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Lake A has a duck population of 30 females and 36 males. Lake B’s duck
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04 May 2021, 03:25
04 May 2021, 03:25
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Lake A has a duck population of 30 females and 36 males. Lake B’s duck population consists of the same ratio of females to males. If 6 new female ducks are added to Lake B and no other ducks are added or removed, the new total duck population of Lake B could be which of the following?
A. 144
B. 105
C. 78
D. 51
E. 36
A. 144
B. 105
C. 78
D. 51
E. 36
Re: Lake A has a duck population of 30 females and 36 males. Lake B’s duck
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04 May 2021, 07:16
04 May 2021, 07:16
1
Solution:-
The ratio of duck's population in lake A=30/36=5/6
Thus, the ratio in lake B=5/6
5x+6x+6=Total population of the duck
11x+6=y
Now, we need a number when subtracted from 6 gives a integer divisible by 11
Only 105 works.
IMO B
The ratio of duck's population in lake A=30/36=5/6
Thus, the ratio in lake B=5/6
5x+6x+6=Total population of the duck
11x+6=y
Now, we need a number when subtracted from 6 gives a integer divisible by 11
Only 105 works.
IMO B
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Re: Lake A has a duck population of 30 females and 36 males. Lake B’s duck
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04 May 2021, 08:23
04 May 2021, 08:23
1
GeminiHeat wrote:
Lake A has a duck population of 30 females and 36 males. Lake B’s duck population consists of the same ratio of females to males. If 6 new female ducks are added to Lake B and no other ducks are added or removed, the new total duck population of Lake B could be which of the following?
A. 144
B. 105
C. 78
D. 51
E. 36
A. 144
B. 105
C. 78
D. 51
E. 36
In Lake B:
\(\frac{F}{M} = \frac{5x}{6x}\)
After adding the female ducks:
\(\frac{F}{M} = \frac{5x+6}{6x}\)
Total Ducks in Lake B = \(11x + 6\)
So, our answer must be a number which is divisible by 11 when 6 is subtracted from it!
Hence, option B
