Adam has some green stickers and blue stickers.
If 116 green stickers are removed, 70% of the stickers will be blue stickers.
If 49 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 + 116 |
14 u |
21 u - 49 |
14 u |
Change |
- 116 |
No change |
+ 49 |
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 - 49 = 6 u + 116
21 u - 6 u = 116 + 49
15 u = 165
1 u = 165 ÷ 15 = 11
Number of green stickers
= 6 u + 116
= 6 x 11 + 116
= 66 + 116
= 182
(b)
Number of blue stickers
= 14 u
= 14 x 11
= 154
Answer(s): (a) 182; (b) 154