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M03#7 [#permalink] New post 02 Jan 2009, 13:33
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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 above products?

(A) 5%
(B) 10%
(C) 15%
(D) 25%
(E) 30%

[Reveal] Spoiler: OA
D

Source: GMAT Club Tests - hardest GMAT questions

Let b_i denote the percentage of buyers who regularly buy i products, and b_x the percentage of buyers who regularly purchase product x . We can construct these equations:
\left{ \begin{eqnarray*} b_1 + 2b_2 + 3b_3 &=& b_m+b_c+b_a = 60\% + 50\% + 35\% = 145\%\\ b_1+b_2+b_3 &=& 100\%\\ b_3 &=& 10\%\\ \end{eqnarray*}

Subtract the second one from the first one:
\begin{eqnarray*} b_2+2b_3 = 45\%\\ b_2 = 25\%\\ \end{eqnarray*}

Alternative explanation:

We can construct the equation:
\begin{eqnarray*} 100% = (60% + 50% + 35%) - x - 2*10%\\ \end{eqnarray*}

In the sum (60%+50%+35%) we count customers that buy two products twice, so we have to extract x . In the same sum (60%+50%+35%) we count customers that buy three products thrice, so we have to extract 2*10%. Therefore, all customers are counted only once.
The correct answer is D.


In the alternative explanation it is stated that customers that buy two products are counted twice, but aren't they counted three times? I don't quite understand.
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Re: M03#7 [#permalink] New post 02 Jan 2009, 16:24
snowy2009 wrote:
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 above products?

(C) 2008 GMAT Club - m03#7

* 5%
* 10%
* 15%
* 25%
* 30%

Let b_i denote the percentage of buyers who regularly buy i products, and b_x the percentage of buyers who regularly purchase product x . We can construct these equations:
\left{ \begin{eqnarray*} b_1 + 2b_2 + 3b_3 &=& b_m+b_c+b_a = 60\% + 50\% + 35\% = 145\%\\ b_1+b_2+b_3 &=& 100\%\\ b_3 &=& 10\%\\ \end{eqnarray*}

Subtract the second one from the first one:
\begin{eqnarray*} b_2+2b_3 = 45\%\\ b_2 = 25\%\\ \end{eqnarray*}

Alternative explanation:

We can construct the equation:
\begin{eqnarray*} 100% = (60% + 50% + 35%) - x - 2*10%\\ \end{eqnarray*}

In the sum (60%+50%+35%) we count customers that buy two products twice, so we have to extract x . In the same sum (60%+50%+35%) we count customers that buy three products thrice, so we have to extract 2*10%. Therefore, all customers are counted only once.
The correct answer is D.

In the alternative explanation it is stated that customers that buy two products are counted twice, but aren't they counted three times? I don't quite understand.



They are counted thrice but should be counted only once. Therefore, they are deducted twice. So now they are counted once.

Hope you got it. :-D
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Re: M03#7 [#permalink] New post 16 Nov 2009, 03:53
In the standard explanation the formula given is:

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.

What is the (3-1) given here? Is it always (3-1) or 2 when there are 3 groups?
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Re: M03#7 [#permalink] New post 17 Jan 2010, 17:26
I could use some help on this one. I have been wracking my brain and I can't seem to understand where the T*(3-1) comes from. I understand how to handle questions with 2 groups, but I haven't been able to figure out the 3 group ones.

Thanks in advance for the help.
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Re: M03#7 [#permalink] New post 02 Feb 2010, 22:29
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snowy2009 wrote:
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 above products?

(C) 2008 GMAT Club - m03#7

* 5%
* 10%
* 15%
* 25%
* 30%


m = 60
c = 50
a = 35
mca = 10
mc+ca+am = ?

total = m+c+a - (mc+ca+am) - 2mca
100 = 60+50+35 - (mc+ca+am) - 2(10)
mc+ca+am = 25
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Re: M03#7 [#permalink] New post 19 Mar 2010, 04:59
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D

This how i solved

60% buy milk
50% buy chicken
35% buy apples

10% buy all three

people who brought not all three:

60-10 = 50% milk
50-10 = 40% chicken
35-10 = 25% apples

total comes to 115% but we have only 90% of people (As 10% had brought all the three)

so, 115-90 = 25%
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Re: M03#7 [#permalink] New post 19 Mar 2010, 06:28
snowy2009 wrote:

In the alternative explanation it is stated that customers that buy two products are counted twice, but aren't they counted three times? I don't quite understand.


They are actually counted thrice, this is why only 2 times it is subtracted. Because we want to keep the amount of all 3 inclusive once only.
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Re: M03#7 [#permalink] New post 19 Mar 2010, 09:11
snowy2009 wrote:
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 above products?

(A) 5%
(B) 10%
(C) 15%
(D) 25%
(E) 30%

[Reveal] Spoiler: OA
D

Source: GMAT Club Tests - hardest GMAT questions

Let b_i denote the percentage of buyers who regularly buy i products, and b_x the percentage of buyers who regularly purchase product x . We can construct these equations:
\left{ \begin{eqnarray*} b_1 + 2b_2 + 3b_3 &=& b_m+b_c+b_a = 60\% + 50\% + 35\% = 145\%\\ b_1+b_2+b_3 &=& 100\%\\ b_3 &=& 10\%\\ \end{eqnarray*}

Subtract the second one from the first one:
\begin{eqnarray*} b_2+2b_3 = 45\%\\ b_2 = 25\%\\ \end{eqnarray*}

Alternative explanation:

We can construct the equation:
\begin{eqnarray*} 100% = (60% + 50% + 35%) - x - 2*10%\\ \end{eqnarray*}

In the sum (60%+50%+35%) we count customers that buy two products twice, so we have to extract x . In the same sum (60%+50%+35%) we count customers that buy three products thrice, so we have to extract 2*10%. Therefore, all customers are counted only once.
The correct answer is D.


In the alternative explanation it is stated that customers that buy two products are counted twice, but aren't they counted three times? I don't quite understand.

x - sum of AnB,BnC,CnA
100 = 60% + 50% + 35% - x + 10
x=55
exactly two products = 55-3(10) = 25 hence D.
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Re: M03#7 [#permalink] New post 19 Mar 2010, 16:49
100 = M+C+A+N-T-2C
N=those who bought none of the products- since the customers buy at least one of the products, N = 0
C = those who bought all three products - 10
T = those who bought exactly two of the products?

100 = 60 + 50 + 35 + 0-T -2*10
100 = 145 - T - 20
=> T = 25%

N.B: where the no of the total products is given we simply multiply by 25% to get the ans is numerical.
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Re: M03#7 [#permalink] New post 07 Jul 2010, 18:19
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I used Venn diagrams with the info below.

Given:

Total Milk = 60
Total Chicken = 50
Total Apples = 35

Milk, Chicken, and Apples = 10

Your Venn diagram now looks as follows:

Milk, Chicken, and Apples = 10

Just Apples and Milk = X
Just Milk and Chicken = Y
Just Chicken and Apples = Z

Just Milk = 60-X-Y-10 = 50-X-Z
Just Chicken = 50-Y-Z-10 = 40-X-Z
Just Apples = 35-X-Z-10 = 25-X-Z

Assume 100 customers (because question gives %'s):

Total = All parts of the Venn diagram added together
100 = 10+X+Y+Z+50-X-Z+40-X-Z+25-X-Z
100 = 125-X-Y-Z
-25 = -X-Y-Z
25 = X+Y+Z

25/100 = 25%
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Re: M03#7 [#permalink] New post 21 Sep 2010, 23:11
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snowy2009 wrote:
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 above products?

(A) 5%
(B) 10%
(C) 15%
(D) 25%
(E) 30%


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.

For more about the formulas for 3 overlapping sets please see my post at: formulae-for-3-overlapping-sets-69014.html?hilit=rather%20memorize

Hope it helps.
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Re: M03#7 [#permalink] New post 23 Mar 2011, 06:58
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M + MC + MA + MAC = 60

C + MC + AC + MAC = 50

A + AC + MA + MAC = 35

MAC = 10


M + C + A + MA + MC + AC + MAC = 100

The question is asking -> MC + MA + AC = ?

M + C + A + 2MC + 2MA + 2AC + 3MAC = 145

=> MC + MA + AC + 2MAC = 45

=> MC + MA + AC = 45 - 2*10 = 25

ANswer - D
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Re: M03#7 [#permalink] New post 23 Mar 2011, 19:45
lionslion wrote:
D

This how i solved

60% buy milk
50% buy chicken
35% buy apples

10% buy all three

people who brought not all three:

60-10 = 50% milk
50-10 = 40% chicken
35-10 = 25% apples

total comes to 115% but we have only 90% of people (As 10% had brought all the three)

so, 115-90 = 25%


Nice explanation!!! I got stucked here -total comes to 115% but we have only 90% of people (As 10% had brought all the three). Thanks!
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Re: M03#7 [#permalink] New post 24 Mar 2011, 03:01
M*=60%
C*=50%
A*=35%
MCA=10%
MA+MC+CA=?

MC* = |C*-(100%-M*)| = |50%-(100%-60%)| = 10%
MC = MC*-MCA = 10%-10% = 0%
MA+CA = A*-MCA = 35%-10% = 25%

MA+MC+CA = 0%+25% = 25%
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Re: M03#7 [#permalink] New post 27 Mar 2012, 21:58
Bunnel man........ lend me your brain for just 4 hours!!!!!

Got C but your approch is definatey best one....

Once again respect!!!
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Re: M03#7 [#permalink] New post 28 Mar 2012, 14:01
I too used Venn diagram.

Since Milk, Chicken and Apples all have 10 common, lets consider the 10 only once. Also since the data is given in percentage, total will be 100

(60 - 10) + (40-10) + (35 -10) + 10 = Total (i.e 100) + Percentage of customers purchasing exact 2 products.

Solving this, Percentage of customers purchasing exact 2 products = 25

Hence answer is D.
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Re: M03#7 [#permalink] New post 28 Mar 2013, 09:28
Bunuel wrote:
snowy2009 wrote:
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 above products?

(A) 5%
(B) 10%
(C) 15%
(D) 25%
(E) 30%


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.

For more about the formulas for 3 overlapping sets please see my post at: formulae-for-3-overlapping-sets-69014.html?hilit=rather%20memorize

Hope it helps.
Hi Bunuel .. the link you provided showed that :

3. To determine the No of persons in exactly two of the sets : P(A n B) + P(A n C) + P(B n C) – 3P(A n B n C)

How can we apply this equation here ? we do not have the intersection of each two sets.

Thanks in advance :)
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Re: M03#7 [#permalink] New post 28 Mar 2013, 10:04
Hi everyone,

I have across the below method which is good in solving venn problems..

Consider
ppl who buy only one product as X
ppl who buy only two products as Y
ppl who buy only three products as Z
ppl who buy no products as N

Assume 100% of ppl to be 100 numbers for simplicity

N is 0 in our case. So X + 2Y + 3Z = 60+50+35
and X + Y + Z + N = 0

From above we get Y + 2Z = 45

Given Z = 10 So Y will be 25.
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Re: M03#7 [#permalink] New post 29 Mar 2013, 23:32
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I used venn diagram. Here is my solution.
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Re: M03#7 [#permalink] New post 31 Oct 2013, 21:11
Why does the normal vent dig formula
n(AuBuC) = n(A)+n(B)+n(C) -n(AnB) -n(AnC)-n(BnC) +n(AnBnC)
Doesn't hold good here. We need to find [n(AnB)+n(AnC)+n(BnC)]
We can straight away find it is
100=60+50+35-x+10
X=55
Here x represents the no. in exactly 2 sets & n(AnBnC) is counted 1 times
Re: M03#7   [#permalink] 31 Oct 2013, 21:11
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