Cathy had a sum of money. She spent 1/5 of it in September, 1/4 of it in October and 120 in November. After spending these amounts of money, she still had 540 left. How much money did she have at first?
Let's denote the amount of money Cathy had at the beginning as "x".
According to the information given:
In September, Cathy spent 1/5 of her money, which is (1/5)x.
In October, she spent 1/4 of her money, which is (1/4)x.
In November, she spent 120.
After spending this money, she still had 540 left, so we can write the equation:
x - (1/5)x - (1/4)x - 120 = 540
To solve this equation, let's simplify and solve for x:
x - (1/5)x - (1/4)x - 120 = 540
Combining like terms, we have:
(20/20)x - (4/20)x - (5/20)x - 120 = 540
(11/20)x - 120 = 540
Adding 120 to both sides, we get:
(11/20)x = 660
Now, multiplying both sides by (20/11), we find:
x = 1200
So, Cathy had 1200 dollars at the beginning.
To find out how much money Cathy had at first, we can work backwards from the amount she had left.
Let's assume the amount of money Cathy had at first is "x".
In September, Cathy spent 1/5 of her money, which is (1/5)x.
So, after September, Cathy had x - (1/5)x = (4/5)x left.
In October, Cathy spent 1/4 of the remaining money, which is (1/4)*(4/5)x = (1/5)x.
So, after October, Cathy had (4/5)x - (1/5)x = (3/5)x left.
In November, Cathy spent 120, so she had (3/5)x - 120 left.
We know she had 540 left after spending these amounts, so we can write the equation:
(3/5)x - 120 = 540
To solve for x, we can add 120 to both sides:
(3/5)x = 660
Then, we can multiply both sides by 5/3 to isolate x:
x = (660 * 5)/3
x = 1100/3
Therefore, Cathy had 1100/3 dollars at first.
1200
Cathy had $X.
x - x/5 - x/4 - 120 = 540.
x - 4x/20 - 5x/20 - 120 = 540,
20x - 4x - 5x - 2400 = 10,800,
11x = 13,200, X = $1200.