+81 Each of n cats has 2n fleas. If two cats (and their fleas) are removed, and three fleas are removed from each remaining cat, the total number of fleas remaining would be half the original total number of fleas. What is the value of n?
The original number of fleas is 2n*n=2n^2. After two cats are removed, there are n-2 cats left. The number of fleas remaining is (n-2)(2n-3)=2n^2-8n+6. We are told that this is half the original number of fleas, so 2n^2-8n+6=2n^2/2=n^2. Solving for n, we get n=3. Therefore, the value of n is 3.
+81 Uh, but when you were multiplying the (n-2)(2n-3), wouldn't it multiply/add to -7n?
Let's start by using algebraic variables to represent the number of cats and fleas.
Let "n" be the number of cats, and "2n" be the number of fleas for each cat. Therefore, the total number of fleas is 2n * n = 2n^2.
When two cats are removed, we are left with (n-2) cats. Each of these cats now has 2n-3 fleas. Therefore, the total number of fleas remaining is (n-2)(2n-3).
According to the problem, this is equal to half the original number of fleas:
(n-2)(2n-3) = (1/2)(2n^2)
Simplifying this equation gives:
2n^2 - 7n + 6 = n^2
Subtracting n^2 from both sides gives:
n^2 - 7n + 6 = 0
Factoring this quadratic equation gives:
(n-6)(n-1) = 0
Therefore, n = 6 or n = 1.
However, if n = 1, then there is only one cat, which contradicts the assumption that we removed two cats. Therefore, the only solution is n = 6.
So there are 6 cats, and each cat has 2n = 12 fleas. Intotal, there are 2n^2 = 72 fleas.
If we remove two cats and three fleas from each remaining cat, we are left with (6-2) = 4 cats, each with (2*6-3) = 9 fleas. Therefore, the total number of fleas remaining is 4*9 = 36.
And indeed, 36 is half of the original number of fleas, which was 72. So n = 6 is the correct solution.
