Difference between revisions of "2024 AIME I Problems/Problem 1"

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)

@@ Line 30: / Line 30: @@
 -Failure.net
+==Solution 2==
+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==
+https://youtu.be/5-CC_-LuCFg
+~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
 ==See also==
 {{AIME box|year=2024|n=I|before=First Problem|num-a=2}}
+[[Category:Introductory Algebra Problems]]
 {{MAA Notice}}

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