swerve wrote:The male alpine rabbits Tzatsek nature reserve has 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
Here's an algebraic solution that uses 2 variables:
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 die, 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
