2024 AMC 10A Problems/Problem 2

The following problem is from both the 2024 AMC 10A #2 and 2024 AMC 12A #2, so both problems redirect to this page.

Solution 2

Alternatively, observe that using $a=10x$ and $b=\frac{y}{100}$ makes the numbers much more closer to each other in terms of magnitude.

Plugging in the new variables: \begin{align*} 69&=15x+8y, \\ 69&=12x+11y. \end{align*}

The solution becomes more obvious in this way, with $15+8=12+11=23$ , and since $23\cdot 3=69$ , we determine that $x=y=3$ .

The question asks us for $4.2a+4000b=42x+40y$ . Since $x=y$ , we have $(40+42)\cdot 3=\boxed{\textbf{(B) }246}$ .

~Edited by Rosiefork

~Education, the Study of Everything

~Thesmartgreekmathdude

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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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 AMC 12 Problems and Solutions