Adam has some green stickers and red stickers.
If 208 green stickers are removed, 25% of the stickers will be green stickers.
If 102 red stickers are removed, 20% 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 + 208 |
3 u |
4 p |
1 p + 102 |
Change |
- 208 |
No change |
No change |
- 102 |
After |
1 u |
3 u |
4 p |
1 p |
(a)
25% =
25100 =
14 20% =
20100 =
15 Scenario 1Fraction of the stickers that are red
= 1 -
14 =
34 Scenario 2Fraction of the stickers that are green
= 1 -
15 =
451 u + 208 = 4 p --- (1)
3 u = 1 p + 102 --- (2)
From (1)
1 u = 4 p - 208 --- (3)
(3)
x 33 u = 12 p - 624 --- (4)
(4) = (2)
12 p - 624 = 1 p + 102
12 p - 1 p = 624 + 102
11 p = 726
1 p = 726 ÷ 11 = 66Number of green stickers = 4 p
= 4 x 66= 264 (b) Number of red stickers = 1 p + 102 = 66 + 102 = 168 Answer(s): (a) 264; (b) 168