Difference between revisions of "2017 AMC 8 Problems/Problem 14"
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<math>\textbf{(A) }89\qquad\textbf{(B) }92\qquad\textbf{(C) }93\qquad\textbf{(D) }96\qquad\textbf{(E) }98</math> | <math>\textbf{(A) }89\qquad\textbf{(B) }92\qquad\textbf{(C) }93\qquad\textbf{(D) }96\qquad\textbf{(E) }98</math> | ||
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==Solution 1== | ==Solution 1== | ||
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Zoe got <math>\frac{90x}{100} + \frac{ax}{100} = \frac{186x}{100}</math> problems right out of <math>2x</math>. Therefore, Zoe got <math>\frac{\frac{186x}{100}}{2x} = \frac{93}{100} = \boxed{\textbf{(C) } 93}</math> percent of the problems correct. | Zoe got <math>\frac{90x}{100} + \frac{ax}{100} = \frac{186x}{100}</math> problems right out of <math>2x</math>. Therefore, Zoe got <math>\frac{\frac{186x}{100}}{2x} = \frac{93}{100} = \boxed{\textbf{(C) } 93}</math> percent of the problems correct. | ||
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==Solution 2== | ==Solution 2== | ||
| − | Assume the total amount of problems is <math>100</math> per half homework assignment, since we are dealing with percentages, and no values. Then, we know that Chloe got <math>80</math> problems correct by herself, and got <math>176</math> problems correct overall. We also know that Zoe had <math>90</math> problems she did alone correct. We can see that the total amount of correct problems Chloe had when Zoe and she did the homework together is <math>176-80=96</math>, which is the total amount of problems she got correct, subtracted by the number of correct problems she did alone. Therefore Zoe has <math>96+90=186</math> problems out of <math>200</math> problems correct. This is <math>\boxed{\textbf{(C) } 93}</math> percent. | + | Assume the total amount of problems is <math>100</math> per half homework assignment, since we are dealing with percentages, and no values. Then, we know that Chloe got <math>80</math> problems correct by herself, and got <math>176</math> problems correct overall. We also know that Zoe had <math>90</math> problems she did alone correct. We can see that the total amount of correct problems Chloe had when Zoe and she did the homework together is <math>176-80=96</math>, which is the total amount of problems she got correct, subtracted by the number of correct problems she did alone. Therefore Zoe has <math>96+90=186</math> problems out of <math>200</math> problems correct. This is <math>\boxed{\textbf{(C) } 93}</math> percent. |
==See Also== | ==See Also== | ||
Revision as of 15:08, 22 June 2019
Contents
Problem 14
Chloe and Zoe are both students in Ms. Demeanor's math class. Last night they each solved half of the problems in their homework assignment alone and then solved the other half together. Chloe had correct answers to only 
of the problems she solved alone, but overall 
of her answers were correct. Zoe had correct answers to 
of the problems she solved alone. What was Zoe's overall percentage of correct answers?
Solution 1
Let the number of questions that they solved alone be 
. Let the percentage of problems they correctly solve together be 
%.
As given, ![\[\frac{80x}{100} + \frac{ax}{100} = \frac{2 \cdot 88x}{100}\]](http://latex.artofproblemsolving.com/6/c/2/6c22dec68f64f92c3368913213f608267d1e4a43.png)
.
Hence, 
.
Zoe got 
problems right out of 
. Therefore, Zoe got 
percent of the problems correct.
Solution 2
Assume the total amount of problems is 
per half homework assignment, since we are dealing with percentages, and no values. Then, we know that Chloe got 
problems correct by herself, and got 
problems correct overall. We also know that Zoe had 
problems she did alone correct. We can see that the total amount of correct problems Chloe had when Zoe and she did the homework together is 
, which is the total amount of problems she got correct, subtracted by the number of correct problems she did alone. Therefore Zoe has 
problems out of 
problems correct. This is 
percent.
See Also
| 2017 AMC 8 (Problems • Answer Key • Resources) | ||
| Preceded by Problem 13 |
Followed by Problem 15 | |
| 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. 
