Solved on Jan 25, 2024

a) How many $1$ -L paint cans needed to cover a $5.2 \times 5.2 \mathrm{~m}$ ceiling if $1$ can covers $11 \mathrm{~m}^{2}$ ? b) What are the max dimensions of a square ceiling that can be painted with $4 \mathrm{L}$ of paint if $1$ can covers $11 \mathrm{~m}^{2}$ ? (to nearest $0.1 \mathrm{~m}$ )

STEP 1

Assumptions
1. A 1-L can of paint covers $11 \mathrm{~m}^{2}$ .
2. The ceiling is square with sides of length $5.2 \mathrm{~m}$ .
3. For part b, we have $4 \mathrm{~L}$ of paint available.
4. We are looking for the maximum dimensions of a square ceiling that can be painted with $4 \mathrm{~L}$ of paint.

STEP 2

First, calculate the area of the ceiling that needs to be painted.
$Area = Side \times Side$

STEP 3

Plug in the given value for the side of the ceiling.
$Area = 5.2 \mathrm{~m} \times 5.2 \mathrm{~m}$

STEP 4

Calculate the area of the ceiling.
$Area = 5.2 \mathrm{~m} \times 5.2 \mathrm{~m} = 27.04 \mathrm{~m}^{2}$

STEP 5

Now, we need to determine how many 1-L cans of paint are required to cover the ceiling area.
$Number\, of\, cans = \frac{Ceiling\, area}{Area\, covered\, by\, one\, can}$

STEP 6

Plug in the values for the ceiling area and the area covered by one can.
$Number\, of\, cans = \frac{27.04 \mathrm{~m}^{2}}{11 \mathrm{~m}^{2}/\text{can}}$

STEP 7

Calculate the number of cans needed.
$Number\, of\, cans = \frac{27.04 \mathrm{~m}^{2}}{11 \mathrm{~m}^{2}/\text{can}} \approx 2.46 \text{ cans}$

STEP 8

Since we cannot buy a fraction of a can, we need to round up to the nearest whole number.
$Number\, of\, cans = \lceil 2.46 \rceil = 3 \text{ cans}$
For part a), you would need 3 cans of paint to cover the ceiling.

STEP 9

Now, for part b, we need to find the maximum dimensions of a square ceiling that can be painted with $4 \mathrm{~L}$ of paint.

STEP 10

First, calculate the total area that can be covered with $4 \mathrm{~L}$ of paint.
$Total\, area\, covered = Number\, of\, liters \times Area\, covered\, by\, one\, liter$

STEP 11

Plug in the given values.
$Total\, area\, covered = 4 \mathrm{~L} \times 11 \mathrm{~m}^{2}/\text{L}$

STEP 12

Calculate the total area that can be covered.
$Total\, area\, covered = 4 \mathrm{~L} \times 11 \mathrm{~m}^{2}/\text{L} = 44 \mathrm{~m}^{2}$

STEP 13

Since we are looking for the dimensions of a square ceiling, the side length can be found by taking the square root of the total area covered.
$Side\, length = \sqrt{Total\, area\, covered}$

STEP 14

Plug in the total area covered.
$Side\, length = \sqrt{44 \mathrm{~m}^{2}}$

STEP 15

Calculate the side length.
$Side\, length = \sqrt{44 \mathrm{~m}^{2}} \approx 6.633 \mathrm{~m}$

STEP 16

Round the side length to the nearest tenth of a meter.
$Side\, length \approx 6.6 \mathrm{~m}$
For part b), the maximum dimensions of a square ceiling you could paint with $4 \mathrm{~L}$ of paint is $6.6 \mathrm{~m}$ by $6.6 \mathrm{~m}$ to the nearest tenth of a meter.

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a) How many $1$ -L paint cans needed to cover a $5.2 \times 5.2 \mathrm{~m}$ ceiling if $1$ can covers $11 \mathrm{~m}^{2}$ ? b) What are the max dimensions of a square ceiling that can be painted with $4 \mathrm{L}$ of paint if $1$ can covers $11 \mathrm{~m}^{2}$ ? (to nearest $0.1 \mathrm{~m}$ )

STEP 1

STEP 2

First, calculate the area of the ceiling that needs to be painted.
$Area = Side \times Side$

STEP 3

Plug in the given value for the side of the ceiling.
$Area = 5.2 \mathrm{~m} \times 5.2 \mathrm{~m}$

STEP 4

Calculate the area of the ceiling.
$Area = 5.2 \mathrm{~m} \times 5.2 \mathrm{~m} = 27.04 \mathrm{~m}^{2}$

STEP 5

Now, we need to determine how many 1-L cans of paint are required to cover the ceiling area.
$Number\, of\, cans = \frac{Ceiling\, area}{Area\, covered\, by\, one\, can}$

STEP 6

Plug in the values for the ceiling area and the area covered by one can.
$Number\, of\, cans = \frac{27.04 \mathrm{~m}^{2}}{11 \mathrm{~m}^{2}/\text{can}}$

STEP 7

Calculate the number of cans needed.
$Number\, of\, cans = \frac{27.04 \mathrm{~m}^{2}}{11 \mathrm{~m}^{2}/\text{can}} \approx 2.46 \text{ cans}$

STEP 8

Since we cannot buy a fraction of a can, we need to round up to the nearest whole number.
$Number\, of\, cans = \lceil 2.46 \rceil = 3 \text{ cans}$
For part a), you would need 3 cans of paint to cover the ceiling.

STEP 9

Now, for part b, we need to find the maximum dimensions of a square ceiling that can be painted with $4 \mathrm{~L}$ of paint.

STEP 10

First, calculate the total area that can be covered with $4 \mathrm{~L}$ of paint.
$Total\, area\, covered = Number\, of\, liters \times Area\, covered\, by\, one\, liter$

STEP 11

Plug in the given values.
$Total\, area\, covered = 4 \mathrm{~L} \times 11 \mathrm{~m}^{2}/\text{L}$

STEP 12

Calculate the total area that can be covered.
$Total\, area\, covered = 4 \mathrm{~L} \times 11 \mathrm{~m}^{2}/\text{L} = 44 \mathrm{~m}^{2}$

STEP 13

Since we are looking for the dimensions of a square ceiling, the side length can be found by taking the square root of the total area covered.
$Side\, length = \sqrt{Total\, area\, covered}$

STEP 14

Plug in the total area covered.
$Side\, length = \sqrt{44 \mathrm{~m}^{2}}$

STEP 15

Calculate the side length.
$Side\, length = \sqrt{44 \mathrm{~m}^{2}} \approx 6.633 \mathrm{~m}$

STEP 16

Round the side length to the nearest tenth of a meter.
$Side\, length \approx 6.6 \mathrm{~m}$
For part b), the maximum dimensions of a square ceiling you could paint with $4 \mathrm{~L}$ of paint is $6.6 \mathrm{~m}$ by $6.6 \mathrm{~m}$ to the nearest tenth of a meter.

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STEP 1

STEP 2

First, calculate the area of the ceiling that needs to be painted.Area=Side×SideArea = Side \times SideArea=Side×Side

STEP 3

Plug in the given value for the side of the ceiling.Area=5.2 m×5.2 mArea = 5.2 \mathrm{~m} \times 5.2 \mathrm{~m}Area=5.2 m×5.2 m

STEP 4

Calculate the area of the ceiling.Area=5.2 m×5.2 m=27.04 m2Area = 5.2 \mathrm{~m} \times 5.2 \mathrm{~m} = 27.04 \mathrm{~m}^{2}Area=5.2 m×5.2 m=27.04 m2

STEP 5

Now, we need to determine how many 1-L cans of paint are required to cover the ceiling area.Number of cans=Ceiling areaArea covered by one canNumber\, of\, cans = \frac{Ceiling\, area}{Area\, covered\, by\, one\, can}Numberofcans=AreacoveredbyonecanCeilingarea​

STEP 6

Plug in the values for the ceiling area and the area covered by one can.Number of cans=27.04 m211 m2/canNumber\, of\, cans = \frac{27.04 \mathrm{~m}^{2}}{11 \mathrm{~m}^{2}/\text{can}}Numberofcans=11 m2/can27.04 m2​

STEP 7

Calculate the number of cans needed.Number of cans=27.04 m211 m2/can≈2.46 cansNumber\, of\, cans = \frac{27.04 \mathrm{~m}^{2}}{11 \mathrm{~m}^{2}/\text{can}} \approx 2.46 \text{ cans}Numberofcans=11 m2/can27.04 m2​≈2.46 cans

STEP 8

Since we cannot buy a fraction of a can, we need to round up to the nearest whole number.Number of cans=⌈2.46⌉=3 cansNumber\, of\, cans = \lceil 2.46 \rceil = 3 \text{ cans}Numberofcans=⌈2.46⌉=3 cansFor part a), you would need 3 cans of paint to cover the ceiling.

STEP 9

Now, for part b, we need to find the maximum dimensions of a square ceiling that can be painted with 4 L4 \mathrm{~L}4 L of paint.

STEP 10

First, calculate the total area that can be covered with 4 L4 \mathrm{~L}4 L of paint.Total area covered=Number of liters×Area covered by one literTotal\, area\, covered = Number\, of\, liters \times Area\, covered\, by\, one\, literTotalareacovered=Numberofliters×Areacoveredbyoneliter

STEP 11

Plug in the given values.Total area covered=4 L×11 m2/LTotal\, area\, covered = 4 \mathrm{~L} \times 11 \mathrm{~m}^{2}/\text{L}Totalareacovered=4 L×11 m2/L

STEP 12

Calculate the total area that can be covered.Total area covered=4 L×11 m2/L=44 m2Total\, area\, covered = 4 \mathrm{~L} \times 11 \mathrm{~m}^{2}/\text{L} = 44 \mathrm{~m}^{2}Totalareacovered=4 L×11 m2/L=44 m2

STEP 13

Since we are looking for the dimensions of a square ceiling, the side length can be found by taking the square root of the total area covered.Side length=Total area coveredSide\, length = \sqrt{Total\, area\, covered}Sidelength=Totalareacovered​

STEP 14

Plug in the total area covered.Side length=44 m2Side\, length = \sqrt{44 \mathrm{~m}^{2}}Sidelength=44 m2​

STEP 15

Calculate the side length.Side length=44 m2≈6.633 mSide\, length = \sqrt{44 \mathrm{~m}^{2}} \approx 6.633 \mathrm{~m}Sidelength=44 m2​≈6.633 m

STEP 16

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First, calculate the area of the ceiling that needs to be painted.
$Area = Side \times Side$

Plug in the given value for the side of the ceiling.
$Area = 5.2 \mathrm{~m} \times 5.2 \mathrm{~m}$

Calculate the area of the ceiling.
$Area = 5.2 \mathrm{~m} \times 5.2 \mathrm{~m} = 27.04 \mathrm{~m}^{2}$

Now, we need to determine how many 1-L cans of paint are required to cover the ceiling area.
$Number\, of\, cans = \frac{Ceiling\, area}{Area\, covered\, by\, one\, can}$

Plug in the values for the ceiling area and the area covered by one can.
$Number\, of\, cans = \frac{27.04 \mathrm{~m}^{2}}{11 \mathrm{~m}^{2}/\text{can}}$

Calculate the number of cans needed.
$Number\, of\, cans = \frac{27.04 \mathrm{~m}^{2}}{11 \mathrm{~m}^{2}/\text{can}} \approx 2.46 \text{ cans}$

Since we cannot buy a fraction of a can, we need to round up to the nearest whole number.
$Number\, of\, cans = \lceil 2.46 \rceil = 3 \text{ cans}$
For part a), you would need 3 cans of paint to cover the ceiling.

Now, for part b, we need to find the maximum dimensions of a square ceiling that can be painted with $4 \mathrm{~L}$ of paint.

First, calculate the total area that can be covered with $4 \mathrm{~L}$ of paint.
$Total\, area\, covered = Number\, of\, liters \times Area\, covered\, by\, one\, liter$

Plug in the given values.
$Total\, area\, covered = 4 \mathrm{~L} \times 11 \mathrm{~m}^{2}/\text{L}$

Calculate the total area that can be covered.
$Total\, area\, covered = 4 \mathrm{~L} \times 11 \mathrm{~m}^{2}/\text{L} = 44 \mathrm{~m}^{2}$

Since we are looking for the dimensions of a square ceiling, the side length can be found by taking the square root of the total area covered.
$Side\, length = \sqrt{Total\, area\, covered}$

Plug in the total area covered.
$Side\, length = \sqrt{44 \mathrm{~m}^{2}}$

Calculate the side length.
$Side\, length = \sqrt{44 \mathrm{~m}^{2}} \approx 6.633 \mathrm{~m}$