Races and Games-JA

Races and Games

(i) Race : A competition in running, riding, swimming, cycling etc.   (ii) Race Course : The plane on which competitions of races are contested.   (iii) Starting Point : The point from which the race starts .   (iv) Winning Point (Post) : The last point at which race is completed.   (v) Dead Heat: A contest of race in which all the competitors set the same time and no one is winner.   (vi) A Game of Hundred : A game in which the players agree that whoever first score hundred points is the winner.   (vii) while A scores 100 points, B scores only 75 points, then we say that ‘In a game of 100 A can give B 25 points’.   (viii) In a race of 100 meters A beats B by 20 meters, means while A runs 100 meters B runs 80 meters.

Question 1 :

In a 100 m race, A can give B 10 m and C 28 m. In the same race B can give C:

A.  18 m

B.  20 m

C.  27 m

D.  9 m

Answer: Option B

Explanation:

A : B = 100 : 90.

A : C = 100 : 72.

frac{B}{A}
frac{A}{C}
frac{90}{100}
frac{100}{72}
frac{90}{72}

B : C = × =  ×  = 

When B runs 90 m, C runs 72 m.

left ( frac{72}{90}times100 right )

When B runs 100 m, C runs m= 80 m.

∴ B can give C 20 m.

Question 2:

A and B take part in 100 m race. A runs at 5 kmph. A gives B a start of 8 m and still beats him by 8 seconds. The speed of B is:

A.   5.15 kmph

B.   4.14 kmph

C.   4.25 kmph

D.   4.4 kmph

Answer: Option B

Explanation:

left ( 5timesfrac{5}{18} right )
frac{25}{18}

A’s speed = m/sec=m/sec.

left ( 100timesfrac{18}{25} right )

Time taken by A to cover 100 m = sec= 72 sec.

∴Time taken by B to cover 92 m = (72 + 8) = 80 sec.

left ( frac{92}{80}timesfrac{18}{5} right )

∴ B’s speed = kmph= 4.14 kmph.

Question 3:

In a 500 m race, the ratio of the speeds of two contestants A and B is 3 : 4. A has a start of 140 m.Then, A wins by:

A.  60 m

B.  40 m

C.  20 m

D.  10 m

Answer: Option C

Explanation:

To reach the winning post A will have to cover a distance of (500 – 140)m, i.e., 360 m.

While A covers 3 m, B covers 4 m.

left ( frac{4}{3}times360 right )

While A covers 360 m, B covers m= 480 m.

Thus, when A reaches the winning post, B covers 480 m and therefore remains 20 m behind.

∴ A wins by 20 m.

Question 4 :

In a 100 m race, A beats B by 10 m and C by 13 m. In a race of 180 m, B will beat C by:

A.  5.4 m

B.  4.5 m

C.  5 m

D.  6 m

Answer : Option D

Explanation:

A : B = 100 : 90.

A : C = 100 : 87.

frac{B}{C}
frac{B}{A}
frac{A}{C}
frac{90}{100}
frac{1000}{87}
frac{30}{29}

 =  ×  = × = 

When B runs 30 m, C runs 29 m.

left ( frac{29}{30}times180 right )

When B runs 180 m, C runsm= 174 m.

∴ B beats C by (180 – 174) m = 6 m.

Question 5 :

At a game of billiards, A can give B 15 points in 60 and A can give C to 20 points in 60. How many points can B give C in a game of 90?

A.  30 points

B.  20 points

C.  10 points

D.  12 points

Answer: Option C

Explanation:

A : B = 60 : 45.

A : C = 60 : 40.

frac{B}{C}
left ( frac{B}{A}timesfrac{A}{C} right )
left ( frac{45}{60}timesfrac{60}{40} right )
frac{45}{40}timesfrac{90}{80}

∴ =  =  == 90 : 80.

∴ B can give C 10 points in a game of 90.

Question 6 :

In a race of 200 m, A can beat B by 31 m and C by 18 m. In a race of 350 m, C will beat B by:

A.  22.75 m

B.  25 m

C.  19.5 m

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D.   m

Answer : Option C

Explanation:

A : B = 200 : 169.

A : C = 200 : 182.

frac{C}{B}
frac{C}{A}timesfrac{A}{B}
frac{182}{200}timesfrac{200}{169}

 =  = = 182 : 169.

When C covers 182 m, B covers 169 m.

left ( frac{169}{182}times350 right )

When C covers 350 m, B covers  m = 325 m.

Therefore, C beats B by (350 – 325) m = 25 m.

Question 7 :

In 100 m race, A covers the distance in 36 seconds and B in 45 seconds. In this race A beats B by:

A.  20 m

B.  25 m

C.  22.5 m

D.  9 m

Answer: Option A

Explanation:

left ( frac{100}{45}times9 right )

Distance covered by B in 9 sec. == 20 m.

∴ A beats B by 20 meters.

Question 8 :

In a game of 100 points, A can give B 20 points and C 28 points. Then, B can give C:

A. 8 points

B. 10 points

C. 14 points

D. 40 points

Answer: Option B

Explanation:

A : B = 100 : 80.

A : C = 100 : 72.

frac{B}{C}
left ( frac{B}{A}timesfrac{A}{C} right )
left ( frac{80}{100}timesfrac{100}{72} right )
frac{10}{9}
frac{100}{90}

∴  = =  =  =  = 100 : 90.

∴ B can give C 10 points.

Question 9 :

In a 200 meters race A beats B by 35 m or 7 seconds. A’s time over the course is:

A. 40 sec

B. 47 sec

C. 33 sec

D. None of these

Answer: Option C

Explanation:

B runs 35 m in 7 sec.

left ( frac{7}{35}times200 right )

∴B covers 200 m in = 40 sec.

B’s time over the course = 40 sec.

∴A’s time over the course (40 – 7) sec = 33 sec.

Question 10:

A can run 22.5 m while B runs 25 m. In a kilometer race B beats A by:

A. 100 m

{111_{9}}^{1}

B.  m

C. 25 m

D. 50 m

Answer: Option A

Explanation:

frac{45}{2}

When B runs 25 m, A runs  m

left ( frac{45}{2}timesfrac{1}{25}times1000 right )

When B runs 1000 m, A runs m = 900 m.

∴ B beats A by 100 m.

Question 11 :

In a 300 m race A beats B by 22.5 m or 6 seconds. B’s time over the course is:

A. 86 sec

B. 80 sec

C. 76 sec

D. None of these

Answer: Option B

Explanation:

frac{45}{2}

B runs   m in 6 sec.

left ( 6timesfrac{45}{2}times300 right )

B covers 300 m in sec = 80 sec.

Question 12:

A runs 1 times as fast as B. If A gives B a start of 80 m, how far must the winning post be so that A and B might reach it at the same time?

A. 200 m

B. 300 m

C. 270 m

D. 160 m

Answer: Option A

Explanation:

frac{5}{3}

Ratio of the speeds of A and B = : 1 = 5 : 3.

Thus, in race of 5 m, A gains 2 m over B.

2 m are gained by A in a race of 5 m.

left ( frac{5}{2}times80 right )

80 m will be gained by A in race of  m = 200 m.

Winning post is 200 m away from the starting point.

Question 13 :

In a 100 m race, A can beat B by 25 m and B can beat C by 4 m. In the same race, A can beat C by:
A. 21 m

B. 26 m

C. 28 m

D. 29 m

Answer : Option C

Explanation:

left ( frac{A}{B}timesfrac{B}{C} right )
left ( frac{100}{75}timesfrac{100}{96} right )
frac{100}{72}

A : B = 100 : 75
B : C = 100 : 96.
∴A : C = = = = 100 : 72.

∴ A beats C by (100 – 72) m = 28 m.