# Word Problem for college math

Here's a true story that contains a puzzle. I've changed the numbers to make it simpler. See if you can solve it: A friend of mine was commissioned to reconstruct the missing pieces of a Japanese raku antique. She was given the dimensions of the missing pieces and had to recreate them exactly to fit with the original. Clay shrinks when you fire it, and the percent of shrinkage for her clay was $14%$. She needed to make a $50$ cm by $50$ $cm^2$, so she added $14%$ to get the size that she should make the piece before firing. $14%$ of $50$ is $7$, so she made a $57$ cm by $57$ $cm^2$. After firing, she got a $49$ cm by $49$ $cm^2$. Thinking she had made a mistake, she rechecked her calculations and made the piece a second time. But the second one turned out exactly like the first. This was really frustrating. Do you see the problem? Can you get the correct size that the piece should be made before firing? So this problem has to do with knowing which number to take the percentage of, when something shrinks or grows. Can you come up with another example when this happens in real life?

• All mathematicians know that if something grows by 10% and then shrinks by 10% (or vice versa), it ends up 1% smaller than when it started. – TonyK Nov 17 '14 at 2:11

$\underbrace{{\small\mbox{Clay}}}_{x}$ shrinks when you fire it, and the percent of shrinkage for her $\underbrace{\mbox{clay was 14%}}_{x' = \frac{86}{100} x}$.
She needed to make a 50 centimeter by 50 centimeter square, $\underbrace{\mbox{so she added 14%}}_{x = \frac{114}{100}x'}$ to get the size that she should make the piece before firing. 14% of 50 is 7, so she made a 57-centimeter by 57-centimeter square. After firing, she got a 49-centimeter by 49-centimeter square. Thinking she had made a mistake, she rechecked her calculations and made the piece a second time. But the second one turned out exactly like the first. This was really frustrating. Do you see the problem? Can you get the correct size that the piece should be made before firing?
Using $x$ for a quantity of untreated clay and $x'$ for a quantity of clay after the firing, we see that she hoped for this canceling of shrinking and extra initial material: $$\frac{86}{100}\left( \frac{114}{100} x' \right) = x'$$ alas $$\frac{86}{100} \frac{114}{100} = \frac{9804}{10000} < 1$$ so this will turn out too small. The right factor would have been $$\frac{100}{86} = \frac{100 \frac{100}{86}}{100} = \frac{116.3}{100} > \frac{114}{100}$$ and the size of the the initial piece would have been $$x = \frac{116.3}{100} 50\mbox{cm} = 58.1 \mbox{cm}$$