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11. Two guns are fired from the same place at an interval of 6 minutes. A person approaching the place observes that 5 minutes 52 seconds have elapsed between the hearings of the sound of the two guns. If the velocity of the sound is 330 m/sec, the man was approaching that place at what speed (in km/h)?

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Solution:
Difference of time
= 6 min - 5 mins. 52 secs.
= 8 secs.
Distance covered by man in 5 mins. 52 secs.
= Distance covered by sound in 8 secs.
= 330 × 8 = 2640 m.
∴ Speed of man
=2640 m5 min. 52 secs.=2640352 m/secs=2640352×185 kmph=27 kmph
12. Running at 54 of his usual speed, an athlete improves his timing by 5 minutes. The time he usually takes to run the same distance is:

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Solution:
When the athlete walks at 54 of his usual speed then he takes 45 of his usual time (AsSpeed∝1Time)    and he saves 5 minutes.
→ Usual time - 45 × usual time = 5 minutes
→ 15 × Usual time = 5 minutes
Usual time = 25 minutes.
13. In a race of 1000m, A can beat B by 100m. In a 400m, B beats C by 40m. In a race of 500m. A will beat C by

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Solution:
When A runs 1000m, B runs 900m.
Hence, when A runs 500m, B runs 450m.
Again, when B runs 400m, C runs 360m.
And, when B runs 450m, C runs = 360 × 450400 = 405m.
Required distance = 500 - 405 = 95 meter.
That means when A runs 500 meter then B can run 450m then C runs 405m.
14. A runs twice as fast as B and B runs thrice as fast as C. The distance covered by C in 72 minutes, will be covered by A in:

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Solution:
Ratio of the speed of A, B and C = 6 : 3 : 1
Then, ratio of time taken;
= 16 : 13 : 1 = 1 : 2 : 6
Hence, time taken by A = 726 = 12 minutes.
15. A motorboat in still water travels at speed of 36 kmph. It goes 56 km upstream in 1 hour 45 minutes. The time taken by it to cover the same distance down the stream will be:

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Solution:
1 hour 45 minutes = 1+4560  = 74 hours.
Speed of the motorboat up-stream,
=DistanceTimeTaken=56km74hours=56×47=32kmph
Let the speed of the current = x kmph
Hence, 36 - x = 32
Or, x = 36 - 32 = 4 kmph
Speed of boat down the stream = 36 + 4 = 40 kmph.
∴ Time taken to cover 56 km at 40 kmph = 5640 = 75 hours
or 1 hours 24 minutes.
16. An athlete runs 200 meters race in 24 seconds. His speed in km/h is

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Solution:
Speed of athlete,=200m24secs=200×185×24=30km/hour
17. If A travels to his school from his house at the speed of 3 km/h, then he reaches the school 5 minutes late. If he travels at the speed of 4 km/h, he reaches the school 5 minutes earlier than school time. The distance of his school from his house is:

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Solution:
Let the distance between school and home be x km.
The difference of time when A goes school to school with these two different speed is 10 min
=1060hourNow,x3−x4=1060Or,x12=16Or,x=126=2km
18. A man starts climbing a 11 m high wall at 5 pm. In each minute he climbs up 1 m but slips down 50 cm. At what time will he climb the wall?

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Solution:
1st Method:
Man climbs 1m and slips down 50 cm (0.5m) in one minute
i.e. he climbs (1 - 0.5 = 0.5 m) in one minute.
But in the last minute he will be climbing 1m as he gets on the top so no slip.
Time taken to climb 11 meter = 100.5+1  = 21 minutes.
He climbs the wall at 5:21 pm

2nd Method (short-cut):
=hightofpole−slippeddistanceclimbeddistance−slippeddistance
=x−zy−z×t=11−0.51−0.5×1=21minutes
19. A monkey climbs a 60 m high pole. In first minute he climbs 6 m and slips down 3 m in the next minute. How much time is required by it to reach the top?

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Solution:
Monkey climbs 6 m in 1st minute and slips down 3 m in next minute
i.e. Monkey climbs 3 meter in 2 minute then he climbs in one minute,
= 32 m.
But in the last minute he climbs 6 m as he gets on the top so there is no slip.
Time required = 2×543+66   = 37 minutes.
20. An ant climbing up a vertical pole ascends 12 meters and slips down 5 meters in every alternate hour. If the pole is 63 meters high how long will it take it to reach the top?

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Solution:
Since it climbs 7 meter in every 2 hr. At the end of 14 hours it would have climbed up to 49 mt. In 15th hour it will reach 61 mt. Then drops back to 56 mt in the 16th hour. It will take another 35 min to travel remaining 7 mt since it will be ascending in the 17th hour. Thus, 16 hr and 35 min.

Detail Explanation:
Given,
Ant ascends 12 meters in first hours and slips down 5 meters in second hour.
This is repeated in every alternate hour.
It means in every two hours Ant ascends (12 - 5 = 7 meters)

Therefore, in ends of 14 hours,
Ant ascends = 14×72  = 49 meters.
So, In next hour i.e 15 hour, Ant will ascends another 12 meter.
Till end of 15th hour, Ant climbed = 49 + 12 = 61 meters. But in 16th hour, Ant will slips down 5 meters (61 - 5 = 56 meters).
Remaining pole to climb,
= 63 - 56
= 7

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