Adam has 9 times as many red stickers as blue stickers.
He buys another 80 blue stickers.
Adam has equal number of red and blue stickers.
- How many more red stickers than blue stickers does Adam have at first?
- How many blue stickers and red stickers does Adam have in the end?
|
Red |
Blue |
Before |
9 x 1 = 9 u |
1 x 1 = 1 u |
Change |
No Change |
+ 80 |
After |
1 × 9 = 9 u |
1 × 9 = 9 u |
(a)
The number of red stickers Adam has at first and in the end remains unchanged.
Make the number of red stickers the same using the LCM of 1 and 9.
LCM of 1 and 9 = 9
Number of blue stickers that Adam buys
= 9 u - 1 u
= 8 u
8 u = 80
1 u = 80 ÷ 8 = 10
Number of more red stickers than blue stickers at first
= 9 u - 1 u
= 8 u
= 8 x 10
= 80
(b)
Number of blue stickers and red stickers in the end
= 9 u + 9 u
= 18 u
= 18 × 10
= 180
Answer(s): (a) 80; (b) 180