In a 500 m race, A reaches the final point in 28 s and B reaches in 35 s. By how much distance does A beat B?
Answer: A
A finishes in 28 seconds. In that time B, who needs 35 seconds for 500 m, covers
\[ 500 \times \frac{28}{35} = 400 \text{ m} \]
\[ \text{A beats B by } 500 - 400 = 100 \text{ m} \]
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In a race of 300 metres, A beats B by 15 metres or 5 seconds. How much time does A take to complete the race?
Answer: C
"By 15 metres or 5 seconds" means B needs 5 seconds for that last 15 m.
\[ v_{B} = \frac{15}{5} = 3 \text{ m/s} \]
\[ t_{B} = \frac{300}{3} = 100 \text{ s} \]
\[ t_{A} = 100 - 5 = 95 \text{ s} \]
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In a game of 100 points, A can give B 20 points and C 28 points. Number of points B can give C in a game of 100 points is
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.
\[ A\,100 : B\,80, \qquad A\,100 : C\,72 \]
\[ B : C = 80 : 72 \]
In a game of 100, when B scores 100:
\[ C = 100 \times \frac{72}{80} = 90 \]
\[ \text{B can give C} = 100 - 90 = 10 \text{ points} \]
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In a 100m race, A can beat B by 25m and B can beat C by 4m. In the same race, A can beat C by
Answer: D
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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A can give B 25 point, A can give C 40 points and B can give C 20 Points. How many points make the game?
Answer: B
Let the game be \( n \) points. The three statements give
\[ A\,n : B\,(n-25), \qquad A\,n : C\,(n-40), \qquad B\,n : C\,(n-20) \]
From the first two, \( B : C = (n-25) : (n-40) \); from the third, \( B : C = n : (n-20) \). Equating:
\[ \frac{n-25}{n-40} = \frac{n}{n-20} \]
\[ (n-25)(n-20) = n(n-40) \]
\[ n^{2} - 45n + 500 = n^{2} - 40n \quad\Longrightarrow\quad 5n = 500 \]
\[ n = 100 \]
The \( n^{2} \) terms cancel, leaving a simple linear equation.
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In a 100 m race, A beats B by 20 m. What is the ratio of their speeds ?
Answer: C
Correct answer: 5 : 4
In the time A covers 100 m, B covers only 80 m, and both run for the same length of time.
\[\text{Speed ratio} = 100 : 80 = 5 : 4\]
Speeds are in the ratio of the distances covered in equal times, which is why the beaten distance is subtracted from the full course.
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A and B start from the same point and travel in the same direction around a circular track 3 km in circumference, If their speed is 200 and 150 metres a minute, when will they first together again at the starting point?
Answer: C
Both start together, so they are next together at the start line when each has completed a whole number of laps.
\[ t_{A} = \frac{3000}{200} = 15 \text{ min}, \qquad t_{B} = \frac{3000}{150} = 20 \text{ min} \]
\[ \text{LCM}(15, 20) = 60 \text{ min} = 1 \text{ hour} \]
They actually meet on the track earlier - every \( \tfrac{3000}{50} = 60 \) minutes as it happens - but the question asks specifically about the starting point, which is what makes the LCM the right tool.
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In a race of 200 metres, A can give a start of 10 metres to B, and C can give a start of 20 metres to A. The start that C can give to B, in the same race, is
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.
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In a game of 100 points, A can give B 20 points and C 28 points. Then, B can give C:
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:
\[ C = 100 \times \frac{72}{80} = 90 \]
\[ \text{B can give C} = 100 - 90 = 10 \text{ points} \]
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At a game of billiards, A can give B 15 points in 60 and A can give C to 20 points in 60. How many points can B give C in a game of 90?
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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