Races & Games of Skill Questions and Answers

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.

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: B

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: 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.