 ###### Kendal Orenstein

Rutger's University
M.Ed., Columbia Teachers College

Kendal founded an academic coaching company in Washington D.C. and teaches in local area schools. In her spare time she loves to explore new places.

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# Solving Sig Figs Involving Multiplication and Division - Concept

Kendal Orenstein ###### Kendal Orenstein

Rutger's University
M.Ed., Columbia Teachers College

Kendal founded an academic coaching company in Washington D.C. and teaches in local area schools. In her spare time she loves to explore new places.

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All right, so let's talk about tips on using the correct number of Sig Figs during calculations involving multiplication and division. So you're back in the lab, and you have these two data points, and you need to multiply them together, to get other data points and to get other measurements, so you can manipulate the data, and get some information.

We've got one measurement that's 21.301g, and you've got another measurement that's 7.2g. You're asked to multiply that. You can put that in your calculator, and you're like I can multiply it and you get 153.362g. You take this to your teacher, and you're all excited. Your teacher looks at you, and says no, this is not correct. This is not significant enough. So now you have to go back, and do this again, and understand why it's not significant enough and give her the most significant answer that you can.

When you go back, we can do this by hand. Let me say 21.301g times 7.2g. We know that the last digit of every measurement is estimated. So this number is estimated, and this number is estimated. We also know that each answer must have only one estimated digit. We're only allowed to have one estimated or doubtful digit. You might hear your teacher say doubtful digit. Either way it's the same thing. Your answer is only allowed to have one. So let's figure out how we can do this.

Any time I work with an estimated digit, my answer is also estimated. So help me out as I multiply this by hand. 2 times 1 is 2. Since I used estimated digits, I'm going to circle it saying this is estimated as well. 2 times 0 is 0, using the estimated 2, so that's also estimated. 2 times 3 is 6, also estimated. 2 times 1 is 2, also estimated. 2 times 2 is 4, also estimated.

Now, going down to the next one. 7 times 1 is 7, also estimated because of the 1. 7 times 0 is 0. 7 times 3 is 21. 7 plus 2 is 9. 7 times 2 is 14. Add this up. Bring down the 2. Estimated 7 plus 0 is 7. Estimated 6 plus 0 is 6. Estimated 2 plus 1 is 3. Estimated 9 plus 4 is 13. Estimated, that 1 is also estimated. 1 plus 4 is 5 estimated and you bring down the 1. Oh my goodness all these numbers are estimated. We're only allowed to have 1. So I have to put a decimal place in. There's three numbers of the decimal in that problem,. So 1, 2, 3, 4 decimal place goes here. But all these numbers are not allowed.

So we can't take 15. It's 153.3672. We can't say 15, it's not even close. So how am I going to make this number one estimated digit? I'm going to change it to 150g is my correct number. Then you go to your teacher and he was like 150, and that's perfect. Your teacher will be very happy with that.

This is crazy. I'm not going to make you do everything like this to get the correct answer. There is a rule you can go by. Let me just explain the rule, and how the rule works.

When multiplying or dividing data, your answer must have the same number of significant digits; sig figs as your measurement with the least amount of sig figs. Looking back at our original data, this has 1, 2, 3, 4, 5 sig figs. Looking at this one, this has two sig figs. 2 is smaller than 5. So our answer must only have two sig figs. A better answer than 150 just getting this in scientific notation and say 1.5 times 10²g.

Hopefully, this helped you to understand why this rule works, and helps you figure out why your answer can't be something the calculator gives you all the time. Because you're smarter than the calculator. Anyway so hope that helped you understanding why we multiply, and divide using this rule.