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

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Bunuel

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11 Jul 2025, 08:00

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Bunuel

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Bunuel

GMAT Club Official Explanation:

1/3 more owls than hawks implies Hawks + 1/3 * Hawks = Owls, which gives 4 * Hawks = 3 * Owls. So:

Owls : Hawks = 4 : 3

3/7 fewer hawks than falcons implies Falcons - 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.

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Let’s assume the wildlife center has 21 falcons (LCM of the denominators). Since hawks are 3/7 fewer than falcons, the number of hawks is 4/7 of 21, which is 12. Owls are 1/3 more than hawks, so the number of owls is 12 plus one-third of 12, which is 16. Therefore, the total number of birds is 21 falcons + 12 hawks + 16 owls = 49 birds. This is the smallest possible total since 21 is the smallest number of falcons that keeps all other bird counts as whole numbers. The correct answer is E, 49.

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11 Jul 2025, 08:17

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Let's define variables:
h = number of hawks
o = number of owls
f= number of falcons
From the first condition: "1/3 more owls than hawks"
0=h+ (1/3)h = (4/3)h
From the second condition: "3/7 fewer hawks than falcons"
h = f- (3/7)f=(4/7)f
Now I can express everything in terms off: h= (4/7)f
0 = (4/3)h = (4/3)(4/7)f = (16/21)f
Since h, o, and f must be integers and at least 1, I need to find the smallest value off such that both (4/f and (16/21)f are integers.
For (4/7)f to be an integer, f must be divisible by 7.
For (16/21)f to be an integer, f must be divisible by 21.
The smallest value off that is divisible by both 7 and 21 is their least common multiple (LCM).
Since 21 = 3 × 7, the LCM of 7 and 21 is 21.
So the smallest possible value for f is 21.
This gives us:
h = (4/7) × 21 = 12
O=(16/21)× 21=16
The total number of birds is: f+h+0=21+12+16=49
Therefore, the smallest possible total number of falcons, hawks, and owls is 49.
The answer is E. 49.

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11 Jul 2025, 08:21

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given
wildlife center can house
1/3 more owls than hawks,
O = h/3 + h = (4/3)h
3/7 fewer hawks than falcons

H= f- (3/7)f = (4/7) f

O = (4/3) h and h = (4/7)f
o/f = 16/21
O = (f*(16/21))

for both o & h to be whole number , dr of 21 of O and that of h 7 ; value of f has to be 21
if f =21
then
o = 16
H= 12
sum of all is 21+16+12 = 49

OPTION E is correct, 49

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Bunuel

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clearly from the options provided and know owls are 1/3 more than hawks and hawks are 3/7 fewer than falcons ie we substitute 25 is not divisible by 3,36 is not divisible by 7, hence the smallest possible total is 42

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11 Jul 2025, 08:26

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Given, O=H*4/3 and H= F*4/7

Simplifying the relation we get,

21O=28H =16F
Diving the equations by LCM of 21,28,16 , we get the ratio.
O:H:F=16:12:21

Hence least possible sum of birds = 16+12+21 = 49

Hence Option E

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

Total= F+H+O = F+(4/7)F+(16/21)F ......(Take LCM)
= (49/21) F
We want smallest possible total number of falcons, hawks, and owls
And the number should be integer, here smallest possible value of F can be same as denominator (21)
Smallest total= 49

E

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

So O in terms of F is :
O = 4/3*4/7*F = 16F/21

min( O + H + F ) = min( [16/21 + 4/7 + 1]F )
So F needs to be a multiple of 21 and since we want the min value thus F=21 itself.

min( O + H + F ) = 16 + 4*3 + 21 = 49

Thus Answer is E : 49

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Ans: E (49)
A wildlife center houses 1/3 more owls than hawks= if hawks are h then o = h +h/3 = 4h/3
3/7 fewer hawks than falcons = h = f-3f/7 = 4f/7

Then o becomes = (4/3)*(4/7)*f = 16/21 f

now the sum =f + 4/7 f + 16/21 f
= 49/21 f

from this the minimum value can be 49 when f = 21.

so the minimum total number of birds is 49

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11 Jul 2025, 08:34

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Given :

1. Three birds in proportion at a wildlife centre: Owls, Hawks & Falcons.
2. Their proportion => Hawks = 4/7 Falcons and Owls = 4/3 Hawks

Asked

The smallest possible total of these three birds.

Solution

Suppose there are 21 Falcons (an integer divisible by 7 and 3),
=> Hawks = 21*4/7 = 12
=> Owls = 12*4/3 =16

Now, do the figures 21, 12, 16 have a common factor- No

Therefore, total smallest possible birds of these types = 21+12+16 = 49.

Option E is the answer.

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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 wildlife center house o owls, h hawks & f falcons.

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

o: h : f = 4/3 : 1 : 7/4 = 16: 12: 21

Smallest possible total number of falcons, hawks and owls = 16 + 12 + 21 = 49

IMO E

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11 Jul 2025, 08:49

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Halk=(1- 3/7) falcon=4/7 falcon
owl=(1+ 1/3)=4/3 halk

falcon: halk : owl = 1 : 4/7 : 16/21

LCM =21

falcon+halk+owl=21+12+16=49

Ans E

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11 Jul 2025, 08:49

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From the question

Owls(o)are more than hawks(h)
o=4/3h (according to question)

Hawks(h) are less than Falcons(f)
h=f-3/7f
h=4/7f

Combining these two equations by replacing h
o=4/3*4/7*f
o=16/21*f
Only if f is 21,will the denominator go away.So f is 21.
o will be 16.
then h will be 12.

Add all those up to get 49

Hence answer is E that is 49

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Let owl, hawk and falcon be O,H,F

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

Smallest number to make all these integers (countable) = 21
F=21
O= 4/3 * 4F/7 = 16F/21 = 16*21/21 = 16
H= 4/7*21 = 12

Total = 21+16+12 =49

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11 Jul 2025, 08:53

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Owl	Hawks	Falcs
4 x 4	3 x 4
	4 x 4	7 x 3

The table above simplifies what is announced, once you get the relation among the qty of Owls and Hawks , and Hawks and Falcons, you use the link to fin the relation among all the birds

hawks are as 12, Owls as 16 and Falcons as 21
We have 49 birds in total as minimun.
Answer is E

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Let f = number of falcons.
Hawks = f-3/7= 4/7f.
Owls= 1/3+h = 4/3
4/7f=16/21f (expressing in terms of f).

Total birds = f+h+o
= f+4/7f+16/21f=49/21f.

So f must be a multiple of 21,the smallest value it can take is =21.

Total birds= (49/ 21 )×21 = 49.
Answer: 49. Option (E).

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11 Jul 2025, 09:02

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Let's say F for falcon, H for hawks, O for owls

As per the question O=H+1/3H=4/3H and F-H=3/4F so, H=4/7F

Now from above we can conclude that from H=4/7F that F is a multiple of 7 and from O=4/3H H is the multiple of 3

For finding H Let's take the possible value of F since F is mutiple of 7 and we have to find min number

So Let's take if F=7 then
H = (4/7) * 7 = 4 & O = (4/3) * H = (4/3) * 4 = 16/3 ( no. of owls or any animals can't be in fraction so F cannot be 7)
If F=14 then
H = (4/7) * 14 = 8 O = (4/3) * H = (4/3) * 8 = 32/3( same as above) It is also not possible
If F=21then
H = (4/7) * 21 = 12 O = (4/3) * H = (4/3) * 12 = 16 . It is Possible
So H+F+O=12+16+21=49 answer is E

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11 Jul 2025, 09:06

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A wildlife center houses 1/3 more owls than hawks
=> O=(1+1/3)H
O=4/3H
O: H= 4:3

and 3/7 fewer hawks than falcons.
=>H=(1-3/7)F
H=4/7F
H: F = 4:7

O:H:F
4:3
4:7
---------
16:12:21

Thus, the combination of the three avian species yields:
16K + 12K + 21K = 49K
When K = 1, representing the minimum value,
16 + 12 + 21 = 49

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Given: Owls = 1/3 (Hawks) + Hawks = 4/3(Hawks)

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

Then owls in tears of falcons = (4/3)*(4/7)*(Falcons) = 16/21(Falcons)

So the total number of birds = Owls + Hawks + Falcons
total = Falcons +4/7*(falcons) + 16/21 (Falcons)
IN order for birds to be integers = the falcons must be= 21.
Therefore the total number of birds(least) = 21 + 12+ 16 = 49. Option E.