Adam has some blue stickers and green stickers.
If 357 blue stickers are removed, 90% of the stickers will be green stickers.
If 147 blue stickers are removed, 20% 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 + 357 |
9 u |
36 u + 147 |
9 u |
Change |
- 357 |
No change |
- 147 |
No change |
After |
1x1 = 1 u |
9x1 = 9 u |
4x9 = 36 u |
1x9 = 9 u |
(a)
90% =
90100 =
91020% =
20100 =
15Scenario 1 Fraction of the stickers that are blue in the end
= 1 -
910 =
110 Scenario 2
Fraction of the stickers that are blue in the end
= 1 -
15 =
45 The number of green stickers remains unchanged in both scenarios.
LCM of 9 and 1 = 9
1 u + 357 = 36 u + 147
36 u - 1 u = 357 - 147
35 u = 210
1 u = 210 ÷ 35 = 6
Number of blue stickers
= 1 u + 357
= 1 x 6 + 357
= 6 + 357
= 363
(b)
Number of green stickers
= 9 u
= 9 x 6
= 54
Answer(s): (a) 363; (b) 54