Adam has some green stickers and blue stickers.
If 65 green stickers are removed, 75% of the stickers will be blue stickers.
If 503 blue stickers are added, 30% of the stickers will be green stickers.
- How many green stickers are there?
- How many blue stickers are there?
|
Scenario 1 |
Scenario 2 |
|
Green |
Blue |
Green |
Blue |
Before |
1 u + 65 |
3 u |
3 p |
7 p - 503 |
Change |
- 65 |
No change |
No change |
+ 503 |
After |
1 u |
3 u |
3 p |
7 p |
(a)
75% =
75100
=
34 30% =
30100 =
310 Scenario 1 Fraction of the stickers that are green in the end
= 1 -
34 =
14 Scenario 2
Fraction of the stickers that are green in the end
= 1 -
310 =
710 1 u + 65 = 3 p --- (1)
3 u = 7 p - 503 --- (2)
(1)
x 3 3 u + 195 = 3 p
3 u = 3 p - 195 --- (3)
(2) = (3)
7 p - 503 = 3 p - 195
7 p - 3 p = 503 - 195
4 p = 308
1 p = 308 ÷ 4 = 77
Number of green stickers
= 3 p
= 3 x 77
= 231
(b)
Number of blue stickers
= 7 p - 503
= 7 x 77 - 503
= 539 - 503
= 36
Answer(s): (a) 231; (b) 36