Adam has some green stickers and red stickers.
If 56 green stickers are removed, 90% of the stickers will be red stickers.
If 84 green stickers are added, 30% of the stickers will be red stickers.
- How many green stickers are there?
- How many red stickers are there?
|
Scenario 1 |
Scenario 2 |
|
Green |
Red |
Green |
Red |
Before |
1 u + 56 |
9 u |
21 u - 84 |
9 u |
Change |
- 56 |
No change |
+ 84 |
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 green in the end
= 1 -
910 =
110 Scenario 2
Fraction of the stickers that are green in the end
= 1 -
310 =
710 The number of red stickers remains unchanged in both scenarios.
LCM of 9 and 3 = 9
21 u - 84 = 1 u + 56
21 u - 1 u = 56 + 84
20 u = 140
1 u = 140 ÷ 20 = 7
Number of green stickers
= 1 u + 56
= 1 x 7 + 56
= 7 + 56
= 63
(b)
Number of red stickers
= 9 u
= 9 x 7
= 63
Answer(s): (a) 63; (b) 63