Time, Speed, and Distance Questions & Answers

Q11 :

A column of men, expanding 250 m in length takes one hour to march through street at the rate of 50 paces a minutes, each pace being 75 cm. Find the lenght of the street.

A

1000 m

B

1750 m

C

2000 m

D

1500 m

View Answer

Correct Answer: C

2000 m

Description:

Distance travelled in one hour

$= 60 times 50 times 75 text{ cm}$

$= frac{60 times 50 times 75}{100} text{ metres } = 2250 text{ metres}$

$therefore text{ Length of the street } = 2250 - 250 = 2000 text{ m}$

Q12 :

A bus left Delhi for Ambala at 50 km/hr and returned over the same route at 40 km/hr. Thus it took 1 hour more on the return trip. The distance between Delhi and Ambala is

A

160 km

B

180 km

C

200 km

D

210 km

View Answer

Correct Answer: C

200 km

Description:

$frac{x}{40} - frac{x}{50} = 1$

$therefore x = 200$

Q13 :

A man goes to a place at the rate of 4 km per hour. He comes back on a bicycle at 16 km per hour. His average speed in km per hour for the entire journey is

A

10

B

8.5

C

5

D

6.4

View Answer

Correct Answer: D

6.4

Description:

$text{Use } frac{2xy}{x + y} = text{ rule }$

Q14 :

A man travelled a distance of 50 km on his first trip. On his second trip, travelling 3 times as fast, he covered 300 km. The ratio of the time taken for the first to that of the second trip is

A

1:1

B

1:2

C

1:3

D

1:6

View Answer

Correct Answer: B

1:2

Description:

Let speed for first trip is $v$ km/hr

$therefore$ The speed for return trip = $3v$ km/hr

Now the times taken for the two trips are

$begin{aligned} & t_1=frac{50}{v} text{ and } t_2 = frac{300}{3v} &therefore frac{t_1}{t_2} = frac{50}{v} times frac{3v}{300} = frac{1}{2} end{aligned}$

Q15 :

A tractor is moving with a speed of 20 km/h, $x$ km ahead of a truck moving with a speed of 35 km/h. If it takes 20 minutes for the truck to overtake the tractor, then $x$ is equal to

A

5 km

B

10 km

C

15 km

D

20 km

View Answer

Correct Answer: A

5 km

Description:

In 20 minutes, the truck covers the distance of $35 times frac{20}{60} text{ km, } = frac{35}{3} text{ km}$
while the tractor covers the distance of
$20 times frac{20}{60} = frac{20}{3 } text{ km }$

$therefore frac{35}{3} = frac{20}{3} + x Rightarrow x = frac{15}{3} = 5$

Q16 :

A person pedals from his house to his office at a speed of $x_1$ km/h and returns by the same route at a speed of $x_2$ km/h. His average speed is

$left(frac{x}{y} right)$

A

$frac{x_1 + x_2}{2}$

B

$frac{x_1 + x_2}{3}$

C

$frac{3}{4} left( frac{x_1x_2}{x_1 + x_2} right)$

D

$frac{2x_1 ;. ;x_2}{x_1 + x_2}$

View Answer

Correct Answer: D

$frac{2x_1 ;. ;x_2}{x_1 + x_2}$

Description:

Let the distance from office to home = y km then time taken for the round trip

$= frac{y}{x_1} + frac{y}{x_2} = frac{yleft(x_1 + x_2right)}{x_1 x_2}$

and total distance travelled in that round trip = $2y$ km

$therefore text{ Average speed } = 2y div frac{yleft(x_1 + x_2right)}{x_1x_2}$

$= frac{2y x_1x_2}{yleft(x_1 + x_2right)} = frac{2x_1x_2}{x_1 + x_2}$

Q17 :

If a train running at 36 km/h takes 2 minutes and 20 seconds to cross a bridge, then the length of the bridge is

A

1.4 km

B

2.2 km

C

3.6 km

D

14 km

View Answer

Correct Answer: A

1.4 km

Description:

Length of the bridge $= frac{36 times 1000}{3600} times 140 = 1400 text{m}$

$therefore 1.4 text{ km}$

Q18 :

If a train takes 1.75 sec to cross a telegraph post and 1.5 sec to overtake a cyclist racing along a road parallel to the track at 10 metres per second, then the length of the train is

A

135 metres

B

125 metres

C

115 metres

D

105 metres

View Answer

Correct Answer: D

105 metres

Description:

Let the length of train be $l$ metres.

Speed be $x$ metre/sec

According to question,

$begin{aligned} & frac{l}{x} = 1.75 Rightarrow l = 1.75 x text{ m } quad ...(i) & text{ and } frac{1}{(x - 10)} = 1.5 Rightarrow l = 1.5 (x - 10) quad ...(ii) & text{From } (i) text{ and } (ii) & 1.75x = 1.5x - 15 & 1.75x - 1.5x = - 15 & therefore quad .25x = - 15 & x = frac{15 times 100}{125} = 60 & therefore quad l = 1.75 times 60 = 105.00 & 105 text{ metres} end{aligned}$

Q19 :

A cyclist cycles non-stop from A to B, a distance of 14 km at a certain average speed. If his average speed reduces by 1 km per hour, he takes $frac{1}{3}$ hour more to cover the same distance. The original average speed cyclist was:

A

5 km/hour

B

6 km/hour

C

7 km/hour

D

8 km/hour

View Answer

Correct Answer: C

7 km/hour

Description:

Let the original average speed $= x$ km/hr.

$therefore text{A.T.S. } dfrac{14}{x - 1} - dfrac{14}{x} = dfrac{1}{3}$

$Rightarrow quad x = 7$

Q20 :

Two trains travel in the same direction at 60 km per hour and 96 km per hour. If the faster train passes a man in the slower train in 20 seconds, then the length of the faster train is:

A

100 m

B

125 m

C

150 m

D

200 m

View Answer

Correct Answer: D

200 m

Description:

Relative speed
= 96 - 60 = 36 km/hr
$= 36 times frac{5}{8}$
= 10 m/sec
As we know
$text{D } = text{ S } times t$
$therefore quad$ Length of faster train = 10 x 20
= 200 m

Time Speed and Distance Questions Answers