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

m (Solution 1)
m (Reverted edits by Richbill (talk) to last revision by Jkibbe)
(Tag: Rollback)
Line 4: Line 4:
  
 
<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>
I promise that you'll never find another like me
 
I know that I'm a handful, baby, uh
 
I know I never think before I jump
 
And you're the kind of guy the ladies want
 
(And there's a lot of cool chicks out there)
 
I know that I went psycho on the phone
 
I never leave well enough alone
 
And trouble's gonna follow where I go
 
(And there's a lot of cool chicks out there)
 
But one of these things is not like the others
 
Like a rainbow with all of the colors
 
Baby doll, when it comes to a lover
 
I promise that you'll never find another like
 
Me-e-e, ooh-ooh-ooh-ooh
 
I'm the only one of me
 
Baby, that's the fun of me
 
Eeh-eeh-eeh, ooh-ooh-ooh-ooh
 
You're the only one of you
 
Baby, that's the fun of you
 
And I promise that nobody's gonna love you like me-e-e
 
I know I tend to make it about me
 
I know you never get just what you see
 
But I will never bore you, baby
 
(And there's a lot of lame guys out there)
 
And when we had that fight out in the rain
 
You ran after me and called my name
 
I never wanna see you walk away
 
(And there's a lot of lame guys out there)
 
'Cause one of these things is not like the others
 
Livin' in winter, I am your summer
 
Baby doll, when it comes to a lover
 
I promise that you'll never find another like
 
Me-e-e, ooh-ooh-ooh-ooh
 
I'm the only one of me
 
Let me keep you company
 
Eeh-eeh-eeh, ooh-ooh-ooh-ooh
 
You're the only one of you
 
Baby, that's the fun of you
 
And I promise that nobody's gonna love you like me-e-e
 
Hey, kids!
 
Spelling is fun!
 
Girl, there ain't no I in "team"
 
But you know there is a "me"
 
Strike the band up, one, two, three
 
I promise that you'll never find another like me
 
Girl, there ain't no I in "team"
 
But you know there is a "me"
 
And you can't spell "awesome" without "me"
 
I promise that you'll never find another like
 
Me-e-e (yeah), ooh-ooh-ooh-ooh (and I want ya, baby)
 
I'm the only one of me (I'm the only one of me)
 
Baby, that's the fun of me (baby, that's the fun of me)
 
Eeh-eeh-eeh, ooh-ooh-ooh-ooh (oh)
 
You're the only one of you (oh)
 
Baby, that's the fun of you
 
And I promise that nobody's gonna love you like me-e-e
 
Girl, there ain't no I in "team" (Ooh-ooh-ooh-ooh)
 
But you know there is a "me"
 
I'm the only one of me (Oh-oh)
 
Baby, that's the fun of me
 
(Eeh-eeh-eeh, ooh-ooh-ooh-ooh)
 
Strike the band up, one, two, three
 
You can't spell "awesome" without "me"
 
You're the only one of you
 
Baby, that's the fun of you
 
And I promise that nobody's gonna love you like me-e-e
 
  
stop reading this
 
stop
 
right now
 
buoi
 
aba-da-anena-ba-NYAH!
 
YA YA
 
 
==Solution 1==
 
==Solution 1==
  
Line 85: Line 13:
  
 
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.
zzzzzzzzzzzzzzzzzzzz
 
  
 
==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 when Zoe and she did the homework together is <math>176-80=96</math>, which is the total amount of problems she got correct, subtracted by the number of correct problems she did alone. 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.Boy I'm really boutta get to yo pickle chin ahh boy, egg head like collard greens head ass boy, oh hell nah boy yo dirt ahh boy stank ahh boy afro head ahh, lip gloss chin ahh boy ugly ahh boi *gagging*
+
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 when Zoe and she did the homework together is <math>176-80=96</math>, which is the total amount of problems she got correct, subtracted by the number of correct problems she did alone. 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 16:08, 22 June 2019

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 when Zoe and she did the homework together is $176-80=96$, which is the total amount of problems she got correct, subtracted by the number of correct problems she did alone. 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
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