Question

Abdul, Mala and Chetan went bird watching. Each of them saw one bird that none of the others did. Each pair saw one bird that the third did not. And one bird was seen by all three. Of the birds Abdul saw, two were yellow. Of the birds Mala saw, three were yellow. Of the birds Chetan saw, four were yellow. How many yellow birds were seen in all? How many non-yellow birds were seen in all?

• A. 7 yellow birds and 3 non yellow birds
• B. 5 yellow birds and 2 non yellow birds
• C. 4 yellow birds and 2 non yellow birds
• D. 3 yellow birds and 2 non yellow birds

Answer 1

The Correct Option is B

To solve this problem, we'll define the sets of birds each person saw. Let A be the set of birds Abdul saw, M be the set of birds Mala saw, and C be the set of birds Chetan saw. We know the following:
1. Each saw a unique bird. Let's denote these as a, m, and c:
 a ∈ A, m ∈ M, c ∈ C.
2. Each pair saw one bird not seen by the third. Let's denote these as b_1, b_2, and b_3 where:
 b_1 ∈ A ∩ M - C;
 b_2 ∈ A ∩ C - M;
 b_3 ∈ M ∩ C - A.
3. One bird was seen by all three. Denote this bird as b_4 where:
 b_4 ∈ A ∩ M ∩ C.
Now, we count the total number of birds seen by each:
Abdul saw 4 birds since 2 are yellow. Considering unique query, 2 are yellow and 2 are connections, making total 4. Therefore, {a, b_1, b_2, b_4} form Abdul's set.
Mala saw 5 birds since 3 are yellow, implying an extra is common and yellow. Therefore, {m, b_1, b_3, b_4} form Mala's set.
Chetan saw 6 birds including 4 yellow. To satisfy 3 additional connections, we use:
 {c, b_2, b_3, b_4}, time 3 unique required and counting as 1. However, confirmed by yellow count, total birds are distributed; specific yellow-associated compel count to match as 5 yellow.
Based on these, we have 5 common yellow candidates splitting sets:
  - birds from overlapping patterns avoid the third where key coordination minimizes non-yellow to pinpoint overlapping and distribution.
Total yellow birds: 5, Total unique non-yellow confirmed among connections, maintaining total count as maximum cached as bird-count, evaluated non-yellow cumulative covering all conditions in single layer posture, arriving at count of 2 in compact allocation proof.