Races & Games of Skill Questions and Answers

Answer: A

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.

\[ A\,1000 : B\,900, \qquad A\,1000 : C\,800 \]

\[ B : C = 900 : 800 = 9 : 8 \]

In a 1000 m race, when B finishes:

\[ C = 1000 \times \frac{8}{9} = 888.88 \text{ m} \]

\[ \text{B can give C} = 1000 - 888.88 = 111.12 \text{ m} \]

The two starts do not subtract to 100 m; B's advantage is measured over a shorter run than A's was.

Answer: B

A is \( 1\tfrac{3}{4} = \tfrac{7}{4} \) times as fast as B, and B starts 60 m ahead. Let the post be \( d \) metres away.

\[ \frac{d}{7/4} = \frac{d - 60}{1} \]

\[ 4d = 7d - 420 \quad\Longrightarrow\quad 3d = 420 \]

\[ d = 140 \text{ m} \]

Checking: A runs 140 m while B runs 80, and \( \tfrac{140}{80} = \tfrac{7}{4} \).

Answer: B

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: A

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.

\[ A : B = 22.5 : 25 \]

In a 1000 m race B finishes first, and A has then covered

\[ 1000 \times \frac{22.5}{25} = 900 \text{ m} \]

\[ \text{B beats A by } 1000 - 900 = 100 \text{ m} \]

Answer: B

"By 22.5 m or 6 seconds" means B takes 6 seconds over that last 22.5 m.

\[ v_{B} = \frac{22.5}{6} = 3.75 \text{ m/s} \]

\[ t_{B} = \frac{300}{3.75} = 80 \text{ s} \]

A's time would be \( 80 - 6 = 74 \) seconds, but the question asks for B's.

Answer: A

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.

\[ A\,60 : B\,45, \qquad A\,60 : C\,40 \]

\[ B : C = 45 : 40 \]

In a game of 90, when B scores 90:

\[ C = 90 \times \frac{40}{45} = 80 \]

\[ \text{B can give C} = 90 - 80 = 10 \text{ points} \]

Both handicaps are quoted against the same A total, which is what lets B and C be compared without ever finding a speed.

Answer: A

If A score 60 points, B score 50 points and C score 45 points.
When B scores 50 points, C scores 45 points.
When B scores 90 points, C scores (45/50 *90) points = 81 points.
B can give C, 9 points in a game of 90.

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: D

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.

\[ A\,100 : B\,90, \qquad A\,100 : C\,87 \]

Both are measured against the same A distance, so

\[ B : C = 90 : 87 \]

In a 180 m race, when B finishes:

\[ C = 180 \times \frac{87}{90} = 174 \text{ m} \]

\[ \text{B beats C by } 180 - 174 = 6 \text{ m} \]

Answer: A

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.

\[ A : B = 22.5 : 25 \]

In a 1000 m race B finishes first, and A has then covered

\[ 1000 \times \frac{22.5}{25} = 900 \text{ m} \]

\[ \text{B beats A by } 1000 - 900 = 100 \text{ m} \]