Difference between revisions of "2017 AMC 8 Problems/Problem 14"

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(Solution 2: I figured to just add a final touch to clarify my brittiant solution, lol 😂)
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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 they did the homework together is <math>176-80=96</math>. Therefore Zoe has <math>96+90=186</math> out of <math>200</math> problems correct. This is <math>\boxed{\textbf{(C) } 93}</math>
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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 they did the homework together is <math>176-80=96</math>, which is the [b]total[/b] amount of problems she got correct, subtracted by the number of problems she did alone, which were correct. 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 12:41, 5 January 2018

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 1

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.

Solution 2

Assume the total amount of problems is $100$ per half homework assignment, since we are dealing with percentages, and no values. Then, we know that Chloe got $80$ problems correct by herself, and got $176$ problems correct overall. We also know that Zoe had $90$ problems she did alone correct. We can see that the total amount of correct problems Chloe had they did the homework together is $176-80=96$, which is the [b]total[/b] amount of problems she got correct, subtracted by the number of problems she did alone, which were correct. Therefore Zoe has $96+90=186$ problems out of $200$ problems correct. This is $\boxed{\textbf{(C) } 93}$ percent.

See Also

2017 AMC 8 (ProblemsAnswer KeyResources)
Preceded by
Problem 13
Followed by
Problem 15
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All AJHSME/AMC 8 Problems and Solutions

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