SOLUTION: A survey of 50 students found that 30 fad cats, 25 had dogs, 5 had white mice, 16 had both dogs and cats, 4 had both dogs and mice, 2 had both cats and mice, and only 1 had all thr

Question 63420: A survey of 50 students found that 30 fad cats, 25 had dogs, 5 had white mice, 16 had both dogs and cats, 4 had both dogs and mice, 2 had both cats and mice, and only 1 had all three kinds of pets. How many students had no pets of these types?

can u please tell me how you found this answer
Answer by Edwin McCravy(20054) (Show Source): You can put this solution on YOUR website!

A survey of 50 students found that 30 had cats, 25 had dogs, 5 had mice, 
16 had both dogs and cats, 4 had both dogs and mice, 2 had both cats and mice,
and only 1 had all three kinds of pets. How many students had no pets of these
types? 

can u please tell me how you found this answer.

You are supposed to use a Venn diagram. But I can't draw a
Venn diagram on here, so I'll have to use algebra instead.

Let X = the number who had no pets of these types

Let C = the number that had cats only, but no dogs or mice.
Let D = the number that had dogs only, but no cats or mice.
Let M = the number that had mice only, but no cats or dogs.

Let CD = the number that had cats and dogs only, but no mice.
Let CM = the number that had cats and mice only, but no dogs.
Let DM = the number that had dogs and mice only, but no cats.

Let CDM = the number that had all three kinds of animals.

>>...50 students...<<

Translation:
eq. 1:    C + D + M + CD + CM + DM + CDM + X = 50 


>>...30 had cats...<<

Translation:
eq. 2:    C + CD + CM + CDM = 30


>>...25 had dogs...<<

Translation:
eq. 3:    D + CD + DM + CDM = 25


>>...5 had mice...<<

Translation:
eq. 4:    M + CM + DM + CDM = 5


>>...16 had both dogs and cats...<<

Translation:
eq. 5:    CD + CDM = 16


>>...4 had both dogs and mice...<< 

Translation:
eq. 6:    DM + CDM = 4 


>>...2 had both cats and mice...<< 

Translation:
eq. 7:    CM + CDM = 2


>>...only 1 had all three kinds of pets...<< 

Translation:
eq. 8:    CDM = 1


So we have 8 equations in 8 unknowns

eq. 1:    C + D + M + CD + CM + DM + CDM + X = 50 
eq. 2:    C + CD + CM + CDM = 30
eq. 3:    D + CD + DM + CDM = 25
eq. 4:    M + CM + DM + CDM = 5
eq. 5:    CD + CDM = 16
eq. 6:    DM + CDM = 4 
eq. 7:    CM + CDM = 2
eq. 8:    CDM = 1

Use equation 8 to substitute 1 for CDM in the other 7 equations,
and you have:

eq. 1:    C + D + M + CD + CM + DM + X = 49 
eq. 2:    C + CD + CM = 29
eq. 3:    D + CD + DM = 24
eq. 4:    M + CM + DM = 4
eq. 5:    CD = 15
eq. 6:    DM = 3 
eq. 7:    CM = 1
eq. 8:    CDM = 1

Use equation 7 to substitute 1 for CM in eqs. 4, 2 and 1,
and you have:

eq. 1:    C + D + M + CD + DM + X = 48 
eq. 2:    C + CD = 28
eq. 3:    D + CD + DM = 24
eq. 4:    M + DM = 3
eq. 5:    CD = 15
eq. 6:    DM = 3 
eq. 7:    CM = 1
eq. 8:    CDM = 1

Use equation 6 to substitute 3 for DM in eqs. 4, 3, and 1,
and you have:

eq. 1:    C + D + M + CD + X = 45 
eq. 2:    C + CD = 28
eq. 3:    D + CD = 21
eq. 4:    M = 0
eq. 5:    CD = 15
eq. 6:    DM = 3 
eq. 7:    CM = 1
eq. 8:    CDM = 1

Use equation 5 to substitute 15 for CD in eqs. 3, 2, and 1,
and you have:

eq. 1:    C + D + M + X = 30 
eq. 2:    C = 13
eq. 3:    D = 6
eq. 4:    M = 0
eq. 5:    CD = 15
eq. 6:    DM = 3 
eq. 7:    CM = 1
eq. 8:    CDM = 1

Use equations 4, 3 and 2 to substitute 0 for M,
6 for D, and 13 for C in eq. 1, and you have

 C + D + M + X = 30
13 + 6 + 0 + X = 30
        19 + X = 30
             X = 11

That's the answer.

Edwin

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