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A reactant $\ce{X}$ is converted into products $\ce{Y}$ and $\ce{Z}$ according to the following first order parallel reactions:

$\ce{X->[k_1]Y}$

$\ce{X->[k_2]Z}$

$\ce{[X]}$ has decreased from $\pu{1.00 M}$ at $t = 0$ to $\pu{0.549 M}$ after $\pu{10 min}$. Calculate the rate constants $k_1$ and $k_2$, knowing that the yields of $\ce{Y}$ and $\ce{Z}$ obtained are $25.0$ and $\pu{75.0 mol\%}$, respectively.

So far, I have the ratio between $k_1$ and $k_2$ $=0.33$, which I found from the yield ratio but I'm unsure what to do with this ratio.

I also have derived the equation $[A] = [A]_0 \cdot e^{-(k_1+ k_2)t}$

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By some simple mathematics, you can reach the following equations (the first of which you have reached):

$[X]=[X]_0\cdot e^{-(k_1+k_2)t}$

$[Y]=\frac{k_1}{k_1+k_2}[X]_0\cdot e^{-k_1 t}$

$[Z]=\frac{k_2}{k_1+k_2}[X]_0\cdot e^{-k_2 t}$

The $1:3$ ratio that you have arrived at is correct, i.e. $\dfrac{k_1}{k_2}=\dfrac{1}{3}$.

Substituting this in the equation for $\ce{X}$, we get

$[X]=[X]_0\cdot e^{-4k_1 t}$

All that is left to do now is to substitute and solve, which leads to the answer:

$(k_1,k_2)\equiv(0.015\ (min)^{-1},0.045\ (min)^{-1})$

There you go!

PS: I've used $[X]_0$ for representing the initial concentration of $\ce{X}$.

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