Adam has some red stickers and green stickers.
If 149 red stickers are removed, 25% of the stickers will be red stickers.
If 158 green stickers are removed, 20% of the stickers will be green stickers.
- How many red stickers are there?
- How many green stickers are there?
|
Scenario 1 |
Scenario 2 |
|
Red |
Green |
Red |
Green |
Before |
1 u + 149 |
3 u |
4 p |
1 p + 158 |
Change |
- 149 |
No change |
No change |
- 158 |
After |
1 u |
3 u |
4 p |
1 p |
(a)
25% =
25100 =
14 20% =
20100 =
15 Scenario 1Fraction of the stickers that are green
= 1 -
14 =
34 Scenario 2Fraction of the stickers that are red
= 1 -
15 =
451 u + 149 = 4 p --- (1)
3 u = 1 p + 158 --- (2)
From (1)
1 u = 4 p - 149 --- (3)
(3)
x 33 u = 12 p - 447 --- (4)
(4) = (2)
12 p - 447 = 1 p + 158
12 p - 1 p = 447 + 158
11 p = 605
1 p = 605 ÷ 11 = 55Number of red stickers = 4 p
= 4 x 55= 220 (b) Number of green stickers = 1 p + 158 = 55 + 158 = 213 Answer(s): (a) 220; (b) 213