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In a group of 100 homeowners, x homeowners had an alarm security syste

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In a group of 100 homeowners, x homeowners had an alarm security syste  [#permalink]

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New post 15 May 2018, 02:35
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In a group of 100 homeowners, x homeowners had an alarm security system and y homeowners had deadbolt locks. If z homeowners had neither an alarm security system nor deadbolt locks, how many homeowners had both an alarm security system and deadbolt locks?


A. 100 – x – y – z

B. 100 – x – y + z

C. x – y – z + 100

D. x + y + z + 100

E. x + y + z – 100

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Re: In a group of 100 homeowners, x homeowners had an alarm security syste  [#permalink]

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New post 15 May 2018, 03:31
Bunuel wrote:
In a group of 100 homeowners, x homeowners had an alarm security system and y homeowners had deadbolt locks. If z homeowners had neither an alarm security system nor deadbolt locks, how many homeowners had both an alarm security system and deadbolt locks?


A. 100 – x – y – z

B. 100 – x – y + z

C. x – y – z + 100

D. x + y + z + 100

E. x + y + z – 100


There can be only 4 catagories

Home with alarm security only + Homes with deadbolt Lock + Home with Both + Home with none = 100%

x = Home with alarm security only+ Home with Both
y = Homes with deadbolt Lock + Home with Both
z = Home with none

x+y+z - Home with Both = 100

So Home with both = x + y + z – 100

Answer: option E
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Re: In a group of 100 homeowners, x homeowners had an alarm security syste  [#permalink]

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New post 15 May 2018, 09:25
GMATinsight wrote:
Bunuel wrote:
In a group of 100 homeowners, x homeowners had an alarm security system and y homeowners had deadbolt locks. If z homeowners had neither an alarm security system nor deadbolt locks, how many homeowners had both an alarm security system and deadbolt locks?


A. 100 – x – y – z

B. 100 – x – y + z

C. x – y – z + 100

D. x + y + z + 100

E. x + y + z – 100


There can be only 4 catagories

Home with alarm security only + Homes with deadbolt Lock + Home with Both + Home with none = 100%

x = Home with alarm security only+ Home with Both
y = Homes with deadbolt Lock + Home with Both
z = Home with none

x+y+z - Home with Both = 100

So Home with both = x + y + z – 100

Answer: option E


I have a question.
Will the equation not become: x+y+z - 2*Home with Both = 100? Why are we counting the "homes with both" only once here?
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Re: In a group of 100 homeowners, x homeowners had an alarm security syste  [#permalink]

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New post 15 May 2018, 10:11
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urvashis09 wrote:
GMATinsight wrote:
Bunuel wrote:
In a group of 100 homeowners, x homeowners had an alarm security system and y homeowners had deadbolt locks. If z homeowners had neither an alarm security system nor deadbolt locks, how many homeowners had both an alarm security system and deadbolt locks?


A. 100 – x – y – z

B. 100 – x – y + z

C. x – y – z + 100

D. x + y + z + 100

E. x + y + z – 100


There can be only 4 catagories

Home with alarm security only + Homes with deadbolt Lock + Home with Both + Home with none = 100%

x = Home with alarm security only+ Home with Both
y = Homes with deadbolt Lock + Home with Both
z = Home with none

x+y+z - Home with Both = 100

So Home with both = x + y + z – 100

Answer: option E


I have a question.
Will the equation not become: x+y+z - 2*Home with Both = 100? Why are we counting the "homes with both" only once here?


Hey urvashis09

Think of it as a Venn diagram where the sum of x and y includes the value of
both(marked in red) twice. So, we need to reduce it once as the total number
of homes will include the home with both facilities once only.

Attachment:
Home_VennDiag.png
Home_VennDiag.png [ 4.02 KiB | Viewed 1020 times ]


Hope this clears your confusion.
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In a group of 100 homeowners, x homeowners had an alarm security syste  [#permalink]

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New post 15 May 2018, 10:59
pushpitkc wrote:
urvashis09 wrote:
GMATinsight wrote:

There can be only 4 catagories

Home with alarm security only + Homes with deadbolt Lock + Home with Both + Home with none = 100%

x = Home with alarm security only+ Home with Both
y = Homes with deadbolt Lock + Home with Both
z = Home with none

x+y+z - Home with Both = 100

So Home with both = x + y + z – 100

Answer: option E


I have a question.
Will the equation not become: x+y+z - 2*Home with Both = 100? Why are we counting the "homes with both" only once here?


Hey urvashis09

Think of it as a Venn diagram where the sum of x and y includes the value of
both(marked in red) twice. So, we need to reduce it once as the total number
of homes will include the home with both facilities once only.

Attachment:
Home_VennDiag.png


Hope this clears your confusion.


Oh, okay so that means subtracting it twice would then just refer to the non-shaded part, which would be wrong. Now I get it! Thank you so much! :-)
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Re: In a group of 100 homeowners, x homeowners had an alarm security syste  [#permalink]

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New post 15 May 2018, 11:13
Alarm: x (not alarm only, therefore could include people with deadbolt)
Deadbolt: y (not deadbolt only, therefore could include people with alarm)
None: z
Both: call it b

Equation: (x - b) + b + (y - b) + z = 100. This can be rewritten as: x + b + y + z = 100. We need b so rewrite as: b = x + y + z - 100. Answer is E.
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Re: In a group of 100 homeowners, x homeowners had an alarm security syste  [#permalink]

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New post 16 May 2018, 10:18
Bunuel wrote:
In a group of 100 homeowners, x homeowners had an alarm security system and y homeowners had deadbolt locks. If z homeowners had neither an alarm security system nor deadbolt locks, how many homeowners had both an alarm security system and deadbolt locks?


A. 100 – x – y – z

B. 100 – x – y + z

C. x – y – z + 100

D. x + y + z + 100

E. x + y + z – 100



The answer should be A. 100 - x - y - z
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Re: In a group of 100 homeowners, x homeowners had an alarm security syste  [#permalink]

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New post 16 May 2018, 10:54
Bunuel wrote:
In a group of 100 homeowners, x homeowners had an alarm security system and y homeowners had deadbolt locks. If z homeowners had neither an alarm security system nor deadbolt locks, how many homeowners had both an alarm security system and deadbolt locks?


A. 100 – x – y – z

B. 100 – x – y + z

C. x – y – z + 100

D. x + y + z + 100

E. x + y + z – 100


We can create the equation:

Total = alarm + deadbolt - both + neither

100 = x + y - both + z

both = x + y + z - 100

Answer: E
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Re: In a group of 100 homeowners, x homeowners had an alarm security syste  [#permalink]

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New post 17 May 2018, 03:59
Bunuel wrote:
In a group of 100 homeowners, x homeowners had an alarm security system and y homeowners had deadbolt locks. If z homeowners had neither an alarm security system nor deadbolt locks, how many homeowners had both an alarm security system and deadbolt locks?


A. 100 – x – y – z

B. 100 – x – y + z

C. x – y – z + 100

D. x + y + z + 100

E. x + y + z – 100


This can be done by using formula :

Total = Group 1 + Group 2 - Both + Neither

Both = Group 1 + Group 2 + Neither - Total

Both = x + y + z - 100

(E)

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In a group of 100 homeowners, x homeowners had an alarm security syste  [#permalink]

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New post 17 May 2018, 10:53
1
Bunuel wrote:
In a group of 100 homeowners, x homeowners had an alarm security system and y homeowners had deadbolt locks. If z homeowners had neither an alarm security system nor deadbolt locks, how many homeowners had both an alarm security system and deadbolt locks?


A. 100 – x – y – z

B. 100 – x – y + z

C. x – y – z + 100

D. x + y + z + 100

E. x + y + z – 100



TOTAL - NEITHER = A + B - EITHER (BOTH)

100 - Z = X + Y - EITHER

EITHER = X + Y - ( 100 - Z )

EITHER = X + Y + Z - 100

ANS = E

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Re: In a group of 100 homeowners, x homeowners had an alarm security syste  [#permalink]

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New post 20 Oct 2019, 06:53
Top Contributor
Bunuel wrote:
In a group of 100 homeowners, x homeowners had an alarm security system and y homeowners had deadbolt locks. If z homeowners had neither an alarm security system nor deadbolt locks, how many homeowners had both an alarm security system and deadbolt locks?


A. 100 – x – y – z

B. 100 – x – y + z

C. x – y – z + 100

D. x + y + z + 100

E. x + y + z – 100


Let's use the Double Matrix Method. This technique can be used for most questions featuring a population in which each member has two characteristics associated with it (aka overlapping sets questions)..

Here, we have a population of homeowners, and the two characteristics are:
- has alarm security system or does NOT have alarm security system
- has deadbolt locks or does NOT have deadbolt locks

In a group of 100 homeowners, x homeowners had an alarm security system and y homeowners had deadbolt locks.
We can set up our matrix as follows:
Image


z homeowners had neither an alarm security system nor deadbolt locks
We get:
Image


Since the two boxes in the BOTTOM ROW must add to 100-y, we know that the missing box must be 100-y-z, since 100-y-z + z = 100-y
Image


Finally, the two boxes in the LEFT-SIDE column must add to x
In other words, ? + (100-y-z) = x
Subtract (100-y-z) from both sides to get: ? = x - (100-y-z)
Image

Take: x - (100-y-z)
Simplify to get: x - 100 + y + z
Rearrange to get: x + y + z - 100

Answer: E

This question type is VERY COMMON on the GMAT, so be sure to master the technique.

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Re: In a group of 100 homeowners, x homeowners had an alarm security syste   [#permalink] 20 Oct 2019, 06:53
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