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Manager
Joined: 02 Jun 2011
Posts: 149

Kudos [?]: 106 [0], given: 11

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17 May 2012, 03:32
2
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Difficulty:

25% (medium)

Question Stats:

79% (01:28) correct 21% (01:18) wrong based on 122 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
[Reveal] Spoiler: OA

Kudos [?]: 106 [0], given: 11

Manager
Affiliations: Project Management Professional (PMP)
Joined: 30 Jun 2011
Posts: 197

Kudos [?]: 83 [0], given: 12

Location: New Delhi, India

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17 May 2012, 05:45
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
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Vaibhav

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Math Expert
Joined: 02 Sep 2009
Posts: 42249

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17 May 2012, 05:48
1
KUDOS
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.

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Kudos [?]: 132540 [1], given: 12324

Manager
Joined: 02 Sep 2014
Posts: 77

Kudos [?]: 58 [0], given: 86

Location: United States
GMAT 1: 700 Q49 V37
GMAT 2: 700 Q47 V40
GMAT 3: 720 Q48 V41
GPA: 3.26
WE: Consulting (Consulting)

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20 Dec 2014, 16:41
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.

Hope this method helps!

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30 Dec 2014, 01:43
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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20 Feb 2017, 04:08
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