Difference between revisions of "2012 AMC 8 Problems/Problem 13"
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| − | You can quickly tell, if you have | + | You can quickly tell, if you have memorized your times tables, that <math>187</math> and <math>143</math> are both multiples of <math>11</math>. Additionally, the difference is 44 cents. That means that there are <math> \boxed{\textbf{(C)}\ 4} </math> more pencils purchased by Sharona because <math>11*4 = 44</math>. |
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==See Also== | ==See Also== | ||
{{AMC8 box|year=2012|num-b=12|num-a=14}} | {{AMC8 box|year=2012|num-b=12|num-a=14}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Latest revision as of 14:36, 28 December 2023
Problem 13
Jamar bought some pencils costing more than a penny each at the school bookstore and paid 
. Sharona bought some of the same pencils and paid 
. How many more pencils did Sharona buy than Jamar?
Solution 1
We assume that the price of the pencils remains constant. Convert and to cents. Since the price of the pencils is more than one penny, we can find the price of one pencil (in cents) by taking the greatest common divisor of 
and 
, which is 
. Therefore, Jamar bought 
pencils and Sharona bought 
pencils. Thus, Sharona bought 
more pencils than Jamar.
Solution 2
We find the difference between 
and 
is 
, which is the extra cost of Sharona's pencils than Jamar's pencils. Because the difference between the amounts of the pencils they bought must be divided evenly by 
, looking into the answers, 
or 
is possibly correct. It gives us the price of each pencil should be 
or 
, respectively. Then we find only 
can be divided evenly by and . So the answer is 
Solution 3 (quick solution)
You can quickly tell, if you have memorized your times tables, that and are both multiples of . Additionally, the difference is 44 cents. That means that there are more pencils purchased by Sharona because 
.
See Also
| 2012 AMC 8 (Problems • Answer Key • Resources) | ||
| Preceded by Problem 12 |
Followed by Problem 14 | |
| 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 AJHSME/AMC 8 Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
