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Missy ate m more crackers than Audrey did, from a box that originally

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LakerFan24
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Missy ate m more crackers than Audrey did, from a box that originally [#permalink] New post   Updated on: 19 Apr 2017, 21:17
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Missy ate m more crackers than Audrey did, from a box that originally contained n crackers. Together, they finished the box. Which of the following represents the number of crackers that Missy ate?

A) \(\frac{(n+m)}{2n}\)

B) \(\frac{(m-n)}{2}\)

C) \(\frac{(n-m)}{2}\)

D) \(\frac{(m+n)}{2}\)

E) \(\frac{(n}{2+m)}\)
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D

Originally posted by LakerFan24 on 19 Apr 2017, 19:23.
Last edited by Bunuel on 19 Apr 2017, 21:17, edited 1 time in total.
Renamed the topic.
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Re: Missy ate m more crackers than Audrey did, from a box that originally [#permalink] New post   19 Apr 2017, 20:10
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LakerFan24 wrote:
Missy ate m more crackers than Audrey did, from a box that originally contained n crackers. Together, they finished the box. Which of the following represents the number of crackers that Missy ate?

A) \(\frac{(n+m)}{2n}\)

B) \(\frac{(m-n)}{2}\)

C) \(\frac{(n-m)}{2}\)

D) \(\frac{(m+n)}{2}\)

E) \(\frac{(n}{2+m)}\)


Hi...

I. Logically..
Since Missy ate more than Audrey and both finished n crackers..
Missy has to eat MORE than half that is >\(\frac{n}{2}\)..
Only D is left..

II. Substitution..
Take total as 10 and M ate 6 and A ate 4...
So m =6-4=2 and n=10..
Substitute in choices and see which gives answer as 6..

A) \(\frac{(n+m)}{2n}....=\frac{10+2}{2*10}\)..No

B) \(\frac{(m-n)}{2}...=\frac{2-10}{2}\)..no

C) \(\frac{(n-m)}{2}...=\frac{10-2}{4}\)..no

D) \(\frac{(m+n)}{2}=\frac{10+2}{2}\).. Yes

E) \(\frac{(n}{2+m)}=10/2+2\)..No

D

III. Algebra..
Let Missy eat x and Audrey eat y..
x-y=m and x+y=n....
Since we are looking for X, add two equations..
So x-y+x+y=m+n......2x=m+n.......x=(m+n)/2
D
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Re: Missy ate m more crackers than Audrey did, from a box that originally [#permalink] New post   19 Apr 2017, 21:05
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LakerFan24 wrote:
Missy ate m more crackers than Audrey did, from a box that originally contained n crackers. Together, they finished the box. Which of the following represents the number of crackers that Missy ate?

A) \(\frac{(n+m)}{2n}\)

B) \(\frac{(m-n)}{2}\)

C) \(\frac{(n-m)}{2}\)

D) \(\frac{(m+n)}{2}\)

E) \(\frac{(n}{2+m)}\)


let a=crackers A ate
a+m=crackers M ate
n=2a+m
a=(n-m)/2
a+m=(m+n)/2
D
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Re: Missy ate m more crackers than Audrey did, from a box that originally [#permalink] New post   26 Jul 2017, 19:52
Missy ate m more crackers than Audrey did, from a box that originally contained n crackers. Together, they finished the box. Which of the following represents the number of crackers that Missy ate?

A) (n+m)2n(n+m)2n

B) (m−n)2(m−n)2

C) (n−m)2(n−m)2

D) (m+n)2(m+n)2

E) (n2+m)

** Sorry I don't want to waste time formatting the A/C, you can find them neater above **


What was key here for me is manipulating the Question into algebra:

M=A+m; A=?; n=total --> This translates to: A+(A+m)=n

- If you got lost: you're basically adding Missy and Audry to get "n", b/c you know they are the only 2 ppl we're talking about who are eating crackers and they finished the box, hence setting them equal to "n".

- This simplifies to: 2A+m=n

Now, let's play around with numbers. If we set n=20, A=9, m=2, M=(9+2) so M=11. NEED TO FIND 11 IN THE A/C
A) 11/20 - wrong;

B) negative # - wrong;

C) 9 - wrong

D) 11 - correct

E) 5 - wrong
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Re: Missy ate m more crackers than Audrey did, from a box that originally [#permalink] New post   22 Jan 2018, 20:48
Picking numbers strategy

Mi = Missy
a = Audrey

I chose n = 5, a = 1 and left with m = 3

(given) Mi = m + a
Mi = 3 + 1
Mi = 4
n = Mi + a

Looking for an answer choice that results in 4

D

LakerFan24 wrote:
Missy ate m more crackers than Audrey did, from a box that originally contained n crackers. Together, they finished the box. Which of the following represents the number of crackers that Missy ate?

A) \(\frac{(n+m)}{2n}\)

B) \(\frac{(m-n)}{2}\)

C) \(\frac{(n-m)}{2}\)

D) \(\frac{(m+n)}{2}\)

E) \(\frac{(n}{2+m)}\)
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Missy ate m more crackers than Audrey did, from a box that originally [#permalink] New post   07 Nov 2019, 04:26
\(S=m+A \implies A=S-m\)

\(A+(m+A)=n\)

\((S-m)+S=n \implies S=\frac{(n+m)}{2}\)
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Re: Missy ate m more crackers than Audrey did, from a box that originally [#permalink] New post   14 Sep 2023, 17:39
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Re: Missy ate m more crackers than Audrey did, from a box that originally [#permalink]
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