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## Arithmetic (all content)

### Unit 6: Lesson 14

Significant figures

# Multiplying and dividing with significant figures

Learn to multiply and divide with significant figures. Created by Sal Khan.

## Want to join the conversation?

• At about , why isn't 10.1 3 sig figs instead of 2?
Because the 0 is sandwiched, it is significant, right? • So if your measurements were 103.323 in by 233. in, what would the answer be (without division by 1.07)? • How do you know how many significant figures to round up to if you have a combination of both multiplication/division and addition/subtraction? • At into the video, the final answer calculated is 114 tiles but I think that 108 tiles are enough to do the job. Below is my solution. Why am I off by 6 tiles?

If you laid out 11 tiles in a row horizontally, that would span 11.77 feet. That means that 0.30 feet are needed for the last horizontal piece. Now, lay out nine rows, which span 9.63 feet, and the last row needs to span 0.47 feet. You have used up 99 tiles so far.

Now, take 3 tiles and cut them into 3 1.07 by 0.30 sections, use those to span the last column. Then, cut 5 tiles each into two 1.07 by 0.47 sections for the last row. Finally, for the last tile, cut it into one 1.07 by 0.47 section and one 1.07 by 0.30 section.

Total tiles used = 99 + 3 + 5 +1 = 108 tiles • I think you might be making a small mistake in the dimensions of your tile. A tile having an area of 1.07 square feet has dimensions of 1.07 feet x 1 foot, meaning that at any one time, it can cover 1.07 feet along one, not both, of length and breadth.

Thus, 9 rows of 11 tiles each may 11.77 feet lengthwise and thus not 9.63 but 9 feet breadthwise, leaving not 0.47 feet to be spanned but 1.1 feet. The left over area at the bottom would be 1.01 x 12.07 = 12.19 sqft. This extra space will mean that you will need 10, not 5 tiles for the last row accounting for 5 of those 6 missing tiles. The last tile is consumed by the 0.1 x 12.07 = 1.207 sqft area at the bottom, accounting for the last missing tile.

9 rows of 11 tiles can also span 9.63 feet breadthwise but only 11 feet lengthwise, thus spanning 0.77 feet less than predicted (leaving 1.07 feet). Thus, you will have to use, not 3 tiles split into threes as predicted, but 11 tiles for the 10.8 sqft space in the last column i.e. 8 tiles more than predicted. The 2 extra tiles in this account are probably due to the fact that I ignored the last tile at the bottom right corner.
• Hey there - can you please do a video on combined operations with significant figures? For example, how would you solve the following problem: (2.756 x 1.20) / (9.5 + 11.28)? Thanks! • Shouldn't the number of tiles be 113 instead of 114?
How can you round up the number of tiles fitting, the are of tiles fitting will be less than or equal to the area of the floor and cant be greater than the area of the floor. • • I think DarkFight is wrong there, at least as far as the 5-5.99 range. If we assume the 5s could have been rounded, then the values they were before hand would have to of been within 4.5 to 5.4 as those are the only values one step of percision greater that could have been rounded to 5. This means the true value could be anywhere from 20.3 to 29.2, which would be one reason why we might want to round to 30 if we could only have 1 significant digit as it is the nearest value of 10 which includes the entire range of potentially more accurate answers.
• I have a similar question as did tiagofischer: from the videos, I understood significant figures as a way to represent the most precise value for a calculation/answer. In the example for the tile-bathroom problem, Mr. Khan rounds up to 114 (). But then why not leave it as 113.9 because it would imply that we were able to measure to the near decimeter as we were when we measured the width of the bathroom (10.1 ft). Thank you and great video!   