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2022 AMC 8 Problems/Problem 15

Revision as of 17:47, 20 February 2022 by S318846 (talk | contribs) (→‎Solution 2)

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?

[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),black); dot((1,1.7),black); dot((1,2),black); dot((1,2.8),black); dot((1.5,2.1),black); dot((1.5,3),black); dot((1.5,3.3),black); dot((1.5,3.75),black); dot((2,2),black); dot((2,2.9),black); dot((2,3),black); dot((2,4),black); dot((2,4.35),black); dot((2,4.8),black); dot((2.5,2.7),black); dot((2.5,3.7),black); dot((2.5,4.2),black); dot((2.5,4.4),black); dot((3,2.5),black); dot((3,3.4),black); dot((3,4.2),black); dot((3.5,3.8),black); dot((3.5,4.5),black); dot((3.5,4.8),black); dot((4,3.9),black); dot((4,5.1),black); dot((4.5,4.75),black); dot((4.5,5),black); dot((5,4.5),black); dot((5,5),black); [/asy]

$\textbf{(A) }1\qquad\textbf{(B) }2\qquad\textbf{(C) }3\qquad\textbf{(D) }4\qquad\textbf{(E) }5$

Solution

[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); draw((0,0)--(6,5),red); dot((1,1.2),black); dot((1,1.7),black); dot((1,2),black); dot((1,2.8),black); dot((1.5,2.1),black); dot((1.5,3),black); dot((1.5,3.3),black); dot((1.5,3.75),black); dot((2,2),black); dot((2,2.9),black); dot((2,3),black); dot((2,4),black); dot((2,4.35),black); dot((2,4.8),black); dot((2.5,2.7),black); dot((2.5,3.7),black); dot((2.5,4.2),black); dot((2.5,4.4),black); dot((3,2.5),blue); dot((3,3.4),black); dot((3,4.2),black); dot((3.5,3.8),black); dot((3.5,4.5),black); dot((3.5,4.8),black); dot((4,3.9),black); dot((4,5.1),black); dot((4.5,4.75),black); dot((4.5,5),black); dot((5,4.5),black); dot((5,5),black); [/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. It is clearly the blue point, which can be found visually. Also, it is the only one with a price per once significantly less than $1$. Finally, we see that the blue point is over the category with a weight of $\boxed{\textbf{(C) } 3}$ 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:

[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),black); dot((1,1.7),black); dot((1,2),black); dot((1,2.8),black); dot((2,2),black); 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),black); dot((4,5.1),black); dot((5,4.5),black); dot((5,5),black); [/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 $\boxed{\textbf{(C) } 3}$ ounces.

~DairyQueenXD (edited by HW73)

Video Solution

https://youtu.be/Ij9pAy6tQSg?t=1305

~Interstigation

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

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