Question 19172: Of 29 students, 11 have cats, 14 have dogs, and 7 have no pets. How many students have 2 pets?
Found 2 solutions by mukhopadhyay, AnlytcPhil:
Answer by mukhopadhyay(490)   (Show Source):
You can put this solution on YOUR website! If 7 students do not have any pets, 22 students have to have at least one pet.
Let x be the number of students having more than one pets (2 in this case).
So, 22-x students have only 1 pet
Total number of pets = 22-x+2x=22+x
Total number of pets from the question=11+14=25
So, 22+x = 25
=>x=3
The answer is: 3 students have 2 pets.
Answer by AnlytcPhil(1806)  (Show Source):
You can put this solution on YOUR website! Of 29 students, 11 have cats, 14 have dogs, and 7 have no pets. How many
students have 2 pets?
There are three methods. Here are all three:
1. Venn diagram method: Draw this with the two sets C and D overlapping
_____________________
| ________ |
| C| ___|_____ |
| |____|___| |D |
| |_________| |
|____________________|
The rectangle C contains the 11 students with cats
The rectangle D contains the 14 students with dogs
The overlaping part contains the students with both cats and dogs
This is the required answer so we put x in the overlapping part
_____________________
| ________ |
| C| ___|_____ |
| | | x | |D |
| ŻŻŻŻ|ŻŻŻ | |
| ŻŻŻŻŻŻŻŻŻ |
ŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻ
Since rectangle C must contain 11 students, the part of C which does not
overlap with D contains 11-x students, so write 11-x in that part of C
_____________________
| ________ |
| C|11-x ___|_____ |
| | | x | |D |
| ŻŻŻŻ|ŻŻŻ | |
| ŻŻŻŻŻŻŻŻŻ |
ŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻ
Since rectangle D must contain 14 students, the part of D which does not
overlap with C contains 14-x students, so write 14-x in that part of C
_____________________
| ________ |
| C|11-x ___|_____ |
| | | x | |D |
| ŻŻŻŻ|ŻŻŻ 14-x | |
| ŻŻŻŻŻŻŻŻŻ |
ŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻ
The part of the large rectangle which is outside the smaller rectangles
contains the number with neither cats nor dogs. This is given to be 7,
so we put 7 in that part:
_____________________
| ________ |
| C|11-x ___|_____ |
| | | x | |D |
| ŻŻŻŻ|ŻŻŻ 14-x | |
| 7 ŻŻŻŻŻŻŻŻŻ |
ŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻ
Now since there are 29 students, we add up all the expressions in the various
parts and equate this to 29:
(11-x) + x + (14-x) + 7 = 29
Solve that equation and get x = 3
---------------------------------------------------------------
2. Set formula method
N(C OR D) = N(C) + N(D) - N(C AND D)
Your book may have "union" for "OR" and "intersection" for "AND"
We are looking for N(C and D) so this will be x
N(C OR D) = N(C) + N(D) - x
We are given N(C) = 11 and N(D) = 14
N(C or D) = 11 + 14 - x
All of the 29 have cats or dogs except for the 7 which have no pets,
so since 29-7 = 22 we substitute 22 for N(C or D)
22 = 11 + 14 - x
Solve this and get x = 3
--------------------------------------------------------
3. System of equations method.
Let x = the number of students with cats and no dogs.
Let y = the number of students with dogs and no cats.
Let z = the number of students with both dogs and cats. This will be
the desired answer
Given 7 = the number of students with no dogs and no cats.
There are 29 students, so
x + y + z + 7 = 29 or, subtracting 7 from both sides
x + y + z = 22
x and z have cats, so
x + z = 11
y and z both have dogs, so
y + z = 14
So we have the system of three equations and three unknowns:
x + y + z = 22
x + z = 11
y + z = 14
x = 8, y = 11, z = 3
So z = 3 = the answer.
Edwin
AnlytcPhil@aol.com
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