Adam has some green stickers and blue stickers.
If 189 green stickers are removed, 80% of the stickers will be blue stickers.
If 53 green stickers are added, 25% 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 + 189 |
4 u |
12 u - 53 |
4 u |
Change |
- 189 |
No change |
+ 53 |
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 green in the end
= 1 -
45 =
15 Scenario 2
Fraction of the stickers that are green in the end
= 1 -
14 =
34 The number of blue stickers remains unchanged in both scenarios.
LCM of 4 and 1 = 4
12 u - 53 = 1 u + 189
12 u - 1 u = 189 + 53
11 u = 242
1 u = 242 ÷ 11 = 22
Number of green stickers
= 1 u + 189
= 1 x 22 + 189
= 22 + 189
= 211
(b)
Number of blue stickers
= 4 u
= 4 x 22
= 88
Answer(s): (a) 211; (b) 88