Adam has some red stickers and green stickers.
If 82 red stickers are removed, 80% of the stickers will be green stickers.
If 586 green stickers are added, 30% of the stickers will be red stickers.
- How many red stickers are there?
- How many green stickers are there?
|
Scenario 1 |
Scenario 2 |
|
Red |
Green |
Red |
Green |
Before |
1 u + 82 |
4 u |
3 p |
7 p - 586 |
Change |
- 82 |
No change |
No change |
+ 586 |
After |
1 u |
4 u |
3 p |
7 p |
(a)
80% =
80100
=
45 30% =
30100 =
310 Scenario 1 Fraction of the stickers that are red in the end
= 1 -
45 =
15 Scenario 2
Fraction of the stickers that are red in the end
= 1 -
310 =
710 1 u + 82 = 3 p --- (1)
4 u = 7 p - 586 --- (2)
(1)
x 4 4 u + 328 = 4 p
4 u = 4 p - 328 --- (3)
(2) = (3)
7 p - 586 = 4 p - 328
7 p - 4 p = 586 - 328
3 p = 258
1 p = 258 ÷ 3 = 86
Number of red stickers
= 3 p
= 3 x 86
= 258
(b)
Number of green stickers
= 7 p - 586
= 7 x 86 - 586
= 602 - 586
= 16
Answer(s): (a) 258; (b) 16