Adam has 9 times as many green stickers as red stickers.
He buys another 120 red stickers.
Adam has equal number of green and red stickers.
- How many more green stickers than red stickers does Adam have at first?
- How many red stickers and green stickers does Adam have in the end?
|
Green |
Red |
Before |
9 x 1 = 9 u |
1 x 1 = 1 u |
Change |
No Change |
+ 120 |
After |
1 × 9 = 9 u |
1 × 9 = 9 u |
(a)
The number of green stickers Adam has at first and in the end remains unchanged.
Make the number of green stickers the same using the LCM of 1 and 9.
LCM of 1 and 9 = 9
Number of red stickers that Adam buys
= 9 u - 1 u
= 8 u
8 u = 120
1 u = 120 ÷ 8 = 15
Number of more green stickers than red stickers at first
= 9 u - 1 u
= 8 u
= 8 x 15
= 120
(b)
Number of red stickers and green stickers in the end
= 9 u + 9 u
= 18 u
= 18 × 15
= 270
Answer(s): (a) 120; (b) 270