Adam has some red stickers and green stickers.
If 56 red stickers are removed, 80% of the stickers will be green stickers.
If 590 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 + 56 |
4 u |
3 p |
7 p - 590 |
Change |
- 56 |
No change |
No change |
+ 590 |
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 + 56 = 3 p --- (1)
4 u = 7 p - 590 --- (2)
(1)
x 4 4 u + 224 = 4 p
4 u = 4 p - 224 --- (3)
(2) = (3)
7 p - 590 = 4 p - 224
7 p - 4 p = 590 - 224
3 p = 366
1 p = 366 ÷ 3 = 122
Number of red stickers
= 3 p
= 3 x 122
= 366
(b)
Number of green stickers
= 7 p - 590
= 7 x 122 - 590
= 854 - 590
= 264
Answer(s): (a) 366; (b) 264