Adam has some blue stickers and green stickers.
If 384 blue stickers are removed, 70% of the stickers will be green stickers.
If 136 blue stickers are added, 30% 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 |
9 u + 384 |
21 u |
49 u - 136 |
21 u |
Change |
- 384 |
No change |
+ 136 |
No change |
After |
3x3 = 9 u |
7x3 = 21 u |
7x7 = 49 u |
3x7 = 21 u |
(a)
70% =
70100 =
71030% =
30100 =
310Scenario 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 -
310 =
710 The number of green stickers remains unchanged in both scenarios.
LCM of 7 and 3 = 21
49 u - 136 = 9 u + 384
49 u - 9 u = 384 + 136
40 u = 520
1 u = 520 ÷ 40 = 13
Number of blue stickers
= 9 u + 384
= 9 x 13 + 384
= 117 + 384
= 501
(b)
Number of green stickers
= 21 u
= 21 x 13
= 273
Answer(s): (a) 501; (b) 273