Life's a race!

Let $v_{A}, v_{B}, v_{C}$ be the respective speeds of runners $A,B,C$ , and let $t_{A}, t_{B}$ be the times runners $A$ and $B$ take to cover the distance $d$ .

Then $d = v_{A}t_{A} = v_{B}t_{A} + 30 = v_{C}t_{A} + 48 \Longrightarrow (v_{B} - v_{C})t_{A} = 18$ ,

and $d = v_{B}t_{B} = v_{C}t_{B} + 20 \Longrightarrow (v_{B} - v_{C})t_{B} = 20$ .

Thus $\dfrac{(v_{B} - v_{C})t_{B}}{v_{B} - v_{C})t_{A}} = \dfrac{20}{18} \Longrightarrow t_{B} = \dfrac{10}{9}t_{A}$ , and so

$d = v_{B}t_{B} = \dfrac{10}{9}v_{B}t_{A} = \dfrac{10}{9}(v_{A}t_{A} - 30)$ . But $d = v_{A}t_{A}$ as well, so

$v_{A}t_{A} = \dfrac{10}{9}(v_{A}t_{A} - 30) \Longrightarrow 9v_{A}t_{A} = 10v_{A}t_{A} - 300 \Longrightarrow d = v_{A}t_{A} = \boxed{300}$ metres.

Assume that running distance is 'L'. During the time B runs (L-30), C runs (L-48). Also by the time B runs L, C runs (L-20). Equating the two ratios and cross-multiplying L(L-48)=(L-20)(L-30). i.e.L=300 is the answer. This is algebra. Still simpler would be to think that when A reaches the end if B runs 30m more she beats C by a further length of 20-18=2m, so in how much run she must have gone ahead of C by 18m. Yes in 9 times 30 length, i.e. 270. So total race length is 30 + 270 = 300.

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When A finishes the race, A-B = 30 and B-C = 18 AND when B finishes the race, B-C = 20 as shown in the Distance v/s Time graph.

The gray triangles are similar, with bases 18 and 20. Hence the finishing times of A and B must be in a ratio 18:20. In fact, we can take them as 9 and 10 in some arbitrary time measuring unit.

That gives us two similar right angle triangles with bases 9 and 10 and heights x and x + 28 respectively.

Hence $\frac{x}{9} = \frac{x+28}{10}$ Solving gives x = 252 and the total distance = x + 18 + 30 = 300

sir use the concept of relative velocity and solve the equation