SOLUTION: i was hoping you could give me the formulas to these 3 problems. 1) how many 4'x 8' sheets of drywall do you need for the walls of a 12' x 20' room, 8ft high? 2) if ceramic til

Algebra ->  Algebra  -> Customizable Word Problem Solvers  -> Unit conversion -> SOLUTION: i was hoping you could give me the formulas to these 3 problems. 1) how many 4'x 8' sheets of drywall do you need for the walls of a 12' x 20' room, 8ft high? 2) if ceramic til      Log On

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Question 206218: i was hoping you could give me the formulas to these 3 problems.
1) how many 4'x 8' sheets of drywall do you need for the walls of a 12' x 20' room, 8ft high?
2) if ceramic tiles are 8" x 8", how many tiles would it take to cover and area of 4' x 6'?
lastly,
3) if a customer buys 20 2x4's 8ft long, at $4.75 each,
how much would the 2x4's cost?
Answer by mickclns(59) About Me  (Show Source):
You can put this solution on YOUR website!
1) Since the room is 8 feet high, you can orient all of your sheets vertically. Each 12' side requires three 4' widths. Each 20' side requires five 4' widths. You need a total of 3 + 5 + 3 + 5 = 16 sheets.
2) First a reminder: to divide by a fraction, multiply by the reciprocal (the flip) of the fraction.
Since 8" = 2/3 foot, in the 4' direction you can fit 4 / (2/3) = 4 x (3/2) = 2 x 3 = 6 tiles. In the same way, in the 6' direction, you can fit 6 / (2/3) = 6 x (3/2) = 3 x 3 = 9 tiles. So, you can tile the area using 9 rows of 6 tiles or 54 tiles.
3) The 2x4 and 8ft numbers are smoke screens, that is, they provide confusion as to what numbers to use. You could equivalently say, "20 things cost $4.75 each, how much would the things cost?" The answer is the product 20 x 4.75. 75 cents is the same as 3 quarters of a dollar -- get it? Actually it's not a joke but a way to simplify multiplication. $4.75 is the same as $4 + $(3/4) so, using the distributive property, 20x$4.75 = 20x($4 + $(3/4) ) = 20x$4 + 20x$(3/4). 1/4 of 20 is 5, so 3/4 of 20 is 15, so $4.75x20 = $80 + $15 = $95 and that's your answer.
- Mick