Races & Games of Skill Questions and Answers

Q: A and B take part in a 100m race. A runs at 5km/hr. A gives B a start of 8m and still beats him by 8 sec. B speed is:

Find A's time first.\[ 5 \text{ km/hr} = 5 \times \frac{5}{18} = \frac{25}{18} \text{ m/s} \]\[ t_{A} = \frac{100}{25/18} = 72 \text{ s} \]B gets an 8 m start, so B runs 92 m, and finishes 8 seconds after A:\[ t_{B} = 72 + 8 = 80 \text{ s} \]\[ v_{B} = \frac{92}{80} = 1.15 \text{ m/s} = 1.15 \times \frac{18}{5} = 4.14 \text{ km/hr} \]The start reduces B's distance to 92 m; the 8 seconds are added to A's time.

Q: A and B take part in 100 m race. A runs at 5 kmph. A gives B a start of 8 m and still beats him by 8 seconds. The speed of B is:

Find A's time first.\[ 5 \text{ km/hr} = 5 \times \frac{5}{18} = \frac{25}{18} \text{ m/s} \]\[ t_{A} = \frac{100}{25/18} = 72 \text{ s} \]B gets an 8 m start, so B runs 92 m, and finishes 8 seconds after A:\[ t_{B} = 72 + 8 = 80 \text{ s} \]\[ v_{B} = \frac{92}{80} = 1.15 \text{ m/s} = 1.15 \times \frac{18}{5} = 4.14 \text{ km/hr} \]The start reduces B's distance to 92 m; the 8 seconds are added to A's time.

Q: In a 100 m race, A can beat B by 25 m and B can beat C by 4 m. In the same race, A can beat C by:

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.

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81.

A and B take part in a 100m race. A runs at 5km/hr. A gives B a start of 8m and still beats him by 8 sec. B speed is:

A.5.15 km/hr
B.4.15 km/hr
C.4.25 km/hr
D.4.14 km/hr

Answer: D

Find A's time first.

\[ 5 \text{ km/hr} = 5 \times \frac{5}{18} = \frac{25}{18} \text{ m/s} \]

\[ t_{A} = \frac{100}{25/18} = 72 \text{ s} \]

B gets an 8 m start, so B runs 92 m, and finishes 8 seconds after A:

\[ t_{B} = 72 + 8 = 80 \text{ s} \]

\[ v_{B} = \frac{92}{80} = 1.15 \text{ m/s} = 1.15 \times \frac{18}{5} = 4.14 \text{ km/hr} \]

The start reduces B's distance to 92 m; the 8 seconds are added to A's time.

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82.

A and B take part in 100 m race. A runs at 5 kmph. A gives B a start of 8 m and still beats him by 8 seconds. The speed of B is:

A.5.15 kmph
B.4.14 kmph
C.4.25 kmph
D.4.4 kmph

Answer: B

Find A's time first.

\[ 5 \text{ km/hr} = 5 \times \frac{5}{18} = \frac{25}{18} \text{ m/s} \]

\[ t_{A} = \frac{100}{25/18} = 72 \text{ s} \]

B gets an 8 m start, so B runs 92 m, and finishes 8 seconds after A:

\[ t_{B} = 72 + 8 = 80 \text{ s} \]

\[ v_{B} = \frac{92}{80} = 1.15 \text{ m/s} = 1.15 \times \frac{18}{5} = 4.14 \text{ km/hr} \]

The start reduces B's distance to 92 m; the 8 seconds are added to A's time.

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83.

In a 100 m race, A can beat B by 25 m and B can beat C by 4 m. In the same race, A can beat C by:

A.21 m
B.26 m
C.28 m
D.29 m

Answer: C

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.

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84.

In 100 m race, A covers the distance in 36 seconds and B in 45 seconds. In this race A beats B by:

A.20m
B.25m
C.22.5m
D.9m

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.

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85.

At a game of billiards, A can give B 15 points in 60 and A can give C 20 points in 60. How many points can B give C in a game of 90 ?

A.30 points
B.20 points
C.10 points
D.12 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.

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86.

In a 100 m race, A beats B by 10 m and C by 13 m. In a race of 180 m, B will beat C by :

A.5.4 m
B.4.5 m
C.5 m
D.6 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:

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

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

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87.

In a 300 m race A Beats B by 22.5 m or 6 seconds. B's time over the course is :

A.86 sec
B.80 sec
C.76 sec
D.None of these.

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.

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