After giving away 20 marbles, joshua had 60% left. How many had he at first.

Please help me solve.

To solve this problem, we'll use the information given:

1. Joshua gave away 20 marbles.
2. After giving away the marbles, he had 60% of the marbles left.

Let's assume that Joshua originally had "x" marbles.

After giving away 20 marbles, Joshua had (100% - 60%) = 40% of the marbles left.

So, 40% of "x" marbles were left after giving away 20 marbles.

To find the value of "x", we can set up the following equation:

40% of x = x × (40/100) = 0.4x

Now, we know that 0.4x = x - 20 (since Joshua gave away 20 marbles).

To solve the equation, we can subtract 0.4x from both sides:

0.4x = x - 20
0.4x - x = -20
-0.6x = -20
x = -20 / -0.6

Simplifying further, we get:

x = 200 / 6 = 33.333...

Therefore, Joshua originally had 33 marbles.

To solve this problem, we need to use the concept of percentages and algebraic equations.

Let's start by understanding what is given in the problem:

1. Joshua had 60% of his original marbles after giving away 20 marbles.

To find out how many marbles Joshua had at first, we can set up an equation.

Let's say the number of marbles Joshua had initially is x.

After giving away 20 marbles, Joshua had 60% of x left, which can be represented as 0.6x.

So, we can write the equation as:

0.6x = x - 20

Now, let's solve the equation to find the value of x:

0.6x = x - 20

To get rid of the decimal point, we can multiply the entire equation by 10:

6x = 10(x - 20)

Expanding the equation:

6x = 10x - 200

Rearranging the equation:

10x - 6x = 200

4x = 200

Dividing both sides of the equation by 4:

x = 200 / 4

x = 50

Therefore, Joshua had initially 50 marbles.

so he must have given away 40%

.40x = 20
x = ...