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A soft-drink producer has done marketing research and found that if it decreases the width of its soda a cans (right

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by BTGmoderatorDC » Tue Feb 02, 2021 6:17 pm
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A soft-drink producer has done marketing research and found that if it decreases the width of its soda a cans (right circular cylinders) by 50%, but keep the height the same, they can still sell the cans for 75% of the original price. By what percent does the price per volume of soda increase to the consumer?

(A) 25%
(B) 50%
(C) 100%
(D) 200%
(E) 300%


OA D

Source: Veritas Prep
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Source: — Problem Solving |

BTGmoderatorDC wrote: ↑
Tue Feb 02, 2021 6:17 pm
A soft-drink producer has done marketing research and found that if it decreases the width of its soda a cans (right circular cylinders) by 50%, but keep the height the same, they can still sell the cans for 75% of the original price. By what percent does the price per volume of soda increase to the consumer?

(A) 25%
(B) 50%
(C) 100%
(D) 200%
(E) 300%


OA D

Source: Veritas Prep
Imagine that the initial can was \(100ml\) \((\pi r^2 h)\), and its price was \(\$100\)

After the proposed reduction in size, the radius will be halved, and the new volume \(=\dfrac{\pi r^2 h}{4}\), which means that the new size \(=\$25\)

for \(25ml,\) the price should be \(\$25,\) however the research proposed \(75\%\) of the original price (\(\$100\)), so the new proposed price \(= \$75\)

This means that the customers will pay \(\$3\) times the original price, i.e \(= 200\%\)
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BTGmoderatorDC wrote: ↑
Tue Feb 02, 2021 6:17 pm
A soft-drink producer has done marketing research and found that if it decreases the width of its soda a cans (right circular cylinders) by 50%, but keep the height the same, they can still sell the cans for 75% of the original price. By what percent does the price per volume of soda increase to the consumer?

(A) 25%
(B) 50%
(C) 100%
(D) 200%
(E) 300%


OA D

Solution:

We can let the original diameter of the circular base of the can (i.e., the width of the can) be 4 (so the radius will be 2) and the height be 10. Therefore, the volume of the original can is V = πr^2h = π x (2^2) x 10 = 40π. We can let the price of this original volume of the can be 40π dollars. Therefore, the original price per volume is 40π/40π = $1.

Now the width of the cans is reduced by 50%, so the new radius is 1. Since the height is the same, the volume of the new can is π x (1^2) x 10 = 10π. Since the new price is 75% of the original price, the new price is 30π dollars, and therefore, the new price per volume is 30π/10π = $3.

Since the original price per volume is $1 and the new price per volume is $3, the increase is $2, which is a 200% increase of the original price.

Answer: D

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