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Ajiteshmathur
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Of the total houses on the market, 75 are brick, 95 have a garden, and Updated on: 03 Jul 2019, 03:43
Originally posted by Ajiteshmathur on 03 Jul 2019, 00:49.
Last edited by Ajiteshmathur on 03 Jul 2019, 03:43, edited 1 time in total.
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65% (hard)

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65% (02:48) correct 35% (02:57) wrong based on 505 sessions

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Of the total houses on the market, 75 are brick, 95 have a garden, and 110 have a porch. If all the houses have at least one of these traits, 50 of the houses have only one of these traits, and 60 of the houses have all three traits, then how may total houses are there on the market ?

A. 110
B. 135
C. 195
D. 280
E. 390

Bunuel VeritasKarishma chetan2u can you please help me out with this one.
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Of the total houses on the market, 75 are brick, 95 have a garden, and 05 Jul 2019, 04:00
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Ajiteshmathur
Of the total houses on the market, 75 are brick, 95 have a garden, and 110 have a porch. If all the houses have at least one of these traits, 50 of the houses have only one of these traits, and 60 of the houses have all three traits, then how may total houses are there on the market ?

A. 110
B. 135
C. 195
D. 280
E. 390

Bunuel VeritasKarishma chetan2u can you please help me out with this one.
Show more

We need the total number of houses.

Attachment:
IMG_1452.jpg
IMG_1452.jpg [ 610.37 KiB | Viewed 7796 times ]
The shaded portion adds up to 50.
a+b+c add up to 'Exactly in 2 sets'.
Total = Exactly in one set + Exactly in 2 sets + Exactly in 3 sets
Total = 50 + Exactly in 2 sets + 60 = 110 + Exactly in 2 sets

Total = 75 + 95 + 110 - Exactly in 2 sets - 2*60 = 160 - Exactly in 2 sets
(because 75, 95 and 110 count "Exactly in 2 sets" twice and "In all 3 sets" thrice. So we subtract the extras out)

110 + Exactly in 2 sets = 160 - Exactly in 2 sets
Exactly in 2 sets = 25
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Of the total houses on the market, 75 are brick, 95 have a garden, and 03 Jul 2019, 01:33
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Most of the set theory questions revolve around 2 equations

Total = I +II + III
Here I = Exactly One
II = Exactly two
III = Exactly three

Hence Union = I + 2 II + 3 III

Here Union = 75 + 95 + 110

Value of I and III is 50 and III is 60 , We get II as 25

We need to find

Total = I +II + III = 50 + 25 + 60 = 135

Hope this is helpful

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Of the total houses on the market, 75 are brick, 95 have a garden, and 03 Jul 2019, 13:48
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Superman249
Most of the set theory questions revolve around 2 equations

Total = I +II + III
Here I = Exactly One
II = Exactly two
III = Exactly three

Hence Union = I + 2 II + 3 III

Here Union = 75 + 95 + 110

Value of I and III is 50 and III is 60 , We get II as 25

We need to find

Total = I +II + III = 50 + 25 + 60 = 135

Hope this is helpful

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Hey! How did you arrive at 25 for II?
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Of the total houses on the market, 75 are brick, 95 have a garden, and 03 Jul 2019, 19:22
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Use equation
Union = I + 2 II + 3 III

Here I= 50
And III = 60

Union = 280

Equate and You get II as 25

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Of the total houses on the market, 75 are brick, 95 have a garden, and 04 Jul 2019, 00:28
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Houses made with Brick (B)=75
Houses with Garden (G)=95
Houses with Porch(P)=110

Let the number of houses with 'only brick' (without porch or garden), 'only garden' and 'only porch' be denoted by 'b', 'g' and 'p' respectively. Given: b+g+p=50.
Let the number of houses with 'both brick and garden', 'both brick and porch', 'both porch and garden' and 'all three (brick, porch and garden)' be denoted by BG, BP, PG and BPG respectively. Given: BPG=60

B=75=b+BP+BG+BPG
BP+BG=15-b.................(I)
Similarly:
BG+PG=35-g.................(II)
BP+PG=50-p..................(III)

Adding LHS and RHS of (I), (II) & (III), we get:
2(BP+BG+PG)=100-(b+p+g)=100-50=50
BP+PG+BG=25

Total number of houses=(b+p+g)+(BP+PG+BG)+BPG=50+25+60=135. ANS:B
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Of the total houses on the market, 75 are brick, 95 have a garden, and 05 Jul 2019, 04:19
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VeritasKarishma
Ajiteshmathur
Of the total houses on the market, 75 are brick, 95 have a garden, and 110 have a porch. If all the houses have at least one of these traits, 50 of the houses have only one of these traits, and 60 of the houses have all three traits, then how may total houses are there on the market ?

A. 110
B. 135
C. 195
D. 280
E. 390

Bunuel VeritasKarishma chetan2u can you please help me out with this one.

We need the total number of houses.

Attachment:
IMG_1452.jpg
The shaded portion adds up to 50.
a+b+c add up to 'Exactly in 2 sets'.
Total = Exactly in one set + Exactly in 2 sets + Exactly in 3 sets
Total = 50 + Exactly in 2 sets + 60 = 110 + Exactly in 2 sets

Total = 75 + 95 + 110 - Exactly in 2 sets - 2*60 = 160 - Exactly in 2 sets
(because 75, 95 and 110 count "Exactly in 2 sets" twice and "In all 3 sets" thrice. So we subtract the extras out)

110 + Exactly in 2 sets = 160 - Exactly in 2 sets
Exactly in 2 sets = 25
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Thank you VeritasKarishma ! The diagram and the explanation helps :)
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Of the total houses on the market, 75 are brick, 95 have a garden, and 05 Jul 2019, 05:22
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Good Question , great Explanation . :thumbup:
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Of the total houses on the market, 75 are brick, 95 have a garden, and 02 Apr 2021, 19:17
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Let:

a = 1 set only

b = 2 sets only

c = all 3 sets

T = total number of unique houses

There is no neither/note category because each house has at least one

T = a + b + c

When we add up all of the Brick, all of the Garden, and all of the Porch:

We will be adding the houses that have exactly two of these TWICE

And the houses that have all three of these THRICE

(75 + 95 + 110) = a + 2b + 3c

280 = a + 2b + 3c


We are given:

50 houses have only 1 of these. This means 50 houses are part of exactly one set.

a = 50

60 houses have all 3.

c = 50

Now plug these into the 2 equations:


T = 50 + b + 60

280 = 50 + 2b + 3(60)

and we have:

T = 110 + b

50 = 2b

Solve for b ——-> b = 25

Then Total number of unique houses:

T = 110 + 25 = 135

135 is the answer

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Of the total houses on the market, 75 are brick, 95 have a garden, and 21 Dec 2025, 10:52
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