Adam has some green stickers and red stickers.
If 31 green stickers are removed, 80% of the stickers will be red stickers.
If 526 red stickers are added, 30% of the stickers will be green stickers.
- How many green stickers are there?
- How many red stickers are there?
|
Scenario 1 |
Scenario 2 |
|
Green |
Red |
Green |
Red |
Before |
1 u + 31 |
4 u |
3 p |
7 p - 526 |
Change |
- 31 |
No change |
No change |
+ 526 |
After |
1 u |
4 u |
3 p |
7 p |
(a)
80% =
80100
=
45 30% =
30100 =
310 Scenario 1 Fraction of the stickers that are green in the end
= 1 -
45 =
15 Scenario 2
Fraction of the stickers that are green in the end
= 1 -
310 =
710 1 u + 31 = 3 p --- (1)
4 u = 7 p - 526 --- (2)
(1)
x 4 4 u + 124 = 4 p
4 u = 4 p - 124 --- (3)
(2) = (3)
7 p - 526 = 4 p - 124
7 p - 4 p = 526 - 124
3 p = 402
1 p = 402 ÷ 3 = 134
Number of green stickers
= 3 p
= 3 x 134
= 402
(b)
Number of red stickers
= 7 p - 526
= 7 x 134 - 526
= 938 - 526
= 412
Answer(s): (a) 402; (b) 412