Q:

Question about significant figures

I was taught that there are two different methods for obtaining results for multiplication/division or addition/subtraction with decimals. For multiplication/division the result will have the least amount of significant figures of the multiplicands or the dividends. For addition/subtraction, the addend or the number being subtracted with the least amount of decimal places will be the amount of decimal places for the result. But what if both multiplication AND addition is being used? I guess I haven't gotten up to that yet. If anyone can help I'd appreciate it, thanks!

A:

I'll answer your question by example. We'll work with three numbers
$$a = 52.4$$
$$b = 0.96$$
$$c = 2.193$$
Here $a$ has three significant figures and one decimal place, $b$ has two significant figures and two decimal places, and $c$ has four significant figures and three decimal places.
If we multiply, the number of significant figures of the result is the same as the multiplicand with the least number of significant figures, so although we have $a \times c = 114.9132$ if we keep all the digits of the result, one of the multiplicands only has three significant figures, so you would write
$$a\times c = 115$$
If you are adding, the number of decimal places of the result is the same as the addend with the least number of decimal places, so although we have $a+c = 54.593$ if we keep all the digits, one of the addends only has one decimal place, so you would write
$$a + c = 54.6$$
If you are adding and subtracting in the same calculation you apply the rules sequentially. So to compute $b + a\times c$, the result of $a\times c$ has three significant figures, and no decimal places. Then when adding $b$, one of the numbers in the sum has two decimal places and the other has none, so the result will have zero decimal places of accuracy. You perform the calculation keeping all the digits and round at the end. Keeping all the digits we have $b + a\times c = 115.8732$. Now rounding that result to zero decimal places gives
$$b + a\times c = 116$$

