2014 AMC 10A Problems/Problem 3
Contents
Problem
Bridget bakes 48 loaves of bread for her bakery. She sells half of them in the morning for each. In the afternoon she sells two thirds of what she has left, and because they are not fresh, she charges only half price. In the late afternoon she sells the remaining loaves at a dollar each. Each loaf costs for her to make. In dollars, what is her profit for the day?
Solution
She first sells one-half of her loaves, or loaves. Each loaf sells for , so her total earnings in the morning is equal to
This leaves 24 loaves left, and Bridget will sell of them for a price of . Thus, her total earnings for the afternoon is
Finally, Bridget will sell the remaining loaves for a dollar each. This is a total of
The total amount of money she makes is equal to .
However, since Bridget spends making each loaf of bread, the total cost to make the bread is equal to .
Her total profit is the amount of money she spent subtracted from the amount of money she made, which is
Solution 2
As we know inherently from the information provided above that B bakes 48 loaves. We know that she sells \frac{1}{2} of the loaves for $2.50 each, and she sells half of them, so we multiply 24 by $2.50, which equals $60. For the two thirds left, or 16 loaves remaining, she sells them for $1.25, so we multiply $1.25 by 16, which equals $20. Finally, she sells the remaining, she sells the loaves at $1 dollar each, for the 8 remaining loaves. We multiply $1 by 8 to get $8, so in total, she makes $88. However, we have to consider the fact that it takes her $0.75 to make a loaf. We multiply that number by the total number of loaves, which is 48, so $0.75 \times 48 = $36. Subtract $36 from $88, we get the final answer of $52 profit.
~Horizon
Video Solution (CREATIVE THINKING)
https://youtu.be/Lr5GAX8vFcU
~Education, the Study of Everything
Video Solution
~savannahsolver
See Also
2014 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 2 |
Followed by Problem 4 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
All AMC 10 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.