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A pizzeria makes pizzas that are shaped as perfect circles, and measur

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A pizzeria makes pizzas that are shaped as perfect circles, and measur  [#permalink]

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17 Apr 2018, 05:24
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45% (medium)

Question Stats:

61% (01:37) correct 39% (01:56) wrong based on 109 sessions

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A pizzeria makes pizzas that are shaped as perfect circles, and measures pizza size by the diameter of a pizza. By surface area, approximately what percent larger is a 16-inch pizza than a 12-inch pizza?

A. 33%
B. 44%
C. 56%
D. 67%
E. 78%

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Re: A pizzeria makes pizzas that are shaped as perfect circles, and measur  [#permalink]

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17 Apr 2018, 05:33
2
Bunuel wrote:
A pizzeria makes pizzas that are shaped as perfect circles, and measures pizza size by the diameter of a pizza. By surface area, approximately what percent larger is a 16-inch pizza than a 12-inch pizza?

A. 33%
B. 44%
C. 56%
D. 67%
E. 78%

Area of 16-inch pizza = pi*8^2 = 64pi

Area of 12-inch- pizza = pi*6^2 = 36pi

16-inch pizza is larger than 12-inch pizza by (64pi-36pi)/36pi = 28pi/36pi = 28/36 = 7/9 = 77.7777% = Approximately 78%

Hence option E = 78% is the answer.
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A pizzeria makes pizzas that are shaped as perfect circles, and measur  [#permalink]

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17 Apr 2018, 09:42
Bunuel wrote:
A pizzeria makes pizzas that are shaped as perfect circles, and measures pizza size by the diameter of a pizza. By surface area, approximately what percent larger is a 16-inch pizza than a 12-inch pizza?

A. 33%
B. 44%
C. 56%
D. 67%
E. 78%

Percent greater than formula

Percent greater than:
$$(\frac{New-Old}{Old}*100)$$ OR $$(\frac{Change}{Original}*100)$$

The 16-inch pizza has $$r=8$$ and
Area of $$\pi r^2=64\pi$$

The 12-inch pizza has $$r=6$$ and
Area $$=\pi r^2=36\pi$$

Area of 16-inch pizza is what percent larger than area of 12-inch pizza?

$$(\frac{64\pi - 36\pi}{36\pi}*100)=(\frac{28\pi}{36\pi}*100)=$$

$$(\frac{7}{9}*100)\approx{.777}*100 \approx{78}$$
percent

Scale factor, k, squared - Typically quicker

If a geometric shape changes by a certain percent or fraction, the multiplier for the percent change is the scale factor, $$k$$

If you find scale factor, you find percent change

Scale factor affects length.
Area = length * length, so
"Percent change" of an increased area =
(scale factor*scale factor) = $$k^2$$

Length = radius = $$r$$
$$k=\frac{r_{2}}{r_{1}}= \frac{8}{6}=\frac{4}{3}$$

Percent change in area here = $$k^2= (\frac{4}{3})^2=\frac{16}{9}\approx{1.7777}\approx{1.78}$$

New area is
$$(1.78-1)=.78=78$$ percent greater than smaller area

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Re: A pizzeria makes pizzas that are shaped as perfect circles, and measur  [#permalink]

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19 Apr 2018, 16:52
Bunuel wrote:
A pizzeria makes pizzas that are shaped as perfect circles, and measures pizza size by the diameter of a pizza. By surface area, approximately what percent larger is a 16-inch pizza than a 12-inch pizza?

A. 33%
B. 44%
C. 56%
D. 67%
E. 78%

A 16-inch pizza has a surface area of π x 8^2 = 64π.

A 12-inch pizza has a surface area of π x 6^2 = 36π.

Now we can determine what percent larger the 16-inch pizza is versus the 12-inch pizza.

(64π - 36π)/36π x 100 = 28/36 x 100 = 7/9 x 100 ≈ 78%.

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Re: A pizzeria makes pizzas that are shaped as perfect circles, and measur  [#permalink]

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29 Nov 2019, 05:35
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Re: A pizzeria makes pizzas that are shaped as perfect circles, and measur   [#permalink] 29 Nov 2019, 05:35
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