# Distance and Time Quiz Set 020

### Question 1

A boy walks along the three edges of an equilateral triangle at average speeds of 6 km/h, 2 km/h and 9 km/h. What is the average speed along the whole journey?

A

\$3{6/7}\$ kmph.

B

\$5{2/3}\$ kmph.

C

\$2{2/9}\$ kmph.

D

\$5{1/3}\$ kmph.

Soln.
Ans: a

Let the edge of the triangle be L. Total distance travelled = 3L. Time taken = \$L/6 + L/2 + L/9\$. So overall average speed = \${3L}/{L/6 + L/2 + L/9}\$ which simplifies to \${27/7}\$, which is same as: \$3{6/7}\$ km/h.

### Question 2

A cyclist travels at a speed of 5km/h. Had he travelled at 10km/h he would have covered a distance of 15 km more. What is the total distance travelled by him?

A

15 km.

B

16 km.

C

14 km.

D

17 km.

Soln.
Ans: a

By the given problem, the time is same for both cases. If the total distance covered by him at 5km/h is x, then, \$x/5 = {x + 15}/10\$. Solving, we obtain x = 15 km.

### Question 3

A city bus has an average speed of 17 km/h if it doesn't stop anywhere. But if it stops in-between the average speed drops to 14 km/h. How many minutes does it stop in 1 hour?

A

\$10{10/17}\$ mins.

B

\$12{5/16}\$ mins.

C

\$8{11/19}\$ mins.

D

\$12{3/19}\$ mins.

Soln.
Ans: a

Due to stoppages, it covers a less distance of 17 - 14 = 3 in one hour. The time taken for that distance would be the wastage due to stopping = \$3/17\$ × 60 = \$10{10/17}\$ mins.

### Question 4

Speeds of A and B are in the ratio 7 : 8. What is the ratio of the times that they will take to cover a distance of 100 km?

A

8 : 7.

B

7 : 8.

C

8 : 100.

D

100 : 7.

Soln.
Ans: a

Let the speeds be 7x and 8x. The times they take to cover 100 km are \$100/{7x}\$ and \$100/{8x}\$. The ratio would be 8 : 7.

### Question 5

A boy walks along the three edges of an equilateral triangle at average speeds of 5 km/h, 8 km/h and 3 km/h. What is the average speed along the whole journey?

A

\$4{44/79}\$ kmph.

B

\$5{49/78}\$ kmph.

C

\$3{38/81}\$ kmph.

D

\$7{10/27}\$ kmph.

Soln.
Ans: a

Let the edge of the triangle be L. Total distance travelled = 3L. Time taken = \$L/5 + L/8 + L/3\$. So overall average speed = \${3L}/{L/5 + L/8 + L/3}\$ which simplifies to \${360/79}\$, which is same as: \$4{44/79}\$ km/h. 