I’m a big advocate of error logs: notebooks in which students analyse their mistakes. I recommend a three-column approach: in the first, write the question, in the second, what went wrong, and in the last, how to do it correctly. Oddly, that’s the format for this post, too.

The question

The densities of three metal alloys, A, B and C, are in the ratio 13:15:21.

1m³ of alloy B has a mass of 8600kg.

Work out the difference between 5m³ of alloy A and 3m³ of alloy C. Give your answer correct to 3 significant figures.

What went wrong

Student tried to split 8600 in the given ratio, asserted that each ‘share’ was about 175kg, so A was 2,280kg and B 3,690kg (to 3sf), giving a difference of 1,410kg.

There are (I think) three errors here:

  • The student has not dealt correctly with the ratio (we’re given the mass of alloy B, not the total mass)
  • The student has not used the volumes of blocks A and C
  • The student has rounded too early.

How to do it right

We’re told that 1m³ of alloy B has a mass of 8600kg. Alloy A is $\frac{13}{15}$ as dense, so 1m³ has a mass $\frac{13}{15}$ as big - $7453 \frac{1}{3}$ kg. Similarly, the mass of 1m³ of alloy C is $\frac{21}{15}$ as large, which is 12,040kg.

The mass of 5kg of alloy A is $37,266 \frac{2}{3}$kg, and 3kg of alloy B gives 36,120kg. The difference is $1,146 \frac{2}{3}$ kg, which is 1,150kg to 3sf.

Notice that rounding to 3sf before subtracting gives an incorrect answer of $37,300 - 36,100 = 1,200$.

Oh, hello, Mathematical Ninja!

“What’s all this?”

“Well, sensei, I can expl… ow.”

“15 units of alloy B represents 8,600kg. 5kg of alloy A is 65 units. 3kg of alloy C is 63 units. The difference is two units, so you need two-fifteenths of 8,600.”

“But, sensei…”

“But me no buts. $8,600 \times \frac{2}{15} = 1,720 \times \frac{2}{3} = \frac{3,440}{3} = 1,146 \frac{2}{3}$kg, directly.”

“Thank you, s… oh, the Mathematical Ninja has left the building.”

“I thought that was a bit unnecessary.”

“Me, too.”

* Edited 2017-07-04 to correct a rounding error and a mistranscription of what the Mathematical Ninja said. Thanks to @ImMisterAl for mentioning that I should have, um, read more carefully. I’ll be watching my back.