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Difference between revisions of "2013 AIME II Problems/Problem 1"

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==Problem 1==
 
Suppose that the measurement of time during the day is converted to the metric system so that each day has <math>10</math> metric hours, and each metric hour has <math>100</math> metric minutes. Digital clocks would then be produced that would read <math>\text{9:99}</math> just before midnight, <math>\text{0:00}</math> at midnight, <math>\text{1:25}</math> at the former <math>\text{3:00}</math> AM, and <math>\text{7:50}</math> at the former <math>\text{6:00}</math> PM. After the conversion, a person who wanted to wake up at the equivalent of the former <math>\text{6:36}</math> AM would set his new digital alarm clock for <math>\text{A:BC}</math>, where <math>\text{A}</math>, <math>\text{B}</math>, and <math>\text{C}</math> are digits. Find <math>100\text{A}+10\text{B}+\text{C}</math>.
 
Suppose that the measurement of time during the day is converted to the metric system so that each day has <math>10</math> metric hours, and each metric hour has <math>100</math> metric minutes. Digital clocks would then be produced that would read <math>\text{9:99}</math> just before midnight, <math>\text{0:00}</math> at midnight, <math>\text{1:25}</math> at the former <math>\text{3:00}</math> AM, and <math>\text{7:50}</math> at the former <math>\text{6:00}</math> PM. After the conversion, a person who wanted to wake up at the equivalent of the former <math>\text{6:36}</math> AM would set his new digital alarm clock for <math>\text{A:BC}</math>, where <math>\text{A}</math>, <math>\text{B}</math>, and <math>\text{C}</math> are digits. Find <math>100\text{A}+10\text{B}+\text{C}</math>.
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==Solution==
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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>.
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==Solution 2==
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First we want to find out what fraction of a day has passed at 6:36 AM. One hour is <math>\frac{1}{24}</math> of a day, and 36 minutes is <math>\frac{36}{60}=\frac{3}{5}</math> of an hour, so at 6:36 AM, <math>6 \cdot \frac{1}{24} + \frac{1}{24} \cdot \frac{3}{5}=\frac{1}{4}+\frac{1}{40}=\frac{11}{40}</math> of a day has passed. Now the metric timing equivalent of <math>\frac{11}{40}</math> of a day is <math>\frac{11}{40}\cdot 1000=275</math> metric minutes, which is equivalent to 2 metric hours and 75 metric minutes, so our answer is <math>\boxed{275}</math> - mathleticguyyy
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== See also ==
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{{AIME box|year=2013|n=II|before=First Problem|num-a=2}}
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{{MAA Notice}}

Latest revision as of 18:03, 3 September 2023

Problem 1

Suppose that the measurement of time during the day is converted to the metric system so that each day has $10$ metric hours, and each metric hour has $100$ metric minutes. Digital clocks would then be produced that would read $\text{9:99}$ just before midnight, $\text{0:00}$ at midnight, $\text{1:25}$ at the former $\text{3:00}$ AM, and $\text{7:50}$ at the former $\text{6:00}$ PM. After the conversion, a person who wanted to wake up at the equivalent of the former $\text{6:36}$ AM would set his new digital alarm clock for $\text{A:BC}$, where $\text{A}$, $\text{B}$, and $\text{C}$ are digits. Find $100\text{A}+10\text{B}+\text{C}$.

Solution

There are $24 \cdot 60=1440$ normal minutes in a day , and $10 \cdot 100=1000$ metric minutes in a day. The ratio of normal to metric minutes in a day is $\frac{1440}{1000}$, which simplifies to $\frac{36}{25}$. This means that every time 36 normal minutes pass, 25 metric minutes pass. From midnight to $\text{6:36}$ AM, $6 \cdot 60+36=396$ normal minutes pass. This can be viewed as $\frac{396}{36}=11$ cycles of 36 normal minutes, so 11 cycles of 25 metric minutes pass. Adding $25 \cdot 11=275$ to $\text{0:00}$ gives $\text{2:75}$, so the answer is $\boxed{275}$.

Solution 2

First we want to find out what fraction of a day has passed at 6:36 AM. One hour is $\frac{1}{24}$ of a day, and 36 minutes is $\frac{36}{60}=\frac{3}{5}$ of an hour, so at 6:36 AM, $6 \cdot \frac{1}{24} + \frac{1}{24} \cdot \frac{3}{5}=\frac{1}{4}+\frac{1}{40}=\frac{11}{40}$ of a day has passed. Now the metric timing equivalent of $\frac{11}{40}$ of a day is $\frac{11}{40}\cdot 1000=275$ metric minutes, which is equivalent to 2 metric hours and 75 metric minutes, so our answer is $\boxed{275}$ - 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

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