Adam has some red stickers and green stickers.
If 52 red stickers are removed, 80% of the stickers will be green stickers.
If 48 red stickers are added, 40% of the stickers will be green stickers.
- How many red stickers are there?
- How many green stickers are there?
|
Scenario 1 |
Scenario 2 |
|
Red |
Green |
Red |
Green |
Before |
1 u + 52 |
4 u |
6 u - 48 |
4 u |
Change |
- 52 |
No change |
+ 48 |
No change |
After |
1x1 = 1 u |
4x1 = 4 u |
3x2 = 6 u |
2x2 = 4 u |
(a)
80% =
80100 =
4540% =
40100 =
25Scenario 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 -
25 =
35 The number of green stickers remains unchanged in both scenarios.
LCM of 4 and 2 = 4
6 u - 48 = 1 u + 52
6 u - 1 u = 52 + 48
5 u = 100
1 u = 100 ÷ 5 = 20
Number of red stickers
= 1 u + 52
= 1 x 20 + 52
= 20 + 52
= 72
(b)
Number of green stickers
= 4 u
= 4 x 20
= 80
Answer(s): (a) 72; (b) 80