelling on bicycle = (9 - x) hr

Distance = Speed * Time, and Total distance = 61 km
So,
4x + 9(9-x) = 61
=> 5x = 20
=> x = 4

So distance traveled on foot = 4(4) = 16 km

Test: Speed, Time & Distance - 2 - Question 2

Walking 6/7th of his usual speed, a man is 12 minutes too late. What is the usual time taken by him to cover that distance?

Detailed Solution for Test: Speed, Time & Distance - 2 - Question 2

Let the distance be x
Let the original speed be y
Time originally taken = distance / speed
That means = x/y
Now the new speed is (6/7)y
And new time is distance /new speed
x/(6/7)y that can be written as 7x/6y
Now new time - original time = 12 min
7x/6y-x/y=12
After subtracting we get
x/y = 12×6
x/y = 72min

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Test: Speed, Time & Distance - 2 - Question 3

A man rides his bicycle 10 km at an average speed of 12 km/hr and again travels 12 km at an average speed of 10 km/hr. What is his average speed for the entire trip approximately?

Detailed Solution for Test: Speed, Time & Distance - 2 - Question 3

Total Distance = 10 +12 = 22 km
Total time = 10/12 + 12/10 = 5/6 + 6/5 = 25/30 + 36/30 = 61/30 hours
Average speed = 22 /(61/30) = 660/61 = 10.8 km/ hour

Test: Speed, Time & Distance - 2 - Question 4

A train can travel 50% faster than a car. Both start from point A at the same time and reach point B 75 kms away from A at the same time. On the way, however, the train lost about 12.5 minutes while stopping at the stations. What is the speed of the car?

Detailed Solution for Test: Speed, Time & Distance - 2 - Question 4

Let the speed of the car be x km/h

So the speed of the train will be 1.5x km/h

According to the question

⇒ 75/x - 75/1.5x = 12.5/60

⇒ (112.5 - 75)/1.5x = 12.5/60

⇒ 37.5/1.5x = 12.5/60

⇒ 1.5x = 37.5 × (60/12.5)

⇒ x = 180/1.5

⇒ x = 120 km/h

∴ The speed of the car is 120 km/h

Test: Speed, Time & Distance - 2 - Question 5

In a flight of 600 km, an aircraft was slowed down due to bad weather. Its average speed for the trip was reduced by 200 km/hr and the time of flight increased by 30 minutes. What is the duration of the flight ?

Detailed Solution for Test: Speed, Time & Distance - 2 - Question 5

option "D"

Test: Speed, Time & Distance - 2 - Question 6

It takes eight hours for a 600 km journey, if 120 km is done by train and the rest by car. It takes 20 minutes more, if 200 km is done by train and the rest by car. What is the ratio of the speed of the train to that of the car?

Detailed Solution for Test: Speed, Time & Distance - 2 - Question 6

Eight hours for a 600 km journey, when 120 km is done by train and 480 km by car.

It takes 20 minutes more if 200 km is done by train and 400 km by car.

Formula used:

Speed = Distance/Time

Calculation:

Let the speed of the train be x km/h

And the speed of the car be y km/h

⇒ 120/x + 480/y = 8

⇒ 120(1/x + 4/y) = 8

⇒ 1/x + 4/y = 1/15     ...i)

In the second condition

⇒ Total time = 8 + 20/60 = 25/3 hr

∴  200/x + 400/y = 25/3

⇒ 200(1/x + 2/y) = 25/3

⇒ 1/x + 2/y = 1/24     ...ii)

After solving equation (i) and (ii)

(By substracting equation 2 from equation 1)

⇒ x = 60 km/h

⇒ y = 80 km/h

Ratio of the speed of train and car is

⇒ 60 : 80

⇒ 3 : 4

∴ The ratio of the speed of train and car is 3 : 4.

Test: Speed, Time & Distance - 2 - Question 7

Brishti went on an 8-hour trip in a car. Before the trip, the car had travelled a total of x km till then, where x is a whole number and is palindromic, i.e., x remains unchanged when its digits are reversed. At the end of the trip, the car had travelled a total of 26862 kms till then, this number again being palindromic. If Brishti never drove at more than 110 kmph, then the greatest possible average speed at which she drove during the rip, in kmph was?

Detailed Solution for Test: Speed, Time & Distance - 2 - Question 7

Total distance travelled at the end of the trip = 26,862.
This includes the distance which car had travelled before the trip and the distance it travels in the 8 hours of the trip.

To maximize speed in 8 hours, we need to maximize distance travelled in 8 hours.
∴ We need to minimize the distance travelled by car before the trip starts, hence we need to minimize x.

Since Brishti drove at less than 110 kmph, hence she must have travelled less than 110 × 8 = 880 kms in the last 8 hours.

∴ She must have travelled more than 26862 – 880 = 25,982 kms before the last 8 hours.
⇒ x > 25,982 and it has to be a palindrome.

Least palindrome greater than 25,982 is 26062.

∴ Car travelled at least 26062 kms before the trip, hence maximum distance travelled during the trip = 26862 – 26062 = 800 kms.

⇒ Maximum average speed for the trip = 800/8 = 100 kmph.

Hence, option (d).

Test: Speed, Time & Distance - 2 - Question 8

Ravi is driving at a speed of 40 km/h on a road. Vijay is 54 meters behind Ravi and driving in the same direction as Ravi. Ashok is driving along the same road from the opposite direction at a speed of 50 km/h and is 225 meters away from Ravi. The speed, in km/h, at which Vijay should drive so that all the three cross each other at the same time, is

Detailed Solution for Test: Speed, Time & Distance - 2 - Question 8

Speed of

Ravi = 40 × 5/18 = 100/9 m/s and

Ashok = 50 × 5/18 = 250/18 m/s

Time taken for them to meet 
⇒ Vijay and Ravi should also meet in 9 secs.

⇒ 9V - 100 = 54
⇒ 9V = 154
⇒ V = 154/9 m/s = 154/9 × 18/5 kmph = 61.6 kmph

Test: Speed, Time & Distance - 2 - Question 9

A boat takes 2 hours to travel downstream a river from port A to port B, and 3 hours to return to port A. Another boat takes a total of 6 hours to travel from port B to port A and return to port B. If the speeds of the boats and the river are constant, then the time, in hours, taken by the slower boat to travel from port A to port B is?

Detailed Solution for Test: Speed, Time & Distance - 2 - Question 9

Let the speeds of river, faster and slower boats be r, f and s km/hr respectively and distance between A and B be 12 kms.

For Faster boat:
⇒ f + r = 12/2 = 6   ...(1)
⇒ f - r = 12/3 = 4   ...(2)

(1) - (2)
⇒ 2r = 6 - 4 = 2
⇒ r = 1

For Slower boat:


⇒ s2 - 1 = 4s
⇒ s2 - 4s - 1 = 0
= 2 ± √5 = 2 + √5 [negative value of s is rejected]
Now, time taken by the slower boat to go from A to B

Hence, option (c).

Test: Speed, Time & Distance - 2 - Question 10

Trains A and B start traveling at the same time towards each other with constant speeds from stations X and Y, respectively. Train A reaches station Y in 10 minutes while train B takes 9 minutes to reach station X after meeting train A. Then the total time taken, in minutes, by train B to travel from station Y to station X  is

Detailed Solution for Test: Speed, Time & Distance - 2 - Question 10

Let the time taken to meet after starting be t mins.

Time taken by A to reach Y after meeting = (10 – t) mins
Time taken by B to reach X after meeting = 9 mins

⇒ t2 = (10 - t) × 9

⇒ t2 +9t – 90 = 0

⇒ (t + 15)(t - 6) = 0

⇒ t = 6 (-15 is rejected)

Total time taken by B to reach Y = 6 + 9 = 15 mins.

Hence, option (b).

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