Adam has some blue stickers and green stickers.
If 290 blue stickers are removed, 70% of the stickers will be green stickers.
If 74 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 + 290 |
7 u |
21 u + 74 |
7 u |
Change |
- 290 |
No change |
- 74 |
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 + 290 = 21 u + 74
21 u - 3 u = 290 - 74
18 u = 216
1 u = 216 ÷ 18 = 12
Number of blue stickers
= 3 u + 290
= 3 x 12 + 290
= 36 + 290
= 326
(b)
Number of green stickers
= 7 u
= 7 x 12
= 84
Answer(s): (a) 326; (b) 84