The number of fish in a tank decreases by x% each year. Given that the number of fish halves in 8 years, work out the value of x. Give your answer to 1dp.

Question

(3marks)

I've tried this question but am not sure of my answer. Could someone please go through it for me, thanks!

Answer 1

Certainly, if you would post your attempt!

Answer 2

dn/dt = -(x/100) n

dn/n = -(x/100) dt

ln n = (-x/100 + C

n = e^(-x/100 + C)t
or
n = c e^-(x/100)t
at t = 0, n = N so c = N
at t = 8, n = N/2
so
c = N
and
N/2 = N e^-8x/100
or
.5 = e^-.08x
ln .5 = -.08 x
-.693 = -.08 x
x = 8.725 %

Answer 3

Sure, so I used the formula Final amount=original amount* (1-x/100)^time.
I rearranged the equation to make x the subject:
X=time√final amount/original amount -1*100
Then I subbed in values:
X=8√50/100 -1 *100
(I used 50 and 100 because 100 would half to a final amount of 50)
I worked out x to be -8.3 to 1dp.
I don't think it should b negative I don't know where I went wrong.

Answer 4

well ok, then your way
so let

(1 - x/100) = z

1/2 = z^8

ln .5 = 8 ln z
so
ln z = -.693/8 = - .0866
then z = e^-.0866 = .917
x/100 = 1-.917 = .083
so
x = 8.3 %

note my way uses continuous rate, yours is yearly change.

Answer 5

The number of fish in a tank decreases by x% each year. Given that the number of fish halves in 8 years,

(1-x/100)^t = (1/2)^(t/8)
t log(1-x/100) = t/8 log(1/2)
log(1-x/100) = log(2^(-1/8))
1 - x/100 = 2^(-1/8)
x/100 = 1 - 2^(-1/8)
x = 100(1 - 2^(-1/8)) = 8.3