Adam has some green stickers and red stickers.
If 50 green stickers are removed, 80% of the stickers will be red stickers.
If 575 red stickers are added, 10% of the stickers will be green stickers.
- How many green stickers are there?
- How many red stickers are there?
|
Scenario 1 |
Scenario 2 |
|
Green |
Red |
Green |
Red |
Before |
1 u + 50 |
4 u |
1 p |
9 p - 575 |
Change |
- 50 |
No change |
No change |
+ 575 |
After |
1 u |
4 u |
1 p |
9 p |
(a)
80% =
80100
=
45 10% =
10100 =
110 Scenario 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 -
110 =
910 1 u + 50 = 1 p --- (1)
4 u = 9 p - 575 --- (2)
(1)
x 4 4 u + 200 = 4 p
4 u = 4 p - 200 --- (3)
(2) = (3)
9 p - 575 = 4 p - 200
9 p - 4 p = 575 - 200
5 p = 375
1 p = 375 ÷ 5 = 75
Number of green stickers
= 1 p
= 1 x 75
= 75
(b)
Number of red stickers
= 9 p - 575
= 9 x 75 - 575
= 675 - 575
= 100
Answer(s): (a) 75; (b) 100