Difference between revisions of "2022 AMC 8 Problems/Problem 15"
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label(scale(0.7)*"$4$", (-0.3,4), black); | label(scale(0.7)*"$4$", (-0.3,4), black); | ||
label(scale(0.7)*"$5$", (-0.3,5), black); | label(scale(0.7)*"$5$", (-0.3,5), black); | ||
− | label(scale(0. | + | label(scale(0.8)*rotate(90)*"Price (dollars)", (-1,3.2), black); |
− | label(scale(0. | + | label(scale(0.8)*"Weight (ounces)", (3.2,-1), black); |
dot((1,1.2),black); | dot((1,1.2),black); | ||
dot((1,1.7),black); | dot((1,1.7),black); | ||
Line 81: | Line 81: | ||
label(scale(0.7)*"$4$", (-0.3,4), black); | label(scale(0.7)*"$4$", (-0.3,4), black); | ||
label(scale(0.7)*"$5$", (-0.3,5), black); | label(scale(0.7)*"$5$", (-0.3,5), black); | ||
− | label(scale(0. | + | label(scale(0.8)*rotate(90)*"Price (dollars)", (-1,3.2), black); |
− | label(scale(0. | + | label(scale(0.8)*"Weight (ounces)", (3.2,-1), black); |
draw((0,0)--(6,5),red); | draw((0,0)--(6,5),red); | ||
Line 125: | Line 125: | ||
</asy> | </asy> | ||
− | We are looking for a black point, that when connected to the origin, yields the lowest slope. The slope represents the price per ounce. | + | We are looking for a black point, that when connected to the origin, yields the lowest slope. The slope represents the price per ounce. We can visually find that the point with the lowest slope is the blue point. Furthermore, it is the only one with a price per ounce significantly less than <math>1</math>. Finally, we see that the blue point is in the category with a weight of <math>\boxed{\textbf{(C) } 3}</math> ounces. |
− | + | ~MathFun1000 | |
− | By answer | + | |
+ | ==Solution 2 (Elimination)== | ||
+ | By the answer choices, we can disregard the points that do not have integer weights. As a result, we obtain the following diagram: | ||
<asy> | <asy> | ||
Line 146: | Line 148: | ||
label(scale(0.7)*"$4$", (-0.3,4), black); | label(scale(0.7)*"$4$", (-0.3,4), black); | ||
label(scale(0.7)*"$5$", (-0.3,5), black); | label(scale(0.7)*"$5$", (-0.3,5), black); | ||
− | label(scale(0. | + | label(scale(0.8)*rotate(90)*"Price (dollars)", (-1,3.2), black); |
− | label(scale(0. | + | label(scale(0.8)*"Weight (ounces)", (3.2,-1), black); |
dot((1,1.2),black); | dot((1,1.2),black); | ||
dot((1,1.7),black); | dot((1,1.7),black); | ||
Line 171: | Line 173: | ||
</asy> | </asy> | ||
− | ~ | + | We then proceed in the same way that we had done in Solution 1. Following the steps, we figure out the blue dot that yields the lowest slope, along with passing the origin. We then can look at the x-axis(in this situation, the weight) and figure out it has <math>\boxed{\textbf{(C) } 3}</math> ounces. |
+ | |||
+ | ~DairyQueenXD (edited by HW73) | ||
+ | |||
+ | ==Solution 3 (Elimination)== | ||
+ | We can find the lowest point in each line (<math>1</math>, <math>2</math>, <math>3</math>, <math>4</math>, or <math>5</math>) and find the price per pound. (Note that we don't need to find the points higher than the points below since we are finding the lowest price per pound.) | ||
+ | |||
+ | <asy> | ||
+ | //diagram by pog | ||
+ | size(5.5cm); | ||
+ | usepackage("mathptmx"); | ||
+ | defaultpen(mediumgray*0.5+gray*0.5+linewidth(0.63)); | ||
+ | add(grid(6,6)); | ||
+ | label(scale(0.7)*"$1$", (1,-0.3), black); | ||
+ | label(scale(0.7)*"$2$", (2,-0.3), black); | ||
+ | label(scale(0.7)*"$3$", (3,-0.3), black); | ||
+ | label(scale(0.7)*"$4$", (4,-0.3), black); | ||
+ | label(scale(0.7)*"$5$", (5,-0.3), black); | ||
+ | label(scale(0.7)*"$1$", (-0.3,1), black); | ||
+ | label(scale(0.7)*"$2$", (-0.3,2), black); | ||
+ | label(scale(0.7)*"$3$", (-0.3,3), black); | ||
+ | label(scale(0.7)*"$4$", (-0.3,4), black); | ||
+ | label(scale(0.7)*"$5$", (-0.3,5), black); | ||
+ | label(scale(0.8)*rotate(90)*"Price (dollars)", (-1,3.2), black); | ||
+ | label(scale(0.8)*"Weight (ounces)", (3.2,-1), black); | ||
+ | dot((1,1.2),red); | ||
+ | dot((1,1.7),black); | ||
+ | dot((1,2),black); | ||
+ | dot((1,2.8),black); | ||
+ | |||
+ | dot((2,2),green); | ||
+ | dot((2,2.9),black); | ||
+ | dot((2,3),black); | ||
+ | dot((2,4),black); | ||
+ | dot((2,4.35),black); | ||
+ | dot((2,4.8),black); | ||
+ | |||
+ | dot((3,2.5),blue); | ||
+ | dot((3,3.4),black); | ||
+ | dot((3,4.2),black); | ||
+ | |||
+ | dot((4,3.9),orange); | ||
+ | dot((4,5.1),black); | ||
+ | |||
+ | dot((5,4.5),purple); | ||
+ | dot((5,5),black); | ||
+ | </asy> | ||
+ | |||
+ | The red dot has a price per pound of something that is larger than <math>1</math>. The green dot has a price per pound of <math>1</math>. The blue dot has a price per pound of something like <math>\frac{2.5}{3}</math>. The orange dot has a price per pound that is less than <math>1</math>, but is very close to it. The purple dot has a price per pound of something like <math>\frac{4.5}{5}</math>. We see that choices <math>\textbf{(A)}</math>, <math>\textbf{(B)}</math>,and <math>\textbf{(D)}</math> are eliminated. Also, <math>\frac{4.5}{5} > \frac{2.5}{3}</math> thus the answer is <math>\boxed{\textbf{(C) } 3}</math>. | ||
+ | |||
+ | ==Video Solution by Math-X (First understand the problem!!!)== | ||
+ | https://youtu.be/oUEa7AjMF2A?si=ZY5Ty2OgPWUtsxgb&t=2374 | ||
+ | |||
+ | ~Math-X | ||
+ | |||
+ | ==Video Solution (CREATIVE THINKING!!!)== | ||
+ | https://youtu.be/rKxwNlKlAW4 | ||
+ | |||
+ | ~Education, the Study of Everything | ||
+ | |||
+ | ==Video Solution== | ||
+ | https://youtu.be/Ij9pAy6tQSg?t=1305 | ||
+ | |||
+ | ~Interstigation | ||
+ | |||
+ | ==Video Solution== | ||
+ | https://youtu.be/p29Fe2dLGs8?t=270 | ||
+ | |||
+ | ~STEMbreezy | ||
+ | |||
+ | ==Video Solution== | ||
+ | https://youtu.be/bNc5j2feJKo | ||
+ | |||
+ | ~savannahsolver | ||
==See Also== | ==See Also== | ||
{{AMC8 box|year=2022|num-b=14|num-a=16}} | {{AMC8 box|year=2022|num-b=14|num-a=16}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 13:49, 23 November 2023
Contents
[hide]Problem
Laszlo went online to shop for black pepper and found thirty different black pepper options varying in weight and price, shown in the scatter plot below. In ounces, what is the weight of the pepper that offers the lowest price per ounce?
Solution
We are looking for a black point, that when connected to the origin, yields the lowest slope. The slope represents the price per ounce. We can visually find that the point with the lowest slope is the blue point. Furthermore, it is the only one with a price per ounce significantly less than . Finally, we see that the blue point is in the category with a weight of ounces.
~MathFun1000
Solution 2 (Elimination)
By the answer choices, we can disregard the points that do not have integer weights. As a result, we obtain the following diagram:
We then proceed in the same way that we had done in Solution 1. Following the steps, we figure out the blue dot that yields the lowest slope, along with passing the origin. We then can look at the x-axis(in this situation, the weight) and figure out it has ounces.
~DairyQueenXD (edited by HW73)
Solution 3 (Elimination)
We can find the lowest point in each line (, , , , or ) and find the price per pound. (Note that we don't need to find the points higher than the points below since we are finding the lowest price per pound.)
The red dot has a price per pound of something that is larger than . The green dot has a price per pound of . The blue dot has a price per pound of something like . The orange dot has a price per pound that is less than , but is very close to it. The purple dot has a price per pound of something like . We see that choices , ,and are eliminated. Also, thus the answer is .
Video Solution by Math-X (First understand the problem!!!)
https://youtu.be/oUEa7AjMF2A?si=ZY5Ty2OgPWUtsxgb&t=2374
~Math-X
Video Solution (CREATIVE THINKING!!!)
~Education, the Study of Everything
Video Solution
https://youtu.be/Ij9pAy6tQSg?t=1305
~Interstigation
Video Solution
https://youtu.be/p29Fe2dLGs8?t=270
~STEMbreezy
Video Solution
~savannahsolver
See Also
2022 AMC 8 (Problems • Answer Key • Resources) | ||
Preceded by Problem 14 |
Followed by Problem 16 | |
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 AJHSME/AMC 8 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.