Difference between revisions of "2014 AMC 10A Problems/Problem 3"
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<math>\dfrac{2}{3}\times 24=16</math> The new price will be <math>\dfrac{1}{2}\textdollar2.50=\textdollar1.25</math>. So <math>\textdollar1.25\times16=\textdollar20.00</math>. This leaves 8 loaves remaining. | <math>\dfrac{2}{3}\times 24=16</math> The new price will be <math>\dfrac{1}{2}\textdollar2.50=\textdollar1.25</math>. So <math>\textdollar1.25\times16=\textdollar20.00</math>. This leaves 8 loaves remaining. | ||
− | We have <math>1\ | + | We have <math>\textdollar 1\times 8=\textdollar 8.00</math>. |
The total amount of money she made for the day is the sum of these amounts, which is <math>60+20+8=\textdollar 88</math>. | The total amount of money she made for the day is the sum of these amounts, which is <math>60+20+8=\textdollar 88</math>. |
Revision as of 20:05, 7 February 2014
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?
$\textbf{(A)}\ 24\qquad\textbf{(B)}\ 36\qquad\textbf{(C)}\ 44\qquad\textbf{(D)}}\ 48\qquad\textbf{(E)}\ 52$ (Error compiling LaTeX. ! Extra }, or forgotten $.)
Solution
of the bread is loaves. . This leaves loaves left.
The new price will be . So . This leaves 8 loaves remaining.
We have .
The total amount of money she made for the day is the sum of these amounts, which is .
The total amount it cost her to make all of the loaves is .
Therefore, her total profit is the amount of money she spent subtracted from the amount of money she made. $88-36=52\implies\boxed{\textbf{(E)}52}}$ (Error compiling LaTeX. ! Extra }, or forgotten $.).
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.