Adam has some green stickers and blue stickers.
If 91 green stickers are removed, 70% of the stickers will be blue stickers.
If 134 green stickers are added, 40% of the stickers will be blue stickers.
- How many green stickers are there?
- How many blue stickers are there?
|
Scenario 1 |
Scenario 2 |
|
Green |
Blue |
Green |
Blue |
Before |
6 u + 91 |
14 u |
21 u - 134 |
14 u |
Change |
- 91 |
No change |
+ 134 |
No change |
After |
3x2 = 6 u |
7x2 = 14 u |
3x7 = 21 u |
2x7 = 14 u |
(a)
70% =
70100 =
71040% =
40100 =
25Scenario 1 Fraction of the stickers that are green in the end
= 1 -
710 =
310 Scenario 2
Fraction of the stickers that are green in the end
= 1 -
25 =
35 The number of blue stickers remains unchanged in both scenarios.
LCM of 7 and 2 = 14
21 u - 134 = 6 u + 91
21 u - 6 u = 91 + 134
15 u = 225
1 u = 225 ÷ 15 = 15
Number of green stickers
= 6 u + 91
= 6 x 15 + 91
= 90 + 91
= 181
(b)
Number of blue stickers
= 14 u
= 14 x 15
= 210
Answer(s): (a) 181; (b) 210