Adam has some blue stickers and green stickers.
If 358 blue stickers are removed, 90% of the stickers will be green stickers.
If 158 blue stickers are removed, 30% 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 + 358 |
9 u |
21 u + 158 |
9 u |
Change |
- 358 |
No change |
- 158 |
No change |
After |
1x1 = 1 u |
9x1 = 9 u |
7x3 = 21 u |
3x3 = 9 u |
(a)
90% =
90100 =
91030% =
30100 =
310Scenario 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 -
310 =
710 The number of green stickers remains unchanged in both scenarios.
LCM of 9 and 3 = 9
1 u + 358 = 21 u + 158
21 u - 1 u = 358 - 158
20 u = 200
1 u = 200 ÷ 20 = 10
Number of blue stickers
= 1 u + 358
= 1 x 10 + 358
= 10 + 358
= 368
(b)
Number of green stickers
= 9 u
= 9 x 10
= 90
Answer(s): (a) 368; (b) 90