Adam has some blue stickers and green stickers.
If 285 blue stickers are removed, 80% of the stickers will be green stickers.
If 87 blue stickers are removed, 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 + 285 |
4 u |
12 u + 87 |
4 u |
Change |
- 285 |
No change |
- 87 |
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
1 u + 285 = 12 u + 87
12 u - 1 u = 285 - 87
11 u = 198
1 u = 198 ÷ 11 = 18
Number of blue stickers
= 1 u + 285
= 1 x 18 + 285
= 18 + 285
= 303
(b)
Number of green stickers
= 4 u
= 4 x 18
= 72
Answer(s): (a) 303; (b) 72