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In a group of 100 homeowners, x homeowners had an alarm security syste
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15 May 2018, 02:35
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E
Difficulty:
35% (medium)
Question Stats:
65% (01:30) correct 35% (01:33) wrong based on 97 sessions
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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?
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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
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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?
Re: In a group of 100 homeowners, x homeowners had an alarm security syste
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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?
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 [ 4.02 KiB | Viewed 1061 times ]
Hope this clears your confusion.
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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!
Re: In a group of 100 homeowners, x homeowners had an alarm security syste
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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
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16 May 2018, 10:18
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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?
Re: In a group of 100 homeowners, x homeowners had an alarm security syste
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16 May 2018, 10:54
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?
Re: In a group of 100 homeowners, x homeowners had an alarm security syste
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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
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17 May 2018, 10:53
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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
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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:
z homeowners had neither an alarm security system nor deadbolt locks We get:
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
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)
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.