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

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iamchinu97

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11 Jul 2025, 09:10

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Ok so we know that
O = (1+1/3)H = 4H/3
H = (1-3/7)F = 4F/7
So F = 7H/4

Now
=> O : H : F
=> 4H/3:H:7H/4
=> 16: 12: 21
So min sum = 16+12+21 = 49

hence Ans E

1111fate

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11 Jul 2025, 09:14

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The Answer is 49
O = H + (1/3)H = (4/3)H
H = F - (3/7)F = (4/7)F
O = (4/3) * (4/7)F
O = (16/21)F

O = (16/21)F
H = (4/7)F
F = F
F must be a multiple of 7 and 21. LCM is 21
Hence F = 21
H = (4/7) * 21 = 4 * 3 = 12
O = (16/21) * 21 = 16
21+12+16=49

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11 Jul 2025, 09:22

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let Falcon be X
Hawk = x-3x/7= 4x/7
owl=4x/7 + 1/3(4x/7) = 16x/21

adding this we get

(21x+12x+16x) / 21 = 49x/21
smallest value of x for which this expression will be an integer is 21

so total sum will be 49 ( option E)

Bunuel

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?

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

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We represent the number of owls, hawks, and falcons as O, H, and F respectively. From the given information, we establish two relationships: owls are 4/3 the number of hawks (O = 3/4 H), and hawks are 4/7 the number of falcons (H = 7/4 F). To find the smallest integer numbers of birds, we express O and H in terms of F. For H to be an integer, F must be a multiple of 7. Substituting H in the O equation, we find that F must also be a multiple of 3 to make O an integer. Therefore, F must be a multiple of both 7 and 3, meaning F is a multiple of 21. Since we need at least one of each bird, the smallest possible value for F is 21. This gives H = 12 and O = 16. Summing these values (21 + 12 + 16) yields the smallest total number of birds, which is 49.

Regards,
Lucas

Bunuel

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11 Jul 2025, 09:46

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Let O be the number of owls, H be the number of hawks, and F be the number of falcons.

From the first statement: "A wildlife center houses 1/3 more owls than hawks"
This means the number of owls is the number of hawks plus one-third of the number of hawks:
O=H+ (1/3)*H = (4/3)*H

From the second statement: "and 3/7 fewer hawks than falcons"
This means the number of hawks is the number of falcons minus three-sevenths of the number of falcons:
H=F− (3/7)*F = (4/7)*F

Now we have a system of two equations:
O= (4/3)*H
H= (4/7)*F

We need to find the smallest possible total number of birds (O+H+F), such that O,H,F are all positive integers (since the center has at least one of each bird).

Let's express O in terms of F by substituting the second equation into the first:
O= (4/3)*{(4/7)*F}
O= (16/21)*F

Now we have the number of owls and hawks in terms of falcons:
O= (16/21)*F
H= (4/7)*F

For O to be an integer, F must be a multiple of 21 (since 16 and 21 share no common factors other than 1).
For H to be an integer, F must be a multiple of 7.

To satisfy both conditions, F must be a common multiple of 21 and 7. The least common multiple (LCM) of 21 and 7 is 21.

So, the smallest possible integer value for F is 21.

Now, let's calculate the corresponding number of hawks and owls:
For F=21:
H= (4/7) ×21=4×3=12 hawks
O= (16/21) ×21=16 owls

We verify that all quantities are positive integers:
Falcons = 21 (at least 1)
Hawks = 12 (at least 1)
Owls = 16 (at least 1)

The smallest possible total number of birds is the sum of these numbers:
Total = F+H+O=21+12+16=49

The final answer is 49

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11 Jul 2025, 09:49

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Because each species count must be a positive integer, the 'one‐third more owls than hawks' condition forces the ratio of owls to hawks to be 4 to 3, and the 'three‐sevenths fewer hawks than falcons' condition forces the ratio of hawks to falcons to be 4:7. When you link those ratios together, owls : hawks : falcons = 16 : 12 : 21, you see that every complete set contains 49 birds in total. Since you need at least one of each, the smallest multiple of that 16 : 12 : 21 pattern is just one copy. this gives 49 birds altogether.

Bunuel

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11 Jul 2025, 09:51

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1/3 of more owls than hawks, falcon

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11 Jul 2025, 10:30

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1. Ratio of owls to Hawks-> 4:3 * 4 = 16:12

Owl	Hawks
1 (1 + 1/3) =4/3	1

2. Ratio of Hawks to Falcons -> 4:7 * 3 = 12:21

Hawks	Falcons
1 (1 - 3/7) =4/7	1

Ratio of owl:Hawk:falcon=> 16:12:21= 16+12+21= 49 (correct answer is E)

Bunuel

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11 Jul 2025, 10:52

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We can say that ,
$O = \frac{4H}{3}$
and, $H = \frac{4F}{7}$ ----> $F = \frac{7H}{4}$

$Total = O + H + F = x = \frac{4H}{3} + H + \frac{7H}{4}$

This gives us,$ x = \frac{49H}{12}$

We need to find a value of x, such that H is an integer (since the number of any birds cannot be a fraction, it has to be an integer)

The only value from the answer choices that works is 49.

This x should be equal to 49.

Answer E.

kvaishvik24

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11 Jul 2025, 10:58

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let:

Hawks = H
Owls = $H × (1 + \frac{1}{3}) => (\frac{4}{3})H$, so H must be a multiple of 3

Falcons = F
Hawks = $F × (1 − \frac{3}{7}) => (\frac{4}{7})F$, so F must be a multiple of 7
H = 4k ( As F=7k is multiple of 7)

For H to be divisible by 3, k must be a multiple of 3. We need to find smallest so:

F =7k = 7 × 3 => 21 falcons
H = 4k = 4 × 3 => 12 hawks
O = (4/3) × 12 => 16 owls

Total = 21 + 12 + 16 = 49.
So the smallest possible total is 49

E) 49.

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11 Jul 2025, 11:05

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let's number of falcons = f

we have 3/7 fewer hawks than falcons

hawks = f - (3/7)*f = (4/7)*f

and 1/3 more owls than hawks

owls = hawks + (1/3)*hawks = (4/3)*hawks = (4/3)*(4/7)*f = (16/21)*f

number of birds will be in whole numbers

to make number of owls in whole number we need minimum 21 falcons

falcons = 21
hawks = 12
owls = 16

total = 49, smallest possible numbers of owls, hawks & falcons

Bunuel

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11 Jul 2025, 11:07

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$O=\frac{4}{3}H$
$H=\frac{4}{7}F$
So, $O=\frac{4}{3}\times\frac{4}{7}F=\frac{16}{21}F$

O:H:F
$=\frac{16}{21}F:\frac{4}{7}F:F$
=16:12:21
So, the smallest possible total number of falcons, hawks, and owls in the wildlife centre is =16+12+21=49
IMO, E

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11 Jul 2025, 11:12

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Let F be no of Falcons
H = (1-3/7)F = 4F/7
O = (1+1/3)H = 4H/3 = 16F/21

So for owl to be a whole number, F must be multiple of 21
=> F minimum = 21
O minimum = 16
H minimum = 12

Total is 49
Hence E is answer

Bunuel

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Bunuel

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Let the Hawks, Falcons and Owls be H, F and Ow respectively.

Ow = H + (1/3)*H = (4/3)*H

H = F - (3/7)*F = (4/7)*F

Thus, F = (7/4)*H

Total = H + (7/4)*H + (4/3)*H

= (12H + 21H + 16H) /12

= 49H/ 12

The minimum value of H = 12.

Then minimum value of Total = (49*12)/12 = 49

Option E

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Lets consider, number of owls = O, number of hawks = H,number of falcons = F

Given data,
a) wildlife center houses 1/3 more owls than hawks => O=H+ 1/3 H = 4/3H => 3O = 4H

b)3/7 fewer hawks than falcons => H=F-3/7F => 7H = 4F

let x be the smallest possible total number
from above equations (a) and (b) , we can observe H is multiple of 12
lets consider, H = 12x

from equation (a) => 3O=4(12x) => 3O=48x => O=16x
from equation (b) => 7(12x)=4F => 84x=4F => F=21x

from above, we got following with considering smallest x can be 1
Falcons (F) = 21 x = 21(1) = 21
Hawks (H) = 12x = 12 ( 1) = 12
Owls (O) = 16x = 16(1) = 16

smallest possible total number of birds = F + H + O = 21 +12+16=49

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Let owls , hawks and falcons be denoted by o,h and f.

o = h + 1/3h = 4/3h
It can be also written as,
h = 3/4o

h = f - 3/7f = 4/7f

From above logic
3/4o = 4/7f

o/f = 16/21

o:f:h = 16:21:12

Since number of a particular bird must be an integer, and sum of ratios is (16+21+12 = 49). So minimum 49 birds can be there. Option E os correct answer.

Bunuel

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11 Jul 2025, 12:07

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

Thus O = 16F/21
Smallest possible total number of birds ?
We need to minimize 49F/21 while keeping all O, H and F as whole number values.
For that the only possibility is F = 21. Thus answer is E, 49.

Bunuel

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The correct answer is choice (E) 49

To answer this question I first wrote out the information given in the question in terms of fractions of each type of bird. This helps establish the numerical relationships. I then used the fractions to determine the least possible whole number of each bird type and then added them all up to get the final answer.

Owls - we know this is 1/3 more than Hawks, so O -> 1+ 1/3 -> 4/3(H)
Hawks - we know this is 3/7 less than Falcons, or the remaining 4/7 (F)

I then tested if 4 Hawks and 7 Falcons could give us a whole number of Owls, but since it needed to be divisible by 3, I needed to try a higher multiple of 7 and apply that same multiple to 4. I tried 21 Falcons (so a multiple of 3) as this would be divisible by 3 as well, then since that was 3 times 7, I also applied it to Hawks to get 3 times 4 = 12 Hawks.

3 x 7 = 21 Falcons
4/7 of that = 3 x 4 = 12 Hawks
4/3 of 12 = 16 Owls

21+12+16 = 49 total birds

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11 Jul 2025, 13:26

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E. 49

Get all to the same unit (hawks)

O=4/3H
H
F=7/4H

Now avoid having parts of birds, you need entire birds. To get that, you have to get to multiply by the LCM (12):

16
12
21

Summing up: 49

Bunuel

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