Arithmetic with Significant Figures

Transcript

0:00Introduction

Hello everyone! Okay, one small puzzle before we start. Take four point two three seven, and multiply it by two point five one. Your calculator proudly shows ten point six three four eight seven. Seven digits! But the numbers you started with had only four significant figures and three significant figures. So tell me, can a calculation ever be more precise than the measurements it came from? Hmm, that does not sound right, does it? Today we find out how many digits you may really keep. And look, in JEE, the options in such questions often differ only in the number of digits. So this one skill decides your marks. Let's begin.

0:45Multiply or divide: count significant figures

So, rule number one, for multiplication and division: the cross and the division sign you see on screen. The answer keeps as many significant figures as the least number in any factor, okay? The weakest measurement decides. Now our puzzle. Four point two three seven, times two point five one. The calculator says ten point six three four eight seven. Four point two three seven has four significant figures, and two point five one has three. So how many can the answer keep? Three, yes! The least one. This double arrow on screen means therefore. We keep three significant figures: ten point six. Not seven digits, just three. Mystery solved: the extra digits were never measured at all, the calculator simply made them up. Clear? Division follows the same rule. Zero point zero five two five, divided by two point one, is zero point zero two five. The zeros in front do not count, remember? So zero point zero five two five has three significant figures, and two point one has only two. So the answer keeps two: zero point zero two five. Really simple.

1:56Add or subtract: count decimal places

Alright, now rule number two, for addition and subtraction: the plus and minus signs on screen. Here we do not count significant figures at all. We count decimal places, the digits after the decimal point. The answer keeps as many decimal places as the least number in any term, okay? Let's add. Four hundred thirty six point three two, plus two hundred twenty seven point two, plus zero point three zero one. The calculator gives six hundred sixty three point eight two one. Now, which term decides? Zero point three zero one has the fewest significant figures, only three, so it decides, right? No, no, careful! That is the multiplication rule. This is the classic JEE trap. For addition, look at decimal places. Two hundred twenty seven point two is known to only one decimal place. So the sum keeps only one decimal place too: six hundred sixty three point eight, okay? Subtraction, same rule. Ten point four six minus two point three is eight point one six. This single arrow means rounds to. Two point three has one decimal place, so eight point one six rounds to eight point two. Got it?

3:10Powers, roots, and when to round

Next, rule number three: powers and roots. Squares, cubes, square roots. The answer keeps the same number of significant figures as the measured quantity, okay? Now, the square root of two point five six. Your calculator shows one point six. So the answer is one point six, right? Hmm, wait. Count the figures first. Two point five six has three significant figures, so its root must also keep three: one point six zero, with that zero written, okay? That zero tells everyone how precise you are. Same way, one point two zero squared is one point four four, three figures again. Now one very important habit, and many students get this wrong. When should you round? After every step, or only at the end? Only at the end, yes, exactly! Round only the final answer, okay? In the middle steps, keep one extra digit, so that early rounding does not spoil your final answer. Clear? Look, like this. Three point four five plus zero point two is three point six five. By the addition rule, this sum may keep only one decimal place. But it is a middle step, so we carry the extra digit. Then three point six five divided by two point zero is one point eight two five. Two point zero has two significant figures, so the final answer is one point eight. See? The extra digit waited quietly in the middle, and we rounded just once, at the end. Got it?

4:43Trap: subtracting nearly equal numbers

Now one exception, and examiners love it. Twelve point three five four, minus twelve point three five one, is zero point zero zero three. Count the significant figures in this answer. Just one! The zeros in front do not count. Each number we started with had five significant figures. After the subtraction, only one survives. So, subtracting nearly equal numbers destroys significant figures, okay? And such a difference gives a large relative error. So when two nearly equal readings are subtracted, be careful: the answer is far less precise than either reading. Keep this in mind, because it comes back in our second example. Remember that!

5:29Example 1: 25.2 x 1374 / 33.3 with correct sig figs

Okay, now your turn. Evaluate twenty five point two, times one thousand three hundred seventy four, divided by thirty three point three, with correct significant figures. Pause the video, and try it first. Done? Okay, let's check. Twenty five point two times one thousand three hundred seventy four is thirty four thousand six hundred twenty four point eight. Divide by thirty three point three, and the calculator gives one thousand thirty nine point seven eight. Only multiplication and division here, so we count significant figures. Twenty five point two has three, one thousand three hundred seventy four has four, and thirty three point three has three. The least is three. So, three figures. Just write one thousand and forty, right? Hmm, careful! Remember the counting rules: trailing zeros in a number without a decimal point are ambiguous. Nobody can tell if that last zero counts. So we use scientific notation: one point zero four, into ten to the power three. Exactly three significant figures. Nice and clean.

6:41Example 2: box and gold pieces, total mass and difference

Now a slightly harder one, JEE level. Read it with me. There is a box of mass two point three kilograms, holding two gold pieces, of twenty point one five grams and twenty point one seven grams. Find the total mass, and the difference of the masses of the two pieces, with correct significant figures. Pause and try. Hint: put everything in the same unit first. Done? Okay, let's check. Kilograms and grams cannot be added directly, so change the grams to kilograms. Twenty point one five grams is zero point zero two zero one five kilograms, and twenty point one seven grams is zero point zero two zero one seven kilograms. Now part a. Add two point three, plus the two small pieces, gives two point three four zero three two kilograms. So the total is two point three four kilograms, right? The gold pieces are measured so precisely, aren't they? No, no, careful! This is the trap. It is an addition, so count decimal places. The box, two point three kilograms, is known to only one decimal place. So the total also keeps one decimal place: two point three kilograms. Very precise gold pieces cannot rescue a rough reading of the box, okay? Now part b, the difference. Twenty point one seven minus twenty point one five: zero point zero two grams. Both readings have two decimal places, so the answer keeps two decimal places too. But count the significant figures in zero point zero two. Just one! Four significant figures in each reading, and only one in the difference. Exactly the exception we saw: nearly equal numbers, and the precision is gone. See how it all connects?

8:29Quick revision: sig figs for x and /, decimals for + and -

So, quick revision. Multiply or divide: keep the least number of significant figures. Add or subtract: keep the least number of decimal places. Powers and roots: keep the figures of the measured value, and round only at the end. And watch out when you subtract nearly equal numbers. Now go and try the practice problems on this page. You will find them easy, I promise. All the best!