# Distance and Time Quiz Set 017

### Question 1

A vehicle travels 50% of its distance at 3 km/h, and the remaining 50% at 7 km/h. What is the total distance, if it travelled for a total duration of 20 hours?

A

84 km.

B

85 km.

C

83 km.

D

86 km.

Soln.
Ans: a

Let 2x be the actual duration of the journey. Then, \$x/3 + x/7 = 20\$. Solving for x we get x = 42, and so, 2x = 84 km.

### Question 2

A boy goes to his school at an average speed of 5 km/h, and returns back at an average speed of 3 km/h. What is the average speed for the to and fro journey?

A

\$3{3/4}\$ km/h.

B

\$6{1/3}\$ km/h.

C

\$1{5/6}\$ km/h.

D

\$4{1/2}\$ km/h.

Soln.
Ans: a

If u and v are the to and fro speeds, then the standard formula is \${2uv}/{u + v}\$ = \${2 × 5 × 3}/{5 + 3}\$ = \${15/4}\$, which is same as: \$3{3/4}\$ km/h.

### Question 3

A school bus covers a certain distance in 4 hours at a speed of 56 km/h. At what speed should it travel to cover the same distance in 7 hours?

A

32 kmph.

B

33 kmph.

C

31 kmph.

D

34 kmph.

Soln.
Ans: a

Let t1, v1 and t2, v2 be the speeds and times for covering the same distance. We have t1 × v1 = t2 × v2. From here we get v2 = \${4 × 56}/7\$ = 32km/h.

### Question 4

An aircraft was on a 1050 km journey. Its engine developed a snag so it had to slow down. It's average speed reduced by 5km/h, as a result of which it reached its destination late by 1 hour. What was the actual duration of the journey?

A

14 hrs.

B

15 hrs.

C

13 hrs.

D

16 hrs.

Soln.
Ans: a

Let x be the actual duration of the journey. Then, \$1050/x - 1050/{x + 1} = 5\$. Which gives 1050 × \$1/{x (x + 1)} = 5\$. Solving for x, or by trying the options one by one, we get x = 14 hours.

### Question 5

A city bus has an average speed of 25 km/h if it doesn't stop anywhere. But if it stops in-between the average speed drops to 17 km/h. How many minutes does it stop in 1 hour?

A

\$19{1/5}\$ mins.

B

\$25{1/4}\$ mins.

C

13 mins.

D

\$15{6/7}\$ mins.

Soln.
Ans: a

Due to stoppages, it covers a less distance of 25 - 17 = 8 in one hour. The time taken for that distance would be the wastage due to stopping = \$8/25\$ × 60 = \$19{1/5}\$ mins.