A baker makes a combination of chocolate chip cookies and peanut butte

Question

A baker makes a combination of chocolate chip cookies and peanut butter cookies for a school bake sale. His recipes only allow him to make chocolate chip cookies in batches of 7 and peanut butter cookies in batches of 6. If he makes exactly 95 cookies for the bake sale, what is the minimum number of chocolate chip cookies that he could make?

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

Show Answer

Official Answer

E

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

BrentGMATPrepNow · Accepted Answer

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

narangvaibhav · Answer

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

Bunuel · Answer

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.

pate13 · Answer

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!

PareshGmat · Answer

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

fskilnik · Answer

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

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

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

{	ext{? = }}{\left( {{	ext{7x}}} ight)_{\,\min }}

{\left( {{	ext{7x}}} ight)_{\,\min }}\,\,\, \Leftrightarrow \,\,\,{x_{\min }}\,\,\,

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

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

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

Regards, 
fskilnik.

ScottTargetTestPrep · Answer

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.

Answer: E

yashbabu · Answer

LCM OF 95 is 5 and 19
so smaller number is 5 or multiple of 5 which is 35

yashbabu · Answer

LCM OF 95 is 5 and 19
so smaller number is 5 or multiple of 5 which is 35

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A baker makes a combination of chocolate chip cookies and peanut butte

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