Adam has 8 times as many red stickers as green stickers.
He buys another 70 green stickers.
Adam has equal number of red and green stickers.
- How many more red stickers than green stickers does Adam have at first?
- How many green stickers and red stickers does Adam have in the end?
|
Red |
Green |
Before |
8 x 1 = 8 u |
1 x 1 = 1 u |
Change |
No Change |
+ 70 |
After |
1 × 8 = 8 u |
1 × 8 = 8 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 8.
LCM of 1 and 8 = 8
Number of green stickers that Adam buys
= 8 u - 1 u
= 7 u
7 u = 70
1 u = 70 ÷ 7 = 10
Number of more red stickers than green stickers at first
= 8 u - 1 u
= 7 u
= 7 x 10
= 70
(b)
Number of green stickers and red stickers in the end
= 8 u + 8 u
= 16 u
= 16 × 10
= 160
Answer(s): (a) 70; (b) 160