2 n^2 fleas to start
2 n^2 - 4 n - 3(n - 2) after removal
n^2 = 2 n^2 - 7 n + 6 ... 0 = n^2 - 7 n + 6
solve the quadratic for n
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?
Please help. I really need help!
3 answers
Thank you. This helped me a lot.
now:
number of cats --- n
number fleas per cat --- 2n
total number of fleas = 2n^2
after removal episode:
number of cats = n-2
fleas removed from those cats = 2(2n) = 4n
fleas removed from remaining cats = 3(n-2) = 3n - 6
fleas remaining = 2n^2 - 4n - (3n-6)
= 2n^2 - 7n + 6
2n^2 - 7n + 6 = (1/2)(2n^2) = n^2
n^2 - 7n + 6 = 0
(n - 1)(n - 6) = 0
n = 1 or n = 6, but we can't have 1 cat and then remove 2
so n = 6
check:
6 cats each with 12 fleas , total fleas = 72
2 cats removed leaves 4 cats, also 24 fleas were removed
also the remaining 4 cats each had 3 fleas removed
= 12
remaining fleas = 72-24-12 = 36
which is half of the original number of fleas
My answer is correct
number of cats --- n
number fleas per cat --- 2n
total number of fleas = 2n^2
after removal episode:
number of cats = n-2
fleas removed from those cats = 2(2n) = 4n
fleas removed from remaining cats = 3(n-2) = 3n - 6
fleas remaining = 2n^2 - 4n - (3n-6)
= 2n^2 - 7n + 6
2n^2 - 7n + 6 = (1/2)(2n^2) = n^2
n^2 - 7n + 6 = 0
(n - 1)(n - 6) = 0
n = 1 or n = 6, but we can't have 1 cat and then remove 2
so n = 6
check:
6 cats each with 12 fleas , total fleas = 72
2 cats removed leaves 4 cats, also 24 fleas were removed
also the remaining 4 cats each had 3 fleas removed
= 12
remaining fleas = 72-24-12 = 36
which is half of the original number of fleas
My answer is correct