Adam has some red stickers and green stickers.
If 156 red stickers are removed, 70% of the stickers will be green stickers.
If 129 red stickers are added, 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 |
6 u + 156 |
14 u |
21 u - 129 |
14 u |
Change |
- 156 |
No change |
+ 129 |
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 red in the end
= 1 -
710 =
310 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 7 and 2 = 14
21 u - 129 = 6 u + 156
21 u - 6 u = 156 + 129
15 u = 285
1 u = 285 ÷ 15 = 19
Number of red stickers
= 6 u + 156
= 6 x 19 + 156
= 114 + 156
= 270
(b)
Number of green stickers
= 14 u
= 14 x 19
= 266
Answer(s): (a) 270; (b) 266