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Difference between revisions of "2017 AMC 8 Problems/Problem 14"
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==Solution==
 
==Solution==
  
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To solve this problem, first substitute the amount of problems as 100. This means that 50 of the problems were solved alone. 80% of 50 is 40. This means that 48 of the problems were solved together. 90% of 50 is 45. This means that 45+48, so (C) percent of Zoe’s problems were correct.
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Let the number of questions that they solved alone be <math>x</math>. Let the percentage of problems they correctly solve together be <math>a</math>%.  
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As given, <cmath>\frac{80x}{100} + \frac{ax}{100} = \frac{2 \cdot 88x}{100}</cmath>.  
  
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~pegasuswa
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Hence, <math>a = 96</math>.
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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.
  
 
==See Also==
 
==See Also==

Revision as of 16:53, 22 November 2017

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 $80\%$ of the problems she solved alone, but overall $88\%$ of her answers were correct. Zoe had correct answers to $90\%$ of the problems she solved alone. What was Zoe's overall percentage of correct answers?

$\textbf{(A) }89\qquad\textbf{(B) }92\qquad\textbf{(C) }93\qquad\textbf{(D) }96\qquad\textbf{(E) }98$

Solution

Let the number of questions that they solved alone be $x$. Let the percentage of problems they correctly solve together be $a$%. As given, \[\frac{80x}{100} + \frac{ax}{100} = \frac{2 \cdot 88x}{100}\].

Hence, $a = 96$.

Zoe got $\frac{90x}{100} + \frac{ax}{100} = \frac{186x}{100}$ problems right out of $2x$. Therefore, Zoe got $\frac{\frac{186x}{100}}{2x} = \frac{93}{100} = \boxed{\textbf{(C) } 93}$ percent of the problems correct.

See Also

2017 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

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