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A certain store will order 25 crates of apples. The apples

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A certain store will order 25 crates of apples. The apples

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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.

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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!

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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
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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.

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Fabio Skilnik :: GMATH method creator ( Math for the GMAT)
English-speakers :: https://www.gmath.net
Portuguese-speakers :: https://www.gmath.com.br

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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 ≈ 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, it’s the least possible number of crates for the variety of apples - Winesap - that has the greatest number of crates.

Alternate Solution:

Let’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

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scott@targettestprep.com



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