Adam has some red stickers and green stickers.
If 176 red stickers are removed, 90% of the stickers will be green stickers.
If 56 red stickers are removed, 30% 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 + 176 |
9 u |
21 u + 56 |
9 u |
Change |
- 176 |
No change |
- 56 |
No change |
After |
1x1 = 1 u |
9x1 = 9 u |
7x3 = 21 u |
3x3 = 9 u |
(a)
90% =
90100 =
91030% =
30100 =
310Scenario 1 Fraction of the stickers that are red in the end
= 1 -
910 =
110 Scenario 2
Fraction of the stickers that are red in the end
= 1 -
310 =
710 The number of green stickers remains unchanged in both scenarios.
LCM of 9 and 3 = 9
1 u + 176 = 21 u + 56
21 u - 1 u = 176 - 56
20 u = 120
1 u = 120 ÷ 20 = 6
Number of red stickers
= 1 u + 176
= 1 x 6 + 176
= 6 + 176
= 182
(b)
Number of green stickers
= 9 u
= 9 x 6
= 54
Answer(s): (a) 182; (b) 54