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

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

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

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

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

option "D"

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.

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).

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

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).

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).