Adam has some blue stickers and green stickers.
If 56 blue stickers are added, 70% of the stickers will be green stickers.
If 386 blue stickers are added, 40% of the stickers will be green stickers.
- How many blue stickers are there?
- How many green stickers are there?
|
Scenario 1 |
Scenario 2 |
|
Blue |
Green |
Blue |
Green |
Before |
6 u - 56 |
14 u |
21 u - 386 |
14 u |
Change |
+ 56 |
No change |
+ 386 |
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 blue in the end
= 1 -
710 =
310 Scenario 2
Fraction of the stickers that are blue in the end
= 1 -
25 =
35 The number of green stickers remains unchanged in both scenarios.
LCM of 7 and 2 = 14
21 u - 386 = 6 u - 56
21 u - 6 u = 386 - 56
15 u = 330
1 u = 330 ÷ 15 = 22
Number of blue stickers
= 6 u - 56
= 6 x 22 - 56
= 132 - 56
= 76
(b)
Number of green stickers
= 14 u
= 14 x 22
= 308
Answer(s): (a) 76; (b) 308