Adam has some blue stickers and green stickers.
If 53 blue stickers are removed, 75% of the stickers will be green stickers.
If 587 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 + 53 |
3 u |
3 p |
7 p - 587 |
Change |
- 53 |
No change |
No change |
+ 587 |
After |
1 u |
3 u |
3 p |
7 p |
(a)
75% =
75100
=
34 30% =
30100 =
310 Scenario 1 Fraction of the stickers that are blue in the end
= 1 -
34 =
14 Scenario 2
Fraction of the stickers that are blue in the end
= 1 -
310 =
710 1 u + 53 = 3 p --- (1)
3 u = 7 p - 587 --- (2)
(1)
x 3 3 u + 159 = 3 p
3 u = 3 p - 159 --- (3)
(2) = (3)
7 p - 587 = 3 p - 159
7 p - 3 p = 587 - 159
4 p = 428
1 p = 428 ÷ 4 = 107
Number of blue stickers
= 3 p
= 3 x 107
= 321
(b)
Number of green stickers
= 7 p - 587
= 7 x 107 - 587
= 749 - 587
= 162
Answer(s): (a) 321; (b) 162