1. Narayan walking at a speed of 20 km/h reaches his college 10 minutes late. Next time he increases his speed by 5 km/h, but finds that he is still late by 4 minutes. What is the distance of his college from his house?

a) 10 km

b) 6 km

c) 12 km

d) 15 km

Explanation: By increasing his speed by 25%, he will reduce his time by 20%. (This corresponds to a 6 minutes drop in his time.)

Hence, his time originally must have been 30 minutes.

Required distance = 20 kmph × 0.5 hours = 10 km.

2. A motor car does a journey in 17.5 hours, covering the first half at 30 km/h and the second half at 40 km/h. Find the distance of the journey?

a) 684 km

b) 600 km

c) 624 km

d) 584 km

Explanation: Let the total distance be 2X.

A_____X km_____M_____X km_____B

Total time taken in the journey = 17.5 hours

Time taken to cover X km at 30 km/h = $$\frac{{\text{X}}}{{30}}$$

Time taken to cover X km at 40 km/h = $$\frac{{\text{X}}}{{40}}$$

$$\eqalign{ & {\frac{X}{{30}}} + {\frac{X}{{40}}} = 17.5 \cr & {\frac{{ {40X + 30X} }}{{1200}}} = 17.5 \cr & 70X = 17.5 \times 1200 \cr & X = 300\,km \cr & {\text{Total}}\,{\text{distance}},\,2x \cr & = 600\,km \cr} $$

3. A car after traveling 18 km from a point A developed some problem in the engine and the speed became $$\frac{4}{5}$$ th of its original speed. As a result, the car reached point B 45 minutes late. If the engine had developed the same problem after traveling 30 km from A, then it would have reached 36 minutes late. The original speed of the car (in km/h) is:

a) 25 kmph

b) 30 kmph

c) 20 kmph

d) 35 kmph

Explanation: He proceeds at $$\frac{4}{5}$$ S where S is his usual speed means $$\frac{1}{5}$$ decrease in speed which will lead to $$\frac{1}{4}$$ increase in time.

Now the main difference comes in those 12km (30 - 18) and the change in difference of time = (45 - 36) min = 9 min.

$$\frac{1}{4}$$ × T = 9 min where T is the time required to cover the distance of (30 - 18) = 12 km

T = 36 min = $$\frac{36}{60}$$ hours = 0.6 hours.

Speed of the car = $$\frac{12}{0.6}$$ = 20 kmph.

4. A hunter fired two shots from the branch of a tree at an interval of 76 seconds. A tiger separating too fast hears the two shots at an interval of 83 seconds. If the velocity of the sound is 1195.2 km/h, then find the speed of the tiger.

a) 112.8 km/h

b) 100.8 km/h

c) 80.16 km/h

d) 85.16 km/h

Explanation: In the case of increasing gap between two objects

$$\frac{{{\text{Speed }}\,{\text{of }}\,{\text{sound}}}}{{{\text{Speed }}\,{\text{of }}\,{\text{tiger}}}} = $$ $$\frac{{{\text{Time }}\,{\text{Utilized}}}}{{{\text{Difference }}\,{\text{in }}\,{\text{time}}}}$$

$$\frac{{1195.2}}{{\text{S}}} = \frac{{83}}{7}$$

S = 100.8 km/h.

5. A man can cross a downstream river by steamer in 40 minutes and same by boat in 1 hour. If the time of crossing Upstream by streamer is 50% more than downstream time by steamer and the time required by boat to cross same river by boat in upstream is 50% more than time required by in downstream. What is the time taken for the man to cross the river downstream by steamer and then return to same place by boat half the way and by steamer the rest of the way?.

a) 85 minutes

b) 115 minutes

c) 120 minutes

d) 125 minutes

Explanation: Downstream (steamer) = 40 min

Downstream (boat) = 60 min

Upstream (steamer) = 60 min

Upstream (boat) = 90 min

Required time = 40 + 30 + 45 = 115 minutes

6. A man can row 15 km/h in still water and he finds that it takes him twice as much time to row up than as to row down the same distance in the river. The speed of current (in km/h) :

a) 6 km/h

b) 6.5 km/h

c) 4.5 km/h

d) 5 km/h

Explanation: Let the speed of the current be x km/h.

$$\frac{{15 + x}}{{15 - x}} = \frac{2}{1}$$

15 + x = 30 - 2x

3x = 30 - 15

3x = 15

x = 5 km/h

7. A boat covers 48 km upstream and 72 km downstream in 12 hours, while it covers 72 km upstream and 48 km downstream in 13 hours The speed of stream is

a) 2 kmph

b) 2.2 kmph

c) 2.5 kmph

d) 4 kmph

Explanation: Let downstream speed be D and Upstream speed be U. Then

$$\eqalign{ & {\frac{{48}}{U} + \frac{{72}}{D}} = 12\,.\,.\,.\,.\,.\left( 1 \right) \cr & {\frac{{72}}{U} + \frac{{48}}{D}} = 13\,.\,.\,.\,.\,.\left( 2 \right) \cr} $$

On solving the equation (1) and (2),

D = 12 kmph

U = 8 kmph

Thus, speed of the current

$$ = \frac{{D - U}}{2} = \frac{4}{2} = 2\,{\text{kmph}}$$

8. A motor boat takes 2 hours to travel a distance of 9 km downstream and it takes 6 hours to travel the same distance against the current. The speed of the boat in still water and that of the current (in km/h) respectively are:

a) 6, 5 km/h

b) 3, 1.5 km/h

c) 8, 5 km/h

d) 9, 3 km/h

Explanation: Downstream speed,

= $$\frac{9}{2}$$ = 4.5 kmph

Upstream speed,

= $$\frac{9}{6}$$ = 1.5 kmph

Speed of boat in still water,

= $$\frac{{4.5 + 1.5}}{2}$$ = 3 km/h

Speed of current = $$\frac{{4.5 - 1.5}}{2}$$ = 1.5 km/h

9. A train met with an accident 120 km from station A. It completed the remaining journey at $$\frac{5}{6}$$ of its previous speed and reached 2 hour late at station B. Had the accident taken place 300 km further, it would have been only 1 hour late. What is the speed of the train?

a) 100 km/h

b) 12 km/h

c) 60 km/h

d) 50 km/h

Explanation: A ____100 km____ P

_{1}________300 km_____ P

_{2}____X______ B

If speed becomes $$\frac{5}{6}$$ then time taken will be $$\frac{6}{5}$$ of original time.

Thus, extra time = 2 hour

$$\frac{1}{5}$$ = 2 hour

Thus unusual time which is require to complete the journey = 5 × 2 = 10 hours.

It means It covers 300 km distance in 5 hours.

Speed = $$\frac{{300}}{5}$$ = 60 km/h

10. A runs $$\frac{7}{4}$$ times as fast as B. If A gives B a start of 300 m, how far must the winning post be if both A and B have to end the race at same time?

a) 1400 m

b) 700 m

c) 350 m

d) 210 m

Explanation: If A runs 7/4 times as fast as B

A : B = 7 : 4

Let x be the constant ratio

A : B = 7x : 4x

The difference is 300m

7x - 4x = 300

3x = 300

x = 100m

The winning post is the distance A has to run

A = 7x = 7(100) = 700 m