Adam has some red stickers and green stickers.
If 147 red stickers are removed, 80% of the stickers will be green stickers.
If 32 red stickers are removed, 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 + 147 |
4 u |
6 u + 32 |
4 u |
Change |
- 147 |
No change |
- 32 |
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
1 u + 147 = 6 u + 32
6 u - 1 u = 147 - 32
5 u = 115
1 u = 115 ÷ 5 = 23
Number of red stickers
= 1 u + 147
= 1 x 23 + 147
= 23 + 147
= 170
(b)
Number of green stickers
= 4 u
= 4 x 23
= 92
Answer(s): (a) 170; (b) 92