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.

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

\[ B : C = 90 : 72 \]

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: 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.

\[ A\,200 : B\,169, \qquad A\,200 : C\,182 \]

\[ C : B = 182 : 169 \]

In a 350 m race, when C finishes:

\[ B = 350 \times \frac{169}{182} = 325 \text{ m} \]

\[ \text{C beats B by } 350 - 325 = 25 \text{ m} \]

Answer: D

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

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: 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

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

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: 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

This is a simple speed time problem. Given conditions:

=>Speed of P= 224/28 = 8m/s

=>Speed of Q=224/32 = 7m/s

=>Difference in time taken = 4 seconds

Therefore, Distance covered by P in that time= 8m/s x 4 seconds = 32 metres