A baker makes chocolate cookies and peanut cookies. His

Question

Source: Manhattan Prep 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 OA is E.

Brent@GMATPrepNow · Answer

BTGmoderatorLU wrote:Source: Manhattan Prep 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 OA is E. 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

fskilnik@GMATH · Answer

BTGmoderatorLU wrote:Source: Manhattan Prep 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.

Scott@TargetTestPrep · Answer

BTGmoderatorLU wrote:Source: Manhattan Prep 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. Answer: E

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