Difference between revisions of "2013 AIME II Problems/Problem 1"
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==Solution== | ==Solution== | ||
| − | There are <math>24 | + | There are <math>24 \cdot 60=1440</math> normal minutes in a day , and <math>10 \cdot 100=1000</math> metric minutes in a day. The ratio of normal to metric minutes in a day is <math>\frac{1440}{1000}</math>, which simplifies to <math>\frac{36}{25}</math>. This means that every time 36 normal minutes pass, 25 metric minutes pass. From midnight to <math>\text{6:36}</math> AM, <math>6 \cdot 60+36=396</math> normal minutes pass. This can be viewed as <math>\frac{396}{36}=11</math> cycles of 36 normal minutes, so 11 cycles of 25 metric minutes pass. Adding <math>25 \cdot 11=275</math> to <math>\text{0:00}</math> gives <math>\text{2:75}</math>, so the answer is <math>\boxed{275}</math>. |
==Solution 2== | ==Solution 2== | ||
Latest revision as of 18:03, 3 September 2023
Contents
Problem 1
Suppose that the measurement of time during the day is converted to the metric system so that each day has 
metric hours, and each metric hour has 
metric minutes. Digital clocks would then be produced that would read 
just before midnight, 
at midnight, 
at the former 
AM, and 
at the former 
PM. After the conversion, a person who wanted to wake up at the equivalent of the former 
AM would set his new digital alarm clock for 
, where 
, 
, and 
are digits. Find 
.
Solution
There are 
normal minutes in a day , and 
metric minutes in a day. The ratio of normal to metric minutes in a day is 
, which simplifies to 
. This means that every time 36 normal minutes pass, 25 metric minutes pass. From midnight to AM, 
normal minutes pass. This can be viewed as 
cycles of 36 normal minutes, so 11 cycles of 25 metric minutes pass. Adding 
to gives 
, so the answer is 
.
Solution 2
First we want to find out what fraction of a day has passed at 6:36 AM. One hour is 
of a day, and 36 minutes is 
of an hour, so at 6:36 AM, 
of a day has passed. Now the metric timing equivalent of 
of a day is 
metric minutes, which is equivalent to 2 metric hours and 75 metric minutes, so our answer is - mathleticguyyy
See also
| 2013 AIME II (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. 
