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Of the 150 houses in a certain development, 60 percent have
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29 May 2017, 02:27

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Of the 150 houses in a certain development, 60 percent have air-conditioning, 50 percent have a sunporch, and 30 percent have a swimming pool. If 5 of the houses have all three of these amenities and 5 have none of them, how many of the houses have exactly two of these amenities?

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15 Nov 2017, 16:22

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ganand wrote:

Of the 150 houses in a certain development, 60 percent have air-conditioning, 50 percent have a sunporch, and 30 percent have a swimming pool. If 5 of the houses have all three of these amenities and 5 have none of them, how many of the houses have exactly two of these amenities?

(A) 10 (B) 45 (C) 50 (D) 55 (E) 65

We can create the following equation:

Total houses = number with air conditioning + number with sunporch + number with pool - number with only two of the three things - 2(number with all three things) + number with none of the three things

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29 May 2017, 05:24

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ganand wrote:

Of the 150 houses in a certain development, 60 percent have air-conditioning, 50 percent have a sunporch, and 30 percent have a swimming pool. If 5 of the houses have all three of these amenities and 5 have none of them, how many of the houses have exactly two of these amenities?

(A) 10 (B) 45 (C) 50 (D) 55 (E) 65

This can be solved using Venn Diagram (refer the attachment) AC = 60% of 150 = 90 Sunporch = 50% of 150 = 75 SP = 30% of 150 = 45

Using the venn diagram we can form the following equations a+b+c+d+e+f+g+h = 150 a+b+c+d+f+g = 140 ----- (1) (as e = h = 5)

Taking 1 circle at a time AC = a+b+d = 85 ---- (2) .. (as e = 5) SunPorch = c+b+f = 70 ---- (3) .. (as e = 5) SP = g+d+f = 40 ---- (4) .. (as e = 5)

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29 May 2017, 07:23

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ganand wrote:

Of the 150 houses in a certain development, 60 percent have air-conditioning, 50 percent have a sunporch, and 30 percent have a swimming pool. If 5 of the houses have all three of these amenities and 5 have none of them, how many of the houses have exactly two of these amenities?

(A) 10 (B) 45 (C) 50 (D) 55 (E) 65

all three = neither =5 = 10/3 % of 150

working formula :- Total = AC + sunporch + swimming pool - Exactly two - 2*all three + neither

100 = 60 + 50 + 30- Exactly two - 2*10/3 + 10/3 100 =140 - Exactly two - 10/3 Exactly two = 40 - 10/3 = (120-10)/3 = 110/3 % 110/3 % of 150 = 110*150/300 =55

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28 Jun 2017, 05:54

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ganand wrote:

Of the 150 houses in a certain development, 60 percent have air-conditioning, 50 percent have a sunporch, and 30 percent have a swimming pool. If 5 of the houses have all three of these amenities and 5 have none of them, how many of the houses have exactly two of these amenities?

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28 Jun 2017, 05:58

hazelnut wrote:

ganand wrote:

Of the 150 houses in a certain development, 60 percent have air-conditioning, 50 percent have a sunporch, and 30 percent have a swimming pool. If 5 of the houses have all three of these amenities and 5 have none of them, how many of the houses have exactly two of these amenities?

Total = A + B + C − (sum of 2−group overlaps) + (all three) + Neither 100 = 60 + 50 + 30 - (x) + (5/150)(100) + (5/150)(100) 100 = 140 + (20/3) - x x = 40 + (20/3) x = 140/3

We need to find exact two. The formula you have listed is for general sum of 2 group overlap. The total you have listed is 100 while in the question it is 150.

Formula for exact two is: Total = A + B + C + Neither - (exact 2) - 2 * (all three)

150=90+75+45+5-10-sum of exact 2. 205-150= sum of exact 2 sum of exact 2=55

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08 May 2018, 11:40

ganand wrote:

Of the 150 houses in a certain development, 60 percent have air-conditioning, 50 percent have a sunporch, and 30 percent have a swimming pool. If 5 of the houses have all three of these amenities and 5 have none of them, how many of the houses have exactly two of these amenities?

here is my solution ...I just need to understand what I did wrong here

60 percent have air-conditioning = 90

50 percent have a sunporch = 75

30 percent have a swimming pool = 45

5 houses has all three amenities

5 houses has non of these amenities

we need to find number of houses that have exactly two of the ameneties X = both

so here is my equation

150 = 90+75+45-(5+x)+5 ( let me explain what i did here, i summed up three groups 90+75+45 minus ( 5 houses with all three feature +houses with only both features ) +neither

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08 May 2018, 12:35

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dave13 wrote:

ganand wrote:

Of the 150 houses in a certain development, 60 percent have air-conditioning, 50 percent have a sunporch, and 30 percent have a swimming pool. If 5 of the houses have all three of these amenities and 5 have none of them, how many of the houses have exactly two of these amenities?

here is my solution ...I just need to understand what I did wrong here

60 percent have air-conditioning = 90

50 percent have a sunporch = 75

30 percent have a swimming pool = 45

5 houses has all three amenities

5 houses has non of these amenities

we need to find number of houses that have exactly two of the ameneties X = both

so here is my equation

150 = 90+75+45-(5+x)+5 ( let me explain what i did here, i summed up three groups 90+75+45 minus ( 5 houses with all three feature +houses with only both features ) +neither

- your issue is addressed in warriorguy 's post HERE I am a fan of Venn diagrams. I am not a fan of Venn diagrams for three overlapping set questions, which typically involve a very specific question.

Try this formula for three overlapping sets:

Total = Group1 + Group 2 + Group 3 - (sum of 2-group overlaps) - 2*(all three) + None

What we need to understand while solving problems like these is that

P(Airconditioner) = 0.6(150) = 90 = P(Only Airconditioner) + A + B + P(All 3) P(Sunporch) = 0.5(150) = 75 = P(Only SunPorch) + B + C + P(All 3) P(SwimmingPool) = 0.3(150) = 45 = P(Only SwimmingPool) + B + C + P(All 3)

P(Total) = P(Only Airconditioner) + P(Only SunPorch) + P(Only SwimmingPool) + (A + B + C) + P(All 3) 150 = 90 + 75 + 45 - (A + B + C) - 2*P(All 3) + 5

Therefore, the houses which have only two amenities are P(Only 2) = A + B + C = 205 - 150 = 55 (Option D)

Hope this helps you!

pushpitkc , generis many thanks for explanation, almost got it appreciate!

pushpitkc , just one question. Didnt you make a typo here (see highlighted part of your solution above or below )

P(Airconditioner) = 0.6(150) = 90 = P(Only Airconditioner) + A + B + P(All 3) P(Sunporch) = 0.5(150) = 75 = P(Only SunPorch) + B + C + P(All 3) P(SwimmingPool) = 0.3(150) = 45 = P(Only SwimmingPool) + B + C + P(All 3)

Lets start from case #1: this one one P(Airconditioner) = 0.6(150) = 90 = P(Only Airconditioner) + A + B + P(All 3) I interpert it as follows: So there are 90 houses with air conditioning among which are houses featuring only Airconditioner, OR both amenties Airconditioner+Porch and Airconditioner + Swimming pool OR three amenites. Am i understanding it correctly ?

If the above mentioned information i understand correctly, then why here P(Sunporch) = 0.5(150) = 75 = P(Only SunPorch) + B + C + P(All 3) are you adding up + B + C + P(All 3) and not +A+C+(All 3) ? As per your diagram Sunporch has intersection with air conditioning and swimming pool. I am just following the pattern as in case #1.

Case # 1 has intersection with A, B and P so you added A + B + P(All 3)

But case # 2 has this intersection +A+C+P but you sum up B + C + P

i would appreaciate if you could expain this

Further to the above questions can you please confirm if the detaled solution based on suggested formula is correct pls see below

You have the eyes of a hawk. You spotted the typo, which I've now edited in my solution for this problem.

dave13 wrote:

Lets start from case #1: this one one P(Airconditioner) = 0.6(150) = 90 = P(Only Airconditioner) + A + B + P(All 3) I interpert it as follows: So there are 90 houses with air conditioning among which are houses featuring only Airconditioner, OR both amenties Airconditioner+Porch and Airconditioner + Swimming pool OR three amenites. Am i understanding it correctly ?

Your understanding is spot on, here! The same line of reasoning for the other 2 aspects of the house - P(Sunporch) and P(SwimmingPool). It was the typo which caused the confusion Sorry about that!

Also, the understanding of the final solution is perfect

Hope this helps you
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Re: Of the 150 houses in a certain development, 60 percent have
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10 May 2018, 00:59

Let a = # of houses with only one b = # of houses with two *We're solving for b. c = # of houses with all three (given: 5) d = # of houses with none (given: 5)

There are 150 houses total: a + b + c + d = 150 substitute c=5 & d=5 ==> a + b = 140. So 140 houses have either one or two amenities.

They gave us 60% + 50% + 30% = 140% so there is some overlap - some houses have multiple amenities.

Because b's are double counted and c's are triple counted, a + 2b + 3c = 150(140%) = 210 Substitute 3c = 3*5 = 15 a + 2b + 15 = 210 From above, a + b = 140. Therefore, (140) + b + 15 = 210 b = 210 - 155 = 55

On this question, while I would still draw the Venn Diagram myself, the key is to know the formula for three-set Venn Diagrams that has been mentioned in this thread:

Total = Group1 + Group2 + Group3 - (sum of 2-group overlaps) - 2*(all three) + None

Then, all we need to do is plug in the numbers we are given (converting percents to numbers) and solve for the sum of 2-group overlaps, because that is what the question is asking us for.

Please let me know if you have any questions, or if you want me to post a video solution!
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Re: Of the 150 houses in a certain development, 60 percent have
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12 Sep 2018, 04:37

Please kindly post a video solution. I cannot seem to understand where 2(5) came from. I seem to get the other workings but not sure about the 2(5).

JeffYin wrote:

If anyone is looking for 3-set Venn Diagram questions to practice on from the Official Guide, I have compiled a list at the following link:

On this question, while I would still draw the Venn Diagram myself, the key is to know the formula for three-set Venn Diagrams that has been mentioned in this thread:

Total = Group1 + Group2 + Group3 - (sum of 2-group overlaps) - 2*(all three) + None

Then, all we need to do is plug in the numbers we are given (converting percents to numbers) and solve for the sum of 2-group overlaps, because that is what the question is asking us for.

Please let me know if you have any questions, or if you want me to post a video solution!

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12 Sep 2018, 11:59

deethompson wrote:

Please kindly post a video solution. I cannot seem to understand where 2(5) came from. I seem to get the other workings but not sure about the 2(5).

Sure, deethompson, I would be happy to post a video solution! I should be able to get to that later this week. In the meantime, I think that the key thing is to understand why we are subtracting some numbers in the first place. We subtract the overlaps because we are double- or triple-counting some houses if we add up all of the houses that have each of the 3 features. We are double-counting those that have exactly 2 of the features, so we need to subtract those once to account for that. We are triple-counting those that have all 3 of the features, so we need to subtract 2X that number of houses to account for that. Because we are told that 5 houses have all 3 features, we need to subtract 2 times that number, or 2*5, to account for this triple-counting.

Let me know if you have any other questions that you want me to address in my video solution!
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