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

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Ved22

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09 Apr 2019, 11:09

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Azedenkae

Winesap = W
McIntosh = M
Rome = R

W>M, W>R

So 'best' case scenario is that M = R, and W must be higher than both.

Let's just make M = n, then R = n, and W = n + x

Total apple crates is 3n + x = 25. n has to be highest possible. In this case n = 8 is highest possible (25/3=8.33...). x = 25 - 24 = 1. W = 8 + 1 = 9.

Answer is C.

This stops you from having to test the answer by substituting values into the equation.

SO GLAD SAME SOLUTION FOLLOWED BY ME TOO

ramgo11

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28 May 2019, 00:09

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chetan2u

Bunuel

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

NEW question from GMAT® Official Guide 2019

(PS01233)

W>M and W>R...

To minimize W, maximii the other two..
If all three are equal, each would be 25/3=8 1/3 so slightly more than 8..
So let the other two be 8, total =2*8=16
W =25-16=9

C

I get that the answer is option C. But why should we consider that M and R would be not 0? Why cannot we assume that R could be 0 (which still makes W>R), and both W and M should equal the 25 crates? In that case, the least number of crates of W would be 13 (since M could be 12).

Thanks!

bawatwr

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08 Jun 2019, 14:57

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Hi,

I have a doubt here. Let's say we have least possible number of W as 9 according to the given answer so remaining distribution would be between R and M, since there is no explicit relation given among them or any constraint of on their quantity we can say that M can be 10 and R would be 6 in that case W<M and hence violates the law of question.

My answer would be 13 for least W as in that case remaining would be 11 and hence one cannot violate the equation of W>M and W>R. Also you can have any value greater than 13 for W and hence maximum possible would be 23 and 1 for each R and M.I am confused here it would be helpful if someone would throw light on this or in case I have misunderstood the question.

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03 Sep 2019, 03:41

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(x-1)+2x=25 –> x=9

colinlin1

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03 Sep 2019, 06:43

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There are 3 categories of apple.
We can divide 25 by 3, get 8 and remainder 1.
So we can get 9.

MathRevolution

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12 Sep 2020, 00:25

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Total crates: 25

=> 3 Varieties: McIntosh, Rome, and Winesap

=> Conditions: Winesap > McIntosh (crates) and Winesap > Rome (crates)

So, the number of crates of Mcintosh can be lesser than those of crates of Rome, or they both can be equal.

=> A rough division of 25 in 3 gives us 9, 8, 8

Winesap: 9; McIntosh: 8 and Rome: 8

Any other combinations will increase the crates of Winesap. For example McIntosh : 8 > Rome 7 will give Winesap = 10.

So, the minimum possible crates for Winseap = 9

Answer C

ABCDEFU

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10 Oct 2020, 06:41

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Wanted to see if my approach is correct:

W + M + R = 25

&, W > M & W>R is what we want!

So, adding the 2 inequalities,
2W > M + R

Now, from the first equation, we have M + R = 25 - W

Put this in the inequality,

2W > 25-W => 3W > 25 => W > 8.something.

Since create has to be an integer and has to be greater than 8, next smallest possible value is 9.

Hence, answer is (C)

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24 Dec 2020, 06:44

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Bunuel

If the 25 crates are equally divided then \(\frac{25}{3}=8^+\); Given that \(W>M, W>R\)

So, if W takes 9 then M and R take 8 each. This will least possible option for W under the given condition.

The answer is \(C\)

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06 May 2021, 12:02

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target find least value which is greater than the other two options of apples i.e Mcintosh & Rome
using answer options
if Winesap is 8 then other two are 7 each ; 25-14 ; 11 which is not valid
if winesapp is 9 then other two are max 8 each ; 25-16 ; 9 correct and matching to option
C is correct

Bunuel

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10 Nov 2022, 04:17

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Bunuel

Given:
M + R + W =25
M < W -------(1)
R < W -------(2)
Adding 1 & 2
M + R < 2W
Adding W both sides
M + R + W < 3W

=> 25 < 3W

W > 25/3
W > 8.33

W = 9

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26 Jul 2024, 09:23

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Valhalla

Let
McIntosh = McI ; Rome = R ; Winesap = W
Given
W > McI ; W > R
W + McI + R = 25

To minimize W , McI and R must be maximize ; Therefore McI = R = (W-1)
Therefore
W + R + McI = 25
w + (w-1) + (w-1) = 25
w = 9

Um why did you put McI and R value as W-1? Why -1 particularly. Cause we could have chosen W+1 and gotten a lesser W value in the end!

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