Paul has two thirds as many postcards and shawn the number of postcards shawn has three fifths of a number of postcard tim has if three boys have 280 postcards altogether how many more postcards does Tim have than Paul
7/8
To solve this problem, let's break it down step by step:
Step 1: Set up equations to represent the information given in the problem.
Let's assume that Tim has T postcards, Paul has P postcards, and Shawn has S postcards.
According to the problem:
1) Paul has two-thirds as many postcards as Tim.
So, P = (2/3)T
2) Shawn has three-fifths as many postcards as Tim.
So, S = (3/5)T
3) The three boys together have 280 postcards.
So, P + S + T = 280
Step 2: Simplify and rearrange the equations.
We can rewrite the first two equations as:
P = (2/3)T
S = (3/5)T
Step 3: Substitute the simplified equations into the third equation.
Now we can substitute the values of P and S into the equation from step 1:
(2/3)T + (3/5)T + T = 280
Step 4: Solve the equation to find the value of T.
Combine the like terms:
[(2/3) + (3/5) + 1] T = 280
To add fractions with different denominators, we need to find a common denominator, which in this case is 15:
[(10/15) + (9/15) + 15/15] T = 280
Now simplify further:
(34/15) T = 280
Multiply both sides of the equation by (15/34) to isolate T:
T = (280 * (15/34)) = 120
Therefore, Tim has 120 postcards.
Step 5: Find the number of postcards Paul has (P).
To find the number of postcards Paul has, substitute the value of T into the equation P = (2/3)T:
P = (2/3) * 120 = 80
Therefore, Paul has 80 postcards.
Step 6: Find the number of postcards Shawn has (S).
To find the number of postcards Shawn has, substitute the value of T into the equation S = (3/5)T:
S = (3/5) * 120 = 72
Therefore, Shawn has 72 postcards.
Step 7: Calculate how many more postcards Tim has than Paul.
The question asks for the difference in the number of postcards between Tim and Paul, so we subtract the number of postcards Paul has (80) from the number Tim has (120):
120 - 80 = 40
So, Tim has 40 more postcards than Paul.
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