souma730 wrote:
The male alpine rabbits of the Tzatzek nature reserve have suffered a disease that killed 90 of them, causing the male to female ratio to drop from 3:2 to 2:3. How many alpine rabbits lived in the reserve before the disease struck?
A. 180
B. 270
C. 360
D. 450
E. 540
One option is to TEST the answer choices. I'll leave that to you.
Here's an algebraic solution.
Let M = the # of male rabbits BEFORE the disease struck.
Let F = the # of female rabbits BEFORE the disease struck.
We're told that the male/female ratio was 3:2 BEFORE the disease.
So, we can write: M/F = 3/2
Cross multiply to get:
2M = 3F
--------------------------
When the disease hits, 90 male rabbits die.
So, M - 90 = the # of male rabbits AFTER the disease struck.
Since no females dies, F = the # of female rabbits AFTER the disease struck.
We're told that the male/female ratio was 2:3 AFTER the disease.
So, we can write: (M - 90)/F = 2/3
Cross multiply to get: 3(M - 90) = 2F
Simplify, to get:
3M - 270 = 2F
--------------------------
We now have two equations:
2M = 3F
3M - 270 = 2F
Multiply the top equation by 2 to get:
4M = 6F
Multiply the bottom equation by 3 to get:
9M - 810 = 6F
Since both equations are set equal to 6F, we can conclude that 4M = 9M - 810
Subtract 9M from both sides to get: -5M = -810
Solve, M = 162
To solve for F, we can use one of the equations we created earlier.
Take
2M = 3F and replace M with 162 to get 2(162) = 3F
Simplify: 324 = 3F
Solve: F = 108
How many alpine rabbits lived in the reserve BEFORE the disease struck?
So, M + F = 162 + 108
=
270
=
B
Cheers,
Brent
