Adam has some blue stickers and green stickers.
If 154 blue stickers are removed, 80% of the stickers will be green stickers.
If 86 blue stickers are added, 20% 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 |
1 u + 154 |
4 u |
16 u - 86 |
4 u |
Change |
- 154 |
No change |
+ 86 |
No change |
After |
1x1 = 1 u |
4x1 = 4 u |
4x4 = 16 u |
1x4 = 4 u |
(a)
80% =
80100 =
4520% =
20100 =
15Scenario 1 Fraction of the stickers that are blue in the end
= 1 -
45 =
15 Scenario 2
Fraction of the stickers that are blue in the end
= 1 -
15 =
45 The number of green stickers remains unchanged in both scenarios.
LCM of 4 and 1 = 4
16 u - 86 = 1 u + 154
16 u - 1 u = 154 + 86
15 u = 240
1 u = 240 ÷ 15 = 16
Number of blue stickers
= 1 u + 154
= 1 x 16 + 154
= 16 + 154
= 170
(b)
Number of green stickers
= 4 u
= 4 x 16
= 64
Answer(s): (a) 170; (b) 64