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61. The driver of an ambulance sees a school bus 40 m ahead of him after 20 seconds, the school bus is 60 meter behind. If the speed of the ambulance is 30 km/h, what is the speed of the school bus?

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Solution:
Relative Speed,
=Total distanceTotal time=60+4020=5m/s=5×185=18kmph
Relative Speed = (speed of ambulance - speed of school bus)
Speed of school bus = speed of ambulance - relative speed.
= 30 - 18
= 12 kmph.
62. A minibus takes 6 hour less to cover 1680 km distance, if its speed is increased by 14 kmph ? What is the usual time of the minibus ?

Discuss
Solution:
Let
Usual Speed = x kmph.
Now,
Speed of bus after increasing the speed = (x + 14)kmph. . . . . . . (1)
A ____________1680km ____________B
In first case,
Time taken to covered the distance 1680 km = 1680x . . . . . . . (2)
In Second Case,
Time Taken to covered the distance 1680 km = 1680x+14
Time difference = 6 Hours.
So,
⇒1680x−1680x+14=6⇒168014x2+14x=6⇒280×14=x2+14x⇒x2+70x−56x−3920=0⇒x(x+70)−56(x+70)=0⇒x=−70,56.hence,speedofminibusis56km/hputx=56inequation(2)T=168056=30hours
63. A man reduces his speed from 20 kmph to 18 kmph. So, he takes 10 minutes more than the normal time. what is the distance traveled by him.

Discuss
Solution:
Usual time = 9 × 10 = 90 min = 32 h
Thus,
Distance traveled = Speed × time = 20×32  = 30 km.

Alternatively,
As the speed decreases from 20 kmph to 18 kmph i.e. 10 % increment in usual time.
10% = 10 min
100% = 100 min.
Now,
Distance traveled by him,
= 10060×18
= 30 km.
64. A certain distance is covered at a certain speed. If half of this distance is covered in double the time, the ratio of the two speed is :

Discuss
Solution:
Let the original speed be S1 and time t1 and distance be D.
Now,
D22t1=S2S2=D4t1andS1=Dt1Thus,S1S2=41=4:1
65. A is twice fast as B and B is thrice fast as C. The journey covered by C in 78 minutes will be covered by A in :

Discuss
Solution:
The ratio of speeds of A, B, C = 6 : 3 : 1
The ratio of time taken by A, B, C = 16 : 13 : 1
To simplify it, we will multiply it by LCM of ratio of speeds given.
Hence, the ratio of time taken by A, B, C = 1 : 2 : 6
[Speed is inversely proportional to time, means if speed increase time decreases. So, ratio of time would be reciprocal of the ratio of speed given. ]
Time taken by C to covered given distance = 78 = 6 × 13
The ratio of time of A and C = 1 : 6
Thus, time taken by A = 13 min.
66. Vinay and Versha run a race with their speed in the ratio of 5 : 3. They prefer to run on a circular track of circumference 1.5 km. What is the distance covered by Vinay when he passes Versha for the seventh time?

Discuss
Solution:
Since, the speeds of Vinay and Versha are in the ratio 5 : 3 i.e. when Vinay covers 5 rounds, then Versa covers 3 rounds, but first time Vinay and Versha meet when Vinay completes {212=2.5}   round and Versha completes 12 round.
For Vinay to pass Versha seventh time, Vinay would have completed,
= 7 × 2.5 rounds
Since, each round is 1.5 km, the distance covered by Vinay is,
= 7 × 2.5 × 1.5
= 26.25 km.
67. When do the two hands of a clock of just after 3 pm make 30°angel between them?

Discuss
Solution:
(90−30)o5.5=(60×2)11=10(1011)=10min54secThus,Requiredtime=3:10:54
68. Two boats go downstream from point X to Y. The faster boat covers the distance from X to Y, 1.5 times as fast as slower boat. It is known that for every hour slower boat lags behinds the faster boat by 8 km. however, if they go upstream, then the faster boat covers the distance from Y to X in half the time as the slower boat. Find the speed of the faster boat in still water?

Discuss
Solution:
Given,
Speed of the faster boat Downstream = 1.5 × speed of the slower boat downstream ----------(1) [Difference in First hour]
Speed of the Faster Boat Downstream = Speed of the slower boat + 8 ------------- (2)
Using Equation (1) and (2), we get
Speed of the faster Boat Downstream = 16 kmph
Now,
Time taken by the faster BoatTime taken by the Slower boat Upstream        = 12
Hence,
Time taken by the faster Boat Upstream = 2 × Time taken by the slower Boat Upstream . . . . . . . (3)
And,
Faster boat's speed upstream - 8 = Slower boat's speed upstream . . . . . . . . (4)Using (4) and (3), we get
Speed of the faster Boat upstream = 8 kmph
Thus,
Speed of the faster Boat in still water = 20 kmph
69. A dog after traveling 50 km meets a swami who counsels him to go slower. He then proceeds at 34 of his former speed and arrives at his destination 35 min late. Had the meeting occurred 24 km further the dog would have reached its destination 25 min late. The speed of dog is:

Discuss
Solution:
He proceeds at 34 S where S is his usual speed means 14 decrease in the speed which will lead to 13 increase in the time. Now the main difference comes in those 24km and the change in difference of time = 35 - 25 m = 10 m.

⇒ 13 × T = 10 where T is the time required to cover the distance of (74 - 50) = 24 km.
T = 30 min = 0.5 hours.

Speed of the dog = 240.5  = 48 kmph.
70. A girl while walking diametrically across a semicircular playground, takes 3 minutes less than if she had kept walking round the circular path from A to B. If she walks 60 metres a minute, what is diameter of the play ground?

Discuss
Solution:
Let the radius be r, then difference in the distance,
πr - 2r = r (π - 2)
= r [(3.14) -2] = 60 × 3
2r = 315 m.[πr = semi perimeter, 2r = diameter]

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