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\frac{p + q}{2}
  2. p2+q2p+q\frac{p^2 + q^2}{p + q}
  3. 2pqp+q\frac{2pq}{p + q}
  4. 2(p2+pq+p2)3(p+q)\frac{2(p^2 + pq + p^2)}{3(p + q)}
  5. p+q+2pq(p+q+2)\frac{p + q + 2pq}{(p + q + 2)}
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 of liquid total, which means pp+1x\frac{p}{p+1}x cc’s are acid and 1p+1x\frac{1}{p+1}x cc’s are water.

The same goes for q:1.qq+1xq:1. \frac{q}{q+1}x is acid and 1q+1x\frac{1}{q+1}x is water.

The total acid is (pp+1(\frac{p}{p+1} + qq+1\frac{q}{q+1})xx = ((p(q+1)+q(p+1))x(p+1)(q+1)(\frac{(p(q + 1)+ q(p + 1))x}{(p + 1)(q + 1)} = (2pq+p+q)x(p+1)(q+1)\frac{(2pq + p + q)x}{(p + 1)(q + 1)}

The total water is (1p+1\frac{1}{p+1} + 1q+1\frac{1}{q+1})xx = (p+1+q+1)x(p+1)(q+1)\frac{(p + 1 + q + 1)x}{(p + 1)(q + 1)} = (p+q+2)x(p+1)(q+1)\frac{(p + q + 2)x}{( p + 1)(q + 1)}

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

It’s getting started that’s hard.