Difference between revisions of "2017 AMC 8 Problems/Problem 14"
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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 <math>80\%</math> of the problems she solved alone, but overall <math>88\%</math> of her answers were correct. Zoe had correct answers to <math>90\%</math> of the problems she solved alone. What was Zoe's overall percentage of correct answers? | 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 <math>80\%</math> of the problems she solved alone, but overall <math>88\%</math> of her answers were correct. Zoe had correct answers to <math>90\%</math> of the problems she solved alone. What was Zoe's overall percentage of correct answers? | ||
<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 | + | Assume the total amount of problems is <math>100</math> per half homework assignment, since we are dealing with percentages, and not 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 correct alone. We can see that the total amount of correct problems Chloe and Zoe did was <math>176-80=96</math>. 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. |
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| + | ==Solution 3== | ||
| + | In the problem, we can see that Chloe solved 80% of the problems she solved alone, but 88% of her answers are correct. If 80 and another number's average is 88, the other number must be 96. Then Zoe solved 90% of the problems she did alone, but 96% of her answers were correct. Then the average of 90 and 96 is <math>\boxed{\textbf{(C) } 93}</math>. | ||
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| + | (Slightly different Solution) | ||
| + | Suppose we said that there was 100 problems in their assignment. Then Chloe had 40 correct and 10 incorrect on her portion, and 48 correct and 2 incorrect on the portion she and Zoe solved. Zoe has 45 correct and 5 incorrect on her portion, and 48 correct and 2 incorrect on the portion that she and Chloe solved. Then, Zoe has 48 + 45 = 93 correct answers out of 100, so the answer is <math>\boxed{\textbf{(C) } 93}</math>. -HW73 | ||
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| + | ==Solution 4== | ||
| + | Let the total number of problems be <math>t</math>. Let the percentage of the number of problems that Chloe and Zoe did together and got right be <math>x</math>. As we can see, Chloe got <math>80</math>% of <math> \frac {1}{2} </math> of the total problems right, hence, <math>{0.80 \cdot \frac{1}{2}t}</math> . We also know that Chloe got <math>88</math>% of <math>t</math> problems right altogether, making it <math>{0.88 \cdot t}</math> total problems right. If we add <math>x</math> to the percentage of correct problems that Chloe solved alone, then that should be equal to the total number of problems that Chloe got right, making the equation: <math>{0.80 \cdot \frac{1}{2}t} + x = {0.88 \cdot t}</math> . Solving that, we get <math>x = 0.48t</math> . We also know that Zoe got <math>90</math>% of <math> \frac {1}{2} </math> of the total problems right, making it <math>{0.90 \cdot \frac{1}{2}t}</math>. We now add that amount to the percentage of problems that Chloe and Zoe got right together, making <math>{0.90 \cdot \frac{1}{2}t}+ 0.48t</math>. Solving that, we get <math>0.93t</math>, which is equal to <math>93</math>%, hence, <math>\boxed{\textbf{(C) } 93}</math> | ||
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| + | -fn106068 | ||
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| + | ==Video Solution== | ||
| + | https://youtu.be/WgoAEitW5D4 | ||
| + | |||
| + | https://youtu.be/1VWcwRNNJoI | ||
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| + | ~savannahsolver | ||
==See Also== | ==See Also== | ||
Revision as of 11:13, 3 March 2022
Problem
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 not 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 correct alone. We can see that the total amount of correct problems Chloe and Zoe did was 
. Therefore Zoe has 
problems out of 
problems correct. This is 
percent.
Solution 3
In the problem, we can see that Chloe solved 80% of the problems she solved alone, but 88% of her answers are correct. If 80 and another number's average is 88, the other number must be 96. Then Zoe solved 90% of the problems she did alone, but 96% of her answers were correct. Then the average of 90 and 96 is .
(Slightly different Solution)
Suppose we said that there was 100 problems in their assignment. Then Chloe had 40 correct and 10 incorrect on her portion, and 48 correct and 2 incorrect on the portion she and Zoe solved. Zoe has 45 correct and 5 incorrect on her portion, and 48 correct and 2 incorrect on the portion that she and Chloe solved. Then, Zoe has 48 + 45 = 93 correct answers out of 100, so the answer is . -HW73
Solution 4
Let the total number of problems be 
. Let the percentage of the number of problems that Chloe and Zoe did together and got right be . As we can see, Chloe got % of 
of the total problems right, hence, 
. We also know that Chloe got 
% of problems right altogether, making it 
total problems right. If we add to the percentage of correct problems that Chloe solved alone, then that should be equal to the total number of problems that Chloe got right, making the equation: 
. Solving that, we get 
. We also know that Zoe got % of of the total problems right, making it 
. We now add that amount to the percentage of problems that Chloe and Zoe got right together, making 
. Solving that, we get 
, which is equal to 
%, hence,
-fn106068
Video Solution
~savannahsolver
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. 
