Difference between revisions of "2020 AMC 8 Problems/Problem 1"

Revision as of 19:54, 19 November 2020

Luka is making lemonade to sell at a school fundraiser. His recipe requires $4$ times as much water as sugar and twice as much sugar as lemon juice. He uses $3$ cups of lemon juice. How many cups of water does he need?

$\textbf{(A) }6 \qquad \textbf{(B) }8 \qquad \textbf{(C) }12 \qquad \textbf{(D) }18 \qquad \textbf{(E) }24$

Solution 1

Luka will need $3\cdot 2=6$ cups of sugar and $6\cdot 4=24$ cups of water. The answer is $\boxed{\textbf{(E) } 24}$ .

Solution 2

Let $W, S,$ and $L$ represent the number of cups of water, sugar, and lemon juice that Luka needs for his recipe, respectively. We are given that $W=4S$ and $S=2L$ . Since $L=3$ , it follows that $S=6$ , which in turn implies that $W=24 \implies\boxed{\textbf{(E) }24}$ .
~junaidmansuri

Solution 3

We have that $\textsf{lemonade} : \textsf{water} : \textsf{lemon juice} = 8 : 2 : 1,$ so we have $3 \cdot 8 = \boxed{\textbf{(E) }24}.$

[pog]

Solution 4

We are given that $4w:s$ and $2s=l$ which we combine to get $8w:2s:l$ . Letting all the variables equal $3$ , we find that the answer is $3\cdot 8=\boxed{\textbf{(E) }24} 24$ .

-franzliszt

Solution 5

Put the numbers in ratios $4w:s$ and $2s:lj$ when w = water, s = sugar, and lj = lemon juice. then since we know there is $3$ cups of lemon juice, do the math. $3\cdot2\cdot4=6\cdot4=24$

~ bsu1

Video Solution

https://youtu.be/FPC792h-mGE

~savannahsolver

2020 AMC 8 (Problems • Answer Key • Resources)
Preceded by First problem	Followed by Problem 2
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.

we just get 4*3*2=24$ which is E -oceanxia

@@ Line 17: / Line 17: @@
 ==Solution 4==
-We are given that <math>4w:s</math> and <math>2s=l</math> which we combine to get <math>8w:2s:l</math>. Letting all the variables equal <math>3</math>, we find that the answer is <math>3\cdot 8=\textbf{(E)}\ 24</math>.
+We are given that <math>4w:s</math> and <math>2s=l</math> which we combine to get <math>8w:2s:l</math>. Letting all the variables equal <math>3</math>, we find that the answer is <math>3\cdot 8=\boxed{\textbf{(E) }24} 24</math>.
 -franzliszt
 ==Solution 5==

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