Difference between revisions of "2023 AMC 12A Problems/Problem 2"
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==Solution 3== | ==Solution 3== | ||
− | + | <math>\frac{P}{3} + \frac{7}{2} R = \frac{3}{4} P + \frac{R}{2}</math> where <math>P</math> is the pizza weight and <math>R</math> is the weight of cup of oranges | |
− | Since oranges weigh 1/ | + | Since oranges weigh <math>\frac{1}{4}</math> pound per cup, the oranges on the LHS weigh <math>\frac{7}{2}</math> cups x <math>\frac{1}{4}</math> pounds/cup = <math>\frac{7}{8}</math> pound, |
− | and those on the RHS weigh 1/ | + | and those on the RHS weigh <math>\frac{1}{2}</math> cup x <math>\frac{1}{4}</math> pounds/cup = <math>\frac{1}{8}</math> pound. |
− | So P/ | + | So <math>\frac{P}{3}</math> + <math>\frac{7}{8}</math> pound = <math>\frac{3}{4} P</math> + <math>\frac{1}{8}</math> pound; <math>\frac{P}{3}</math> + <math>\frac{3}{4}</math> pound = <math>\frac{3}{4} P</math>. |
− | Multiplying both sides by | + | Multiplying both sides by <math>\text{lcm}(3,4) = 12</math>, we have |
− | 4P + 9 | + | <math>4P + 9 = 9P</math>; <math>5P = 9</math>; <math>P</math> = weight of a large pizza = <math>\frac{9}{5}</math> pounds = <math>\boxed{\textbf{(A)}1 \frac{4}{5}}</math> pounds. |
~Dilip | ~Dilip | ||
+ | ~<math>\LaTeX</math> by A_MatheMagician | ||
==Video Solution by Math-X (First understand the problem!!!)== | ==Video Solution by Math-X (First understand the problem!!!)== |
Revision as of 11:44, 11 November 2023
- The following problem is from both the 2023 AMC 10A #2 and 2023 AMC 12A #2, so both problems redirect to this page.
Contents
[hide]Problem
The weight of of a large pizza together with cups of orange slices is the same as the weight of of a large pizza together with cup of orange slices. A cup of orange slices weighs of a pound. What is the weight, in pounds, of a large pizza?
Solution 1 (Substitution)
Use a system of equations. Let be the weight of a pizza and be the weight of a cup of orange slices. We have Rearranging, we get Plugging in pounds for by the given gives
~ItsMeNoobieboy ~walmartbrian
Solution 2
Let: be the weight of a pizza. be the weight of a cup of orange.
From the problem, we know that .
Write the equation below:
Solving for :
~d_code
Solution 3
where is the pizza weight and is the weight of cup of oranges Since oranges weigh pound per cup, the oranges on the LHS weigh cups x pounds/cup = pound, and those on the RHS weigh cup x pounds/cup = pound.
So + pound = + pound; + pound = .
Multiplying both sides by , we have ; ; = weight of a large pizza = pounds = pounds.
~Dilip ~ by A_MatheMagician
Video Solution by Math-X (First understand the problem!!!)
https://youtu.be/cMgngeSmFCY?si=ULlMU09VdlpsRW3n&t=205
Video Solution
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
See Also
2023 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 1 |
Followed by Problem 3 | |
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 |
2023 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 1 |
Followed by Problem 3 |
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 12 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.