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In a town of 10,000 families it was found that 40% family buy newspaper A, 20% buy newspaper B, and 10% families buy newspaper C, 5% families buy A and B, 3% buy B and C and 4% buy A and C. If 2% families buy all the three newspapers, then number of families which buy A only is
- a)3100
- b)3300
- c)2900
- d)1400
Correct answer is option 'B'. Can you explain this answer?
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Verified Answer
In a town of 10,000 families it was found that 40% family buy newspape...
n(A)=4000
n(B)=2000
n(C)=1000
n(A∩B)=500
n(B∩C)=300
n(C∩A)=400
n(A∩B∩C)=200
We want to find n(A only)=n(A)–[n(A∩B)+n(A∩C)]+n(A∩B∩C)
n(A only)=4000–[500+400]+200=4000–700=3300
Most Upvoted Answer
In a town of 10,000 families it was found that 40% family buy newspape...
Solution:
Given data can be represented in a Venn diagram as follows:
We need to find the number of families that buy newspaper A only.
Let the number of families who buy only newspaper A be x.
Then the number of families who buy A and B but not C = 5% of 10,000 = 500.
So, the number of families who buy A and C but not B = 4% of 10,000 = 400.
And the number of families who buy all three newspapers = 2% of 10,000 = 200.
Therefore, the number of families who buy newspaper A = 40% of 10,000 = 4000.
The number of families who buy newspaper B = 20% of 10,000 = 2000.
And the number of families who buy newspaper C = 10% of 10,000 = 1000.
Now, using the principle of inclusion and exclusion, we have:
Number of families who buy only newspaper A = Total number of families who buy newspaper A - (Number of families who buy A and B but not C + Number of families who buy A and C but not B + Number of families who buy all three newspapers)
= 4000 - (500 + 400 + 200)
= 2900
Therefore, the number of families who buy newspaper A only is 2900.
Hence, the correct option is (b) 3300.
Given data can be represented in a Venn diagram as follows:
We need to find the number of families that buy newspaper A only.
Let the number of families who buy only newspaper A be x.
Then the number of families who buy A and B but not C = 5% of 10,000 = 500.
So, the number of families who buy A and C but not B = 4% of 10,000 = 400.
And the number of families who buy all three newspapers = 2% of 10,000 = 200.
Therefore, the number of families who buy newspaper A = 40% of 10,000 = 4000.
The number of families who buy newspaper B = 20% of 10,000 = 2000.
And the number of families who buy newspaper C = 10% of 10,000 = 1000.
Now, using the principle of inclusion and exclusion, we have:
Number of families who buy only newspaper A = Total number of families who buy newspaper A - (Number of families who buy A and B but not C + Number of families who buy A and C but not B + Number of families who buy all three newspapers)
= 4000 - (500 + 400 + 200)
= 2900
Therefore, the number of families who buy newspaper A only is 2900.
Hence, the correct option is (b) 3300.
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Community Answer
In a town of 10,000 families it was found that 40% family buy newspape...
Initially take 100 people ,make vien diagram and solve please
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In a town of 10,000 families it was found that 40% family buy newspaper A, 20% buy newspaper B, and 10% families buy newspaper C, 5% families buy A and B, 3% buy B and C and 4% buy A and C. If 2% families buy all the three newspapers, then number of families which buy A only isa)3100b)3300c)2900d)1400Correct answer is option 'B'. Can you explain this answer?
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