Races & Games of Skill Questions and Answers

Answer: B

A took 220 seconds. Let A's time be T. Then B takes T + 50 and C takes T + 80 seconds. When A finishes the 1500 m, C is 400 m behind, so C has run 1100 m in T seconds. C's speed is 1100/T, and he runs 1500 m in T + 80 seconds: \[\frac{1500}{T + 80} = \frac{1100}{T}\] \[1500T = 1100T + 88000\] \[T = 220\] Check: C covers 1100 m in 220 s, 5 m per second, and 1500 m in 300 s, which is 220 + 80. The other options give a speed that does not match this.

Answer: C

The race is 78 m long. A is 4% faster, so speeds are in the ratio 104 : 100. B is given a 3 m lead, so B runs (L - 3) while A runs L in the same time: \[ \frac{L}{104} = \frac{L - 3}{100} \] \[ 100L = 104L - 312 \Rightarrow L = 78 \]

74, 76 and 80 do not satisfy this equation.

Answer: D

Anil beats Pawan by 434 m. Pawan's speed is 924/66 = 14 m/s. In the 35 seconds Anil takes, Pawan covers 14 x 35 = 490 m, so he is 924 - 490 = 434 m behind when Anil finishes. 431, 432 and 433 m do not follow from these speeds. Anil's speed is 924/35 = 26.4 m/s. The distance is the same, so the time ratio (35 : 66) is the inverse of the speed ratio. Alternatively, Pawan runs 35/66 of the race in Anil's time, which is 924 x 35/66 = 490 m.

Answer: D

A takes 920 seconds.

Let A take t seconds. When A gives B a 40 m start the race is a dead heat, so B runs 460 m in t seconds. With a 30 m start B runs 470 m, which takes 20 seconds more than t. At B's speed 460/t, the time for 470 m is 470t/460.

\[ \frac{470t}{460}=t+20 \]

\[ \frac{10t}{460}=20 \Rightarrow t=920 \]

325, 417 and 266 seconds do not satisfy these equations.

The key idea is that the 10 m extra headstart changes the result from B finishing 20 s behind to a dead heat, so B covers 10 m in 20 s at the speed 460 / t. This gives B's speed of 0.5 m/s, and the 460 m B covers takes 920 seconds, which is A's time over 500 m.

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.