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GMAT Club Olympics 2025 (Day 10): A wildlife center houses 1/3 more

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Shin0099

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11 Jul 2025, 13:50

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Bunuel

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To answer this question, we just need to put the numbers into relation and bruteforce the question.

Hawk = 1
Owls = Hawk + 1/3 = 4/3
Falcons = (Hawk - 3/7)^-1 = 7/4

Now, we are looking for the LCM.
That is 12.

12*(1+4/3+7/4) = 49

Therefore, the lowest value possible in that case is 49 and the answer is E)

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1/3 more owls than hawk

Hawk : Owl
H : H+1/3*H
H:4H/3
3H:4H
3:4

For every 3 hawks they will have 4 owls.

3/7 fewer hawks than falcons.

This means
Hawk : Falcon
F-3/7*F : F
4/7 F : F
4F: 7F

For every 4 haws they have 7 falcons

Their total ratio is

	Owl	Hawk	Falcon
O:H	4	3
H:F		4	7
O:H:F	4*4	3*4	7*3
	16	12	21

this is the smallest ratio we can find so minimum number of total animals can be

16+12+21=49

Answer E. 49

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let us assume owls-o, hawks-h, falcons-f
o=(1+1/3)*h =>3o=4h
h=(1-3/7)*f => 7h=4f
to find the ratio 0:h:f
21o=28h=16f
LCM(21,28,16)=336
hence o:h:f=16:12:21
thus minimum total no of above birds=16+12+21=49
hence E

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"A wildlife center houses 1/3 more owls than hawks, and 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?"

Let O = Owl, H = Hawks, F = Falcon

F = X

3/7 fewer hawks than falcons:
H = (4/7)* X

1/3 more owls than hawks:
O = (4/3) * ( (4/7) * X ) = (16/21) * X

Ratio
X [F] : 4*X/7 [H] : 16*X/21 [O]

LCM(1, 7, 21) = 21

21*X/21 [F] : 12*X/21 [H] : 16*X/21 [O]
21*X [F] : 12*X [H] : 16*X [O]

GCF( 12, 16, 21 ) = 1

Then, we can not have a lower amount of each than this one.

Therefore the smallest possible total number of falcons, hawks, and owls it could have is 12 + 16 + 21 = 49

Answer = E. 49

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Topic(s)- Fractions to Ratios, Least Common Multiples
Strategy- Ratio Tables
Variable(s)- Owls = "o"; Hawks = "h"; Falcons = "f"; 1st Ratio Multiplier = "x"; 2nd Ratio Multiplier = "y"; 3rd Ratio Multiplier = "z"; Scale Factor = "SF"

1) Think of the fractions in terms of ratios
i) h = (3o/3) + (1o/3) = (4o/3)
(h/o) = (4/3)
i.e. 4 parts h to 3 parts o is 7 parts total
ii) h = (7f/7) - (3f/7) = (4f/7)
(h/f) = (4/7)
i.e. 4 parts h to 7 parts f is 11 parts total

2) Ratio Table
i) Determine the Scale factors to combine each phrase into a 3-part ratio

SF	o	h	Sum(o+h)	SF	h	f	Sum(h+f)
(4z/x)	4x	3x	7x	(3z/y)	4y	7y	11y
--	(4z/x)*4x =16z	(4z/x)*3x =12z	--	--	(3z/y)*4y =12z	(3z/y)*7y =21z	--

O:H:F = 16z : 12z : 21z
ii) Total = 16z + 12z + 21z = 49z

Answer: E

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For easy calculation, we assume 21 for the falcons. So, the number of hawks is $\frac{4}{7}*21=12$, and the number of owls is $\frac{4}{3}*12=16$
So the ratio of owls to hawks to falcons is 16:12:21
Since we can't divide all the numbers in the ratio by any factor, they are the smallest possible numbers for each one.
The question asks for the smallest total number, which is: 16+12+21=49

The answer is E.

Bunuel

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A wildlife center houses 1/3 more owls than hawks, and 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?

Fractions have numerators 3 and 7 => both primes => LCM > 21

Take 21 as the potential number for the falcons , this would mean that there are 21-3/7= 12 Hawks, which in that case would mean that there are 12 x 4/3 = 16 owls.

21 + 12 + 16 = 49

Answer E

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A wildlife center houses 1/3 more owls than hawks (h + h * 1/3 = o), and 3/7 fewer hawks than falcons (f - f * 3/7 = h). 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?

h = f * 4/7; Falcon must be a multiple of 7.

o = h * 4/3; Hawk must be a multiple of 3.

Since h must be a multiple of 3, the expression f * 4/7 must be a multiple of 3, and hence f must also be a multiple of 3;

Min f = 7*3 = 21;

Then h = f * 4/7 = 12;

Then o = h * 4/3 = 16;

Total = 49

A. 28
B. 36
C. 42
D. 45
E. 49

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A wildlife center houses 1/3 more owls than hawks, and 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?
Let no. of Owls be O, Hawks be H and Falcons be F
Owls,O=H+1/3H= 4/3 H
Hawks,H= F-3/7F= 4/7F
Owls= 4/3*4/7= 16/21F
Also O=>1 F=>1 H=>1
For owls to be whole no greater than 1
F should be multiple of 21
For minimum owls Falcons=21
Owls = 16 so Hawks =12
O=16,F=21 H=12
Sum= 49 Answer should be E.

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Bunuel

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H = Hawks
O= Owls
F = Falcons

O = H + 1/3H
= 4/3H

H = F- 3/7F
= 4/7F

=> O = 4/3*4/7F
= 16/21F

Total: F+ 4/7F+16/21F = 49F/21
Now, for the minimum possible value, F=21 and F has to be a multiple of 21 to make the number of falcons an integer.

Total: 49
ANSWER: E

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4/3 h= owls
4/7f=h
f=7h/4

Ratio of O:H:F= 4/3h:h:7h/4
= 14:12:21

Minimum number would be =16+12+21=49.

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Owls =$ H+ \frac{1}{3}H = \frac{4}{3}H $
Hawks = $H = F - \frac{3}{7}F = \frac{4}{7}F$
Owls =$ H+ \frac{1}{3}H = \frac{4}{3}H = \frac{4}{3} * \frac{4}{7}F = \frac{16}{21}F $

Total = $ \frac{16}{21}F + \frac{4}{7}F + F = \frac{49}{21}F$

While 49/21 can be reduced to 7/3, for the number of owls and hawks to be a whole number, F needs to be a multiple of 21. If inot, then number of owls = 16/21 of falcons will be a decimal, and not an integer,
So F = 21 , is the minimum multiple of 21.
then O = 16
H = 12
Total = 49

So minimum is 49
Answer is E

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Now looking at total denominator ie. If falcons are 7x then owls are 16/3x and hawks are 4k. Now sum should be a positive integer.
If we take 3 as x we get 21+16+12=49

Bunuel

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Correct Answer: Option E 49

Let's assume,
Owl= O
Hawks= H
Falcon= F

Here we will covert all the data in terms of falcon,
Falcons= F
Hawks= 4/7 F (As hawks are 3/7 less than falcons)

For owl firstly calculate 1/3 from hawks and then we will add it,

4/7 *1/3= 4/21F this will now add into hawks as 4/21F is more than hawks.
4/21 F + 4/7 F= 4+12/21 F
Owl = 16/21 F

So now,
Falcons= F
Hawks= 4/7 F
Owl = 16/21 F

Now we will add all three and checks through options:

= F + 4/7 F + 16/21 F
= (21F + 12F + 16F )/21
= 49 F/21

We can check that option E is divisible by 49 and we try to check that,

Falcon= 21
Owl= 16
Hawks= 12

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From the above fractions ($\frac{1}{3}$, $\frac{3}{7}$) calculate LCM of denominators = 21

Now let's assume Number of Falcons = $21x$ (for easy calculations)

Given :
Hawks are $\frac{3}{7}$ fewer than falcons,meaning hawks are $\frac{4}{7}$ of falcons
= $\frac{4}{7}$*$21x$= $12x$
Owls are $\frac{1}{3}$ more than hawks meaning owls are $\frac{4}{3}$ of hawks = $\frac{4}{3}$*$12x$= $16x$

Total birds = $21x+12x+16x = 49x$
We want smallest possible value, so put $x=1$
Then total = 49

Option E

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Owls, Hawks, Falcons are $o, h, f$

$o=\frac{4}{3}h$;

$h=(1-\frac{3}{7})f=\frac{4}{7}f$; $f=\frac{7}{4}h$

Total birds $=o+h+f=h(\frac{4}{3}+1+\frac{7}{4})=\frac{16+12+21}{12}h=\frac{49h}{12}$

Since the no of birds must be an integer, the min possible total number of birds is when $h=12$

$Total =49$

Answer: E

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Bunuel

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O = H + 1/3 H = 4/3 H

H = F - 3/7 F = 4/7 F

F = 7/4 H

Total = H + 7/4H + 4/3H

LCM = 12

H = 12
F = 21
O = 16

Total = 49

Option E

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Let the number of Hawks be H
Number of Owl O= (4/3)H
Number of Falcon F -> H = (4/7)F
-> F = (7/4)H
The total number -> H + O + F = H + 4H/3 + 7H/4
= H*(12+16+21)
= 49H/12
Using options only multiples of 49 gives whole number to number of hawks, since they want minimum take first multiple = 49
Hence H = 12, O = 16, F = 21

Bunuel

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1/3 more owls than hawks => ratio between owls and hawks is O:H=4:3 = 16:12
3/7 fewer hawks than falcons => ratio between falcons and hawks is F:H=7:4 = 21:12
So we have O:H:F=16:12:21. The smallese possible total number of falcons, hawks, and owls is 16+12+21= 49

Bunuel

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