Answer:
Approximately 1.0% of the remaining water leaked out in the second hour.
Explanation:
If 20% of the water leaked out in the first hour, then 80% of the water remained in the barrel at the end of the first hour.
To find out what percent of the remaining water leaked out in the second hour, we need to first calculate the fraction of the original amount of water that remains after the first hour. If we let W be the original amount of water in the barrel, then the amount of water that remains after the first hour is:
W1 = W(1 - 0.20) = 0.8W
where W1 is the amount of water remaining after the first hour.
The amount of water that leaks out in the second hour is not given, so let's call it x. Then, the amount of water remaining in the barrel after the second hour is:
W2 = W1(1 - x)
Substituting the expression for W1 we found above, we get:
W2 = 0.8W(1 - x)
We want to find the percent of the remaining water that leaked out in the second hour, which is equal to the fraction (x/W2) multiplied by 100%. So we have:
(x/W2) x 100% = (x / (0.8W(1 - x))) x 100%
We're given that this percentage is equal to some value, say P. So we can write:
P = (x / (0.8W(1 - x))) x 100%
Solving for x:
x = (0.8W(1 - x) / 100%) x P
x = 0.008W(1 - x)P
Expanding and simplifying:
x = 0.008WP - 0.008xP
1.008xP = 0.008WP
x = (0.008WP) / 1.008P
x = 0.00794W
Therefore, the percent of the remaining water that leaked out in the second hour is:
(x/W2) x 100% = (0.00794W / (0.8W(1 - 0.00794))) x 100% ≈ 1.0%
So approximately 1.0% of the remaining water leaked out in the second hour.