Aaron, Bert and Carol had some stickers. Bert had 200 more stickers than Aaron. The ratio of the number of stickers Bert had to the number of stickers Carol had was 4:3. After Bert gave 1/8 of his stickers to Aaron, he had 60 fewer stickers than Aaron. How many stickers did Carol have at first?
Let's say Aaron had A stickers at first.
Since Bert had 200 more stickers than Aaron, Bert had B = A + 200 stickers at first.
The ratio of the number of stickers Bert had to the number of stickers Carol had was 4:3, so Carol had C = (3/7) * (A + 200) stickers at first.
After Bert gave 1/8 of his stickers to Aaron, Bert had B - B/8 = 7B/8 stickers left.
Since Bert had 60 fewer stickers than Aaron, we have 7B/8 = A - 60.
Substituting A + 200 for B, we have 7(A + 200)/8 = A - 60.
Simplifying the right side, we have 7A + 1400/8 = A - 60.
Combining like terms on the left side, we have 7A + 1400 = 8A - 480.
Subtracting 7A from both sides, we have 1400 = A - 480.
Adding 480 to both sides, we have 1880 = A.
Therefore, Aaron had 1880 stickers at first.
Since Carol had (3/7) * (A + 200) = 3/7 * 1880 + 600 = 1020 + 600 = 1620 stickers at first,
So the answer is 1620
A==Aaron, B==Bert, C==Carol
B==A + 200
B / C==4 / 3
A + 1/8[A + 200]==What Aaron has
7/8[A + 200] ==A + 1/8[A + 200] - 60, solve for A
A==840 stickers - what Aaron had at first
840 + 200 ==1040 stickers - what Bert had at first
1040 / C ==4 / 3, solve for C
C ==780 stickers - what Carol had at first.
