Adam has some green stickers and red stickers.
If 272 green stickers are removed, 70% of the stickers will be red stickers.
If 167 green stickers are removed, 40% 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 |
6 u + 272 |
14 u |
21 u + 167 |
14 u |
Change |
- 272 |
No change |
- 167 |
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 red stickers remains unchanged in both scenarios.
LCM of 7 and 2 = 14
6 u + 272 = 21 u + 167
21 u - 6 u = 272 - 167
15 u = 105
1 u = 105 ÷ 15 = 7
Number of green stickers
= 6 u + 272
= 6 x 7 + 272
= 42 + 272
= 314
(b)
Number of red stickers
= 14 u
= 14 x 7
= 98
Answer(s): (a) 314; (b) 98