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31 Jul 2025, 12:32
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A wildlife center houses \(\frac{1}{3}\) more owls than hawks, and \(\frac{3}{7}\) fewer hawks than falcons. If the center has at least one of each bird, what is the smallest possible total number of falcons, hawks, and owls it could have?
A. 28
B. 36
C. 42
D. 45
E. 49
A. 28
B. 36
C. 42
D. 45
E. 49
31 Jul 2025, 12:32
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Official Solution:
A wildlife center houses \(\frac{1}{3}\) more owls than hawks, and \(\frac{3}{7}\) fewer hawks than falcons. If the center has at least one of each bird, what is the smallest possible total number of falcons, hawks, and owls it could have?
A. 28
B. 36
C. 42
D. 45
E. 49
\(\frac{1}{3}\) more owls than hawks implies \(Hawks + \frac{1}{3} * Hawks = Owls\), which gives \(4 * Hawks = 3 * Owls\). So:
Owls : Hawks = 4 : 3
\(\frac{3}{7}\) fewer hawks than falcons implies \(Falcons - \frac{3}{7} * Falcons = Hawks\), which gives \(4 * Falcons = 7 * Hawks\). So:
Falcons : Hawks = 7 : 4
Adjust the ratios so that the number corresponding to Hawks matches by multiplying the first ratio by 4 and the second by 3:
Owls : Hawks = 16 : 12
Falcons : Hawks = 21 : 12
So, Owls : Hawks : Falcons = 16 : 12 : 21
Therefore, the minimum number of birds the center could have is 16 + 12 + 21 = 49.
Answer: E
A wildlife center houses \(\frac{1}{3}\) more owls than hawks, and \(\frac{3}{7}\) fewer hawks than falcons. If the center has at least one of each bird, what is the smallest possible total number of falcons, hawks, and owls it could have?
A. 28
B. 36
C. 42
D. 45
E. 49
\(\frac{1}{3}\) more owls than hawks implies \(Hawks + \frac{1}{3} * Hawks = Owls\), which gives \(4 * Hawks = 3 * Owls\). So:
Owls : Hawks = 4 : 3
\(\frac{3}{7}\) fewer hawks than falcons implies \(Falcons - \frac{3}{7} * Falcons = Hawks\), which gives \(4 * Falcons = 7 * Hawks\). So:
Falcons : Hawks = 7 : 4
Adjust the ratios so that the number corresponding to Hawks matches by multiplying the first ratio by 4 and the second by 3:
Owls : Hawks = 16 : 12
Falcons : Hawks = 21 : 12
So, Owls : Hawks : Falcons = 16 : 12 : 21
Therefore, the minimum number of birds the center could have is 16 + 12 + 21 = 49.
Answer: E
08 Sep 2025, 05:25
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I did not quite understand the solution. When you have 4 hawks = 3 owls how did you go from that to owls : hawks = 4:3 in the next line?
Probably missing something but don't quite follow. Thank you.
Probably missing something but don't quite follow. Thank you.
