Difference between revisions of "2023 AMC 12A Problems/Problem 8"
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<cmath>\frac{ax+33}{a+3}=x+2\qquad (2)</cmath> | <cmath>\frac{ax+33}{a+3}=x+2\qquad (2)</cmath> | ||
− | Multiplying <math>(1)</math> by <math>(a+1)</math> and solving, we get: | + | Multiplying equation <math>(1)</math> by <math>(a+1)</math> and solving, we get: |
<cmath>ax+11=ax+a+x+1</cmath> | <cmath>ax+11=ax+a+x+1</cmath> | ||
<cmath>11=a+x+1</cmath> | <cmath>11=a+x+1</cmath> | ||
<cmath>a+x=10\qquad (3)</cmath> | <cmath>a+x=10\qquad (3)</cmath> | ||
− | Multiplying <math>(2)</math> by <math>(a+3)</math> and solving, we get: | + | Multiplying equation <math>(2)</math> by <math>(a+3)</math> and solving, we get: |
<cmath>ax+33=ax+2a+3x+6</cmath> | <cmath>ax+33=ax+2a+3x+6</cmath> | ||
<cmath>33=2a+3x+6</cmath> | <cmath>33=2a+3x+6</cmath> | ||
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~Dilip ~megaboy6679 (latex) | ~Dilip ~megaboy6679 (latex) | ||
− | ==Solution 3== | + | ==Solution 3 (do this if you are bored)== |
Let <math>s</math> represent the sum of Maureen's test scores previously and <math>t</math> be the number of scores taken previously. | Let <math>s</math> represent the sum of Maureen's test scores previously and <math>t</math> be the number of scores taken previously. | ||
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We can use the first equation to write <math>s</math> in terms of <math>t</math>. | We can use the first equation to write <math>s</math> in terms of <math>t</math>. | ||
− | We then substitute this into the second equation: <math>\frac{-t^2+10t+33}{t+3} = \frac{-t^2+ | + | We then substitute this into the second equation: <math>\frac{-t^2+10t+33}{t+3} = \frac{-t^2+10t}{t}+2</math> |
− | From here, we solve for t, | + | From here, we solve for t: multiply both sides by (<math>t</math>) and then (<math>t+3</math>), combining like terms to get <math>t^2-3t=0</math>. Factorize to get <math>t=0</math> or <math>t=3</math>, and therefore <math>t=3</math> (makes sense for the problem). |
We substitute this to get <math>s=21</math>. | We substitute this to get <math>s=21</math>. | ||
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Therefore, the solution to the problem is <math>\frac{21}{3}=</math> <math>\boxed{\textbf{(D) }7}</math> | Therefore, the solution to the problem is <math>\frac{21}{3}=</math> <math>\boxed{\textbf{(D) }7}</math> | ||
− | ~milquetoast | + | ~milquetoast ~the_eaglercraft_grinder |
==Solution 4 (Trial and Error)== | ==Solution 4 (Trial and Error)== | ||
Let's consider all the answer choices. If the average is <math>8</math>, then, we can assume that all her test choices were <math>8</math>. We can see that she must have gotten <math>8</math> twice, in order for another score of <math>11</math> to bring her average up by one. However, adding three <math>11</math>'s will not bring her score up to 10. Continuing this process for the answer choices, we see that the answer is <math>\boxed{\textbf{(D) }7}</math> | Let's consider all the answer choices. If the average is <math>8</math>, then, we can assume that all her test choices were <math>8</math>. We can see that she must have gotten <math>8</math> twice, in order for another score of <math>11</math> to bring her average up by one. However, adding three <math>11</math>'s will not bring her score up to 10. Continuing this process for the answer choices, we see that the answer is <math>\boxed{\textbf{(D) }7}</math> | ||
~andliu766 | ~andliu766 | ||
+ | |||
+ | ==Solution 5== | ||
+ | Let <math>n</math> be the number of existing quizzes. So after one more test, score <math>11</math> has <math>n+1</math> extra points to distribute to <math>n+1</math> quizzes. Also, after three more quizzes, there will be <math>3(n+1)</math> extra points to distribute to the <math>n+3</math> quizzes. So <math>3n+3=2(n+3)</math>. This means <math>n=3</math>. <math>n+1</math> extra points means original mean (average) is <math>7</math> | ||
+ | |||
+ | ~Pratima | ||
==Video Solution by Power Solve (easy to digest!)== | ==Video Solution by Power Solve (easy to digest!)== | ||
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~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com) | ~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com) | ||
+ | |||
Latest revision as of 19:15, 30 July 2024
- The following problem is from both the 2023 AMC 10A #10 and 2023 AMC 12A #8, so both problems redirect to this page.
Contents
- 1 Problem
- 2 Solution 1
- 3 Solution 2 (Variation on Solution 1)
- 4 Solution 3 (do this if you are bored)
- 5 Solution 4 (Trial and Error)
- 6 Solution 5
- 7 Video Solution by Power Solve (easy to digest!)
- 8 Video Solution (🚀 Just 3 min 🚀)
- 9 Video Solution by Math-X (First understand the problem!!!)
- 10 Video Solution by CosineMethod [🔥Fast and Easy🔥]
- 11 Video Solution
- 12 See Also
Problem
Maureen is keeping track of the mean of her quiz scores this semester. If Maureen scores an on the next quiz, her mean will increase by
. If she scores an
on each of the next three quizzes, her mean will increase by
. What is the mean of her quiz scores currently?
Solution 1
Let represent the amount of tests taken previously and
the mean of the scores taken previously.
We can write the following equations:
Multiplying equation by
and solving, we get:
Multiplying equation by
and solving, we get:
Solving the system of equations for and
, we find that
and
.
~walmartbrian ~Shontai ~andyluo ~megaboy6679
Solution 2 (Variation on Solution 1)
Suppose Maureen took tests with an average of
.
If she takes another test, her new average is
Cross-multiplying: , so
.
If she takes more tests, her new average is
Cross-multiplying: , so
.
But can also be written as
. Therefore
~Dilip ~megaboy6679 (latex)
Solution 3 (do this if you are bored)
Let represent the sum of Maureen's test scores previously and
be the number of scores taken previously.
So, and
We can use the first equation to write in terms of
.
We then substitute this into the second equation:
From here, we solve for t: multiply both sides by () and then (
), combining like terms to get
. Factorize to get
or
, and therefore
(makes sense for the problem).
We substitute this to get .
Therefore, the solution to the problem is
~milquetoast ~the_eaglercraft_grinder
Solution 4 (Trial and Error)
Let's consider all the answer choices. If the average is , then, we can assume that all her test choices were
. We can see that she must have gotten
twice, in order for another score of
to bring her average up by one. However, adding three
's will not bring her score up to 10. Continuing this process for the answer choices, we see that the answer is
~andliu766
Solution 5
Let be the number of existing quizzes. So after one more test, score
has
extra points to distribute to
quizzes. Also, after three more quizzes, there will be
extra points to distribute to the
quizzes. So
. This means
.
extra points means original mean (average) is
~Pratima
Video Solution by Power Solve (easy to digest!)
https://www.youtube.com/watch?v=ZUgo3-BKt30
Video Solution (🚀 Just 3 min 🚀)
~Education, the Study of Everything
Video Solution by Math-X (First understand the problem!!!)
https://youtu.be/cMgngeSmFCY?si=MHL95YihFdxKROrU&t=2280 ~Math-X
Video Solution by CosineMethod [🔥Fast and Easy🔥]
https://www.youtube.com/watch?v=1QJ_BQOfZtg
Video Solution
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
See Also
2023 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 9 |
Followed by Problem 11 | |
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 7 |
Followed by Problem 9 |
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.