Adam has 10 times as many red stickers as blue stickers.
He buys another 90 blue stickers.
Adam has equal number of red and blue stickers.
- How many more red stickers than blue stickers does Adam have at first?
- How many blue stickers and red stickers does Adam have in the end?
|
Red |
Blue |
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 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 10.
LCM of 1 and 10 = 10
Number of blue stickers that Adam buys
= 10 u - 1 u
= 9 u
9 u = 90
1 u = 90 ÷ 9 = 10
Number of more red stickers than blue stickers at first
= 10 u - 1 u
= 9 u
= 9 x 10
= 90
(b)
Number of blue stickers and red stickers in the end
= 10 u + 10 u
= 20 u
= 20 × 10
= 200
Answer(s): (a) 90; (b) 200