Races & Games of Skill Questions and Answers

Answer: D

In a 100 meters race, when Karan runs 100 meters, Arjun runs 90 meters. Since time is constant, the ratio of distances is equal to the ratio of speeds. So, the ratio of speeds of Karan and Arjun is 100:90 = 10:9. In the second case, Karan will have to run 110 meters to complete the race. In this case, the ratio of distances travelled will be equal to the ratio of speeds 10:9. So, when Karan runs 110 meters, Arjun will run (110 x 9)/10 = 99 meters, i.e., he needs to cover 1 meter to complete the race. In other words, Karan beats Arjun by 1 meter.

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.

\[ \text{A } 60 : \text{B } 50, \qquad \text{A } 60 : \text{C } 45 \]

Both are stated against the same A score of 60, so B and C compare directly:

\[ B : C = 50 : 45 \]

In a game of 90, when B scores 90:

\[ C = 90 \times \frac{45}{50} = 81 \]

\[ \text{B can give C} = 90 - 81 = 9 \text{ points} \]

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.

\[ A\,100 : B\,90, \qquad A\,100 : C\,72 \]

\[ B : C = 90 : 72 \]

In a 100 m race, when B finishes:

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

\[ \text{B can give C} = 100 - 80 = 20 \text{ m} \]

The difference of the two handicaps, \( 28 - 10 = 18 \), is not the answer, because B's lead is measured over a shorter effective distance.

Answer: C

Convert both times to seconds.

\[ A = 4 \text{ min } 54 \text{ s} = 294 \text{ s}, \qquad B = 300 \text{ s} \]

For a dead heat, B must finish in A's 294 seconds. In that time B normally covers

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

\[ \text{start A can give} = 1000 - 980 = 20 \text{ m} \]

The 6-second gap corresponds to exactly the 20 m B would still have to run.

Answer: C

Given that, A can win B by 50 m.
That is, When A runs 200 m then B can run (200 - 50 = )150 m.
And, B can win C by 8 m means, When B runs 200 m then C can run (200 - 8 =) 192 m.
Therefore, A:B = 200:150 and B:C = 200:192
We have to find A: C.
A:C = A/C = A/B x B/C = (200 / 150) x ( 200 / 192) = (4 / 3) x (200 / 192) = 200 / (3 x 48) = 200 /1 44
That is, A:C = 200:144.
Hence, A wins C by 200-144 m = 56 m.

Answer: C

A is \( 2\tfrac{1}{3} = \tfrac{7}{3} \) times as fast as B, and B starts 80 m ahead. Let the course be \( d \) metres.

\[ \frac{d}{7/3} = \frac{d - 80}{1} \]

\[ 3d = 7d - 560 \quad\Longrightarrow\quad 4d = 560 \]

\[ d = 140 \text{ m} \]

Checking: A runs 140 m while B runs 60, and \( \tfrac{140}{60} = \tfrac{7}{3} \).

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

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.

\[ A\,100 : B\,90, \qquad A\,100 : C\,87 \]

Both are measured against the same A distance, so

\[ B : C = 90 : 87 \]

In a 180 m race, when B finishes:

\[ C = 180 \times \frac{87}{90} = 174 \text{ m} \]

\[ \text{B beats C by } 180 - 174 = 6 \text{ m} \]

Answer: B

A finishes in 20 seconds. In that time B, who needs 25 seconds for 80 m, covers

\[ 80 \times \frac{20}{25} = 64 \text{ m} \]

\[ \text{A beats B by } 80 - 64 = 16 \text{ m} \]

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.