Adam has 10 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 |
10 x 1 = 10 u |
1 x 1 = 1 u |
Change |
No Change |
+ 90 |
After |
1 × 10 = 10 u |
1 × 10 = 10 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 10.
LCM of 1 and 10 = 10
Number of red stickers that Adam buys
= 10 u - 1 u
= 9 u
9 u = 90
1 u = 90 ÷ 9 = 10
Number of more blue stickers than red stickers at first
= 10 u - 1 u
= 9 u
= 9 x 10
= 90
(b)
Number of red stickers and blue stickers in the end
= 10 u + 10 u
= 20 u
= 20 × 10
= 200
Answer(s): (a) 90; (b) 200