Adam has some red stickers and green stickers.
If 12 red stickers are removed, 80% of the stickers will be green stickers.
If 528 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 + 12 |
4 u |
3 p |
7 p - 528 |
Change |
- 12 |
No change |
No change |
+ 528 |
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 + 12 = 3 p --- (1)
4 u = 7 p - 528 --- (2)
(1)
x 4 4 u + 48 = 4 p
4 u = 4 p - 48 --- (3)
(2) = (3)
7 p - 528 = 4 p - 48
7 p - 4 p = 528 - 48
3 p = 480
1 p = 480 ÷ 3 = 160
Number of red stickers
= 3 p
= 3 x 160
= 480
(b)
Number of green stickers
= 7 p - 528
= 7 x 160 - 528
= 1120 - 528
= 592
Answer(s): (a) 480; (b) 592