Adam has some blue stickers and green stickers.
If 388 blue stickers are removed, 70% of the stickers will be green stickers.
If 46 blue stickers are removed, 25% 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 |
3 u + 388 |
7 u |
21 u + 46 |
7 u |
Change |
- 388 |
No change |
- 46 |
No change |
After |
3x1 = 3 u |
7x1 = 7 u |
3x7 = 21 u |
1x7 = 7 u |
(a)
70% =
70100 =
71025% =
25100 =
14Scenario 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 -
14 =
34 The number of green stickers remains unchanged in both scenarios.
LCM of 7 and 1 = 7
3 u + 388 = 21 u + 46
21 u - 3 u = 388 - 46
18 u = 342
1 u = 342 ÷ 18 = 19
Number of blue stickers
= 3 u + 388
= 3 x 19 + 388
= 57 + 388
= 445
(b)
Number of green stickers
= 7 u
= 7 x 19
= 133
Answer(s): (a) 445; (b) 133