john poured 2 1 / 2 liters of water into a tank. then he poured out 3 2 / 5 liters of water from the tank , leaving 4 1 over 5 liters of water in the tank how much water was in the tank at first?
4 1 over 5 = 4 1/5?
Let x = amount in tank at first
x + 2.5 - 3.4 = 4.2
Solve for x.
To find out how much water was in the tank at first, we need to perform the following steps:
Step 1: Convert mixed numbers to fractions.
2 1/2 liters = 5/2 liters
3 2/5 liters = 17/5 liters
4 1/5 liters = 21/5 liters
Step 2: Create an equation to represent the situation.
Let x represent the amount of water in the tank at first.
x - (17/5) = 21/5
Step 3: Solve the equation.
We can simplify the equation by multiplying through by 5 to eliminate the denominators.
5x - 17 = 21
Add 17 to both sides of the equation.
5x = 21 + 17
5x = 38
Divide both sides of the equation by 5.
x = 38/5
Step 4: Simplify the fraction.
x = 7 3/5 liters
Therefore, there were 7 3/5 liters of water in the tank at first.
To find out how much water was in the tank at first, we need to perform some calculations.
First, let's convert the mixed numbers into improper fractions:
2 1/2 liters = (2 x 2) + 1 / 2 = 5/2 liters
3 2/5 liters = (3 x 5) + 2 / 5 = 17/5 liters
4 1/5 liters = (4 x 5) + 1 / 5 = 21/5 liters
Now, we can set up the equation:
Initial amount - Amount poured out = Remaining amount
Let's call the initial amount of water in the tank x:
x - 17/5 = 21/5
To solve for x, we need to isolate the variable. Adding 17/5 to both sides of the equation:
x = 21/5 + 17/5
x = 38/5
Therefore, the initial amount of water in the tank was 38/5 liters, or 7 3/5 liters.