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| {{duplicate|[[2025 AMC 10A Problems/Problem 1|2025 AMC 10A #1]] and [[2025 AMC 12A Problems/Problem 1|2025 AMC 12A #1]]}}
| | #redirect [[2025 AMC 12A Problems/Problem 1]] |
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| == Problem ==
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| Andy and Betsy both live in Mathville. Andy leaves Mathville on his bicycle at <imath>1{:}30</imath>, traveling due northat a steady <imath>8</imath> mile per hour. Betsy leaves on her bicycle from the same point at <imath>2{:}30</imath>, traveling due east at a steady <imath>12</imath> miles per hour. At what time will they be exactly the same distance from their common starting point?
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| <imath>\textbf{(A) } {3:30}\qquad\textbf{(B) } {3:45}\qquad\textbf{(C) } {4:00}\qquad\textbf{(D) } {4:15}\qquad\textbf{(E) } {4:30}</imath>
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| == Solution 1 ==
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| We can see that Betsy travels 1 hour after Andy started. We have <imath>lcm(8, 12)=24</imath>. Now we can find the total time Andy has taken once Betsy catches up: <imath>\frac{24}{8} = 3 \text{ hours}</imath>
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| So the answer is <imath>1{:}30 + 3{:}00 = \boxed{\textbf{(E) } 4{:}30}</imath>
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| ~Boywithnuke(Goal: 10 followers)
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| ~minor edits by ChickensEatGrass
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| == Solution 2 ==
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| <imath>h</imath> hours after Betsy left, Andy has traveled <imath>8(h+1)</imath> miles, and Betsy has traveled <imath>12h</imath> miles. We are told these are equal, so <imath>8h+8=12h</imath>. Solving, we get <imath>h=2</imath>, so Andy and Betsy will be exactly the same distance from their common starting point two hours after Betsy leaves, or <imath>\text{(E) }4:30</imath>.
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| ~mithu542
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| ==Solution 3 (bash)==
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| We can use all the answer choices that we are given.
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| Let's use casework for each of the answers:
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| At 3:30, Andy will have gone <imath>2\cdot8=16</imath> miles. Betsy will have gone <imath>1\cdot12=12</imath> miles.
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| At 3:45, we can just add the distance each of them goes in 1/4 hours. Therefore, Andy goes 18 miles, and Betsy goes 15 miles.
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| At 4:00, we see from the same logic that Andy has gone 20 miles, and Betsy has gone 18 miles.
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| At 4:15 we see that Andy has gone 22 miles, and Betsy has gone 21 miles.
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| At E, 4:30, we see that both Andy and Betsy have gone 24 miles.
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| Now we see that <imath>\text{(E) }4:30</imath> is the correct answer.
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| ~vgarg
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| ==Solution 4==
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| We can see that at <imath>2:30</imath>, Andy will be <imath>8</imath> miles ahead. For every hour that they both travel, Betsy will gain <imath>4</imath> miles on Andy. Therefore, it will take <imath>2</imath> more hours for Betsy to catch up, and they will be at the same point at <imath>\text{(E) }4:30</imath>.
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| ~vinceS
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| ==Solution 5==
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| The distance Andy travels can be represented by <imath>8x</imath> and Betsy with the equation <imath>12(x-1).</imath> The solution to this is <imath>x = 3,</imath> so the answer is <imath>1:30</imath> plus <imath>3</imath> hours or <imath>\boxed{\text{(E) }4:30}</imath>.
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| ~minor LaTeX edits by zoyashaikh
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| ==See Also==
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| {{AMC10 box|year=2025|ab=A|before=First Problem|num-a=2}}
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| {{AMC12 box|year=2025|ab=A|before=First Problem|num-a=2}}
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| * [[AMC 10]]
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| * [[AMC 10 Problems and Solutions]]
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| * [[Mathematics competitions]]
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| * [[Mathematics competition resources]]
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| {{MAA Notice}}
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