Races & Games of Skill Questions and Answers

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.

\[ A\,100 : B\,80, \qquad A\,100 : C\,72 \]

\[ B : C = 80 : 72 \]

In a game of 100, when B scores 100:

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

\[ \text{B can give C} = 100 - 90 = 10 \text{ points} \]

Answer: A

"A can give B 10 m".
In a 100 m race, while A starts from the starting point, B starts 10 meters ahead from the same starting point at the same time.
"A can give C 28 m"
In a 100 m race, while A starts from the starting point, C starts 28 meters ahead from the same starting point at the same time.
i.e., in a 100-10=90 m race, B can give C 28-10=18 m
In a 100 m race, B can give C 18/90×100 = 20 m

Answer: C

In a game of 100 points, A scores 100 points, B scores (100-20)=80 points and C scores (100-28)=72 points
i.e., when B scores 80 points, C scores 72 points
When B scores 100 points, C scores 72/80×100 = 90 points
i.e., In a game of 100 points, B can give C (100-90)=10 points

Answer: A

A finishes in 28 seconds. In that time B, who needs 35 seconds for 500 m, covers

\[ 500 \times \frac{28}{35} = 400 \text{ m} \]

\[ \text{A beats B by } 500 - 400 = 100 \text{ m} \]

Answer: C

"By 15 metres or 5 seconds" means B needs 5 seconds for that last 15 m.

\[ v_{B} = \frac{15}{5} = 3 \text{ m/s} \]

\[ t_{B} = \frac{300}{3} = 100 \text{ s} \]

\[ t_{A} = 100 - 5 = 95 \text{ s} \]

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.

\[ A\,100 : B\,80, \qquad A\,100 : C\,72 \]

\[ B : C = 80 : 72 \]

In a game of 100, when B scores 100:

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

\[ \text{B can give C} = 100 - 90 = 10 \text{ points} \]

Answer: D

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: B

Let the game be \( n \) points. The three statements give

\[ A\,n : B\,(n-25), \qquad A\,n : C\,(n-40), \qquad B\,n : C\,(n-20) \]

From the first two, \( B : C = (n-25) : (n-40) \); from the third, \( B : C = n : (n-20) \). Equating:

\[ \frac{n-25}{n-40} = \frac{n}{n-20} \]

\[ (n-25)(n-20) = n(n-40) \]

\[ n^{2} - 45n + 500 = n^{2} - 40n \quad\Longrightarrow\quad 5n = 500 \]

\[ n = 100 \]

The \( n^{2} \) terms cancel, leaving a simple linear equation.

Answer: C

Correct answer: 5 : 4

In the time A covers 100 m, B covers only 80 m, and both run for the same length of time.

\[\text{Speed ratio} = 100 : 80 = 5 : 4\]

Speeds are in the ratio of the distances covered in equal times, which is why the beaten distance is subtracted from the full course.

Answer: C

Both start together, so they are next together at the start line when each has completed a whole number of laps.

\[ t_{A} = \frac{3000}{200} = 15 \text{ min}, \qquad t_{B} = \frac{3000}{150} = 20 \text{ min} \]

\[ \text{LCM}(15, 20) = 60 \text{ min} = 1 \text{ hour} \]

They actually meet on the track earlier - every \( \tfrac{3000}{50} = 60 \) minutes as it happens - but the question asks specifically about the starting point, which is what makes the LCM the right tool.