1. 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?

a) 10 kmph

b) 12 kmph

c) 15 kmph

d) 22 kmph

Explanation: Relative Speed,

$$\eqalign{ & = \frac{{{\text{Total }}\,{\text{distance}}}}{{{\text{Total time}}}} \cr & = \frac{{60 + 40}}{{20}} \cr & = 5\,\,{\text{m/s}} \cr & = \frac{{5 \times 18}}{5} \cr & = 18\,\,{\text{kmph}} \cr} $$

Relative Speed = (speed of ambulance - speed of school bus)

Speed of school bus = speed of ambulance - relative speed

= 30 - 18

= 12 kmph

2. 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 ?

a) 15 hours

b) 24 hours

c) 25 hours

d) 30 hours

Explanation: Let Usual Speed = x kmph

Speed of bus after increasing the speed = (x + 14)kmph

A ____________1680km ____________B

In first case, Time taken to covered the distance 1680 km = $$\frac{{1680}}{x}$$ . . . . . (1)

In Second Case, Time Taken to covered the distance 1680 km = $$\frac{{1680}}{{x + 14}}$$ . . . . . (2)

Time difference = 6 Hours.

So,

$$\eqalign{ & {\frac{{1680}}{x} - \frac{{1680}}{{ {x + 14} }}} = 6 \cr & 1680 {\frac{{14}}{{ {{x^2} + 14x} }}} = 6 \cr & 280 \times 14 = {x^2} + 14x \cr & {x^2} + 70x - 56x - 3920 = 0 \cr & x\left( {x + 70} \right) - 56\left( {x + 70} \right) = 0 \cr & x = - 70,56. \cr & {\text{hence,}}\,{\text{speed}}\,{\text{of}}\,{\text{minibus}}\,{\text{is}}\,56\,km/h \cr & {\text{put}}\,x = 56\,{\text{in}}\,{\text{equation}}\,(2) \cr & T = \frac{{1680}}{{56}} \cr & \,\,\,\,\,\, = 30\,{\text{hours}} \cr} $$

3. 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.

a) 30 km

b) 25 km

c) 50 km

d) 36 km

Explanation: Usual time = 9 × 10 = 90 min = $$\frac{3}{2}$$ h

Distance traveled = Speed × time = $$20 \times \frac{3}{2}$$ = 30 km

4. 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 :

a) 1 : 16

b) 4 : 1

c) 2 : 1

d) 1 : 4

Explanation: Let the original speed be S

_{1}and time t

_{1}and distance be D.

$$\eqalign{ & \frac{{ {\frac{D}{2}} }}{{2{t_1}}} = {S_2} \cr & {S_2} = \frac{D}{{4{t_1}}}\,{\text{and}}\,{S_1} = \frac{D}{{{t_1}}} \cr & \frac{{{S_1}}}{{{S_2}}} = \frac{4}{1} = 4:1 \cr} $$

5. 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 :

a) 13 min

b) 15.5 min

c) 17 min

d) 12 min

Explanation: The ratio of speeds of A, B, C = 6 : 3 : 1

The ratio of time taken by A, B, C = $$\frac{1}{6}$$ : $$\frac{1}{3}$$ : 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.

6. A man goes to the fair with his son and dog. Unfortunately man misses his son which he realizes 20 minutes later. The son comes back towards his home at the speed of 20 m/min and man follows him at 40 m/min. The dog runs to son and comes back to the man to show him the direction of his son. He keeps moving to and fro at 60 m/min between son and father, till the man meets the son. What is the distance traveled by the dog in the direction of the son?

a) 800 m

b) 1675 m

c) 848 m

d) 1000 m

Explanation: In 20 minutes the difference between man and son,

= 20 × 20

= 400 m.

Distance traveled by dog when he goes towards the child,

= $$400 \times \frac{{60}}{{40}}$$

= 600 m and time required is 10 minutes.

In 10 min remaining distance between man and child,

= 400 - (20 × 10)

= 200 m.

Time taken by dog to meet the man,

= $$\frac{{200}}{{100}}$$

= 2 min

(Relative speed of dog with child is 40 m/min and same with man is 100 m/min.)

Remaining distance in 2 min,

= 200 - (2 × 20),

= 160 m.

Now, the time taken by dog to meet the child again,

= $$\frac{{160}}{{40}}$$

= 4 min.

In 4 min he covers = 4 × 60 = 240 m.

Now, remaining distance in 4 min = 160 - (4 × 20) = 80 m.

Time required by dog to meet the man once again = $$\frac{{80}}{{100}}$$ = 0.8 min.

Now remaining distance = 80 - (0.8 × 20) = 64 m.

Now, time taken by dog to meet the child again = (64/40) x (8/5) min

Distance travelled by dog = (8/5) x 60 = 96 m

Thus, we can observe that every next time dog just go 2/5th of previous distance to meet the child in the direction of child.

So, we can calculate the total distance covered by dog in the direction of child with the help of GP formula.

Here, first term (a) = 600 and common ratio (r) = 2/5

Sum of the infinite GP = a/(1 - r)

= 600/(1 - 2/5) = (600 x 5)/3 = 1000 m

7. A tiger is 50 of its own leaps behinds a deer. The tiger takes 5 leaps and per minutes to the deer's 4. If the tiger and the deer cover 8 m and 5 m per leap respectively, what distance will the tiger have to run before it caches the deer?

a) 600 m

b) 700 m

c) 800 m

d) 1000 m

Explanation: Speed of tiger = 40 m/min

Speed of deer = 20 m/min.

Relative speed = 40 - 20 = 20 m/min.

Initial difference in distance = 50 × 8 = 400 m

Time take to catch = $$\frac{{400}}{{20}}$$ = 20 min.

Distance traveled in 20 min,

= 20 × 40

= 800 m

8. A candle of 6 cm long burns at the rate of 5 cm in 5 hour and an another candle 8 cm long burns at the rate of 6 cm in 4h. What is the time required to each candle to remain of equal lengths after burning for some hours, when they starts to burn simultaneously with uniform rate of burning?

a) 1 hours

b) 1.5 hours

c) 2 hours

d) 4 hours

Explanation: (6 - x) = (8 - 1.5x)

x = 4 cm.

It will take 4 hours to burn it in such a way that they will remain equal in lengths.

9. A man walking at the speed of 4 km/hr,cross a square field diagonally in 3 minutes.The area of the field is?

a) 20000 sqm

b) 25000 sqm

c) 18000 sqm

d) 19000 sqm

Explanation: Speed of man,

= 4 kmph = $$\frac{{4 \times 5}}{{18}}$$ m/sec

In 3 min (180 sec) man will go = $$\frac{{20 \times 180}}{8}$$ = 200 m.

That means the diagonal of the square field = 200 m.

Diagonal of square,

= Side of Square × $$\sqrt 2 $$ = 200 m.

Side of Square = $$\frac{{200}}{{\sqrt 2 }}$$

Area of Square = Side

^{2}

Area = $$\frac{{200 \times 200}}{2}$$ = 20000 sqm.

10. A train approaches a tunnel AB. Inside the tunnel a cat located at a point i.e. $$\frac{5}{{12}}$$ of the distance AB measured from the entrance A. When the train whistles the Cat runs. If the cat moves to the exit B, the train catches the cat exactly the exit. The speed of the train is greater than the speed of the cat by what order ?

a) 1 : 6

b) 3 : 5

c) 6 : 1

d) 5 : 4

Explanation:Train(T)__________ A_____5k____CAT__________B

T-------x------------------12k--------------------------

Let the speed of train be u and the speed of Cat be v and train whistles at a point T, X km away from A. Let AB = 12k and Cat was 5k distance away from A. Time was constant for both, then

$$\eqalign{ & \Rightarrow \frac{v}{u} = \frac{x}{{5k}} = \frac{{ {x + 12k} }}{{7k}} \cr & 7x = 5\left( {x + 12k} \right) \cr & \frac{x}{k} = \frac{{30}}{1} \cr & {\text{Thus}}, \cr & \Rightarrow \frac{u}{v} = \frac{{30}}{5} = \frac{6}{1} \cr} $$

so, 6 : 1