Adam has some green stickers and red stickers.
If 52 green stickers are added, 70% of the stickers will be red stickers.
If 307 green stickers are added, 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 - 52 |
14 u |
21 u - 307 |
14 u |
Change |
+ 52 |
No change |
+ 307 |
No change |
After |
3x2 = 6 u |
7x2 = 14 u |
3x7 = 21 u |
2x7 = 14 u |
(a)
70% =
70100 =
71040% =
40100 =
25 Scenario 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
21 u - 307 = 6 u - 52
21 u - 6 u = 307 - 52
15 u = 255
1 u = 255 ÷ 15 = 17
Number of green stickers
= 6 u - 52
= 6 x 17 - 52
= 102 - 52
= 50
(b)
Number of red stickers
= 14 u
= 14 x 17
= 238
Answer(s): (a) 50; (b) 238