Why are there only 2 significant figures in the answer?

Question

In the problem $\frac{8.01-7.50}{3.002}$

Why would the answer be $0.17$ and not $0.170$? My least amount of sig figs is $3$ in the original equation. The only thing I can come up with is in the intermediate step.$8.01-7.50= 0.51$ exactly, which only has $2$ significant figures. Does the intermediate step really count in determining significant figures? Thank you. :)

Answer 1

It is often regarded as good practice to give $1$ fewer sig. figs. than in the given numbers. So you are right, the least number of sig figs in the original equation is $3$ therefore give $2$ in the final answer.

(This has nothing to do with the intermediate step you mention.)

To see why this is regarded as best practice you could look at what the original numbers might mean. For example they might have been rounded from $8.014,7.495,3.0015$.

Your calculation with these numbers gives $0.173$ to $3$ sig figs . So $0.17$ to 2 sig figs gives a more honest degree of accuracy.

Answer 2

Unless the 0 in $7.50$ is an exact figure, you can't deem it significant.

The significant figures (also known as the significant digits and decimal places) of a number are digits that carry meaning contributing to its measurement resolution. This includes all digits except:[1]

All leading zeros. For example, "013" has two significant figures: 1 and 3; Trailing zeros when they are merely placeholders to indicate the scale of the number (exact rules are explained at identifying significant figures); and Spurious digits introduced, for example, by calculations carried out to greater precision than that of the original data, or measurements reported to a greater precision than the equipment supports.

So unless the 0 is showing accuracy of a scale or an exact measurement ( like there being exactly 453592370 micrograms to a lb by definition) , there could be rounding inaccuracies ( like quoting a nutrition label 64 g of sugar is 21% of a 2000 Calorie diet, or 25 grams sugar to 8% for another, if they round from half percents the range on the first gives $128\over 41$ to $128\over 43$ the second gives $10\over 3$ to $50 \over 17$ (that's all assuming grams are completely accurate)).

So arguably, you could have as few as two significant figures, or the 0 could be from rounding.

Source wikipedia on significant digits.