2140.71 – Beakers, Ratio of Acid to Water

Two identical beakers are filled with acid solutions, the ratio of the volume of acid to the volume of water being p:1p:1 in one beaker and q:1q:1 in the other beaker. If the entire contents of the two beakers are mixed together, the ratio of the volume of acid to the volume of water is:

  1. p+q2\dfrac{p + q}{2}

  2. p2+q2p+q\dfrac{p^2 + q^2}{p + q}

  3. 2pqp+q\dfrac{2pq}{p + q}

  4. 2(p2+pq+p2)3(p+q)\dfrac{2(p^2 + pq + p^2)}{3(p + q)}

  5. p+q+2pq(p+q+2)\dfrac{p + q + 2pq}{(p + q + 2)}

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Solution

This looks nasty, but here we go.

If the ratio of acid to water is p:1p:1, then if the beaker contains xx cc's of liquid total, which means (pp+1)×x\left(\dfrac{p}{p+1}\right) \times x cc’s are acid and (1p+1)×x\left(\dfrac{1}{p+1}\right) \times x cc’s are water. The same goes for q:1q:1. (qq+1)×x\left(\dfrac{q}{q+1}\right) \times x is acid and (1q+1)×x\left(\dfrac{1}{q+1}\right) \times x is water.

The total acid is:

(pp+1+qq+1)x=(p(q+1)+q(p+1))x(p+1)(q+1)=(2pq+p+q)x(p+1)(q+1)\begin{align*}\left(\dfrac{p}{p+1} + \dfrac{q}{q+1}\right)x &= \dfrac{(p(q + 1)+ q(p + 1))x}{(p + 1)(q + 1)} \\[1em]&= \dfrac{(2pq + p + q)x}{(p + 1)(q + 1)}\end{align*}

The total water is:

(1p+1+1q+1)x=(p+1+q+1)x(p+1)(q+1)=(p+q+2)x(p+1)(q+1)\begin{align*}\left(\dfrac{1}{p+1} + \dfrac{1}{q+1}\right)x &= \dfrac{(p + 1 + q + 1)x}{(p + 1)(q + 1)} \\[1em]&= \dfrac{(p + q + 2)x}{( p + 1)(q + 1)}\end{align*}

So acid/water = (2pq+p+q)x(p+1)(q+1)÷(p+q+2)x(p+1)(q+1)=2pq+p+qp+q+2\dfrac{(2pq + p + q)x}{(p + 1)(q + 1)} \div \dfrac{(p + q + 2)x}{( p + 1)(q + 1)} = \dfrac{2pq + p + q}{p + q + 2}, which is e.

It’s getting started that’s hard.