Races & Games of Skill Questions and Answers

Answer: A

Let the winning post be \( d \) metres away. A runs \( d \), B runs \( d - 50 \), and they finish together.

\[ \frac{d}{1.5} = \frac{d - 50}{1} \]

\[ 2d = 3d - 150 \quad\Longrightarrow\quad d = 150 \text{ m} \]

Checking: A covers 150 m while B covers 100, and \( \tfrac{150}{100} = 1.5 \).

Answer: D

A can give B a start of 15 seconds in a km race.
B takes 4 minutes to run a km. i.e 1000/4= 250 m/min = 250/60 m/sec
Therefore, B will cover a distance of = 62.5 meters in 15 seconds.
The start that A can give B in a km race therefore, is 62.5 meters, the distance that B run in 15 seconds.
Hence in a 2 km race, A can give B a start of 62.5 * 2 = 125 m or 30 seconds.

Answer: B

Correct answer: 20 m

A finishes in 294 seconds. In that time B, who needs 300 seconds for the full kilometre, covers

\[1000 \times \frac{294}{300} = 980 \text{ m}\]

So A beats B by \(1000 - 980 = 20\) m, which corresponds to the 6-second gap.

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: D

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.

The two handicaps are quoted over different totals, so scale both to a common basis.

\[ A\,80 : B\,64 \ \Rightarrow\ B = \tfrac{4}{5}A \]

\[ A\,90 : C\,75 \ \Rightarrow\ C = \tfrac{5}{6}A \]

\[ \frac{C}{B} = \frac{5/6}{4/5} = \frac{25}{24} \]

C is the stronger, so in a game of 50, when C reaches 50:

\[ B = 50 \times \frac{24}{25} = 48 \]

\[ \text{C can give B} = 50 - 48 = 2 \text{ points} \]

The two handicaps cannot be compared directly, since one is over 80 and the other over 90.

Answer: C

A gives B a start of 25 meters and still wins the race by 10 seconds.
Alternatively, if A gives B a start of 45 meters, then the race ends in a dead heat.
Therefore, the additional 20 meters start given to B compensates for the 10 seconds.i.e., B runs 20 meters in 10 seconds.
Hence, B will take 100 seconds to run 200 meters.
We know that A gives B a start of 45 meters. B will take 22.5 seconds to run this 45 meters as B runs 20 meters in 10 seconds or at the speed of 2 m/s.
Hence, A will take 22.5 seconds lesser than B or 100 - 22.5 = 77.5 seconds to complete the race.

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.

\[ A\,200 : B\,169, \qquad A\,200 : C\,182 \]

\[ C : B = 182 : 169 \]

In a 350 m race, when C finishes:

\[ B = 350 \times \frac{169}{182} = 325 \text{ m} \]

\[ \text{C beats B by } 350 - 325 = 25 \text{ m} \]

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

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

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

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: 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.

\[ A\,60 : B\,40, \qquad A\,60 : C\,30 \]

\[ B : C = 40 : 30 = 4 : 3 \]

In a game of 100, when B scores 100:

\[ C = 100 \times \frac{3}{4} = 75 \]

\[ \text{B can give C} = 100 - 75 = 25 \text{ points} \]