Question 1

In a 100 m race, Kamal defeats Bimal by 5 seconds. If the speed of Kamal is 18 Km/h, then the speed of Bimal is :

Accepted Answer

Time taken by Kamal to run 100 m = 100 / 18 x 5 / 18 = 20 s Therefore, time taken by Bimal to run 100 m = 20 + 5 = 25 s Hence, Bimal's speed = 100 / 25 = 4 m/sec = 4 x 18 Km/h = 14.4 Km/h

Question 2

In a 100 meter race, A runs at 5Km/h. A gives B a start of 8 meter and still beats him by 8 sec. Find out the speed of B.

Accepted Answer

Time taken by A to cover 100 meters = 100 ÷ (5 * 5 / 18) = 72 seconds. ∴ B covers (100 - 8) or, 92 meters in (72 + 8) or, 80 seconds. ∴ Speed of B = ( 92 / 80 ) * ( 16 / 5 ) = 4.14 Km/h.

Question 3

In a Km race A beats B by 100 m and C by 200 m. By how many can B beat C in a race of 1350 m?

Accepted Answer

While Arun's 1000 m, B runs 1000 - 100 = 900 m and C runs 1000 - 200 = 800 m. ∴ While B runs 900 m, C runs 800 m. ∴ While B runs 1350 m; C runs 800 / 900 × 1350 = 1200 m ∴ B can beat C by 1350 - 1200 = 150 m.

Question 4

A can run a distance of 1 Km in 3 minutes 10 seconds and B can run the same distance in 3 minutes 20 seconds. If they start together, by what distance will A beat B?

Accepted Answer

According to question, A can run a distance of 1 Km in 3 minutes 10 seconds and B can run the same distance in 3 minutes 20 seconds. A beats B by 10 sec. So distance covered by B in 10 sec. = (1000 *10 / 200) m = 50 m ∴ A beat B by 50 m.

Question 5

In a game of billiards, A can give B 20 points in the game of 120 points and he can give C 30 points in the game of 120 points, How many points can B give C in a game of 90?

Accepted Answer

If A scores 120 points, then B scores 100 points and C scores 90 points. When B scores 100 points, then C scores 90 points. When B scores 90 points than C scores (90/100) x 90 points = 81 points ∴ B can give C, 9 points in a game of 90.

Question 6

Arun and Bhaskar start from place P at 6:00 am and 7:30 am, respectively and run in the same direction. Arun and Bhaskar run at 8 km/h and 12 km/h. respectively. Bhaskar overtakes Arun at?

Accepted Answer

Distance between Arun and Bhaskar at 7:30 am = 8 x 11/2 = 12 km Time taken by Bhaskar in covering a distance of 12 km = 12/(12 - 8) = 3 h ∴ Required time = 10 : 30 am

Question 7

A, B and C walk 1 km in 5 min, 8 min and 10 min, respectively, C starts walking from a points at a certain time, B starts from the same point 1 min later and A starts from the same point 2 min later than C, Then, A meet B and C at times?

Accepted Answer

A walks 1 km in 5 min. Then, distance covered by A in 1 min = 1000/5 = 200 m B walks 1 km in 8 min. Then, distance covered by B in 1 min = 1000/8 = 125 m C walks 1 km in 10 min. Then, distance covered by C in 1 min = 1000/101 = 100 m Let A meets B and C in x and y min, respectively. Then, according to the question, Distance covered by C in(x + 2) min = Distance covered by A in x min ⇒ 100(x + 2) = 200 x ⇒100 x + 200 = 200 x ⇒ 200 = 100 x ∴ x = 200/100 = 2 min Now, for A and B Distance covered by B in (y + 1) min = Distance covered by A in y min ⇒ 125(y + 1) = 200y ⇒ 125y + 125 = 200y ⇒ 125 = 200y - 125y ⇒ 125 = 75y ∴ y = 125/75 = 5/3 min.

Question 8

A 10 km race is organised at 800 m circular race course. P and Q are the contestants of the race. If the ratio of the speed of P and Q is 5 : 4, how many times will the winner overtake the loser?

Accepted Answer

Speed of P : Speed of Q = 5 : 4 Time taken by P to cover 5 rounds = Time taken by Q to cover 4 rounds Distance covered by P in 5 rounds = 5 x (800/1000) = 4 km Distance covered by Q in 4 rounds = 4 x (800/1000) = 16/5 km In 5 rounds, P will overtake Q every time. It means that after covering 4 km, P will overtake Q one time . ∴ After covering 10 km P will overtake Q, = (1/4) x 10 = 21/2 times

Question 9

A runs 1 2/ 3 times as fast as B. If A gives B a start of 40 m, how far must the winning post be, so that A and B might reach it at the same time?

Accepted Answer

Ratio of the speeds of A and B = 5 : 3 Thus, in a race of 5m, A gains 2 m over B. 2 m are gained by A in a race of 5 m. 40 m will be gained by A in a race of (5/2) x 40 m = 100 m ∴ Winning post is 100 m away from the starting point.

Question 10

In a 20 m race, A runs at a speed of 2 m/s. If A gives B a start of 10 m and still beats him by 5 s, what will be the speed of B?

Accepted Answer

Time taken by A to cover 200 m = 200/2 = 100 s ∴ B covers (200 - 10) = 190 m in (100 + 5) = 105 s ∴ B' s speed = 190/105 = 38/21 = 117/21 m/s

Question 11

Shantanu started cycling along the boundaries of a square field ABCD from corner point A. After 1/2 h, he reached the corner point C, diagonally opposite to point A. If his speed was 16 km/h. Find the area of the field?

Accepted Answer

Distance = (AB + BC) = 2L = 16 x 1/2 = 8 km ∴ L = 8/2 = 4 km ∴ Area of the field = 4 x 4 = 16 sq km.

Question 12

In a race of 800 meter, A can beat B by 40 meter. In a race of 500 meter, B can beat C by 5 meter. In a race of 200 meter, A will beat C by

Accepted Answer

In a 800 meter race, A covers 800 meter while B cover 760 meter. ∴ A covers 200 m while B will cover 760 x 200 / 800 = 190 m Again, B covers 500 m while C 495 m ∴ B covers 190 m while C will cover = 495 x 190 / 500 = 188.10 m Since, A covers 200 m while C 188.10 m Hence, A will beat C by (200 – 188.10) = 11.9 m

Question 13

Three men Ram, Karan and Mohan run around an oval garden from a fixed post at the rates of 5, 5 1⁄ 7, 5 1⁄ 3 per hour. They come back to the fixed post in 3 3⁄ 4 hours. Find the greatest length, the race course can have.

Accepted Answer

Total number of rounds made by Ram in 15 / 4 hrs = 5 × 15 / 4 = 75 / 4 ∴ Total number of rounds made by Karan in 15 / 4 hrs = 36 / 7 * 15 / 4 = 135 / 7 Total number of rounds made by Mohan in 15 / 4 hrs = 16 / 3 * 15 / 4 = 20 H.C.F. of 75 / 4 , 135 / 7 and 2 = 5 / 28

Question 14

A and B start from the same point and travel in the same direction round a circular track 3 km in circumference. If their speed be 200 and 150 metres a minute, when will they first together again at the starting point?

Accepted Answer

Time taken by A in one round = 3000 / 200 = 15 min. Time taken by B in one round = 3000 / 150 = 20 min L.C.M. of 15 min and 20 min = 60 min = 1 hr. Hence, they will meet first together at the starting point after 1 hour. 2nd Method to solve this Let us assume after T minutes both A and B will meet first time together at starting points. According to question, Distance covered by A in T minutes = 200 x T = 200T Distance covered by B in T minutes = 150 x T = 150T 200T - 150T = 3000 As we know (3 KM = 3000 meters) ⇒ 50T = 3000 ⇒ 5T = 300 ⇒ T = 300/5 = 60 Minutes. As we know ( 60 minutes = 1 hours) ⇒ T = 1 Hours

Question 15

A can beat B by 5 meters in 100 m and B can beat C by 15 meters in 200 m race. A can beat C in 400 m race by:

Accepted Answer

∴ A runs 100 m while B runs 95 m ∴ A runs 400 m while B runs 95 x 400 / 100 = 380 m Again, B runs 200 m while C runs 185 m ∴ B runs 380 m while C runs 185 x 200 / 380 = 3515 / 10 = 351.5 m Hence, A runs 400 m while C runs 351.5 m So, A can beat C by (400 – 351.5) = 48.5 m

Question 16

At a game of billiards A can give B 6 points in 50 and he can give C 13 points in 65, how many points can B give C in a game of 55?

Accepted Answer

In a game of 50, If A scores 50 then B scores 44. ∴ A scores 65 while B 44 * 65 / 50 = 57.2 points Now, B scores 57.2 points while C 52 ∴ B scores 55 points while C = 52 * 55 / 57.2 = 50 Hence, in game of 55 points B can give C (55 – 50) = 5 points

Question 17

In a race of 200 metres, B can give a start of 10 metres to A and C can give a start of 20 metres to B. The start that C can give to A, in the same race, is

Accepted Answer

B covers 200 m while A 190 m B covers 180 m while A 190 x 180 / 200 = 171 m Hence, C covers 200 m while A = 171 m So, C can give to A (200 – 171) = 29 m start.

Question 18

A can run 500m in 30 seconds and B in 35 seconds. How many meter start can A give to B in a Km race so that the race may end in a dead-heat?

Accepted Answer

Time taken by A to run 1 Km = 30 × 2 = 60 seconds. Time taken by B to run 1 Km = 35 × 2 = 70 seconds. ∴ A can give B a start of (70 - 60) = 10 seconds. In 35 sec B runs 500 m ∴ In 10 sec B runs = 500 × 10 / 35 = 1000 / 7 = 1426⁄7 meters So, A can give B a start of 1426⁄7 meters in a Km race.

Question 19

A's speed is 1 3⁄ 8 times of B's. In a race A gives B a start of 300 m. How long should the race course be so that both reach the winning post simultaneously?

Accepted Answer

A's speed : B's speed = 11⁄2 : 1 = 3⁄2 : 1 = 3 : 2 It means that in a race of 3 m, A gains (3 - 2). = 1 m over B. 1 m is gained by A in a race of 3 m. ∴ 300 m is gained by A in a race of = 3⁄1 × 300 = 900 m.

Question 20

In a Km race A can beat B by 80 m and B can beat C by 60 m. In the same race, A can beat C by :

Accepted Answer

While A runs 1000 m, B runs 1000 - 80 = 920 m and While B runs 1000 m, C runs 1000 - 60 =940 m. ∴ While B runs 920 m; C runs 940 / 1000 × 920 = 4324 / 5 m ∴ While B runs 920 m; C runs 940 × 920 / 1000 = 4324 / 5 m ∴ While A runs 1000 m, C runs 4324 / 5 m ∴ A can beat C by 1000 - 4324 / 5 = 676 / 5 = 1351⁄5 m.

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