, 2012

Initially, the bowl contains 320 g of white sugar and 0 g of brown sugar.

Mixture Y contains $(320-x)$ g of white sugar and $x$ g of brown sugar.

When Mixture Z (the final mixture) is formed, there is still 320 g of sugar in the bowl.

Since we are told that the ratio of the mass of white sugar to the mass of brown sugar is $49:15$, then the mass of white sugar in Mixture Z is $\frac{49}{49+15}\cdot 320 = \frac{49}{64}\cdot 320 = 49 \cdot 5 = 245$ g and the mass of brown sugar in Mixture Z is $320 - 245 = 75$ g.

In order to determine the value of $x$ (and hence determine the values of $w$ and $b$), we need to determine the mass of each kind of sugar in Mixture Z in terms of $x$.

Recall that Mixture Y consists of $(320-x)$ g of white sugar and $x$ g of brown sugar, which are thoroughly mixed together.

Because Mixture Y is thoroughly mixed, then each gram of Mixture Y consists of $\dfrac{320-x}{320}$ g of white sugar and $\dfrac{x}{320}$ g of brown sugar.

To form Mixture Z, $x$ g of Mixture Y are removed.

This amount of Mixture Y that is removed contains $x \cdot \dfrac{x}{320} = \dfrac{x^2}{320}$ g of brown sugar.

Mixture Z is made by removing $x$ g of Mixture Y (which contains $\dfrac{x^2}{320}$ g of brown sugar),then adding $x$ g of brown sugar.

Thus the mass of brown sugar, in g, in Mixture Z is $x - \dfrac{x^2}{320} + x$.

Since Mixture Z includes 75 g of brown sugar, then $$\begin{aligned} 2x - \dfrac{x^2}{320} & = & 75\\ 0 & = & x^2 - 2(320)x + 75(320) \\ 0 & = & x^2 - 640x + 24000 \\ 0 & = & (x-40)(x-600)\end{aligned}$$ Therefore, $x = 40$ or $x=600$.

Since the initial mixture consists of 320 g of sugar, then $x<320$, so $x=40$.

This tells us that Mixture Y consists of $320-40=280$ g of white sugar and 40 g of brown sugar. The ratio of these masses is $280:40$, which equals $7:1$ in lowest terms. Thus, $w=7$ and $b=1$.

Therefore, $x+w+b=40+7+1=48$.