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

(Solution 1)
(Solution 1)
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Lastly, <math>s + \frac{1}{2} = 3</math> kilometers per hour, so
 
Lastly, <math>s + \frac{1}{2} = 3</math> kilometers per hour, so
  
<math>\frac{9}{3} + 0.4 = 3.4</math> hours, or <math>204</math> minutes
+
<math>\frac{9}{3} + 0.4 = 3.4</math> hours, or <math>\framebox{204}</math> minutes
  
 
-Failure.net
 
-Failure.net

Revision as of 13:22, 2 February 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