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A baker makes chocolate cookies and peanut cookies. His

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A baker makes chocolate cookies and peanut cookies. His [#permalink] New post 17 May 2012, 02:32
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A baker makes chocolate cookies and peanut cookies. His recipes allow him to make chocolate cookie in batches of 7 and peanut cookies in batches of 6. If he makes exactly 95 cookies, what is the minimum number of chocolate chip cookies he makes?

A. 7
B. 14
C. 21
D. 28
E. 35
[Reveal] Spoiler: OA
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Re: Algebra [#permalink] New post 17 May 2012, 04:45
kashishh wrote:
A baker makes chocolate cookies and peanut cookies. His recipes allow him to make chocolate cookie in batches of 7 and peanut cookies in batches of 6. i f he makes exactly 95 cookies, what is the minimum number of chocolate chip cookies he makes?
A. 7
B. 14
C. 21
D. 28
E. 35


7C+6P=95
We need to maximize P to minimize C so that the eq is also satisfied
Try substitution for C & P to solve so that eqn is satisfied

The least value of C for which equation gets satisfied is 5
i.e. 7*5+6*10=35+60=95
Hence E is the answer
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Re: A baker makes chocolate cookies and peanut cookies. His [#permalink] New post 17 May 2012, 04:48
Expert's post
kashishh wrote:
A baker makes chocolate cookies and peanut cookies. His recipes allow him to make chocolate cookie in batches of 7 and peanut cookies in batches of 6. If he makes exactly 95 cookies, what is the minimum number of chocolate chip cookies he makes?

A. 7
B. 14
C. 21
D. 28
E. 35


Say x is the number of chocolate cookies and y is the number of peanut cookies Bob makes. Notice that since chocolate cookies are in batches of 7 and peanut cookies are in batches of 6 then x must be a multiple of 7 and y must be a multiple of 6.

Given: x+y=95 --> y=95-x. We want to minimize x, so we need to find the minimum value of a multiple of 7 (x) that must be subtracted from 95 to get a multiple of 6 (y). The minimum value turns out to be 35=5*7: 95-35=60=6*10.

Answer: E.
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A baker makes chocolate cookies and peanut cookies. His [#permalink] New post 20 Dec 2014, 15:41
kashishh wrote:
A baker makes chocolate cookies and peanut cookies. His recipes allow him to make chocolate cookie in batches of 7 and peanut cookies in batches of 6. If he makes exactly 95 cookies, what is the minimum number of chocolate chip cookies he makes?

A. 7
B. 14
C. 21
D. 28
E. 35


The first thing I noticed is that the answer choice must be odd. Since 6 is even, the quantity of peanut butter cookies must be even, so in order to add to 95, which is odd, the quantity of chocolate cookies must be odd. That eliminates answer choices B and D.

To find the answer, I started with the lowest odd answer choice, subtracted that number from 95, and found if it was divisible by 6.

A) 95 - 7 = 88. 88 is not divisible by 6 because 8+8 is not divisible by 3. Eliminate A.
C) 95 - 21 = 74. 74 is not divisible by 6 because 7+4 is not divisible by 3. Eliminate C.

This leaves only answer choice E remaining.

The answer is E.

Hope this method helps!
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Re: A baker makes chocolate cookies and peanut cookies. His [#permalink] New post 30 Dec 2014, 00:43
We require to check divisibility of (95 - OA) by 6

95 is not divisible by 6, however 95+1 = 96 is divisibly by 6

So, adding 1 to the OA, only 35+1 = 36 is divisible by 6

Answer = E
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Re: A baker makes chocolate cookies and peanut cookies. His   [#permalink] 30 Dec 2014, 00:43
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