Races & Games of Skill Questions and Answers

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: C

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: D

Find A's time first.

\[ 5 \text{ km/hr} = 5 \times \frac{5}{18} = \frac{25}{18} \text{ m/s} \]

\[ t_{A} = \frac{100}{25/18} = 72 \text{ s} \]

B gets an 8 m start, so B runs 92 m, and finishes 8 seconds after A:

\[ t_{B} = 72 + 8 = 80 \text{ s} \]

\[ v_{B} = \frac{92}{80} = 1.15 \text{ m/s} = 1.15 \times \frac{18}{5} = 4.14 \text{ km/hr} \]

The start reduces B's distance to 92 m; the 8 seconds are added to A's time.

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

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

"By 35 m or 7 seconds" means B needs 7 seconds to cover that last 35 m.

\[ v_{B} = \frac{35}{7} = 5 \text{ m/s} \]

\[ t_{B} = \frac{200}{5} = 40 \text{ s} \]

\[ t_{A} = 40 - 7 = 33 \text{ s} \]

The pairing of a distance with a time is what gives the loser's speed, and everything else follows from it.

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.

\[ A : B = 22.5 : 25 \]

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: 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: 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.