T-Run-CTVT

This solution assumes that runner A is consistently running faster than runner B, who is consistently running faster than runner C, at the very least during the final portions of the race; if that is not the case then the correct answer should be that there is insufficient information to answer the question.

Given the above assumption, we can say that towards the end of the race, the gap between A and B is steadily increasing, as is the gap between B and C. By the time B reaches the finish line, the gap between B and C is 20m; however this means that when A reaches the finish line, 10m ahead of B, the gap between B and C has not yet become 20m, and therefore A will beat C by less than 30m.

Let $x$ be the distance of the lap, and $t_1$ the time of the first race, where $A$ is against $B$ . Since $B$ 's distance is $10$ m less than the lap, $B$ 's speed is $\frac{x - 10}{t_1}$ .

In the second race, where $B$ is against $C$ , $B$ finishes the race at the same speed as before, and so the time $t_2 = \frac{xt_1}{x - 10}$ . Since C's distance is $20$ m less than the lap, $C$ 's speed is $\frac{(x - 20)(x - 10)}{xt_1}$ .

In the third race, where $A$ is against $C$ , $A$ will finish in the same time as before, so the time is $t_1$ . $C$ 's distance would then be $\frac{(x - 20)(x - 10)}{xt_1} \cdot t_1 = \frac{(x - 20)(x - 10)}{x}$ . $C$ 's distance behind $A$ would be $x - \frac{(x - 20)(x - 10)}{x} = 30 - \frac{200}{x}$ , which is necessarily less than $30$ m for all positive values of $x$ (and a positive value for all values of $x$ greater than $\frac{20}{3}$ , which $x$ must be since $B$ beats $C$ by $20$ m).

Relevant wiki: Speed, Distance, and Time

Let $v_a, v_b, v_c$ be the velocities of runners $a, b, c$ , and let $L$ be the length of track lap, and $x$ the unknown distance. Then we have the following equations

$v_b \dfrac{L}{v_a} = L-10$

$v_c \dfrac{L}{v_b} = L-20$

$v_c \dfrac{L}{v_a} = L-x$

That's $3$ equations for $4$ unknowns, so we're not going to have an unique $x$ . However, multiplying togther the first two equations and comparing with the 3rd, we have

$v_c \dfrac{L}{v_a}L = (L-10)(L-20)=(L-x)L$

After some algebraic ado, we end up with

$x=\dfrac{30L-200}{L}$

which means $x <30$ for any finite $L$

In a race with all the runners, when A reaches the finish line, B has yet to achieve the distance of 20 meters to C, because he's still 10m away from the finishing line, therefore if the distance between B and C is less than 20m, the distance between A and C is less then 30m.

Let $V_A, V_B, V_C$ be the speeds of runners A, B & C respectively, $D$ be the length of track lap & $x$ be the distance by which A beats C then

Race between the runners A & B, \frac{D}{V_A}=\frac{D-10}{V_B}\tag1

Race between the runners B & C, \frac{D}{V_B}=\frac{D-20}{V_C}\tag2

Race between the runners A & C, \frac{D}{V_A}=\frac{D-x}{V_C}\tag3

Equating (1) & (3), we get \frac{D-10}{V_B}=\frac{D-x}{V_C}\tag4

Now, subtracting (4) from (2), we get $\color{#D61F06}{\frac{10}{V_B}=\frac{x-20}{V_C}}$

But runner B beats C that means $V_B>V_C \implies \color{#D61F06}{10}>\color{#D61F06}{x-20}$ $\implies x<30$

hence runner A beats C by a distance $x<30 \ m$

Imagine that B was also running in the last run. When the runner A crosses the finish line, B is behind him 10 meters. We know that when B finishes his race, C will be 20 meters behind B. But A finishes the race earlier than B and so, there is not enough time to B build his 20 meters advantage over C. So, the gap between A and C must be 10 meters plus a number inferior to 20 meters, i. e., the answer is less than 30 meters.

This relies on the fact that speeds are constant and the gap between the same runners is proportional to the time since the race started.

Suppose they run on a 100m run. When A hits the 100m mark B will be at 90m. When B hits 100m, C will be at 80m. This problem can be solved as a simple problem involving proportions i.e. A:B = 100:90 and B:C = 100:80 so A:C= 100/90 * 100/80 = 100:72. In other words when A is on the 100m mark C is on the 72m mark, or 28m behind.

Since, in the question there is no information about how many metres race it is, let me assume it as 100 metres (you'll come to know why 100; it makes our job easier). We know that $Speed$ = $\frac{distance}{Time taken}$ . And since speeds are constant, we get $\frac{a}{b}$ = $\frac{100}{90}$ and $\frac{b}{c}$ = $\frac{100}{80}$ where $a$ , $b$ and $c$ are the speeds of A, B and C, respectively. Multiplying both, we get $\frac{a}{c}$ = $\frac{100}{72}$ . This means that the runners A and C will be 28 metres apart if they run the lap together. Hence, the option 1.

Assume runner A runs 100m in 10s => 10m/s. Therefore, runner B runs 100m - 10m = 90m in 10s => 9m/s

Runner B then runs 100m while going at 9m/s => $\frac{100m}{9m/s}$ = 11.11s

Therefore runner C runs 100m - 20m = 80m in 11.11s => $\frac{80m}{11.11m/s}$ = 7.2m/s

So, since runner A runs 100m in 10s, runner C will run $x$ meters in 10s at 7.2m/s => $x$ = 7.2m/s * 10s = 72m, which is less than 30m from runner A.

Assume A is running 100 meters "100% fast". So B is only 90% fast, but still C is 20% slower than B. So C runs 80% of 90% = 72% fast. So he is only 28% slower than A. You can easily see that the inequality holds true for different runway lengths, because it comes from the product of relative speeds.

Assume B run together. Distance of B and C is increasing over time. We know when B finished, B beats C 20m. But when A finished race, B isn't approach finish line. So distance between C and B $k$ less than 20m. $10+k<30$ .So answer is A.

Here is a solution without math, Consider A,B and C running together, sure A will win the race and when he just completes the race, B will be 10 meter behind A as given. C will be behind B and since B has greater velocity than C the distance between them is increasing with time. When B finishes the race C will be 20 meters behind B as given but at any instant before this, distance between them is less than 20(because distance between them is increasing). So 10 and something less than 20 gives something less than 30.

A beats C by exactly 28 meters.

• Imagine total distance from start to end is 100 meters.
• In the 1st race, A finish race in 1 second - A's velocity is 100 meters per second; B's velocity is 90meters per second.
• In the 2nd race, B runs 100 meters in 100/90 seconds.C runs 80 meters in 100/90 seconds. Therefore, C's velocity is (80*90)/100 =72 meters per second
• In the 3rd race, A finish race in 1 second and B runs 72 meters in 1 second.

It take the length of the entire Lap( at constant speeds) for the gap of 10 m to develop between A & B and similarly for B&C( gap of 20 m). Now, Imagine all three run at the same time. When A finishes, B will be 10 m behind. C will be behind B but with a distance < 20 m.

$d =$ race distance

$v_A, v_B, v_C =$ speeds of A, B, and C respectively

$t_A, t_B, t_C =$ respective times it takes A, B, and C to finish the race

$t_B = t_A + \dfrac{10}{v_B}$

$t_C = t_B + \dfrac{20}{v_C} = t_A + \dfrac{10}{v_B} + \dfrac{20}{v_C} < t_A + \dfrac{10}{v_C} + \dfrac{20}{v_C} = t_A + \dfrac{30}{v_C}$

Since $\dfrac{30}{v_C}$ is the time it takes C to run $30$ m, A beats C by less than $30$ m.

Assume all speeds are uniform from start to finish. Let $t_A$ be time for runner A to finish and $t_B$ for runner B to finish. Then, using V for speeds, $t_A(V_A-V_B) = 10$ and $t_B(V_B-V_C) = 20$ . Do a little algebra to get $t_A(V_A-V_C) = 10 + (\frac{t_A}{t_B}$ ) 20. $\frac{t_A}{t_B}$ < 1, so less than 30.

Imagine C running behind B running behind A. When A reaches the finish and the distance between A and B is 10 m, the distance between B and C is less than 20 m. That's because B is faster than C and he has to keep running to reach the finish to increase his lead over B for the full 20 m.

We need to assume that the runners move at a constant rate, so that B = ,9A and C = .8B. Then C = .72A. For convenience, assume the track is 100 m. When A is at the finish line, C is at 72, or 28 m. behind. Ed Gray

Imagine that in the third race, all three runners are visible. From diagram 1, we know that runner B would be 10 m behind runner A. However, we cannot assume that runner C would be 20 m behind runner B. Because C is slower than B, the gap between the runners is increasing the whole time. If B is 20 m ahead at the finish line, B is less than 20 m ahead before the finish line. Therefore, in the third race, the distance between B and C is less than 20 m. Runner A would be less than 30 m ahead of runner C.