Adam has 5 times as many blue stickers as red stickers.
He buys another 60 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 |
5 x 1 = 5 u |
1 x 1 = 1 u |
Change |
No Change |
+ 60 |
After |
1 × 5 = 5 u |
1 × 5 = 5 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 5.
LCM of 1 and 5 = 5
Number of red stickers that Adam buys
= 5 u - 1 u
= 4 u
4 u = 60
1 u = 60 ÷ 4 = 15
Number of more blue stickers than red stickers at first
= 5 u - 1 u
= 4 u
= 4 x 15
= 60
(b)
Number of red stickers and blue stickers in the end
= 5 u + 5 u
= 10 u
= 10 × 15
= 150
Answer(s): (a) 60; (b) 150