Difference between revisions of "2017 AMC 8 Problems/Problem 14"
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− | ==Problem | + | ==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 <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 | + | Assume the total amount of problems is <math>100</math> per half homework assignment since we are dealing with percentages, 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 correctly alone. We can see that the total amount of correct problems Chloe and Zoe did together was <math>176-80=96</math>. Therefore, Zoe did <math>96+90=186</math> problems out of <math>200</math> problems correctly. This is <math>\boxed{\textbf{(C) } 93}</math> 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 <math>\boxed{\textbf{(C) } 93}</math>. | ||
+ | |||
+ | ==Solution 4== | ||
+ | (Slightly different Solution) | ||
+ | Suppose we said that there were <math>100</math> problems in their assignment. Then, Chloe had <math>40</math> correct and <math>10</math> incorrect on her portion, and | ||
+ | <math>48</math> correct and <math>2</math> incorrect on the portion she and Zoe solved. Zoe has <math>45</math> correct and <math>5</math> incorrect on her portion, and <math>48</math> correct and <math>2</math> incorrect on the portion that she and Chloe solved. Then, Zoe has <math>48 + 45 = 93</math> correct answers out of <math>100</math>, so the answer is <math>\boxed{\textbf{(C) } 93}</math>. | ||
+ | |||
+ | -HW73 | ||
+ | |||
+ | ==Solution 5== | ||
+ | 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>. | ||
+ | |||
+ | -fn106068 | ||
+ | |||
+ | ==Video Solution (CREATIVE THINKING + ANALYSIS!!!)== | ||
+ | https://youtu.be/6Am5gaLa4JQ | ||
+ | |||
+ | ~Education, the Study of Everything | ||
+ | |||
+ | ==Video Solution== | ||
+ | https://youtu.be/WgoAEitW5D4 | ||
+ | |||
+ | https://youtu.be/1VWcwRNNJoI | ||
+ | |||
+ | ~savannahsolver | ||
==See Also== | ==See Also== |
Latest revision as of 13:49, 14 January 2024
Contents
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,
.
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, 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 correctly alone. We can see that the total amount of correct problems Chloe and Zoe did together was
. Therefore, Zoe did
problems out of
problems correctly. 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 .
Solution 4
(Slightly different Solution)
Suppose we said that there were problems in their assignment. Then, Chloe had
correct and
incorrect on her portion, and
correct and
incorrect on the portion she and Zoe solved. Zoe has
correct and
incorrect on her portion, and
correct and
incorrect on the portion that she and Chloe solved. Then, Zoe has
correct answers out of
, so the answer is
.
-HW73
Solution 5
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 (CREATIVE THINKING + ANALYSIS!!!)
~Education, the Study of Everything
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.