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2017 AMC 8 Problems/Problem 14

Revision as of 23:50, 18 June 2019 by Richbill (talk | contribs) (→‎Problem 14)

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?

$\textbf{(A) }89\qquad\textbf{(B) }92\qquad\textbf{(C) }93\qquad\textbf{(D) }96\qquad\textbf{(E) }98$ 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

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. zzzzzzzzzzzzzzzzzzzz

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 (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

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