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100 points for $49 worth of Veritas practice GMATs FREE VERITAS PRACTICE GMAT EXAMS Earn 10 Points Per Post Earn 10 Points Per Thanks Earn 10 Points Per Upvote ## A certain store will order 25 crates of apples. The apples tagged by: BTGmoderatorLU ##### This topic has 4 expert replies and 0 member replies ### Top Member ## A certain store will order 25 crates of apples. The apples ## Timer 00:00 ## Your Answer A B C D E ## Global Stats Difficult Source: Official Guide A certain store will order 25 crates of apples. The apples will be of three different varieties-McIntosh, Rome, and Winesap-and each crate will contain apples of only one variety. If the store is to order more crates of Winesap than crates of McIntosh and more crates of Winesap than crates of Rome, what is the least possible number of crates of Winesap that the store will order? A. 7 B. 8 C. 9 D. 10 E. 11 The OA is C. ### GMAT/MBA Expert GMAT Instructor Joined 25 May 2010 Posted: 15348 messages Followed by: 1864 members Upvotes: 13060 GMAT Score: 790 BTGmoderatorLU wrote: Source: Official Guide A certain store will order 25 crates of apples. The apples will be of three different varieties-McIntosh, Rome, and Winesap-and each crate will contain apples of only one variety. If the store is to order more crates of Winesap than crates of McIntosh and more crates of Winesap than crates of Rome, what is the least possible number of crates of Winesap that the store will order? A. 7 B. 8 C. 9 D. 10 E. 11 Let W = Winesap crates, M = McIntosh crates, and R = Roma crates. Since a total of 25 crates are ordered, we get: W + M + R = 25. To MINIMIZE the value of W, we must MAXIMIZE the values of M and R. Since the store is to order more crates of Winesap than crates of McIntosh and more crates of Winesap than crates of Rome, W>M and W>R. Thus, M and R will be maximized as follows: M = W-1 (implying that the number of McIntosh crates is 1 less than the number of Winesap crates) R = W-1 (implying that the number of Roma crates is 1 less than the number of Winesap crates) Substituting M=W-1 and R=W-1 into W+M+R = 25, we get; W + (W-1) + (W-1) = 25 3W - 2 = 25 3W = 27 W = 9. The correct answer is C. Alternatively, we can PLUG IN THE ANSWERS, which represent the value of W. When the correct answer is plugged in, W+M+R = 25. Since the question stem asks for the least possible value of W, start with the smallest answer choice. A: 7 If W=7, then the greatest possible values for M and R are M=6 and R=6. In this case, W+R+M = 7+6+6 = 19. The sum is too small. Eliminate A. B: 8 If W=8, then the greatest possible values for M and R are M=7 and R=7. In this case, W+R+M = 8+7+7 = 22. The sum is too small. Eliminate B. C: 9 If W=9, then the greatest possible values for M and R are M=8 and R=8. In this case, W+R+M = 9+8+8 = 25. Success! _________________ Mitch Hunt Private Tutor for the GMAT and GRE GMATGuruNY@gmail.com If you find one of my posts helpful, please take a moment to click on the "UPVOTE" icon. Available for tutoring in NYC and long-distance. For more information, please email me at GMATGuruNY@gmail.com. Student Review #1 Student Review #2 Student Review #3 Free GMAT Practice Test How can you improve your test score if you don't know your baseline score? Take a free online practice exam. Get started on achieving your dream score today! Sign up now. ### GMAT/MBA Expert GMAT Instructor Joined 22 Aug 2016 Posted: 1975 messages Followed by: 30 members Upvotes: 470 BTGmoderatorLU wrote: Source: Official Guide A certain store will order 25 crates of apples. The apples will be of three different varieties-McIntosh, Rome, and Winesap-and each crate will contain apples of only one variety. If the store is to order more crates of Winesap than crates of McIntosh and more crates of Winesap than crates of Rome, what is the least possible number of crates of Winesap that the store will order? A. 7 B. 8 C. 9 D. 10 E. 11 The OA is C. Say the # of crates of McIntosh, Rome, and Winesap are x, y and z, respectively. Thus, x + y + z = 25 such that x > y and x > z. Note that there is no comparison given between y and z. Since we have to find out the least possible value of x, let's assume that x = y = z, thus from x + y + z = 25, we have x = y = z = 8.33 Let's try with the nearest greater value of 8.33 for x and the nearest smaller value of 8.33 for y and z, we have x = 9, y = z = 8. This works since 9 + 8 + 8 = 25. The correct answer: C Hope this helps! -Jay _________________ Manhattan Review GMAT Prep Locations: Manhattan Review Bangalore | Hyderabad GMAT Prep | Bangalore GMAT Courses | Mehdipatnam GRE Prep | and many more... Schedule your free consultation with an experienced GMAT Prep Advisor! Click here. ### GMAT/MBA Expert GMAT Instructor Joined 09 Oct 2010 Posted: 1449 messages Followed by: 32 members Upvotes: 59 BTGmoderatorLU wrote: Source: Official Guide A certain store will order 25 crates of apples. The apples will be of three different varieties-McIntosh, Rome, and Winesap-and each crate will contain apples of only one variety. If the store is to order more crates of Winesap than crates of McIntosh and more crates of Winesap than crates of Rome, what is the least possible number of crates of Winesap that the store will order? A. 7 B. 8 C. 9 D. 10 E. 11 $$?\,\,\, = \,\,\,\min \left( W \right)$$ $$M + R + W = 25\,\,\left( * \right)$$ $$\left\{ \matrix{ \,W > M \ge 1\,\,\,{\rm{ints}} \hfill \cr \,W > R \ge 1\,\,\,{\rm{ints}} \hfill \cr} \right.\,\,\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\left\{ \matrix{ \,W = M + k\,,\,\,\,\,\,M,k\,\, \ge 1\,\,\,{\rm{ints}} \hfill \cr \,W = R + j\,,\,\,\,\,\,R,j\,\, \ge 1\,\,\,{\rm{ints}} \hfill \cr} \right.\,\,\,\,\,\left( {**} \right)$$ $$\left( * \right) \cap \left( {**} \right)\,\,\,\,\, \Rightarrow \,\,\,\,\,M + R + {{\left( {M + R + k + j} \right)} \over 2} = 25\,\,\,\,\,\, \Rightarrow \,\,\,\,\,{3 \over 2}\left( {M + R} \right) + {{k + j} \over 2} = 25$$ $$\min \left( W \right)\,\,\,\,\,\mathop \Leftrightarrow \limits^{\left( ** \right)} \,\,\,\,\,\min \left( {k + j} \right)\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,k + j = 1 + 1 = 2\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,M + R = {2 \over 3}\left( {24} \right) = 16$$ $$?\,\,\,\mathop = \limits^{\left( {**} \right)} \,\,\,\,{{\left( {M + R} \right) + \left( {k + j} \right)} \over 2}\,\,\, = \,\,\,{{16 + 2} \over 2}\,\,\, = \,\,\,9$$ This solution follows the notations and rationale taught in the GMATH method. Regards, Fabio. _________________ Fabio Skilnik :: GMATH method creator ( Math for the GMAT) English-speakers :: https://www.gmath.net Portuguese-speakers :: https://www.gmath.com.br ### GMAT/MBA Expert GMAT Instructor Joined 25 Apr 2015 Posted: 2791 messages Followed by: 18 members Upvotes: 43 BTGmoderatorLU wrote: Source: Official Guide A certain store will order 25 crates of apples. The apples will be of three different varieties-McIntosh, Rome, and Winesap-and each crate will contain apples of only one variety. If the store is to order more crates of Winesap than crates of McIntosh and more crates of Winesap than crates of Rome, what is the least possible number of crates of Winesap that the store will order? A. 7 B. 8 C. 9 D. 10 E. 11 The OA is C. On average, each variety of apples is approximately 25/3 â‰ˆ 8 crates. So we can have 9 crates of Winesap and 8 crates of Rome and 8 crates of McIntosh, for a total of 25 crates. Since 9 is the closest number to the average, itâ€™s the least possible number of crates for the variety of apples - Winesap - that has the greatest number of crates. Alternate Solution: Letâ€™s try each answer choice, starting from the smallest. Answer Choice A: 7 Winesap crates If there are 7 Winesap crates, then there are 25 - 7 = 18 crates of McIntosh and Rome crates. If there are 18 crates of McIntosh and Rome crates combined, then either one of these crates has to be more than the number of Winesap crates (which is 7). This is because, if both the number of McIntosh and Rome crates are less than 7, then there can be at most 12 remaining crates, but we have 18. Answer Choice B: 8 Winesap crates Similar to the above discussion, there are 25 - 8 = 17 crates of McIntosh and Rome crates. Again, if one of these crates is less than 8, then the other one will definitely be greater than 8. Answer Choice C: 9 Winesap crates In this case, there are 25 - 9 = 16 crates of McIntosh and Rome crates. In this case, we observe that the store could have ordered 8 crates of the McIntosh and Rome apples each, so it is possible for the store to have ordered 9 Winesap crates. Since we are looking for the smallest value, this is the correct value. Answer: C _________________ Scott Woodbury-Stewart Founder and CEO scott@targettestprep.com See why Target Test Prep is rated 5 out of 5 stars on BEAT the GMAT. Read our reviews • Award-winning private GMAT tutoring Register now and save up to$200

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