Adam has some green stickers and blue stickers.
If 288 green stickers are removed, 25% of the stickers will be green stickers.
If 192 blue stickers are removed, 20% 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 + 288 |
3 u |
4 p |
1 p + 192 |
Change |
- 288 |
No change |
No change |
- 192 |
After |
1 u |
3 u |
4 p |
1 p |
(a)
25% =
25100 =
14 20% =
20100 =
15 Scenario 1Fraction of the stickers that are blue
= 1 -
14 =
34 Scenario 2Fraction of the stickers that are green
= 1 -
15 =
451 u + 288 = 4 p --- (1)
3 u = 1 p + 192 --- (2)
From (1)
1 u = 4 p - 288 --- (3)
(3)
x 33 u = 12 p - 864 --- (4)
(4) = (2)
12 p - 864 = 1 p + 192
12 p - 1 p = 864 + 192
11 p = 1056
1 p = 1056 ÷ 11 = 96Number of green stickers = 4 p
= 4 x 96= 384 (b) Number of blue stickers = 1 p + 192 = 96 + 192 = 288 Answer(s): (a) 384; (b) 288