Asked by Anonymous
The floor of a rectangular room is to be tiled with 1/3 (fraction) foot square tiles along a 10 5/8 (fraction) foot wall. How many tiles will be needed along the wall?
Would I multiply 10 5/8 times 1/3 to get the answer?
Would I multiply 10 5/8 times 1/3 to get the answer?
Answers
Answered by
jim
Let's estimate the problem first.
Handy to check our thinking.
There are 3 tiles per foot, right?
And there are a bit over 10 feet to tile, at 3 tiles per foot, so our answer is going to be near 30 tiles.
If you multiplied 1/3 by 10, you'd get about 3 as an answer, so something is going wrong with that.
Suppose it was a 9-foot wall, and you were using 3-foot tiles. How many 3-foot tiles would you need to cover 9 feet?
You need to divide the space into the width of the tile.
This shows that you don't want to multiply; you need to divide. And dividing by 1/3 is the same as multiplying by 3. Let's look at that:
10 5/8 = 85/8
(85/8) * 3 = 255/8 = 31 7/8
A possible little catch here is: can you order 7/8 of a tile? In a real life situation, you'd buy 32, and just throw away the last piece.
Handy to check our thinking.
There are 3 tiles per foot, right?
And there are a bit over 10 feet to tile, at 3 tiles per foot, so our answer is going to be near 30 tiles.
If you multiplied 1/3 by 10, you'd get about 3 as an answer, so something is going wrong with that.
Suppose it was a 9-foot wall, and you were using 3-foot tiles. How many 3-foot tiles would you need to cover 9 feet?
You need to divide the space into the width of the tile.
This shows that you don't want to multiply; you need to divide. And dividing by 1/3 is the same as multiplying by 3. Let's look at that:
10 5/8 = 85/8
(85/8) * 3 = 255/8 = 31 7/8
A possible little catch here is: can you order 7/8 of a tile? In a real life situation, you'd buy 32, and just throw away the last piece.
Answered by
Anonymous
That is the part I was missing. Forgetting you can multiply be three instead of divide by 1/3. You certainly make problems make sense, thank you for your help.
Answered by
jim
You're welcome!
Answered by
Bryan
thank you :)
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