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# A baker makes chocolate cookies and peanut cookies. His

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Posts: 125
A baker makes chocolate cookies and peanut cookies. His  [#permalink]

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17 May 2012, 02:32
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5
00:00

Difficulty:

35% (medium)

Question Stats:

77% (02:05) correct 23% (02:01) wrong based on 183 sessions

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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
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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]

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17 May 2012, 04:48
1
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.

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A baker makes chocolate cookies and peanut cookies. His  [#permalink]

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20 Dec 2014, 15:41
1
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]

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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

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Re: A baker makes chocolate cookies and peanut cookies. His  [#permalink]

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04 Sep 2018, 14:13
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Top Contributor
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

We're looking for the smallest possible number of chocolate chip cookies.
So, let's start by testing answer choice A

A) If we make 7 chocolate chip cookies, then the remaining 88 cookies are peanut cookies.
We're told that peanut cookies are baked in batches of 6
However, 88 is NOT divisible by 6, which means there cannot be 88 peanut cookies.
ELIMINATE A

B) If we make 14 chocolate chip cookies, then the remaining 81 cookies are peanut cookies.
However, 81 is NOT divisible by 6, which means there cannot be 81 peanut cookies.
ELIMINATE B

C) If we make 21 chocolate chip cookies, then the remaining 74 cookies are peanut cookies.
However, 74 is NOT divisible by 6, which means there cannot be 74 peanut cookies.
ELIMINATE C

D) If we make 28 chocolate chip cookies, then the remaining 67 cookies are peanut cookies.
However, 67 is NOT divisible by 6, which means there cannot be 67 peanut cookies.
ELIMINATE D

By the process of elimination, the correct answer is E

Cheers,
Brent
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Re: A baker makes chocolate cookies and peanut cookies. His  [#permalink]

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04 Sep 2018, 18:22
GMATPrepNow 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

$$x \geqslant 1\,\,\,{\text{choco}}\,\,{\text{batches}}{\text{,}}\,\,\,\,\,{\text{7}}\,\,{\text{choco/batch}}$$

$$y \geqslant 1\,\,\,{\text{pean}}\,\,{\text{batches}}{\text{,}}\,\,\,\,\,{\text{6}}\,\,{\text{pean/batch}}$$

$$7x + 6y = 95\,\,\,\,\,\left( * \right)$$

$${\text{? = }}{\left( {{\text{7x}}} \right)_{\,\min }}$$

$${\left( {{\text{7x}}} \right)_{\,\min }}\,\,\, \Leftrightarrow \,\,\,{x_{\min }}\,\,\,$$

$${\left( {multiple\,\,of\,\,6} \right)_{\max }} = {\left( {6y} \right)_{\max }}\mathop = \limits^{\left( * \right)} \,\,95 - 7x$$

$$\begin{gathered} x = 1\,\,\, \Rightarrow \,\,\,95 - 7x = 88\,\,{\text{not}}\,\,{\text{divisible}}\,\,{\text{by}}\,\,3\, \hfill \\ x = 2\,\,\, \Rightarrow \,\,\,95 - 7x = {\text{odd}}\,\,\left( {{\text{not}}\,\,{\text{divisible}}\,\,{\text{by}}\,\,2} \right) \hfill \\ x = 3\,\,\, \Rightarrow \,\,\,95 - 7x = 74\,\,{\text{not}}\,\,{\text{divisible}}\,\,{\text{by}}\,\,3 \hfill \\ x = 4\,\,\, \Rightarrow \,\,\,95 - 7x = {\text{odd}}\,\,\left( {{\text{not}}\,\,{\text{divisible}}\,\,{\text{by}}\,\,2} \right) \hfill \\ x = 5\,\,\, \Rightarrow \,\,\,95 - \boxed{7x = 35} = {\text{60}}\,\,\underline {{\text{divisible}}\,\,{\text{by}}\,\,6!} \hfill \\ \end{gathered}$$

The above follows the notations and rationale taught in the GMATH method.

Regards,
fskilnik.
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Re: A baker makes chocolate cookies and peanut cookies. His  [#permalink]

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07 Sep 2018, 16:12
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

We can let c = the number of batches of chocolate chip cookies made and p = the number of batches of peanut cookies made and create the equation:

7c + 6p = 95

7c = 95 - 6p

c = (95 - 6p)/7

In order for c to be an integer, we need (95 - 6p) to be a multiple of 7.

Let’s start with the largest value of p and work our way down.

When p is 15, we have:

c = 5/7... this does not work

When p is 14, we have:

c = 11/7… this does not work

When p is 13, we have:

c = 17/7... this does not work

When p is 12, we have:

c = 23/7… this does not work

When p is 11, we have:

c = 29/7... this does not work

When p is 10, we have:

c = 35/7 = 5... this works!

So, the minimum number of batches of chocolate chip cookies is 5, and, thus, the minimum number of chocolate chip cookies is 5 x 7 = 35.

Alternate Solution:

Let’s test each answer choice, starting with the smallest value:

Answer Choice A: c = 7

If c = 7, then the number of peanut butter cookies is 95 - 7 = 88; however, since 88 is not divisible by 6, c = 7 is not possible.

Answer Choice B: c = 14

If c = 14, then the number of peanut butter cookies is 95 - 14 = 81; however, since 81 is not divisible by 6, c = 14 is not possible.

Answer Choice C: c = 21

If c = 21, then the number of peanut butter cookies is 95 - 21 = 74; however, since 74 is not divisible by 6, c = 21 is not possible.

Answer Choice D: c = 28

If c = 28, then the number of peanut butter cookies is 95 - 28 = 67; however, since 67 is not divisible by 6, c = 28 is not possible.

Since we eliminated every other answer choice, we know by this point that the correct answer is E; however, let’s verify this as an exercise:

Answer Choice E: c = 35

If c = 35, then the number of peanut butter cookies is 95 - 35 = 60, which is a possible value since 60 is divisible by 6.

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Re: A baker makes chocolate cookies and peanut cookies. His   [#permalink] 07 Sep 2018, 16:12
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# A baker makes chocolate cookies and peanut cookies. His

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