Adam has some blue stickers and green stickers.
If 51 blue stickers are added, 70% of the stickers will be green stickers.
If 186 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 - 51 |
14 u |
21 u - 186 |
14 u |
Change |
+ 51 |
No change |
+ 186 |
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 - 186 = 6 u - 51
21 u - 6 u = 186 - 51
15 u = 135
1 u = 135 ÷ 15 = 9
Number of blue stickers
= 6 u - 51
= 6 x 9 - 51
= 54 - 51
= 3
(b)
Number of green stickers
= 14 u
= 14 x 9
= 126
Answer(s): (a) 3; (b) 126