In a 200 metres race A beats B by 35 m or 7 seconds. A's time over the course is:
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.
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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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In a 200m race, if A gives B a start of 25 meters, then A wins the race by 10 seconds. Alternatively, if A gives B a start of 45 meters the race ends in a dead heat. How long does A take to run 200m?
Answer: C
A gives B a start of 25 meters and still wins the race by 10 seconds.
Alternatively, if A gives B a start of 45 meters, then the race ends in a dead heat.
Therefore, the additional 20 meters start given to B compensates for the 10 seconds.i.e., B runs 20 meters in 10 seconds.
Hence, B will take 100 seconds to run 200 meters.
We know that A gives B a start of 45 meters. B will take 22.5 seconds to run this 45 meters as B runs 20 meters in 10 seconds or at the speed of 2 m/s.
Hence, A will take 22.5 seconds lesser than B or 100 - 22.5 = 77.5 seconds to complete the race.
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In a race of 200 m, A can beat B by 31 m and C by 18 m. In a race of 350 m, C will beat B by :
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:
\[ B = 350 \times \frac{169}{182} = 325 \text{ m} \]
\[ \text{C beats B by } 350 - 325 = 25 \text{ m} \]
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A can run 22.5 m while B runs 25 m. In a kilometre race B beats A by :
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.
\[ A : B = 22.5 : 25 \]
In a 1000 m race B finishes first, and A has then covered
\[ 1000 \times \frac{22.5}{25} = 900 \text{ m} \]
\[ \text{B beats A by } 1000 - 900 = 100 \text{ m} \]
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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 :
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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In a game of billiards, A can give B 20 points in 60 and he can give C 30 points in 60. How many points can B give C in a game of 100?
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.
\[ A\,60 : B\,40, \qquad A\,60 : C\,30 \]
\[ B : C = 40 : 30 = 4 : 3 \]
In a game of 100, when B scores 100:
\[ C = 100 \times \frac{3}{4} = 75 \]
\[ \text{B can give C} = 100 - 75 = 25 \text{ points} \]
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In a 100 m race, P can give Q 20 m and R 32 m. In the same race Q can give R:
Answer: C
P : Q = 100 : 80.
P : R = 100 : 68.
Q : R = Q/P * P/R
= (80/100) * (100/68)
= 80/68
When Q runs 80 m, R runs 68 m.
When Q runs 100 m, R runs ((68/80) * 100)m = 85m
Q can give R 15 m.
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In a 550 m race, the ratio of the speeds of two contestants P and Q is 3 : 4. P has a start of 170 m. Then, P wins by:
Answer: B
To reach the winning post A will have to cover a distance of (550 - 170)m, i.e., 380 m.
While P covers 3 m, Q covers 4 m. (380 * (4/3)) = 506
While P covers 440 m, Q covers
Thus, when P reaches the winning post, Q covers 506 m and therefore remains 44 m behind.
P wins by 44 m.
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In a kilometer race, A beats B by 50 meters or 10 seconds. What time does A take to complete the race?
Answer: B
"By 50 m or 10 seconds" means B needs 10 seconds for that final 50 m.
\[ v_{B} = \frac{50}{10} = 5 \text{ m/s} \]
\[ t_{B} = \frac{1000}{5} = 200 \text{ s} \]
\[ t_{A} = 200 - 10 = 190 \text{ s} \]
The question asks for A's time, so the 10 seconds must be subtracted from B's.
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