+0  
 
0
51
2
avatar

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?

 Jun 24, 2023
 #1
avatar
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

 Jun 24, 2023
 #2
avatar
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.

 Jun 25, 2023

0 Online Users