Difference between revisions of "2024 AMC 10A Problems/Problem 2"

2024 AMC 10A (Problems • Answer Key • Resources)
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All AMC 10 Problems and Solutions

2024 AMC 12A (Problems • Answer Key • Resources)
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All AMC 12 Problems and Solutions

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 Problem 2. https://www.youtube.com/watch?v=ZAYDcjOSTvk
-==Solution 2==
+Problem 2. https://www.youtube.com/watch?v=ZAYDcjOSTvk
-Alternatively, observe that using <math>a=10x</math> and <math>b=\frac{y}{100}</math> makes the numbers much more closer to each other in terms of magnitude.
-Plugging in the new variables:
-\begin{align*}
-&=15x+8y, \\
-&=12x+11y.
-\end{align*}
-The solution becomes more obvious in this way, with <math>15+8=12+11=23</math>, and since <math>23\cdot 3=69</math>, we determine that <math>x=y=3</math>.
-The question asks us for <math>4.2a+4000b=42x+40y</math>. Since <math>x=y</math>, we have <math>(40+42)\cdot 3=\boxed{\textbf{(B) }246}</math>.
-~Edited by Rosiefork
 == Video Solution by Math from my desk ==

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