Difference between revisions of "2023 AMC 12B Problems/Problem 18"
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<cmath>y_2 = y_1 + 18, \quad z_2 = z_1 + 18</cmath> | <cmath>y_2 = y_1 + 18, \quad z_2 = z_1 + 18</cmath> | ||
− | <cmath> y = \frac{ y_1 n_1 + y_2 n_2 }{ n_1 + n_2} = \frac{ y_1 n_1 + (y_1 + 18) n_2 }{ n_1 + n_2} = y_1 + \frac{18 | + | <cmath> y = \frac{ y_1 n_1 + y_2 n_2 }{ n_1 + n_2} = \frac{ y_1 n_1 + (y_1 + 18) n_2 }{ n_1 + n_2} = y_1 + \frac{18 n_2 }{ n_1 + n_2}</cmath> |
<cmath> z = \frac{ z_1 k_1 + z_2 k_2 }{ k_1 + k_2} = \frac{ z_1 k_1 + (z_1 + 18) k_2 }{ k_1 + k_2} = z_1 = \frac{ 18 k_2 }{ k_1 + k_2}</cmath> | <cmath> z = \frac{ z_1 k_1 + z_2 k_2 }{ k_1 + k_2} = \frac{ z_1 k_1 + (z_1 + 18) k_2 }{ k_1 + k_2} = z_1 = \frac{ 18 k_2 }{ k_1 + k_2}</cmath> | ||
− | <cmath>y - z = y_1 + \frac{18 | + | <cmath>y - z = y_1 + \frac{18 n_2 }{ n_1 + n_2} - z_1 - \frac{ 18 k_2 }{ k_1 + k_2} = y_1 - z_1 + 18 \left( \frac{n_2 }{ n_1 + n_2} - \frac{k_2 }{ k_1 + k_2} \right) = 3 + 18 \left( \frac{n_2 }{ n_1 + n_2} - \frac{k_2 }{ k_1 + k_2} \right)</cmath> |
− | <math>\frac{ | + | <math>\frac{n_2 }{ n_1 + n_2}</math> at most is <math>1</math>, <math>\frac{k_2 }{ k_1 + k_2}</math> is at least <math>0</math>, meaning that <math>\frac{n_2 }{ n_1 + n_2} - \frac{k_2 }{ k_1 + k_2}</math> is at most <math>1</math>. |
Therefore, <cmath>y - z \le 3 + 18 = 21</cmath> | Therefore, <cmath>y - z \le 3 + 18 = 21</cmath> | ||
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~[https://artofproblemsolving.com/wiki/index.php/User:Isabelchen isabelchen] | ~[https://artofproblemsolving.com/wiki/index.php/User:Isabelchen isabelchen] | ||
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==Solution 4== | ==Solution 4== | ||
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Denote <math>y_1</math> and <math>y_2</math> as the quiz averages of Yolanda in the <math>1</math>st and <math>2</math>nd semesters, respectively. Similarly, denote <math>z_1</math> and <math>z_2</math> as the quiz averages of Zelda in the <math>1</math>st and <math>2</math>nd semesters. | Denote <math>y_1</math> and <math>y_2</math> as the quiz averages of Yolanda in the <math>1</math>st and <math>2</math>nd semesters, respectively. Similarly, denote <math>z_1</math> and <math>z_2</math> as the quiz averages of Zelda in the <math>1</math>st and <math>2</math>nd semesters. | ||
− | + | We have <math>y_1 = z_1 + 3</math>, so <math>y_1 - 3 = z_1</math>. We also know that <math>y_2 = 18 +y_1 = 3 + z_2</math>, implying <math>z_2 = 15 + y_1</math>. | |
The average quiz scores for both students must lie between the averages of each semester, i.e <math>\mathrm{Avg_{y}}\in[y_1, 18+y_1],</math> and <math>\mathrm{Avg_{z}}\in[y_1-3, y_1+15].</math> Since <math>z_2>z_1</math>, <math>y_2 > y_1</math>, and <math>y_1>z_1</math>, we have <math>\min\{\mathrm{Avg}\} = \min\{\mathrm{Avg}_z\} = z_1</math> and <math>\max\{\mathrm{Avg}\} = \max\{\mathrm{Avg}_y\} = y_2</math>. Therefore the maximum difference between the two yearly averages is | The average quiz scores for both students must lie between the averages of each semester, i.e <math>\mathrm{Avg_{y}}\in[y_1, 18+y_1],</math> and <math>\mathrm{Avg_{z}}\in[y_1-3, y_1+15].</math> Since <math>z_2>z_1</math>, <math>y_2 > y_1</math>, and <math>y_1>z_1</math>, we have <math>\min\{\mathrm{Avg}\} = \min\{\mathrm{Avg}_z\} = z_1</math> and <math>\max\{\mathrm{Avg}\} = \max\{\mathrm{Avg}_y\} = y_2</math>. Therefore the maximum difference between the two yearly averages is | ||
− | <cmath>|y_2 - z_1| = |(y_1 + 18)-(y_1-3)| = 21 < 22 | + | <cmath>|y_2 - z_1| = |(y_1 + 18)-(y_1-3)| = 21 < 22.</cmath> |
− | + | Therefore, <math>\boxed{\textbf{(A)}~\text{Yolanda's quiz average for the academic year was 22 points higher than Zelda's}}</math> is not possible. | |
-Benedict T (countmath1) | -Benedict T (countmath1) |
Latest revision as of 20:27, 23 May 2024
Contents
Problem
Last academic year Yolanda and Zelda took different courses that did not necessarily administer the same number of quizzes during each of the two semesters. Yolanda's average on all the quizzes she took during the first semester was points higher than Zelda's average on all the quizzes she took during the first semester. Yolanda's average on all the quizzes she took during the second semester was
points higher than her average for the first semester and was again
points higher than Zelda's average on all the quizzes Zelda took during her second semester. Which one of the following statements cannot possibly be true?
Yolanda's quiz average for the academic year was
points higher than Zelda's.
Zelda's quiz average for the academic year was higher than Yolanda's.
Yolanda's quiz average for the academic year was
points higher than Zelda's.
Zelda's quiz average for the academic year equaled Yolanda's.
If Zelda had scored
points higher on each quiz she took, then she would have had the same average for the academic year as Yolanda.
Solution 1
Denote by the average of person with initial
in semester
Thus,
,
,
.
Denote by the average of person with initial
in the full year.
Thus,
can be any number in
and
can be any number in
.
Therefore, the impossible solution is
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
Solution 2
We can use process of elimination by finding possible solutions to answer choices. Let and
be the number of quizzes Yolanda took in the first and second semesters, respectively. Define
and
similarly for Zelda.
Answer choice B is satisfied by .
Answer choice C and E are both satisfied by .
Answer choice D is satisfied by .
Therefore the answer is
~cantalon
Solution 3 (Algebra)
Let Yolanda's average for semester be
, the number of quizzes Yolanda took in semester
be
, Zelda's average for semester
be
, the number of quizzes Zelda took in semester
be
, Yolanda's average for semester
be
, the number of quizzes Yolanda took in semester
be
, Zelda's average for semester
be
, the number of quizzes Zelda took in semester
be
, Yolanda's average for the entire year be
, Zelda's average for the entire year be
.
From the problem we know that
at most is
,
is at least
, meaning that
is at most
.
Therefore,
Hence, is not possible.
Solution 4
Denote and
as the quiz averages of Yolanda in the
st and
nd semesters, respectively. Similarly, denote
and
as the quiz averages of Zelda in the
st and
nd semesters.
We have , so
. We also know that
, implying
.
The average quiz scores for both students must lie between the averages of each semester, i.e and
Since
,
, and
, we have
and
. Therefore the maximum difference between the two yearly averages is
Therefore, is not possible.
-Benedict T (countmath1)
Video Solution 1 by OmegaLearn
Video Solution 2
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
See Also
2023 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 17 |
Followed by Problem 19 |
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.