Adam has some green stickers and blue stickers.
If 66 green stickers are removed, 80% of the stickers will be blue stickers.
If 44 green stickers are added, 40% of the stickers will be blue stickers.
- How many green stickers are there?
- How many blue stickers are there?
|
Scenario 1 |
Scenario 2 |
|
Green |
Blue |
Green |
Blue |
Before |
1 u + 66 |
4 u |
6 u - 44 |
4 u |
Change |
- 66 |
No change |
+ 44 |
No change |
After |
1x1 = 1 u |
4x1 = 4 u |
3x2 = 6 u |
2x2 = 4 u |
(a)
80% =
80100 =
4540% =
40100 =
25Scenario 1 Fraction of the stickers that are green in the end
= 1 -
45 =
15 Scenario 2
Fraction of the stickers that are green in the end
= 1 -
25 =
35 The number of blue stickers remains unchanged in both scenarios.
LCM of 4 and 2 = 4
6 u - 44 = 1 u + 66
6 u - 1 u = 66 + 44
5 u = 110
1 u = 110 ÷ 5 = 22
Number of green stickers
= 1 u + 66
= 1 x 22 + 66
= 22 + 66
= 88
(b)
Number of blue stickers
= 4 u
= 4 x 22
= 88
Answer(s): (a) 88; (b) 88