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

Question

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

(PS01233)

chetan2u · Accepted Answer

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

avigutman · Answer

Video solution from Quant Reasoning: Subscribe for more: https://www.youtube.com/QuantReasoning? ... irmation=1 https://www.youtube.com/watch?v=sEWl1gjT1bI

CrackverbalGMAT · Answer

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)

It's given 
 W > M  and  W > R

#Approach 1: 
In order to find the least possible number of crates of Winesap (W)that the store will order, Values of M and R should be maximum .

Let's say that no of crates of Winesap, W =  x

Max possible value of M =  x-1  Its given that W> M ,also the no of crates should be an integer.
Max possible value of R =  x-1   W > R

Total no of crates ordered in the store = 25

W + M + R = 25

x + x-1 + x-1 = 25

3x -2 = 25

3x = 27

x = 9

So the least possible number of crates of Winesap = 9

Option C

#Approach 2 : Analyzing the answer options

Let's start with Option C:9
i.e W =9 .
Max possible value of M = 8 (W> M)
Max possible value of R = 8 (W> R)
W + M + R =25
9 + 8 + 8 = 25
9,8,8 combination will satisfy all the conditions. As of now we cannot confirm that 9 is the least possible option but we can eliminate Option D and E as we don't need to look for any number greater than 9 since you are asked to find the least possible value.

Let's Check option B to see if any values less than 9 is a possible option.
Option B: 8
i.e W =8.
Max possible value of M = 7 (W> M)
Max possible value of R = 7 (W> R)
W + M + R =25
8 + 7 + 7 = 22
Since the sum of the crates is less than 25 , we can confirm that least possible value of W should be greater than 8.
Option B and A are eliminated.

Only option left is C.

Thanks,
  Clifin J Francis,
GMAT SME

ScottTargetTestPrep · Answer

On average, each variety of apple 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 McIntosh crates

If there are 7 McIntosh crates, then there are 25 - 7 = 18 crates of Winesap and Rome crates. If there are 18 crates of Winesap and Rome crates combined, then either one of these crates has to be more than the number of McIntosh crates (which is 7). This is because, if both the number of Winesap and Rome crates are less than 7, then there can be at most 12 remaining crates, but we have 18.

Answer Choice B: 8 McIntosh crates

Similar to the above discussion, there are 25 - 8 = 17 crates of Winesap 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 McIntosh crates

In this case, there are 25 - 9 = 16 crates of Winesap and Rome crates. In this case, we observe that the store could have ordered 8 crates of the Winesap and Rome apples each, so it is possible for the store to have ordered 9 McIntosh crates. Since we are looking for the smallest value, this is the value we are looking for.

Answer: 9 (C)

GMATinsight · Answer

Answer: Option C

Video solution by  GMATinsight

https://www.youtube.com/watch?v=28LAGgDeoLM

pikolo2510 · Answer

Given :-
W>M
W>R

Solution: -
so possible combinations are
W>M>R or
W>R>M or
W>R=M

Since we want to find out the least possible W , we need to max R and M

Go by options

If W = 10
then R + M = 15
R or M can be 7 or 8 OR 8 or 7

We need not check W=11 since we know W=10 is possible

If W = 9
then R + M = 16
R or M can be 8 each , Hence this possible and we cant discard W=10 as a possibility

lets check for W = 8
then R + M = 17
then R or M has to be 9 or 8 which is not possible

Hence answer should be C

@Experts - looking for a shorter solution. thanks

Azedenkae · Answer

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.

chippoochan · Answer

The problem given 3 types of apples which are
McIntosh(M), Rome(R), and Winesap(W)
and the amount of each type of crates are given that
W>M and W>R
From the problem, total number of ordered crates is 25 crates
to minimize the possible number of W means that we have to maximize the number of M and R
So I solve this problem by dividing total number of crates by 3 --->  25/3=8.33 
since the number of crates must be integers and W must be more than R and M, so I rounded up the number of W and round-down the number of M and R.
The minimum possible value of W is then 9 with R and M are 8 each (9+8+8=25)
so my answer is "C"

Valhalla · Answer

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

dave13 · Answer

here is a shortcut to problem but first of all forget the math

Distrubue simply values proprtionally yet keep in mind W must be bigger let M, R and W be respectively 8+8+9 = 25 i think in all MAX/MIN the best approach is at first to distribute value proportionally initially if possible, if is it not possible then the remainder add to the value we need to maximize

From there you can easily see a bigger picture quickly

lfc · Answer

just use the options,

if w=7, at max m and r can be 6...but 7+6+6 is not equal to 25.
Keep going until you get C as the answer

gracie · Answer

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)

A. 7
 
B. 8
 
C. 9
 
D. 10
 
E. 11

let x=number of M 
let x=number of R 
let x+1=number of W
2x+(x+1)=25
x=8
x+1=9
C

fskilnik · Answer

?\,\,\, = \,\,\,\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.

nust2017 · Answer

Hi everyone. Would be really glad if someone can help on this. Why did we not consider option D or E?

I get that we choose C, because 9 perfectly fits in for W, and we divided R and M to be 8 each, but my question, and the point that I want clarity on is why didn't we choose option D or E, since in both the options, it does tick the boxes for the main criteria of the question which W> M, and W> R, so assuming W is 10, M can be 8, and R can be 7, or assuming W is 11, then M can be and 7, and R can be 7.

Just puzzled about this, and would really appreciate clarity. Thanks in advance

fskilnik · Answer

Hi  nust2017  ,

what is the  LEAST POSSIBLE  number of crates of Winesap that the store will order?

Regards,
Fabio.

nust2017 · Answer

Thanks a lot. That really helps. Appreciate it mate.

fskilnik · Answer

I am glad I could help.

See you in other posts!

Regards,
Fabio.

ShiqiangHAN · Answer

1) W+M+R=25;
2) W>M;
3) W>R;
2)+3) => 2W>M+R=25-W
=> 2W>25-W 
=> W>25/3, Wmin=9

Totyanik12 · Answer

We need not check W=11 since we know W=10 is possible

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