# Problems on Time and Distance TNPSC Questions - 1

Question: 1

A monkey climbing up a pole ascends 6 metres and slips 3 metres in alternate minutes. If the pole is 60 metres high, how long will it take the monkey to reach the top?

(A) 31 min

(B) 35 min

(C) 37 min

(D) 39 min

Ans: C

Net height ascended in 2 min = (6 - 3) = 3m.

Net height ascended in 36 min = \$\$({3}/{2} × 36)\$\$m = 54 m.

In the 37th min, the monkey ascends 6 m and reaches the top

Hence, total time taken = 37 minutes.

Question: 2

Which of the following trains is the fastest?

(A) 25 m/sec

(B) 90 km/hr

(C) 1500 m/min

(D) None of these

Ans: D

25 m/sec = \$\$(25 × {18}/{5})\$\$km/hr = 90 km/hr.

And, 25 m/sec = (25 × 60) m/min = 1500 m/min.

So, all the three speeds are equal.

Question: 3

A train leaves Delhi at 4.10 P.M. and reaches Aligarh at 7.25 P.M. The average speed of the train is 40 km/hr. What is the distance from Delhi to Aligarh?

(A) 110 km

(B) 120 km

(C) 130 km

(D) 140 km

Ans: C

Time taken = 3 hrs 15 min = \$\$3{1}/{4}\$\$hrs = \$\${13}/{4}\$\$hrs.

∴ Required distance = \$\$(40 × {13}/{4})\$\$ = 130 km.

Question: 4

The speeds of three cars are in the ratio 2 : 3 : 4. The ratio of the times taken by these cars to travel the same distance is

(A) 2 : 3 : 4

(B) 4 : 3 : 2

(C) 6 : 4 : 3

(D) 6 : 2 : 2

Ans: C

Ratio of speeds = 2 : 3 : 4.

∴ Ratio of times taken = \$\${1}/{2} : {1}/{3} : {1}/{4}\$\$ = 6 : 4 : 3.

Question: 5

In track meets both 100 yards and 100 metres are used as distances. By how many metres is 100 metres longer than 100 yards?

(A) 8.56 m

(B) 0.856 m

(C) 0.0856 m

(D) 1 m

Ans: A

1 yard = 0.9144 m ⇔ 100 yards = (100 × 0.9144) m = 91.44 m.

∴ Required difference = (100 - 91.44) m = 8.56 m.

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