Adam has some blue stickers and green stickers.
If 44 blue stickers are removed, 80% of the stickers will be green stickers.
If 66 blue stickers are added, 25% of the stickers will be green stickers.
- How many blue stickers are there?
- How many green stickers are there?
|
Scenario 1 |
Scenario 2 |
|
Blue |
Green |
Blue |
Green |
Before |
1 u + 44 |
4 u |
12 u - 66 |
4 u |
Change |
- 44 |
No change |
+ 66 |
No change |
After |
1x1 = 1 u |
4x1 = 4 u |
3x4 = 12 u |
1x4 = 4 u |
(a)
80% =
80100 =
4525% =
25100 =
14Scenario 1 Fraction of the stickers that are blue in the end
= 1 -
45 =
15 Scenario 2
Fraction of the stickers that are blue in the end
= 1 -
14 =
34 The number of green stickers remains unchanged in both scenarios.
LCM of 4 and 1 = 4
12 u - 66 = 1 u + 44
12 u - 1 u = 44 + 66
11 u = 110
1 u = 110 ÷ 11 = 10
Number of blue stickers
= 1 u + 44
= 1 x 10 + 44
= 10 + 44
= 54
(b)
Number of green stickers
= 4 u
= 4 x 10
= 40
Answer(s): (a) 54; (b) 40