In a 500 m race the speeds of A and B are in the ratio 3 : 4, and A is given a start of 140 m. By how much does A win ?
Answer: B
Correct answer: 20 m
With a 140 m start, A has only \(500 - 140 = 360\) m to run. In the time A covers that,
\[B \text{ covers } 360 \times \frac{4}{3} = 480 \text{ m}\]
B still has \(500 - 480 = 20\) m to go when A finishes, so A wins by 20 m.
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In a game of 100 points, A can give B 20 points and A can give C 28 points. How many points can B give C in a game of 100 ?
Answer: B
Correct answer: 10
Giving 20 points means that when A scores 100, B scores 80; likewise C scores 72.
\[B : C = 80 : 72\]
Scaling so that B scores 100, C scores \(100 \times \tfrac{72}{80} = 90\). So B can give C 10 points.
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In a game of Snooker, A can Give B 16 Points in 80 and A can Give C 15 point in 90. How many points can C give B in a game of 50?
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.
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P can run 100m in 20 second and Q in 25 seconds. P beats Q by
Answer: B
P finishes in 20 seconds. In that time Q, who needs 25 seconds for 100 m, covers
\[ 100 \times \frac{20}{25} = 80 \text{ m} \]
\[ \text{P beats Q by } 100 - 80 = 20 \text{ m} \]
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In a 2 Km race, P can give Q 200 m and R 560 m. In the same race, Q can give R
Answer: A
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.
\[ P\,2000 : Q\,1800, \qquad P\,2000 : R\,1440 \]
\[ Q : R = 1800 : 1440 = 5 : 4 \]
In a 2000 m race, when Q finishes:
\[ R = 2000 \times \frac{4}{5} = 1600 \text{ m} \]
\[ \text{Q can give R} = 2000 - 1600 = 400 \text{ m} \]
The difference of the starts, \( 560 - 200 = 360 \), is not the answer - Q's advantage is measured over a shorter effective run.
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A runs 1.5 Times as fast as B can. If A gives B a start of 50 m, how far must the winning post be in order that A and B reach at the same time
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 \( \tfrac{150}{100} = 1.5 \).
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P and Q run a Kilometers and P win by 1 minutes. P and R run a kilometre and P wins by 375 m. Q and R run a kilometre and Q wins By 30 Seconds. Find the time taken by each runner to run a kilometre.
Answer: A
Let the three times over a kilometre be \( p \), \( q \) and \( r \).
P beats Q by 1 minute, and Q beats R by 30 seconds:
\[ q = p + 60, \qquad r = q + 30 = p + 90 \]
P beats R by 375 m, so R covers only 625 m in P's time:
\[ \frac{625}{1000} = \frac{p}{r} \quad\Longrightarrow\quad r = 1.6p \]
\[ 1.6p = p + 90 \quad\Longrightarrow\quad 0.6p = 90 \quad\Longrightarrow\quad p = 150 \]
\[ q = 210 \text{ s}, \qquad r = 240 \text{ s} \]
Checking: R covers \( 1000 \times \tfrac{150}{240} = 625 \) m in P's 150 seconds, exactly 375 m short.
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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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At a game of billiards, A can give B 12 points in a game of 40 and A can give C 10 points in a game of 50. How many points can C give B in a game of 80?
Answer: A
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 handicaps are over different totals, so scale both against A.
\[ A\,40 : B\,28 \ \Rightarrow\ B = \tfrac{7}{10}A \]
\[ A\,50 : C\,40 \ \Rightarrow\ C = \tfrac{8}{10}A \]
\[ B : C = 7 : 8 \]
C is the stronger, so in a game of 80, when C reaches 80:
\[ B = 80 \times \frac{7}{8} = 70 \]
\[ \text{C can give B} = 80 - 70 = 10 \text{ points} \]
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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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