Question

There are currently 6,210 fish in a lake. If the number of fish in the lake increased by 15 percent during the last year and 20 percent during the year before that, how many more fish are in the lake currently than in the last two years ago?

Answer 1

1710 fish

Explanation:

Let the number of fish in the lake last year be 'x'

The equation becomes-
x + 15x/100=6210 (This is because 15% of x is added in x to become
6210... hope you understand)

x + 15x/100 = 6210
115x/100 = 6210
23x/20 = 6210
x = 6210 * 20/23
x = 270 * 20
x = 5400

Therefore- The number of fish last year = 5400


Now, let the number of fish in the lake 2 years ago be 'y'

The equation becomes-
y + 20y/100 = 5400 (Same logic as last time)

y + 20y/100 = 5400
120y/100 = 5400
24y/20 = 5400
6y/5 = 5400
y = 5400 * 5/6
y = 900 * 5
y = 4500

Therefore- Fish in the lake two years ago = 4500 fish

Difference between current and two years ago = 6210-4500
= 1710 fish

Hope you understood the answer and I hope it helps you :)

Answer 2

I’m pretty sure the answer should be 1,710 if not i’m sorry