Orozco Slack and Vincent Lyle folded paper balls over two days. On Saturday, Orozco Slack folded 45 paper balls more than Vincent Lyle. On Sunday, Orozco Slack folded another 20 paper balls and Vincent Lyle folded another 21. At the end of the two days, Orozco Slack made 7/10 of the total number of balls. What was the number of paper balls that Orozco Slack folded?
Let's start by using variables to represent the number of paper balls each person folded on Saturday:
- Let x be the number of paper balls Vincent Lyle folded on Saturday
- Then Orozco Slack folded 45 more, which is x + 45
Together, they folded x + (x + 45) = 2x + 45 paper balls on Saturday.
On Sunday, Orozco Slack folded another 20, bringing his total to x + 45 + 20 = x + 65. Vincent Lyle folded another 21, bringing his total to x + 21.
In total, they folded (2x + 45) + (x + 65) + (x + 21) = 4x + 131 paper balls over two days.
We know that Orozco Slack folded 7/10 of the total, so:
- 7/10(4x + 131) = x + 65
Let's solve for x:
- 28x + 917 = 10x + 650
- 18x = 267
- x ≈ 14.83
Since we can't have a fraction of a paper ball, let's round up to the nearest whole number and say that Vincent Lyle folded 15 paper balls on Saturday.
Then Orozco Slack folded 45 more, which is 60 paper balls on Saturday. On Sunday, Orozco Slack folded another 20, which is a total of 80. Vincent Lyle folded 21 more, which is a total of 36.
Together, they folded 80 + 36 = 116 paper balls over two days. Orozco Slack folded 7/10 of these, which is:
- 7/10(116) = 81.2
Rounding to the nearest whole number, we get that Orozco Slack folded 81 paper balls over two days.