Adam has 6 times as many blue stickers as red stickers.
He buys another 80 red stickers.
Adam has equal number of blue and red stickers.
- How many more blue stickers than red stickers does Adam have at first?
- How many red stickers and blue stickers does Adam have in the end?
|
Blue |
Red |
Before |
6 x 1 = 6 u |
1 x 1 = 1 u |
Change |
No Change |
+ 80 |
After |
1 × 6 = 6 u |
1 × 6 = 6 u |
(a)
The number of blue stickers Adam has at first and in the end remains unchanged.
Make the number of blue stickers the same using the LCM of 1 and 6.
LCM of 1 and 6 = 6
Number of red stickers that Adam buys
= 6 u - 1 u
= 5 u
5 u = 80
1 u = 80 ÷ 5 = 16
Number of more blue stickers than red stickers at first
= 6 u - 1 u
= 5 u
= 5 x 16
= 80
(b)
Number of red stickers and blue stickers in the end
= 6 u + 6 u
= 12 u
= 12 × 16
= 192
Answer(s): (a) 80; (b) 192