Difference between revisions of "2003 AMC 10B Problems/Problem 22"
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==Problem== | ==Problem== | ||
| − | A clock chimes once at <math>30</math> minutes past each hour and chimes on the hour according to the hour. For example, at <math>1 \text{PM}</math> there is one chime and at noon and midnight there are twelve chimes. Starting at <math>11:15 \text{AM}</math> on <math>\text{February 26, 2003},</math> on what date will the <math>2003\text{rd}</math> chime occur? | + | A clock chimes once at <math>30</math> minutes past each hour and chimes on the hour according to the hour. For example, at <math>1 \text{PM}</math> there is one chime and at noon and midnight there are twelve chimes. Starting at <math>11:15 \text{AM}</math> on <math>\text{February 26, 2003},</math> on what date will the <math>2003^{\text{rd}}</math> chime occur? |
<math>\textbf{(A) } \text{March 8} \qquad\textbf{(B) } \text{March 9} \qquad\textbf{(C) } \text{March 10} \qquad\textbf{(D) } \text{March 20} \qquad\textbf{(E) } \text{March 21}</math> | <math>\textbf{(A) } \text{March 8} \qquad\textbf{(B) } \text{March 9} \qquad\textbf{(C) } \text{March 10} \qquad\textbf{(D) } \text{March 20} \qquad\textbf{(E) } \text{March 21}</math> | ||
==Solution== | ==Solution== | ||
| − | First, find how many chimes will have already happened before midnight (the beginning of the day) of <math>\text{February 27, 2003}.</math> <math>13</math> half-hours have passed, and the number of chimes according to the hour is <math>1+2+3+\cdots+12.</math> The total number of chimes is <math>13+78=91</math> | + | First, find how many chimes will have already happened before midnight (the beginning of the day) of <math>\text{February 27, 2003}.</math> <math>13</math> half-hours have passed, and the number of chimes according to the hour is <math>1+2+3+\cdots+12.</math> The total number of chimes is <math>13+78=91.</math> |
Every day, there will be <math>24</math> half-hours and <math>2(1+2+3+\cdots+12)</math> chimes according to the arrow, resulting in <math>24+156=180</math> total chimes. | Every day, there will be <math>24</math> half-hours and <math>2(1+2+3+\cdots+12)</math> chimes according to the arrow, resulting in <math>24+156=180</math> total chimes. | ||
| − | On <math>\text{February | + | On <math>\text{February 26},</math> the number of chimes that still need to occur is <math>2003-91=1912.</math> <math>1912 \div 180=10 \text{R}112.</math> Rounding up, it is <math>11</math> days past <math>\text{February 26},</math> which is <math>\boxed{\textbf{(B) \ } \text{March 9}}</math> |
==See Also== | ==See Also== | ||
{{AMC10 box|year=2003|ab=B|num-b=21|num-a=23}} | {{AMC10 box|year=2003|ab=B|num-b=21|num-a=23}} | ||
| + | {{MAA Notice}} | ||
Latest revision as of 00:07, 8 September 2022
Problem
A clock chimes once at 
minutes past each hour and chimes on the hour according to the hour. For example, at 
there is one chime and at noon and midnight there are twelve chimes. Starting at 
on 
on what date will the 
chime occur?
Solution
First, find how many chimes will have already happened before midnight (the beginning of the day) of 
half-hours have passed, and the number of chimes according to the hour is 
The total number of chimes is 
Every day, there will be 
half-hours and 
chimes according to the arrow, resulting in 
total chimes.
On 
the number of chimes that still need to occur is 
Rounding up, it is 
days past which is 
See Also
| 2003 AMC 10B (Problems • Answer Key • Resources) | ||
| Preceded by Problem 21 |
Followed by Problem 23 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
| All AMC 10 Problems and Solutions | ||
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