Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

(→‎Video Solution)
(→‎Solution 4)
 
(14 intermediate revisions by 5 users not shown)
Line 1: Line 1:
 
==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>
Line 16: Line 16:
 
==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 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.
+
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==
 
==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>.
+
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>.
  
−
(Slightly different Solution)
+
-HW73
−
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
 
  
−
==Solution 4==
+
==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>
+
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
 
-fn106068
 +
 +
==Video Solution (CREATIVE THINKING + ANALYSIS!!!)==
 +
https://youtu.be/6Am5gaLa4JQ
 +
 +
~Education, the Study of Everything
  
 
==Video Solution==
 
==Video Solution==

Latest revision as of 12:49, 14 January 2024

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 $80\%$ 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, not 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 correctly alone. We can see that the total amount of correct problems Chloe and Zoe did together was $176-80=96$. Therefore, Zoe did $96+90=186$ problems out of $200$ problems correctly. This is $\boxed{\textbf{(C) } 93}$ 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 $\boxed{\textbf{(C) } 93}$.

Solution 4

(Slightly different Solution) Suppose we said that there were $100$ 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 $\boxed{\textbf{(C) } 93}$.

-HW73

Solution 5

Let the total number of problems be $t$. Let the percentage of the number of problems that Chloe and Zoe did together and got right be $x$. As we can see, Chloe got $80$% of $\frac {1}{2}$ of the total problems right, hence, ${0.80 \cdot \frac{1}{2}t}$ . We also know that Chloe got $88$% of $t$ problems right altogether, making it ${0.88 \cdot t}$ total problems right. If we add $x$ 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: ${0.80 \cdot \frac{1}{2}t} + x = {0.88 \cdot t}$. Solving that, we get $x = 0.48t$. We also know that Zoe got $90$% of $\frac {1}{2}$ of the total problems right, making it ${0.90 \cdot \frac{1}{2}t}$. We now add that amount to the percentage of problems that Chloe and Zoe got right together, making ${0.90 \cdot \frac{1}{2}t}+ 0.48t$. Solving that, we get $0.93t$, which is equal to $93$%; hence, $\boxed{\textbf{(C) } 93}$.

-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

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. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2017_AMC_8_Problems/Problem_14&oldid=210721"