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Difference between revisions of "2024 AIME I Problems/Problem 1"
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==Solution 2==  
 
==Solution 2==  
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The amount of hours spent while walking on the first travel is <math>\frac{240-t}{6}</math>. Thus, we have the equation <math>(240-t)(s) = 540</math>, and by the same logic, the second equation yields <math>(144-t)(s+2) = 540</math>. We have <math>240s-st = 540</math>, and <math>288+144s-2t-st = 540</math>. We subtract the two equations to get <math>96s+2t-288 = 0</math>, so we have <math>48s+t = 144</math>, so <math>t = 144-48s</math>, and now we have <math>(96+48s)(s) = 540</math>. The numerator of <math>s</math> must evenly divide 540, however, <math>s</math> must be less than 3. We can guess that <math>s = 2.5</math>. Now, <math>2.5+0.5 = 3</math>. Taking <math>\frac{9}{3} = 3</math>, we find that it will take three hours for the 9 kilometers to be traveled. The t minutes spent at the coffeeshop can be written as <math>144-48(2.5)</math>, so t = 24. <math>180 + 24 = 204</math>. -sepehr2010
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The amount of hours spent while walking on the first travel is <math>\frac{240-t}{60}</math>. Thus, we have the equation <math>(240-t)(s) = 540</math>, and by the same logic, the second equation yields <math>(144-t)(s+2) = 540</math>. We have <math>240s-st = 540</math>, and <math>288+144s-2t-st = 540</math>. We subtract the two equations to get <math>96s+2t-288 = 0</math>, so we have <math>48s+t = 144</math>, so <math>t = 144-48s</math>, and now we have <math>(96+48s)(s) = 540</math>. The numerator of <math>s</math> must evenly divide 540, however, <math>s</math> must be less than 3. We can guess that <math>s = 2.5</math>. Now, <math>2.5+0.5 = 3</math>. Taking <math>\frac{9}{3} = 3</math>, we find that it will take three hours for the 9 kilometers to be traveled. The t minutes spent at the coffeeshop can be written as <math>144-48(2.5)</math>, so t = 24. <math>180 + 24 = 204</math>. -sepehr2010
  
 
==Video Solution==
 
==Video Solution==
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==See also==
 
==See also==
 
{{AIME box|year=2024|n=I|before=First Problem|num-a=2}}
 
{{AIME box|year=2024|n=I|before=First Problem|num-a=2}}
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[[Category:Introductory Algebra Problems]]
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 13:22, 1 August 2024

Problem

Every morning Aya goes for a $9$-kilometer-long walk and stops at a coffee shop afterwards. When she walks at a constant speed of $s$ kilometers per hour, the walk takes her 4 hours, including $t$ minutes spent in the coffee shop. When she walks $s+2$ kilometers per hour, the walk takes her 2 hours and 24 minutes, including $t$ minutes spent in the coffee shop. Suppose Aya walks at $s+\frac{1}{2}$ kilometers per hour. Find the number of minutes the walk takes her, including the $t$ minutes spent in the coffee shop.

Solution 1

$\frac{9}{s} + t = 4$ in hours and $\frac{9}{s+2} + t = 2.4$ in hours.

Subtracting the second equation from the first, we get,

$\frac{9}{s} - \frac{9}{s+2} = 1.6$

Multiplying by $(s)(s+2)$, we get

$9s+18-9s=18=1.6s^{2} + 3.2s$

Multiplying by 5/2 on both sides, we get

$0 = 4s^{2} + 8s - 45$

Factoring gives us

$(2s-5)(2s+9) = 0$, of which the solution we want is $s=2.5$.

Substituting this back to the first equation, we can find that $t = 0.4$ hours.

Lastly, $s + \frac{1}{2} = 3$ kilometers per hour, so

$\frac{9}{3} + 0.4 = 3.4$ hours, or $\framebox{204}$ minutes

-Failure.net

Solution 2

The amount of hours spent while walking on the first travel is $\frac{240-t}{60}$. Thus, we have the equation $(240-t)(s) = 540$, and by the same logic, the second equation yields $(144-t)(s+2) = 540$. We have $240s-st = 540$, and $288+144s-2t-st = 540$. We subtract the two equations to get $96s+2t-288 = 0$, so we have $48s+t = 144$, so $t = 144-48s$, and now we have $(96+48s)(s) = 540$. The numerator of $s$ must evenly divide 540, however, $s$ must be less than 3. We can guess that $s = 2.5$. Now, $2.5+0.5 = 3$. Taking $\frac{9}{3} = 3$, we find that it will take three hours for the 9 kilometers to be traveled. The t minutes spent at the coffeeshop can be written as $144-48(2.5)$, so t = 24. $180 + 24 = 204$. -sepehr2010

Video Solution

https://youtu.be/5-CC_-LuCFg

~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)


See also

2024 AIME I (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
All AIME Problems and Solutions

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