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

(Solution 2)
(Solution 2)
Line 8: Line 8:
 
==Solution 2==
 
==Solution 2==
 
Let <math>W, S,</math> and <math>L</math> represent the number of cups of water, sugar, and lemon juice that Luka needs for his recipe, respectively. We are given that <math>W=4S</math> and <math>S=2L</math>. Since <math>L=3</math>, it follows that <math>S=6</math>, which in turn implies that <math>W=24 \implies\boxed{\textbf{(E) }24}</math>.<br>
 
Let <math>W, S,</math> and <math>L</math> represent the number of cups of water, sugar, and lemon juice that Luka needs for his recipe, respectively. We are given that <math>W=4S</math> and <math>S=2L</math>. Since <math>L=3</math>, it follows that <math>S=6</math>, which in turn implies that <math>W=24 \implies\boxed{\textbf{(E) }24}</math>.<br>
~jmansuri
+
~ junaidmansuri
  
 
==See also==
 
==See also==
 
{{AMC8 box|year=2020|before=First problem|num-a=2}}
 
{{AMC8 box|year=2020|before=First problem|num-a=2}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 06:23, 18 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

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

See also

2020 AMC 8 (ProblemsAnswer KeyResources)
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. AMC logo.png