Difference between revisions of "2023 AMC 12A Problems/Problem 2"
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{{duplicate|[[2023 AMC 10A Problems/Problem 2|2023 AMC 10A #2]] and [[2023 AMC 12A Problems/Problem 2|2023 AMC 12A #2]]}} | {{duplicate|[[2023 AMC 10A Problems/Problem 2|2023 AMC 10A #2]] and [[2023 AMC 12A Problems/Problem 2|2023 AMC 12A #2]]}} | ||
− | ==Problem== | + | ==Problem 2== |
The weight of <math>\frac{1}{3}</math> of a large pizza together with <math>3 \frac{1}{2}</math> cups of orange slices is the same as the weight of <math>\frac{3}{4}</math> of a large pizza together with <math>\frac{1}{2}</math> cup of orange slices. A cup of orange slices weighs <math>\frac{1}{4}</math> of a pound. What is the weight, in pounds, of a large pizza? | The weight of <math>\frac{1}{3}</math> of a large pizza together with <math>3 \frac{1}{2}</math> cups of orange slices is the same as the weight of <math>\frac{3}{4}</math> of a large pizza together with <math>\frac{1}{2}</math> cup of orange slices. A cup of orange slices weighs <math>\frac{1}{4}</math> of a pound. What is the weight, in pounds, of a large pizza? | ||
<math>\textbf{(A) }1\frac{4}{5}\qquad\textbf{(B) }2\qquad\textbf{(C) }2\frac{2}{5}\qquad\textbf{(D) }3\qquad\textbf{(E) }3\frac{3}{5}</math> | <math>\textbf{(A) }1\frac{4}{5}\qquad\textbf{(B) }2\qquad\textbf{(C) }2\frac{2}{5}\qquad\textbf{(D) }3\qquad\textbf{(E) }3\frac{3}{5}</math> | ||
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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 (easy to digest) by Power Solve== | ||
+ | https://www.youtube.com/watch?v=8huvzWTtgaU | ||
+ | |||
+ | ==Video Solution (🚀 Just 1 min 🚀)== | ||
+ | https://youtu.be/7ADHHpSNMsE | ||
+ | |||
+ | ~Education, the Study of Everything | ||
==Video Solution by Math-X (First understand the problem!!!)== | ==Video Solution by Math-X (First understand the problem!!!)== | ||
https://youtu.be/cMgngeSmFCY?si=ULlMU09VdlpsRW3n&t=205 | https://youtu.be/cMgngeSmFCY?si=ULlMU09VdlpsRW3n&t=205 | ||
+ | |||
+ | == Video Solution by CosineMethod [🔥Fast and Easy🔥]== | ||
+ | |||
+ | https://www.youtube.com/watch?v=CpboCxGBcWY | ||
==Video Solution== | ==Video Solution== | ||
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~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com) | ~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com) | ||
+ | |||
+ | |||
+ | ==Video Solution (Easy to Understand) by DR.GOOGLE (YT: Pablo's Math)== | ||
+ | https://youtu.be/N5cHw8ODT_I | ||
==See Also== | ==See Also== |
Latest revision as of 07:36, 26 August 2024
- The following problem is from both the 2023 AMC 10A #2 and 2023 AMC 12A #2, so both problems redirect to this page.
Contents
- 1 Problem 2
- 2 Solution 1 (Substitution)
- 3 Solution 2
- 4 Solution 3
- 5 Video Solution (easy to digest) by Power Solve
- 6 Video Solution (🚀 Just 1 min 🚀)
- 7 Video Solution by Math-X (First understand the problem!!!)
- 8 Video Solution by CosineMethod [🔥Fast and Easy🔥]
- 9 Video Solution
- 10 Video Solution (Easy to Understand) by DR.GOOGLE (YT: Pablo's Math)
- 11 See Also
Problem 2
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 (easy to digest) by Power Solve
https://www.youtube.com/watch?v=8huvzWTtgaU
Video Solution (🚀 Just 1 min 🚀)
~Education, the Study of Everything
Video Solution by Math-X (First understand the problem!!!)
https://youtu.be/cMgngeSmFCY?si=ULlMU09VdlpsRW3n&t=205
Video Solution by CosineMethod [🔥Fast and Easy🔥]
https://www.youtube.com/watch?v=CpboCxGBcWY
Video Solution
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
Video Solution (Easy to Understand) by DR.GOOGLE (YT: Pablo's Math)
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.