An algebra problem by Prathamesh Kanbaskar

Let C be the crossing point of the two cyclicts.

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For cyclist A we know that,

$AC$ : time = $t$ minutes

$BC$ : time = $16$ minutes

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For cyclist B we know that,

$BC$ : $time = t$ minutes

$CA$ : $time = 25$ minutes

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Using the formula for both cyclists: $speed = \frac{distance}{time}$

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Cyclist A: $speed = \frac{AC}{t} = \frac{AC+BC}{t+16}$ $\frac{AC}{BC} = \frac{t}{16}$

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Cyclist B: $speed = \frac{BC}{t} = \frac{AC+BC}{t+25}$ $\frac{AC}{BC} = \frac{25}{t}$

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Thus solving for $t$ we obtain, $t^{2} = 25 \times 16 = 400$ $t = 20$

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Time for A to reach B is $t + 16$ which is $\boxed{16+20=36}$ .

X = Rate of speed of cyclist "1"
Y = Rate of speed of cyclist "2"
T = Time elapsed (in minutes) when cyclists cross paths.
D = Distance from A to B

X · T = Y · 25 mins

=> X = D / (T + 16), and Y = D / (T + 25)

So, substitute and re-express as:

=> (D · T) / (T + 16) = (D · 25) / (T + 25)

=> (D · T)(T + 25) = (D · 25)(T + 16)

=> D · T² = 400 · D

=> T = 20 mins

20 mins + 16 mins = 36 minutes for cyclist "1" to travel from A to B

Let the speeds of cyclist from $A$ and $B$ be $v_A$ and $v_B$ respectively, the distance between $A$ and $B$ be $D$ , the time after they started and distance from $A$ where the cyclists crosses each other be $t$ and $x$ respectively. Then, we have:

$\begin{cases} v_A t = x & ...(1) \\ v_B t = D - x & ...(2) \\ v_A (t+16) = D &...(3) \\ v_B (t+25) = D &...(4) \end{cases}$

$\Rightarrow \begin{cases} (1)+(2): & D = (v_A + v_B ) t &...(5) \\ (3)+(4): & 2D = (v_A+v_B)t + 16v_A+25v_B \\ & \Rightarrow D = 16v_A+25v_B & ...(6) \end{cases}$

From $\dfrac {(3)}{(4)}: \quad \dfrac{v_A}{v_B} = \dfrac{t+25}{t+16}$

From $(5) = (6):$

$\begin{aligned} (v_A + v_B ) t & = 16v_A+25v_B \\ \dfrac{v_A}{v_B} t+ t & = 16\dfrac{v_A}{v_B}+25 \\ \dfrac{v_A}{v_B} (t - 16 ) & = 25 - t \\ \dfrac{v_A}{v_B} & = \dfrac{25 - t}{t-16} \\ \dfrac{t+25}{t+16} & = \dfrac{25 - t}{t-16} \\ (t+25)(t-16) & = (25-t)(t+16) \\ t^2 + 9t - 400 & = -t^2 + 9t + 400 \\ 2t^2 & = 800 \\ t^2 & = 400 \\ t & = 20 \end{aligned}$

Therefore, the time taken for the first cyclist to reach $B$ from $A$ is: $t+16 = 20 + 16 = \boxed{36}$