## Speed, Time and Distance Questions and Answers Part-6

1. Two trains starting at the same time from two-stations 300 km apart and going in opposite directions, cross each other at a distance of 160 km from one of them. The ratio of their speeds is :
a) 7 : 9
b) 16 : 20
c) 8 : 7
d) 8 : 12

Explanation: A → ___160________Meeting____140_____← B
Ratio of their distance covered = $$\frac{{160}}{{140}}$$ = 8 : 7
Ratio of their speeds = 8 : 7 [When time is constant Speed is proportional to Distance covered]

2. Two runner start running together for a certain distance, one at 5 km/h and another at 8 km/h. The former arrives one and half an hour before the latter. The distance in Km is :
a) 12 km
b) 20 km
c) 25 km
d) 36 km

Explanation:
\eqalign{ & \text{Let }x\text{ be the distance} \cr & \frac{x}{5} - \frac{x}{8} = \frac{3}{2} \cr & \frac{8x - 5x}{40} = \frac{3}{2} \cr & \frac{3x}{40} = \frac{3}{2} \cr & x = 20\,{\text{km}} \cr}

3. A railway passenger counts the telegraph poles on the rail road as he passes them. The telegraph poles are at a distance of 50 metres. What will be his count in 4 hours, if the speed of the train is 45 km per hour.
a) 600
b) 2500
c) 3600
d) 5000

Explanation: In train he travels = 4 × 45 = 180 km = 180000 m
The no. of poles,
= $$\frac{{180000}}{{50}}$$   = 3600 [Poles are at the distance of 50 m.]

4. Without stoppage, a train travels a certain distance with an average speed of 60 km/h, and with stoppage it covers the same distance with an average speed of 40 km/h. On an average, how many minutes per hour does train stop during the journey ?
a) 20 min/hour
b) 15 min/hour
c) 10 min/hour
d) 12 min/hour

Explanation: Since, the train travels at 60 km/h, it's speed per minute is 1 km per minute.
If it's speed with stoppage is 40 km/h, it will travel 40 minutes per hour i.e. train stops 20 min/hour.

5. A train without stopping travels 60 km/h and with stoppage 40 km/h. What is the time taken for stoppage on a route 300 km?
a) 10 hours
b) 20 hours
c) 5 hours
d) 2.5 hours

Explanation: Since, the train travels at 60 km/h, it's speed per minute is 1 km per minute. Hence, if it's speed with stoppage is 40 km/h, it will travel 40 minutes per hour i.e. train stops 20 min per hour.
Time taken to travel 300 km with stoppage
$$= \frac{{300}}{{40}} = 7.5\,\,{\text{hours}}$$
Time taken for stoppage as it stops for 20 min per hour
= 7 × 20 + 10
= 140 + 10
= 150 minutes.
= $$\frac{{150}}{{60}}$$  hours = 2.5 hours.

6. A person can row a boat d km upstream and the same distance downstream in 5 hours 15 minutes. Also, he can row the boat 2d km upstream in 7 hours. How long will it take to row the same distance 2d km downstream?
a) $$\frac{{3}}{{2}}$$ hours
b) 7 hours
c) $$\frac{{29}}{{4}}$$ hours
d) $$\frac{{7}}{{2}}$$ hours

Explanation: Let the speeds of boat and stream was $$s$$ and $$v$$ km/hr respectively
Actual Speed Downstream = $$\left(s + v\right)$$  km/hr
Actual Speed upstream = $$\left(s - v\right)$$  km/hr
\eqalign{ & \frac{d}{{s + v}} + \frac{d}{{s - v}} = 5\,{\text{hr}}{\text{.}}\,15\,{\text{min}}{\text{.}} \cr & \Rightarrow \frac{d}{{s + v}} + \frac{d}{{s - v}} = \frac{{21}}{4}\,.\,....\left( 1 \right) \cr & {\text{and}} \cr & \frac{{2d}}{{s - v}} = 7 \cr & \Rightarrow \frac{d}{{s - v}} = \frac{7}{2}\,.....\left( 2 \right) \cr & {\text{By equation }}\left( 1 \right) - \left( 2 \right), \cr & \frac{d}{{s + v}} = \frac{{21}}{4} - \frac{7}{2} \cr & \Rightarrow \frac{d}{{s + v}} = \frac{{21 - 14}}{4} \cr & \Rightarrow \frac{d}{{s + v}} = \frac{7}{4} \cr & \Rightarrow \frac{{2d}}{{s + v}} = \frac{7}{2} \cr & \cr}
Hence, he takes $$\frac{{7}}{{2}}$$ hours to row 2d km distance downstream

7. If a man runs at 6 kmph from his house, he misses the train at the station by 8 min. If he runs at 10 kmph, he reaches 7 min before the departure of the train. What is the distance of the station from his house? (in Km).
a) $$4\frac{3}{4}$$ km
b) $$3\frac{1}{2}$$ km
c) $$4\frac{1}{4}$$ km
d) $$3\frac{3}{4}$$ km

Explanation: Let the distance of the station from the house of the person = x km
\eqalign{ & {\text{Difference}}\,{\text{of}}\,{\text{time}} \cr & = 8 + 7 \cr & = 15\,{\text{minutes}} \cr & = \frac{1}{4}\,hr \cr & {\text{Time}} = \frac{{{\text{Distance}}}}{{{\text{Speed}}}} \cr & \frac{x}{6} - \frac{x}{{10}} = \frac{1}{4} \cr & \frac{{10x - 6x}}{{60}} = \frac{1}{4} \cr & \frac{{2x}}{{30}} = \frac{1}{4} \cr & \Rightarrow x = \frac{{15}}{4} = 3\frac{3}{4}km \cr}

8. A train running at a speed of 54 km/hr crosses a platform in 30 seconds. The platform is renovated and its length is doubled. Now, the same train running at same speed crosses the platform in 46 seconds. Find the length of the train.
a) 180 metres
b) 200 metres
c) 210 metres
d) 240 metres

Explanation: Let length of the Platform is X m and Train is Y m.
Speed of the train = 54 kmph = $$\frac{{54 \times 5}}{{18}}$$ = 15 m/sec.
To cross the platform, train needs to travel (X + Y) m in 30 sec.
\eqalign{ & {\text{Speed}} = \frac{{{\text{Distance}}}}{{{\text{Time}}}} \cr & 15 = \frac{{{\text{X}} + {\text{Y}}}}{{30}} \cr & {\text{X}} + {\text{Y}} = 450\,.\,.\,.\,.\,.\,.\,.\left( 1 \right) \cr}
Now Platform is renovated and its length is doubled. So, train need to travel (2X + Y) m to cross the platform.
\eqalign{ & {\text{Speed}} = \frac{{{\text{Distance}}}}{{{\text{Time}}}} \cr & 15 = \frac{{{\text{2X}} + {\text{Y}}}}{{46}} \cr & {\text{2X}} + {\text{Y}} = 690\,.\,.\,.\,.\,.\,.\,.\,.\left( 2 \right) \cr}
Multiplying equation (1) by (2)
2X + 2Y = 900 ---------- (3)
Now, equation (2) - (3)
2X + Y - 2X - 2Y = 690 - 900
- Y = - 210
Y = 210
Length of the train = 210 metres

9. A person crosses a 600 m long street in 5 minutes. What is his speed in km per hour?
a) 3.6 km/hr
b) 7.2 km/hr
c) 8.4 km/hr
d) 10 km/hr

\eqalign{ & {\text{Speed}} = {\frac{{600}}{{5 \times 60}}} {\text{ m/sec}} = 2{\text{m/sec}} \cr & {\text{Converting}}\,{\text{m/sec}}\,{\text{to}}\,{\text{km/hr}} \cr & = {2 \times \frac{{18}}{5}} {\text{ km/hr}} \cr & = 7.2\,{\text{ km/hr}} \cr}
10. An aeroplane covers a certain distance at a speed of 240 kmph in 5 hours. To cover the same distance in $$1\frac{2}{3}$$ hours, it must travel at a speed of:
\eqalign{ & {\text{Distance}} = {240 \times 5} = 1200\,{\text{km}} \cr & {\text{Speed}} = \frac{{{\text{Distance}}}}{{{\text{Time}}}} \cr & {\text{Speed}} = \frac{{1200}}{{ {\frac{5}{3}} }}\,{\text{ km/hr}}\, \cr & \left[ {{\text{We}}\,{\text{can}}\,{\text{write}}\,1\frac{2}{3}\,{\text{hours}}\,{\text{as}}\,\frac{5}{3}\,{\text{hours}}} \right] \cr & {\text{Required}}\,{\text{speed}} \cr & = {1200 \times \frac{3}{5}} \,{\text{ km/hr}} \cr & = 720\,{\text{ km/hr}} \cr}