Adam has some red stickers and green stickers.
If 208 red stickers are removed, 70% of the stickers will be green stickers.
If 92 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 + 208 |
14 u |
21 u - 92 |
14 u |
Change |
- 208 |
No change |
+ 92 |
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 - 92 = 6 u + 208
21 u - 6 u = 208 + 92
15 u = 300
1 u = 300 ÷ 15 = 20
Number of red stickers
= 6 u + 208
= 6 x 20 + 208
= 120 + 208
= 328
(b)
Number of green stickers
= 14 u
= 14 x 20
= 280
Answer(s): (a) 328; (b) 280