Peter had 3/4 as much money as James at first. After receiving $12 each, Peter had 3/5 as much money as James. How much money did Peter have at first?
p=3/4 J
p+12= 3/5(j+12)
Can you solve for P and J?
To solve this problem, let's break it down into steps:
Step 1: Let's assume that James had x dollars at first.
Step 2: According to the problem, Peter had 3/4 as much money as James at first. Therefore, Peter had (3/4) * x dollars initially.
Step 3: After receiving $12 each, Peter had 3/5 as much money as James. This means that Peter had (3/5) * (x + 12) dollars.
Step 4: Now, we can set up an equation with the information from steps 2 and 3: (3/4) * x = (3/5) * (x + 12).
Step 5: To eliminate the fractions, we can multiply both sides of the equation by the least common multiple (LCM) of the denominators, which is 20. This gives us: 20 * (3/4) * x = 20 * (3/5) * (x + 12).
Simplifying, we get: 15x = 12(x + 12).
Step 6: Distributing on the right side of the equation, we have: 15x = 12x + 144.
Step 7: Subtracting 12x from both sides gives us: 3x = 144.
Step 8: Finally, dividing both sides by 3 gives us: x = 48.
Therefore, James had $48 at first. And since Peter had 3/4 as much money as James at first, Peter had (3/4) * $48 = $36 initially.
So, Peter had $36 at first.