Two Indian tourists in the US cycled towards each other,one from point A and the other from point B. The first tourist left point A $6$ hrs later than the second left point B, and it turned out on their meeting that he had traveled $12$ km less than the second tourist. After their meeting, they kept cycling with the same speed, and the first tourist arrived at B $8$ hours later and the second arrived at A $9$ hrs later. Find the speed of faster tourist.

$\bf\text{options}$ a.)$4$ km/h $\quad$ b.) $6$ km/h $\quad$ c.) $9$ km/h $\quad$ d.) $2$ km/h

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let they meet at point X as shown in the diagram.

From the question i concluded that first cyclist is fastest .

let the time and distance taken by first cyclist be $(t-6)$ hrs and $(m)$ km to travel upto point X

let the time and distance taken by second cyclist be $(t)$ hrs $(m+12)$ km to travel upto point X

i used the formula that
if two persons starting from point x and y with speed $s_1$ ans $s_2$ coming towards each other from opposite direction after meeting each other take time $t_1$ and $t_2 $ to reach points y and x respectively then

$\large \frac{s_1}{s_2}=\sqrt{\frac{t_2}{t_1}}$

and i concluded.

$\large \dfrac{\frac{m}{t-6}}{\frac{m+12}{t}}=\sqrt{\frac{9}{8}}$

now stucked for two unknowns.


2 Answers 2


Your $t+6$ should be $t-6$ as A left after B. We can then write an equation for each that says they traveled at constant speed $$\frac m{t-6}=\frac {m+12}8 \\\frac {m+12}t=\frac m9\\\frac m{m+12}=\frac {t-6}8\\\frac m{m+12}=\frac 9t\\\frac {t-6}8=\frac 9t\\t^2-6t=72$$ and you have a quadratic to solve

  • $\begingroup$ i got $t=12$ from your equation how should i proceed now. $\endgroup$
    – R K
    Commented Mar 6, 2015 at 15:54
  • $\begingroup$ That is correct. Now you can solve for $m$, say by plugging $t=12$ into the third equation. A's speed is $\frac m{t-6}$ $\endgroup$ Commented Mar 6, 2015 at 15:55
  • $\begingroup$ thanks i got A's speed =$6$ km/h, got $m=36$ $\endgroup$
    – R K
    Commented Mar 6, 2015 at 16:01
  • $\begingroup$ That is correct. $\endgroup$ Commented Mar 6, 2015 at 16:04

Let $a$ be the speed of $A$, $b$ be the speed of $B$, $t$ is the time needed by $A$ to meet $B$ (hence the time used by $B$ to meet $A$ is $t+6$, this is slightly different from the definition in the figure), and $m$ is the distance of the meeting from $A$'s starting point, then

Formula (1): $at = m$

Formula (2): $b(t+6) = m + 12$

Formula (3): $8a = m + 12$

Formula (4): $9b = m$

(1) and (4) gives: $at = 9b$, hence $t = \frac{9b}{a}$ ... [Formula (5)]

(2) and (3) gives: $8a = b(t+6)$, hence $t = \frac{8a}{b} - 6$ ... [Formula (6)]

(1) and (2) gives: $b(t+6) = at + 12$, hence $t = \frac{12-6b}{b-a}$ ... [Formula (7)]

(5) and (7) gives: $\frac{9b}{a} = \frac{12-6b}{b-a}$, hence $9b(b-a) = (12-6b)a$

Simpifying gives: $a = \frac{9b^2}{12+3b}$ ... [Formula (8)]

(5) and (6) gives: $\frac{9b}{a} = \frac{8a}{b} - 6$, hence $8a^2 - 6ab - 9b^2 = 0$.

Factorization gives $(4a+3b)(2a-3b) = 0$, meaning $a = \frac{-3b}{4}$ or $ a = \frac{3b}{2}$.

The $a = \frac{-3b}{4}$ is dropped since $a, b$ (speed) must be both positive.

Now with $ a = \frac{3b}{2}$, substitute to Formula (8):

$ \frac{3b}{2} = \frac{9b^2}{12+3b} $

This gives $36b + 9b^2 = 18b^2$, meaning $b = 0$ (rejected) or $b = 4$.

Hence $a = 6$ (km/h) is the answer.


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