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?

Guest Jun 24, 2023

#1**0 **

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

Guest Jun 24, 2023

#2**0 **

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.**

Guest Jun 25, 2023