Problem Solving

Hi,

following is the question that i came across in veritas prep"

In the city of San Durango, 60 people own cats, dogs, or rabbits. If 30 people owned cats, 40 owned dogs, 10 owned rabbits, and 12 owned exactly two of the three types of pet, how many people owned all three?

a)2
b)4
c)8
d)12
e)32

The answer used the following logic:

Sum of totals in three individual categories - (sum of items in two categories) - (2 * number of items in all three categories) = Total

however what i know is that it should be

Sum of totals in three individual categories - (sum of items in two categories) - ( number of items in all three categories) = Total

as the formula is n(AuBuC)=n(A)+n(B)+n(C)+n (A common b common c)- [n(a common b)+n (b common c)n + n (c common a)]

am i right in my understanding or am i missing anything??

manasgoswami1 wrote: In the city of San Durango, 60 people own cats, dogs, or rabbits. If 30 people owned cats, 40 owned dogs, 10 owned rabbits, and 12 owned exactly two of the three types of pet, how many people owned all three?

a)2
b)4
c)8
d)12
e)32

The vast majority of problems with 3 overlapping groups are easily solved with the following formula:

T = A + B + C - (AB + AC + BC) - 2(ABC)

The big idea with overlapping group problems is to SUBTRACT THE OVERLAPS.
When we add together everyone in A, everyone in B, and everyone in C:
Those in exactly 2 of the groups (AB+AC+BC) are counted twice, so they need to be subtracted from the total ONCE.
Those in all 3 groups (ABC) are counted 3 times, so they need to be subtracted from the total TWICE.
By subtracting the overlaps, we ensure that no one is overcounted.

In the problem above:
T = 60
A = cat owners = 30.
B = dog owners = 40.
C = rabbit owners = 10.
The number who own exactly 2 of the 3 types of pet = AB + AC + BC = 12.
The number who own all 3 types of pet = ABC = x.

Plugging these values into the formula, we get:
60 = 30 + 40 + 10 - 12 - 2(x)
-8 = -2x
x=4.

The correct answer is B.

Other problems that can be solved with this formula:
https://www.beatthegmat.com/3-overlappin ... 49650.html
https://www.beatthegmat.com/og-13-178-vi ... 11188.html

Hi Mitch,

there are conflicting theories. if you scroll down on https://webserver.ignou.ac.in/virtualcam ... IT%201.htm

then you see that multiplication by 2 is not there and the explanation too is given using venn diagrams.

Request your help in explaining this anomaly.

regards

Manas

correcting and reiterating my understanding

however what i know is that it should be

Sum of totals in three individual categories - (sum of items in two categories) + ( number of items in all three categories) = Total

as the formula is n(AuBuC)=n(A)+n(B)+n(C)+n (A common b common c)- [n(a common b)+n (b common c)n + n (c common a)]

Hi Mitch,

the following links add to my confusion even more.
Please suggest

https://www.beatthegmat.com/mba/2011/01/ ... set-theory
https://www.sciencehq.com/math-formulas/ ... mulas.html
https://gmat-maths.blogspot.in/2008/05/s ... mulas.html

manasgoswami1 wrote:Hi Mitch,

there are conflicting theories. if you scroll down on https://webserver.ignou.ac.in/virtualcam ... IT%201.htm

then you see that multiplication by 2 is not there and the explanation too is given using venn diagrams.

Request your help in explaining this anomaly.

regards

Manas

Both formulas are correct, but they account for the overlaps differently.
I discuss the difference in my 2nd post here:
https://www.beatthegmat.com/sets-t148362.html

Hi Mitch,

why are we dividing by 2 and why the answer is not 8?

manasgoswami1 wrote:Hi Mitch,

why are we dividing by 2 and why the answer is not 8?

For 3 overlapping groups A, B and C, here are all of the different groupings:

Those in exactly ONE group:
Only A
Only B
Only C

Those in exactly TWO groups:
AB
AC
BC

Those in ALL THREE groups:
ABC

Here is the formula that I typically use:

T = A + B + C - (AB + AC + BC) - 2(ABC).

Let's examine the logic behind the formula.

The values in red represent EVERY element in A, EVERY element in B, and EVERY element in C.
Thus:
A = (Only A + AB + AC + ABC).
B = (Only B + AB + BC + ABC).
C = (Only C + AC + BC + ABC).
Adding together these values, we get:
T = A + B + C = (Only A + Only B + Only C) + (AB + AC + BC) + (AB + AC + BC) + (ABC + ABC + ABC).

Notice that the values in BLUE have been double-counted.
Thus, they must be subtracted from the total ONCE:
T = A + B + C - (AB + AC + BC).

Notice that the value in GREEN has been triple-counted.
Thus, it must be subtracted from the total TWICE:
T = A + B + C - (AB + AC + BC) - 2(ABC).

The result is the following formula:
T = A + B + C - (AB + AC + BC) - 2(ABC).

In the problem above:
T = 60.
A = cat owners = 30.
B = dog owners = 40.
C = rabbit owners = 10.
The number who own exactly 2 types of pet = AB + AC + BC = 12.
The number who own all 3 types of pet = ABC = x.

Plugging these values into the formula, we get:
60 = 30 + 40 + 10 - 12 - 2(x)
-8 = -2x
x=4.