Adam has 7 times as many blue stickers as red stickers.
He buys another 90 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 |
7 x 1 = 7 u |
1 x 1 = 1 u |
Change |
No Change |
+ 90 |
After |
1 × 7 = 7 u |
1 × 7 = 7 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 7.
LCM of 1 and 7 = 7
Number of red stickers that Adam buys
= 7 u - 1 u
= 6 u
6 u = 90
1 u = 90 ÷ 6 = 15
Number of more blue stickers than red stickers at first
= 7 u - 1 u
= 6 u
= 6 x 15
= 90
(b)
Number of red stickers and blue stickers in the end
= 7 u + 7 u
= 14 u
= 14 × 15
= 210
Answer(s): (a) 90; (b) 210