Difference between revisions of "2024 AIME I Problems/Problem 1"
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==Solution 2== | ==Solution 2== | ||
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 | 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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| + | ==Video Solution== | ||
| + | |||
| + | https://youtu.be/5-CC_-LuCFg | ||
| + | |||
| + | ~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com) | ||
| + | |||
==See also== | ==See also== | ||
Revision as of 20:14, 2 February 2024
Problem
Every morning Aya goes for a 
-kilometer-long walk and stops at a coffee shop afterwards. When she walks at a constant speed of 
kilometers per hour, the walk takes her 4 hours, including 
minutes spent in the coffee shop. When she walks 
kilometers per hour, the walk takes her 2 hours and 24 minutes, including minutes spent in the coffee shop. Suppose Aya walks at 
kilometers per hour. Find the number of minutes the walk takes her, including the minutes spent in the coffee shop.
Solution 1
in hours and 
in hours.
Subtracting the second equation from the first, we get,
Multiplying by 
, we get
Multiplying by 5/2 on both sides, we get
Factoring gives us
, of which the solution we want is 
.
Substituting this back to the first equation, we can find that 
hours.
Lastly, 
kilometers per hour, so
hours, or 
minutes
-Failure.net
Solution 2
The amount of hours spent while walking on the first travel is 
. Thus, we have the equation 
, and by the same logic, the second equation yields 
. We have 
, and 
. We subtract the two equations to get 
, so we have 
, so 
, and now we have 
. The numerator of must evenly divide 540, however, must be less than 3. We can guess that 
. Now, 
. Taking 
, 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 
, so t = 24. 
. -sepehr2010
Video Solution
~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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
