Races & Games of Skill Questions and Answers

Answer: B

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.

Chain the two statements through A.

\[ A\,200 : B\,190, \qquad C\,200 : A\,180 \]

When C runs 200, A runs 180; and when A runs 180, B runs

\[ 180 \times \frac{190}{200} = 171 \text{ m} \]

\[ \text{C can give B} = 200 - 171 = 29 \text{ m} \]

Adding the two starts gives 30 m, which is close but wrong: B's 10 m deficit is measured over 200 m, and here B only runs 180.

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: 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.

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

\[ B : C = 45 : 40 \]

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's speed is \( \tfrac{5}{3} \) of B's, and B starts 80 m ahead. Let the post be \( d \) metres away.

They arrive together, so their times are equal:

\[ \frac{d}{5/3} = \frac{d - 80}{1} \]

\[ \frac{3d}{5} = d - 80 \quad\Longrightarrow\quad 3d = 5d - 400 \]

\[ 2d = 400 \quad\Longrightarrow\quad d = 200 \text{ m} \]

Checking: A runs 200 m while B runs 120, and \( \tfrac{200}{120} = \tfrac{5}{3} \).

Answer: A

Let X be the length of the course

according to the question,
A beats B by 15 metres and C by 29 metres

Applying this,
If A runs X metres, B runs (X-15) and C runs ( X-29) metres.

B and C together run, B runs X metres and C runs (X-15)

=> If B runs 1 metre, C runs X-15/X

=> B runs X-15 m, C runs (X-15/x) * (X-15)

APPLYING THIS,
X-29 = (X-15)(X-15) / X

Solving,

X² -29X= X² - 30 X + 225
X = 225

Thus, length of the course is 225 metres.

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.

\[ X : Y = 48 : 42 \]

In a 1000 m race X finishes first, and Y has then covered

\[ 1000 \times \frac{42}{48} = 875 \text{ m} \]

\[ \text{X beats Y by } 1000 - 875 = 125 \text{ m} \]

Answer: B

Chain the two handicaps, both stated in games of 80.

\[ P\,80 : Q\,64, \qquad R\,80 : P\,60 \]

When R scores 80, P scores 60; and when P scores 60, Q scores

\[ 60 \times \frac{64}{80} = 48 \]

\[ R : Q = 80 : 48 = 5 : 3 \]

In a game of 150, when R scores 150:

\[ Q = 150 \times \frac{3}{5} = 90 \]

\[ \text{R can give Q} = 150 - 90 = 60 \text{ points} \]

The handicaps must be chained through P, not added: 16 and 20 do not combine to 36.

Answer: B

A : B = 200 : 169.
A : C = 200 : 182.
C/A = (C/A x A/B) = (182/200 x 200/169) = 182 : 169.
When C covers 182 m, B covers 169 m.
When C covers 350 m, B covers (169/182 x 350) m = 325 m.
Therefore, C beats B by (350 - 325) m = 25 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: D

Convert both times to seconds.

\[ A = 4\tfrac{1}{2} \text{ min} = 270 \text{ s}, \qquad B = 5 \text{ min} = 300 \text{ s} \]

For a dead heat, B must finish in A's 270 seconds, in which B covers

\[ 1000 \times \frac{270}{300} = 900 \text{ m} \]

\[ \text{start A can give} = 1000 - 900 = 100 \text{ m} \]