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17 Apr 2018, 05:24
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E
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
35%
(medium)
Question Stats:
64% (01:38) correct
36%
(01:55)
wrong
based on 214
sessions
History
Date
Time
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Not Attempted Yet
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. 33%
B. 44%
C. 56%
D. 67%
E. 78%
17 Apr 2018, 05:33
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Bunuel
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. 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.
17 Apr 2018, 09:42
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Bunuel
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 formulaA. 33%
B. 44%
C. 56%
D. 67%
E. 78%
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
Answer E
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
Answer E
