Races & Games of Skill Questions and Answers

Answer: B

Correct answer: 10

Giving 20 points means that when A scores 100, B scores 80; likewise C scores 72.

\[B : C = 80 : 72\]

Scaling so that B scores 100, C scores \(100 \times \tfrac{72}{80} = 90\). So B can give C 10 points.

Answer: B

In these problems the two runners are compared by the distances they cover in the same time, so their speeds are in the ratio of those distances.

\[ A\,100 : B\,90, \qquad A\,100 : C\,72 \]

\[ B : C = 90 : 72 \]

In a 100 m race, when B finishes:

\[ C = 100 \times \frac{72}{90} = 80 \text{ m} \]

\[ \text{B can give C} = 100 - 80 = 20 \text{ m} \]

The difference of the two handicaps, \( 28 - 10 = 18 \), is not the answer, because B's lead is measured over a shorter effective distance.

Answer: A

In these problems the two players are compared by the scores they reach in the same time, so their rates are in the ratio of those scores.

\[ A\,400 : B\,380, \qquad A\,400 : C\,361 \]

\[ B : C = 380 : 361 = 20 : 19 \]

In a 400 m race, when B finishes:

\[ C = 400 \times \frac{19}{20} = 380 \text{ m} \]

\[ \text{B can give C} = 400 - 380 = 20 \text{ m} \]

The difference of the starts, \( 39 - 20 = 19 \), is close but not the answer, since C's deficit is now measured over B's 400 m rather than A's.

Answer: C

Chain the two comparisons through B.

\[ A\,100 : B\,75, \qquad B\,100 : C\,96 \]

When A runs 100, B runs 75; and when B runs 75, C runs

\[ 75 \times \frac{96}{100} = 72 \text{ m} \]

\[ \text{A beats C by } 100 - 72 = 28 \text{ m} \]

The two margins do not simply add to 29 - B's 4 m lead over C is measured over 100 m, but B only runs 75 m in this race.

Answer: A

A finishes in 36 seconds. In that time B, who needs 45 seconds for the full 100 m, covers

\[ 100 \times \frac{36}{45} = 80 \text{ m} \]

\[ \text{A beats B by } 100 - 80 = 20 \text{ m} \]

The 9-second difference is B's time for the remaining 20 m, at \( \tfrac{100}{45} = 2.22 \) m/s.

Answer: C

A starts 140 m ahead, so A has only 360 m to cover while B has the full 500 m.

Speeds are in the ratio 3 : 4, so in the time A covers 360 m:

\[ \text{B covers } 360 \times \frac{4}{3} = 480 \text{ m} \]

\[ \text{A wins by } 500 - 480 = 20 \text{ m} \]

The head start is subtracted from A's distance, not added to B's.

Answer: B

Here, greatest have to cover a distance of ( 500-140) = 360 m to reach the winning point while B 500 m Now, A covers 3m while B covers 4 m
. . A covers 360 m while B covers 4/3 x 360 = 480 m
Hence, B will be ( 500- 480) = 20 m behind the winning post.
. . A wins by 20 min.

Answer: B

In these problems the two runners are compared by the distances they cover in the same time, so their speeds are in the ratio of those distances.

\[ A : B = 20 : 25 \]

In a 1000 m race B finishes first, and in that time A covers

\[ 1000 \times \frac{20}{25} = 800 \text{ m} \]

\[ \text{B beats A by } 1000 - 800 = 200 \text{ m} \]

Answer: A

In these problems the two players are compared by the scores they reach in the same time, so their rates are in the ratio of those scores.

\[ P\,2000 : Q\,1800, \qquad P\,2000 : R\,1440 \]

\[ Q : R = 1800 : 1440 = 5 : 4 \]

In a 2000 m race, when Q finishes:

\[ R = 2000 \times \frac{4}{5} = 1600 \text{ m} \]

\[ \text{Q can give R} = 2000 - 1600 = 400 \text{ m} \]

The difference of the starts, \( 560 - 200 = 360 \), is not the answer - Q's advantage is measured over a shorter effective run.

Answer: A

Let the three times over a kilometre be \( p \), \( q \) and \( r \).

P beats Q by 1 minute, and Q beats R by 30 seconds:

\[ q = p + 60, \qquad r = q + 30 = p + 90 \]

P beats R by 375 m, so R covers only 625 m in P's time:

\[ \frac{625}{1000} = \frac{p}{r} \quad\Longrightarrow\quad r = 1.6p \]

\[ 1.6p = p + 90 \quad\Longrightarrow\quad 0.6p = 90 \quad\Longrightarrow\quad p = 150 \]

\[ q = 210 \text{ s}, \qquad r = 240 \text{ s} \]

Checking: R covers \( 1000 \times \tfrac{150}{240} = 625 \) m in P's 150 seconds, exactly 375 m short.