Difference between revisions of "2023 AMC 12A Problems/Problem 2"
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==Video Solution by Math-X (First understand the problem!!!)== | ==Video Solution by Math-X (First understand the problem!!!)== | ||
− | https://youtu.be/cMgngeSmFCY?si= | + | https://youtu.be/cMgngeSmFCY?si=ULlMU09VdlpsRW3n&t=205 |
==See Also== | ==See Also== |
Revision as of 14:53, 10 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
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
gives
~ItsMeNoobieboy
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
P/3 + 7/2 R = 3/4 P + R/2 where P = pizza weight and R = weight of cup of oranges
Since oranges weigh 1/4 pound per cup, the oranges on the LHS weigh 7/2 cups x 1/4 pounds/cup = 7/8 pound, and those on the RHS weigh 1/2 cup x 1/4 pounds/cup = 1/8 pound.
So P/3 + 7/8 pound = 3/4 P + 1/8 pound; P/3 + 3/4 pound = 3/4 P.
Multiplying both sides by LCM (3,4) = 12, we have
4P + 9# = 9P; 5P = 9#; P = weight of a large pizza = 9/5 pounds = pounds.
~Dilip
Video Solution by Math-X (First understand the problem!!!)
https://youtu.be/cMgngeSmFCY?si=ULlMU09VdlpsRW3n&t=205
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.