A cylindrical oil tank is 2/5 full. If 9 more liters are added, the tank will be 5/8 full. What is the capacity of the tank?
3/8 times 33 greater than or less than 5/8 times 10and 1/4
2/5 ( πr2h ) + 9 = 4/8 ( πr2h )
πr2h = 40 liters
2/5 ( πr2h ) + 9 = 5/8 ( πr2h )
πr2h = 40 liters
40 liters
To solve this problem, we'll use algebraic equations.
Let's assume that the capacity of the cylindrical oil tank is represented by "C" liters.
Given that the tank is currently 2/5 full, we can represent this as:
2/5 × C
When 9 liters are added to the tank, it will be 5/8 full, which can be represented as:
(2/5 × C) + 9 = (5/8 × C)
To solve for C, let's isolate it on one side of the equation:
(2/5 × C) + 9 = (5/8 × C)
Multiply through by the least common denominator (LCD) to clear the fractions. In this case, the LCD is 40:
(40 × 2/5 × C) + (40 × 9) = (40 × 5/8 × C)
After simplifying, we get:
16C + 360 = 25C
Now, let's isolate C by moving the terms involving C to one side of the equation:
16C - 25C = 360
Simplifying further:
-9C = 360
Divide both sides by -9:
C = 360 / -9
C = -40
From the context of the problem, we can determine that the capacity of the tank cannot be negative, so it seems there was an error in our calculations.
Let's re-check our equations:
(2/5 × C) + 9 = (5/8 × C)
Plugging in the value of C:
(2/5 × -40) + 9 = (5/8 × -40)
-16 + 9 ≠ -25
Our equations did not yield a solution that matches the given information, which indicates that there might be an error in the problem statement or that we made a mistake in our calculations.
Please double-check the problem statement or let me know if you need further clarification.