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M03-07

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Joined: 02 Sep 2009
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16 Sep 2014, 00:19
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45% (medium)

Question Stats:

64% (01:55) correct 36% (02:18) wrong based on 211 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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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}

$$100=60+50+35-x-2*10+0$$ --&gt; $$x=25$$.

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17 Mar 2015, 11:47
I think this question is good and helpful.
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Joined: 01 Jul 2015
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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
[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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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
[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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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
[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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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
[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

Also check:

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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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10 Mar 2018, 14:23
2
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).

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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09 Aug 2018, 22:16
I think this is a high-quality question and I agree with explanation.
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06 Jun 2019, 09:31
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I solved it slightly differently and find this approach more intuitive than applying a specific formula (mind you got it right at the third try but that is just on my carelessness). Hope this helps:

With questions like these, I find it useful to make a Venn Diagram to help in visualizing and summarizing all the data.

So we are given the info represented in the Venn Diagram, and we are also told that all customers buy AT LEAST one of the three items, so we know that the three should sum up to 100. Further the deductions are just the customers who buy a combination of any two of the three items or all three items in case of the 10 from each of the three items.

Given all this info all we need to do now is to simply add up all these terms and find the value of x + y + z , which represents the proportion of customers that but exactly two of the three items.

So we have the equation:

[60 - x - y -10] + [35 - x - z - 10] + [50 - y - z - 10] + x + y + z + 10 = 100

i.e. [proportion of those that buy only milk] + [proportion of those that buy only apples] + [proportion of those that only buy chicken] + customers that buy milk and apples + those that buy milk and chicken + those that buy chicken and apples + those that buy all three = 100 (because all %ages of customers surveyed should total 100)

When you simplify the above expression you will eventually arrive at:

x + y + z = 25

I know it seems a bit long winded but for me it is the most intuitive way of solving the problem.
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06 Jun 2019, 10:26
aliakberza wrote:
I solved it slightly differently and find this approach more intuitive than applying a specific formula (mind you got it right at the third try but that is just on my carelessness). Hope this helps:

With questions like these, I find it useful to make a Venn Diagram to help in visualizing and summarizing all the data.

So we are given the info represented in the Venn Diagram, and we are also told that all customers buy AT LEAST one of the three items, so we know that the three should sum up to 100. Further the deductions are just the customers who buy a combination of any two of the three items or all three items in case of the 10 from each of the three items.

Given all this info all we need to do now is to simply add up all these terms and find the value of x + y + z , which represents the proportion of customers that but exactly two of the three items.

So we have the equation:

[60 - x - y -10] + [35 - x - z - 10] + [50 - y - z - 10] + x + y + z + 10 = 100

i.e. [proportion of those that buy only milk] + [proportion of those that buy only apples] + [proportion of those that only buy chicken] + customers that buy milk and apples + those that buy milk and chicken + those that buy chicken and apples + those that buy all three = 100 (because all %ages of customers surveyed should total 100)

When you simplify the above expression you will eventually arrive at:

x + y + z = 25

I know it seems a bit long winded but for me it is the most intuitive way of solving the problem.

Thank you for this detailed explanation...this is the best way to see the solution in my opinion
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07 Jun 2019, 22:47
Bunuel 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 products listed above?

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

Let total = 100%

3 category Venn diagram formula

—> Total - neither = A + B + C - (Exactly 2 category) - 2(Exactly 3 category)

buy at least one of the following products —> Neither = 0

—> 100 - 0 = 60 + 50 + 35 - (Exactly 2 category) - 2*10
—> Exactly 2 category = 145 - 20 - 100 = 25

So Option D

Pls Hit kudos if you like the solution

Posted from my mobile device
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11 Jun 2019, 20:55
I think this is a high-quality question and I agree with explanation.
Re M03-07   [#permalink] 11 Jun 2019, 20:55
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