Races & Games of Skill Questions and Answers

Answer: B

A took 220 seconds. Let A's time be T. Then B takes T + 50 and C takes T + 80 seconds. When A finishes the 1500 m, C is 400 m behind, so C has run 1100 m in T seconds. C's speed is 1100/T, and he runs 1500 m in T + 80 seconds: \[\frac{1500}{T + 80} = \frac{1100}{T}\] \[1500T = 1100T + 88000\] \[T = 220\] Check: C covers 1100 m in 220 s, 5 m per second, and 1500 m in 300 s, which is 220 + 80. The other options give a speed that does not match this.

Answer: C

The race is 78 m long. A is 4% faster, so speeds are in the ratio 104 : 100. B is given a 3 m lead, so B runs (L - 3) while A runs L in the same time: \[ \frac{L}{104} = \frac{L - 3}{100} \] \[ 100L = 104L - 312 \Rightarrow L = 78 \]

74, 76 and 80 do not satisfy this equation.

Answer: D

Anil beats Pawan by 434 m. Pawan's speed is 924/66 = 14 m/s. In the 35 seconds Anil takes, Pawan covers 14 x 35 = 490 m, so he is 924 - 490 = 434 m behind when Anil finishes. 431, 432 and 433 m do not follow from these speeds. Anil's speed is 924/35 = 26.4 m/s. The distance is the same, so the time ratio (35 : 66) is the inverse of the speed ratio. Alternatively, Pawan runs 35/66 of the race in Anil's time, which is 924 x 35/66 = 490 m.

Answer: D

A takes 920 seconds.

Let A take t seconds. When A gives B a 40 m start the race is a dead heat, so B runs 460 m in t seconds. With a 30 m start B runs 470 m, which takes 20 seconds more than t. At B's speed 460/t, the time for 470 m is 470t/460.

\[ \frac{470t}{460}=t+20 \]

\[ \frac{10t}{460}=20 \Rightarrow t=920 \]

325, 417 and 266 seconds do not satisfy these equations.

The key idea is that the 10 m extra headstart changes the result from B finishing 20 s behind to a dead heat, so B covers 10 m in 20 s at the speed 460 / t. This gives B's speed of 0.5 m/s, and the 460 m B covers takes 920 seconds, which is A's time over 500 m.

Answer: A

The gap is 20 m.

When Q starts, P still has 500 - 140 = 360 m to run. In the same time Q covers distance in the ratio 4 : 3:

\[ \text{Q's distance} = 360 \times \frac{4}{3} = 480\ \text{m} \]

So when P finishes at 500 m, Q is at 480 m, 20 m behind. P does win, since Q would need 500 m.

140 is only P's head start. A gap of 40 or 60 m would need a different speed ratio, since with 3 : 4 Q gains exactly 120 m on P over P's last 360 m.

Answer: C

They meet at the starting point after 60 seconds.

A completes a lap in 600/30 = 20 s and B in 600/20 = 30 s. Both are at the start together whenever the time is a multiple of both lap times:

\[ \text{LCM}(20, 30) = 60\ \text{s} \]

In 60 s A runs 3 laps and B 2, so both are back at the start. 40, 50 and 70 seconds are not multiples of both lap times; at 40 s, for example, A is at the start but B is two-thirds of the way round.

Answer: B

Bimal's speed is 14.4 kmph.

Kamal runs at 18 km/h = 5 m/s, so he finishes 100 m in 20 s. Bimal takes 5 s longer:

\[ \begin{aligned} v_{B} &= \frac{100}{25} = 4\ \text{m/s} \\ &= 4 \times \frac{18}{5} = 14.4\ \text{km/h} \end{aligned} \]

15.4, 14.5 and 14 km/h correspond to times of about 23.4, 24.8 and 25.7 seconds, none of which is 25.

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.

\[ \text{A } 60 : \text{B } 50, \qquad \text{A } 60 : \text{C } 45 \]

Both are stated against the same A score of 60, so B and C compare directly:

\[ B : C = 50 : 45 \]

In a game of 90, when B scores 90:

\[ C = 90 \times \frac{45}{50} = 81 \]

\[ \text{B can give C} = 90 - 81 = 9 \text{ points} \]

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.

\[ A : B = 20 : 25 \]

In a 1000 m race B finishes first, and in that time A covers

\[ 1000 \times \frac{20}{25} = 800 \text{ m} \]

\[ \text{B beats A by } 1000 - 800 = 200 \text{ m} \]

Answer: C

Convert both times to seconds.

\[ A = 4 \text{ min } 54 \text{ s} = 294 \text{ s}, \qquad B = 300 \text{ s} \]

For a dead heat, B must finish in A's 294 seconds. In that time B normally covers

\[ 1000 \times \frac{294}{300} = 980 \text{ m} \]

\[ \text{start A can give} = 1000 - 980 = 20 \text{ m} \]

The 6-second gap corresponds to exactly the 20 m B would still have to run.