Adam has some blue stickers and green stickers.
If 51 blue stickers are removed, 80% of the stickers will be green stickers.
If 546 green stickers are added, 30% of the stickers will be blue stickers.
- How many blue stickers are there?
- How many green stickers are there?
|
Scenario 1 |
Scenario 2 |
|
Blue |
Green |
Blue |
Green |
Before |
1 u + 51 |
4 u |
3 p |
7 p - 546 |
Change |
- 51 |
No change |
No change |
+ 546 |
After |
1 u |
4 u |
3 p |
7 p |
(a)
80% =
80100
=
45 30% =
30100 =
310 Scenario 1 Fraction of the stickers that are blue in the end
= 1 -
45 =
15 Scenario 2
Fraction of the stickers that are blue in the end
= 1 -
310 =
710 1 u + 51 = 3 p --- (1)
4 u = 7 p - 546 --- (2)
(1)
x 4 4 u + 204 = 4 p
4 u = 4 p - 204 --- (3)
(2) = (3)
7 p - 546 = 4 p - 204
7 p - 4 p = 546 - 204
3 p = 342
1 p = 342 ÷ 3 = 114
Number of blue stickers
= 3 p
= 3 x 114
= 342
(b)
Number of green stickers
= 7 p - 546
= 7 x 114 - 546
= 798 - 546
= 252
Answer(s): (a) 342; (b) 252