Adam has 6 times as many red stickers as blue stickers.
He buys another 50 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 |
6 x 1 = 6 u |
1 x 1 = 1 u |
Change |
No Change |
+ 50 |
After |
1 × 6 = 6 u |
1 × 6 = 6 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 6.
LCM of 1 and 6 = 6
Number of blue stickers that Adam buys
= 6 u - 1 u
= 5 u
5 u = 50
1 u = 50 ÷ 5 = 10
Number of more red stickers than blue stickers at first
= 6 u - 1 u
= 5 u
= 5 x 10
= 50
(b)
Number of blue stickers and red stickers in the end
= 6 u + 6 u
= 12 u
= 12 × 10
= 120
Answer(s): (a) 50; (b) 120