carcass wrote:
On a certain day, a bakery produced a batch of rolls at a total production cost of $ 300. On that day, \(\frac{4}{5}\) of the rolls in the batch were sold, each at a price that was 50 percent greater than the average (arithmetic mean) production cost per roll. The remaining rolls in the batch were sold the next day, each at a price that was 20 percent less than the price of the day before. What was the bakery's profit on this batch of rolls?
A. $ 150
B. $ 144
C. $ 132
D. $ 108
E. $ 90
Method I: assign a value to the total number of rollsBecause all the rolls are sold, and because they are all sold at equal prices on their respective days, we can use any number of rolls we want to calculate
Profit as (Total Revenue - Total Cost).
The profit is the same for 300 rolls that cost $300 as it is for 30 rolls that cost $300.
Let # of rolls = 30
Cost per roll = $10
Day 1: \(\frac{4}{5}\) of 30, or 24 rolls, are sold at 50 percent above cost -->
Day 1, PRICE: 50% above cost is (1.5 * $10) = $15 each
Day 1, Total Revenue: (24 * $15) = $360
Day 2: remaining 6 rolls are sold at 80 percent of Day 1's price (of $15) -->
Day 2's price is (.8 * $15)= $12 each
Day 2, Total Revenue: (6 * $12) = $72
Total Revenue for both days: ($360 + $72) = $432
(TR - TC) = PROFIT
PROFIT = ($432 - $300) = $132
Answer C
Method II - Algebra\(xy = 300\) = Total cost, where \(x\) is unit cost of rolls and \(y\) is # of rolls
Day 1 = \(\frac{4}{5}y * \frac{3}{2}x = \frac{6}{5}xy\)
Day 2 = \(\frac{1}{5}y * (\frac{4}{5}*\frac{3}{2})x\)
\(= (\frac{1}{5}y*\frac{12}{10}x)=\frac{12}{50}xy=\frac{6}{25}xy\)
Day 1 + Day 2: \((\frac{6}{5}xy + \frac{6}{25}xy)= \frac{36}{25}xy\)
\(xy = 300\), so
Total revenue = \((\frac{36}{25})($300) = $432\)
Total cost = \($300\)
Profit = (TR - TC)
Profit = \($132\)
Answer C
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