Adam has some green stickers and red stickers.
If 215 green stickers are removed, 25% of the stickers will be green stickers.
If 265 red stickers are removed, 10% of the stickers will be red stickers.
- How many green stickers are there?
- How many red stickers are there?
|
Scenario 1 |
Scenario 2 |
|
Green |
Red |
Green |
Red |
Before |
1 u + 215 |
3 u |
9 p |
1 p + 265 |
Change |
- 215 |
No change |
No change |
- 265 |
After |
1 u |
3 u |
9 p |
1 p |
(a)
25% =
25100 =
14 10% =
10100 =
110 Scenario 1Fraction of the stickers that are red
= 1 -
14 =
34 Scenario 2Fraction of the stickers that are green
= 1 -
110 =
9101 u + 215 = 9 p --- (1)
3 u = 1 p + 265 --- (2)
From (1)
1 u = 9 p - 215 --- (3)
(3)
x 33 u = 27 p - 645 --- (4)
(4) = (2)
27 p - 645 = 1 p + 265
27 p - 1 p = 645 + 265
26 p = 910
1 p = 910 ÷ 26 = 35Number of green stickers = 9 p
= 9 x 35= 315 (b) Number of red stickers = 1 p + 265 = 35 + 265 = 300 Answer(s): (a) 315; (b) 300