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.