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M03-07

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M03-07  [#permalink]

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New post 16 Sep 2014, 00:19
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A
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D
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  45% (medium)

Question Stats:

66% (01:37) correct 34% (01:35) wrong based on 128 sessions

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Foodmart customers regularly buy at least one of the following products: milk, chicken, or apples. 60% of shoppers buy milk, 50% buy chicken, and 35% buy apples. If 10% of the customers buy all 3 products, what percentage of Foodmart customers purchase exactly 2 of the products listed above?

A. 5%
B. 10%
C. 15%
D. 25%
E. 30%

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Re M03-07  [#permalink]

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New post 16 Sep 2014, 00:19
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Official Solution:

Foodmart customers regularly buy at least one of the following products: milk, chicken, or apples. 60% of shoppers buy milk, 50% buy chicken, and 35% buy apples. If 10% of the customers buy all 3 products, what percentage of Foodmart customers purchase exactly 2 of the products listed above?

A. 5%
B. 10%
C. 15%
D. 25%
E. 30%


Even though this is not a typical 2-group problem with overlapping members, we can still apply the group formula: \(Total = G_1 + G_2 + N - B\), where \(G_1, G_2\) are group 1 and group 2, \(N\) is neither, and \(B\) is both. Because we have more than 2 groups, we need to adjust the formula to reflect customers purchasing 3 products and therefore being members of 3 groups, and being counted as 3 distinct customers. The formula needs to be modified as follows: \(Total = G_1 + G_2 + G_3 + N - B - T * (3 - 1)\), where \(T\) is members of three groups and \(B\) is members of only two groups.
\(100\% = 60\% + 50\% + 35\% - B - 2 * 10\%\)
\(100\% = 145\% - B - 20\%\)
\(100\% = 125\% - B\)
\(B = 25\%\)

There is no neither group because all customers purchase at least one product.

Alternative Explanation:

Assume there are 100 individual buyers. Let \(C_i\) represent the customers who regularly buy 1, 2, or 3 products, where \(i\) is the number of products. Multiply \(C_2\) by 2 and \(C_3\) by 3 to accurately represent the number of times these buyers were counted since they have purchased chicken and apples or milk, chicken, and apples but in reality are an individual buyer counted multiple times. Construct the following equations:
\(C_1 + 2 * C_2 + 3 * C_3 = M + C + A =\)
\(= 60\% + 50\% + 35\% =\)
\(= 145\%\)
\(C_1 + C_2 + C_3 = 100\%\)
\(C_3 = 10\%\)

Subtract the second equation from the first one:
\(C_2 + 2 * C_3 = 45\%\)
\(C_2 + 2 * 10\% = 45\%\)
\(C_2 = 25\%\)

Alternative Explanation 2:

Apply the formula for 3 overlapping sets:

100%={customers who buy milk}+{customers who buy chicken}+{customers who buy apples} - {customer who buy exactly 2 products} - 2*{customers who by exactly 3 products}+{customers who buy neither of the products}

\(100=60+50+35-x-2*10+0\) --> \(x=25\).


Answer: D
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Re M03-07  [#permalink]

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New post 17 Mar 2015, 11:47
I think this question is good and helpful.
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Re M03-07  [#permalink]

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New post 29 Aug 2015, 23:30
I think this is a high-quality question and I don't agree with the explanation. I think the answer is 55%.
n(AUBUC) = n(A) + n(B) + n(C) -n(AnB)-n(BnC)-n(CnA) + n(AnBnC)
Thus
100 = 60+50+35 -x +10
Therefore, x = 55%
Alternate,
through venn diagram,
if n(AnB) = x; n(BnC) = y; n(CnA) = z
then adding each one individually;
[60-x-(y-10)+50-x-(z-10)+35-z-(y-10)] + [x-10+y-10+z-10]+ [10] = 100
will result in x+y+z = 55%
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Re: M03-07  [#permalink]

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New post 30 Aug 2015, 00:59
srkatikala wrote:
I think this is a high-quality question and I don't agree with the explanation. I think the answer is 55%.
n(AUBUC) = n(A) + n(B) + n(C) -n(AnB)-n(BnC)-n(CnA) + n(AnBnC)
Thus
100 = 60+50+35 -x +10
Therefore, x = 55%
Alternate,
through venn diagram,
if n(AnB) = x; n(BnC) = y; n(CnA) = z
then adding each one individually;
[60-x-(y-10)+50-x-(z-10)+35-z-(y-10)] + [x-10+y-10+z-10]+ [10] = 100
will result in x+y+z = 55%


The formula you are using is not correct. It should be:

{Total} = {Group 1} + {Group 2} + {Group 3} - {those in 2 of the groups} - 2*{those in all 3 groups}
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Collection of Questions:
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Re: M03-07  [#permalink]

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New post 02 Nov 2015, 10:08
Bunuel wrote:
srkatikala wrote:
I think this is a high-quality question and I don't agree with the explanation. I think the answer is 55%.
n(AUBUC) = n(A) + n(B) + n(C) -n(AnB)-n(BnC)-n(CnA) + n(AnBnC)
Thus
100 = 60+50+35 -x +10
Therefore, x = 55%
Alternate,
through venn diagram,
if n(AnB) = x; n(BnC) = y; n(CnA) = z
then adding each one individually;
[60-x-(y-10)+50-x-(z-10)+35-z-(y-10)] + [x-10+y-10+z-10]+ [10] = 100
will result in x+y+z = 55%


The formula you are using is not correct. It should be:

{Total} = {Group 1} + {Group 2} + {Group 3} - {those in 2 of the groups} - 2*{those in all 3 groups}


Could you explain why we multiply the last part by two please? I can't get the logic behind that
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Re: M03-07  [#permalink]

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New post 02 Nov 2015, 22:06
Icecream87 wrote:
Bunuel wrote:
srkatikala wrote:
I think this is a high-quality question and I don't agree with the explanation. I think the answer is 55%.
n(AUBUC) = n(A) + n(B) + n(C) -n(AnB)-n(BnC)-n(CnA) + n(AnBnC)
Thus
100 = 60+50+35 -x +10
Therefore, x = 55%
Alternate,
through venn diagram,
if n(AnB) = x; n(BnC) = y; n(CnA) = z
then adding each one individually;
[60-x-(y-10)+50-x-(z-10)+35-z-(y-10)] + [x-10+y-10+z-10]+ [10] = 100
will result in x+y+z = 55%


The formula you are using is not correct. It should be:

{Total} = {Group 1} + {Group 2} + {Group 3} - {those in 2 of the groups} - 2*{those in all 3 groups}


Could you explain why we multiply the last part by two please? I can't get the logic behind that


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Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


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Re: M03-07  [#permalink]

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New post 03 Nov 2015, 14:25
Great! Thank you Bunuel. I really need to learn my way through this website. So many useful materials.
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Re: M03-07  [#permalink]

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New post 10 Mar 2018, 14:23
1
Good question but on the easier end of 700 Qs. I'd consider more 600 level

Quick way to solve -

Add up all the percentages (145%)
Find up much over you are (45%)
Swap in people for percent - you have 100 people and 145 "counts". Question states that everyone buys at least one product so you can discount the possibility of some number of people buying 0 products (always important to consider this possibility)

10 people buy all 3 products so are counted 3 times - so these 10 people give you 30 "counts". Subtract 30 from your total counts (145) and 10 from your total number of people (100)
The remaining 90 people give you 115 counts. You have 25 counts extra so you know that 25 people are getting counted twice (ie buy 2 products).

Add it all up
1*65 + 2*25+ 3*10 = 65 + 50+ 3*10 = 145 (which is what we want)
65 + 25 + 10 = 100 (which is what we want)
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Re M03-07  [#permalink]

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New post 09 Aug 2018, 22:16
I think this is a high-quality question and I agree with explanation.
Re M03-07 &nbs [#permalink] 09 Aug 2018, 22:16
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