Daily Math Puzzle: 2026-08-24
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2026-08-24
A wildlife hospital has a mixed population of recovering owls and hawks. Currently, there are 12 more owls than hawks. If half of the owls and a third of the hawks are released after successful rehabilitation, there will then be 8 more hawks than owls remaining in the hospital. How many birds were initially in the hospital?
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Solution
180 — Let O be the initial number of owls and H be the initial number of hawks.
1. From the first statement, 'there are 12 more owls than hawks':
O = H + 12
2. After rehabilitation and release:
Half of the owls are released, so remaining owls = O - O/2 = O/2.
A third of the hawks are released, so remaining hawks = H - H/3 = 2H/3.
3. From the third statement, 'there will then be 8 more hawks than owls remaining':
Remaining Hawks = Remaining Owls + 8
2H/3 = O/2 + 8
4. Now, substitute the first equation (O = H + 12) into the third equation:
2H/3 = (H + 12)/2 + 8
5. To eliminate fractions, multiply the entire equation by the least common multiple of 3 and 2, which is 6:
6 * (2H/3) = 6 * ((H + 12)/2) + 6 * 8
4H = 3(H + 12) + 48
4H = 3H + 36 + 48
4H = 3H + 84
6. Solve for H:
4H - 3H = 84
H = 84 (initial number of hawks)
7. Solve for O using O = H + 12:
O = 84 + 12
O = 96 (initial number of owls)
8. The question asks for the *total* number of birds initially in the hospital:
Total Initial Birds = O + H = 96 + 84 = 180.
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