Adam has 7 times as many red stickers as green stickers.
He buys another 60 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 |
7 x 1 = 7 u |
1 x 1 = 1 u |
Change |
No Change |
+ 60 |
After |
1 × 7 = 7 u |
1 × 7 = 7 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 7.
LCM of 1 and 7 = 7
Number of green stickers that Adam buys
= 7 u - 1 u
= 6 u
6 u = 60
1 u = 60 ÷ 6 = 10
Number of more red stickers than green stickers at first
= 7 u - 1 u
= 6 u
= 6 x 10
= 60
(b)
Number of green stickers and red stickers in the end
= 7 u + 7 u
= 14 u
= 14 × 10
= 140
Answer(s): (a) 60; (b) 140