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A person orders a 12-inch circular pizza online. The restaurant calls her back and says that they ran out of 12-inch pizzas and instead offers the fol
Question

A person orders a 12-inch circular pizza online. The restaurant calls her back and says that they ran out of 12-inch pizzas and instead offers the following choices in circular pizzas. Which of them gives the best value for her money?

A.

Six 4-inch pizzas

B.

Four 6-inch pizzas

C.

Seven 3-iinch pizzas

D.

Five 5-inch pizzas

Correct option is B

Given:
Original order: One 12-inch circular pizza
Restaurant offers alternate choices:
A. Six 4-inch pizzas
B. Four 6-inch pizzas
C. Seven 3-inch pizzas
D. Five 5-inch pizzas
All pizzas are circular

We are to determine which option gives the maximum total area, as the value for money is based on the amount of pizza (area).

Formula Used:
Area of a circle = Area of a circle: A=πr2=π(d2)2=πd24\text{Area of a circle: } A = \pi r^2 = \pi \left(\frac{d}{2}\right)^2 = \frac{\pi d^2}{4} \\

Solution:
Let’s calculate total area for each option using​​

Original 12-inch pizza:A=π×1224=144π4=36πOption A: Six 4-inch pizzasTotal Area=6×π×424=6×16π4=6×4π=24πOption B: Four 6-inch pizzasTotal Area=4×π×624=4×36π4=4×9π=36πOption C: Seven 3-inch pizzasTotal Area=7×π×324=7×9π4=63π4=15.75πOption D: Five 5-inch pizzasTotal Area=5×π×524=5×25π4=125π4=31.25πOption B gives the maximum area equal to the original: 36π\text{Original 12-inch pizza:} \\A = \frac{\pi \times 12^2}{4} = \frac{144\pi}{4} = 36\pi \\\text{Option A: Six 4-inch pizzas} \\\text{Total Area} = 6 \times \frac{\pi \times 4^2}{4} = 6 \times \frac{16\pi}{4} = 6 \times 4\pi = 24\pi \\\text{Option B: Four 6-inch pizzas} \\\text{Total Area} = 4 \times \frac{\pi \times 6^2}{4} = 4 \times \frac{36\pi}{4} = 4 \times 9\pi = 36\pi \\\text{Option C: Seven 3-inch pizzas} \\\text{Total Area} = 7 \times \frac{\pi \times 3^2}{4} = 7 \times \frac{9\pi}{4} = \frac{63\pi}{4} = 15.75\pi \\\text{Option D: Five 5-inch pizzas} \\\text{Total Area} = 5 \times \frac{\pi \times 5^2}{4} = 5 \times \frac{25\pi}{4} = \frac{125\pi}{4} = 31.25\pi \\\textbf{} \\\text{Option B gives the maximum area equal to the original: } \boxed{36\pi} \\​​

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