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In a town of 10000 families, it was found that 40% families buy newspaper A, 20% families 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 the number of families that 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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Most Upvoted Answer
In a town of 10000 families, it was found that 40% families buy newspa...
To solve this problem, we can use the principle of inclusion-exclusion. We will break down the problem into various categories and then calculate the number of families in each category.
Let's start by identifying the given information:
- Total number of families = 10,000
- Percentage of families buying newspaper A = 40%
- Percentage of families buying newspaper B = 20%
- Percentage of families buying newspaper C = 10%
- Percentage of families buying both A and B = 5%
- Percentage of families buying both B and C = 3%
- Percentage of families buying both A and C = 4%
- Percentage of families buying all three newspapers = 2%
Now, let's calculate the number of families in each category.
Number of families buying newspaper A = 40% of 10,000 = 4,000
Number of families buying newspaper B = 20% of 10,000 = 2,000
Number of families buying newspaper C = 10% of 10,000 = 1,000
Number of families buying both A and B = 5% of 10,000 = 500
Number of families buying both B and C = 3% of 10,000 = 300
Number of families buying both A and C = 4% of 10,000 = 400
Number of families buying all three newspapers = 2% of 10,000 = 200
Now, let's calculate the number of families buying only newspaper A.
Number of families buying only A = Number of families buying A - Number of families buying both A and B - Number of families buying both A and C + Number of families buying all three newspapers
Number of families buying only A = 4,000 - 500 - 400 + 200 = 3,300
Therefore, the number of families that buy newspaper A only is 3,300.
Hence, the correct answer is option B) 3,300.
Let's start by identifying the given information:
- Total number of families = 10,000
- Percentage of families buying newspaper A = 40%
- Percentage of families buying newspaper B = 20%
- Percentage of families buying newspaper C = 10%
- Percentage of families buying both A and B = 5%
- Percentage of families buying both B and C = 3%
- Percentage of families buying both A and C = 4%
- Percentage of families buying all three newspapers = 2%
Now, let's calculate the number of families in each category.
Number of families buying newspaper A = 40% of 10,000 = 4,000
Number of families buying newspaper B = 20% of 10,000 = 2,000
Number of families buying newspaper C = 10% of 10,000 = 1,000
Number of families buying both A and B = 5% of 10,000 = 500
Number of families buying both B and C = 3% of 10,000 = 300
Number of families buying both A and C = 4% of 10,000 = 400
Number of families buying all three newspapers = 2% of 10,000 = 200
Now, let's calculate the number of families buying only newspaper A.
Number of families buying only A = Number of families buying A - Number of families buying both A and B - Number of families buying both A and C + Number of families buying all three newspapers
Number of families buying only A = 4,000 - 500 - 400 + 200 = 3,300
Therefore, the number of families that buy newspaper A only is 3,300.
Hence, the correct answer is option B) 3,300.
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Community Answer
In a town of 10000 families, it was found that 40% families buy newspa...
Let A, B and C denote the sets of families buying newspapers A, B and C, respectively.
n(A) = 4000, n(B) = 2000 and n(C) = 1000
Also, n(A ∩ B) = 500, n(B ∩ C) = 300 and n(C ∩ A) = 400
n(A ∩ B ∩ C) = 200
Number of families who buy A only = Number of families who buy A - (Number of families who buy A and B, A and C and (A and B and C))
= 4000 - (300 + 200 + 200) = 3300
Alternative method:
We have, N = 10000, n(A) = 40% of 10000 = 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 have to find
= n(A) - n((A ∩ B) ∪ (A ∩ C))
= n(A) - {n(A ∩ B) + n(A ∩ C) - n(A ∩ B ∩ C)}
= 4000 - (500 + 400 - 200) = 3300
n(A) = 4000, n(B) = 2000 and n(C) = 1000
Also, n(A ∩ B) = 500, n(B ∩ C) = 300 and n(C ∩ A) = 400
n(A ∩ B ∩ C) = 200
Number of families who buy A only = Number of families who buy A - (Number of families who buy A and B, A and C and (A and B and C))
= 4000 - (300 + 200 + 200) = 3300
Alternative method:
We have, N = 10000, n(A) = 40% of 10000 = 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 have to find
= n(A) - n((A ∩ B) ∪ (A ∩ C))
= n(A) - {n(A ∩ B) + n(A ∩ C) - n(A ∩ B ∩ C)}
= 4000 - (500 + 400 - 200) = 3300
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In a town of 10000 families, it was found that 40% families buy newspaper A, 20% families 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 the number of families that buy A only isa)3100b)3300c)2900d)1400Correct answer is option 'B'. Can you explain this answer?
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