Difference between revisions of "2012 AMC 8 Problems/Problem 13"
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==Solution 2== | ==Solution 2== | ||
− | We find the difference between <math>1.43</math> and <math>1.87</math> is <math>1.87-1.43 = 0.44</math> | + | We find the difference between <math>1.43</math> and <math>1.87</math> is <math>1.87-1.43 = 0.44</math>, 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 <math>0.44</math>, looking into the answers, <math>2</math> or <math>4</math> is possibly correct. It gives us the price of each pencil should be <math>0.44/2=0.22</math> or <math>0.44/4=0.11</math>, respectively. Then we find only <math>0.11</math> can be divided evenly by <math>1.43</math> and <math>1.87</math>. So the answer is <math> \boxed{\textbf{(C)}\ 4} </math> ---LarryFlora |
==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}} |
Revision as of 13:55, 21 August 2021
Contents
Problem
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 ---LarryFlora
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.