Adam has some green stickers and blue stickers.
If 22 green stickers are added, 80% of the stickers will be blue stickers.
If 382 green stickers are added, 20% of the stickers will be blue stickers.
- How many green stickers are there?
- How many blue stickers are there?
|
Scenario 1 |
Scenario 2 |
|
Green |
Blue |
Green |
Blue |
Before |
1 u - 22 |
4 u |
16 u - 382 |
4 u |
Change |
+ 22 |
No change |
+ 382 |
No change |
After |
1x1 = 1 u |
4x1 = 4 u |
4x4 = 16 u |
1x4 = 4 u |
(a)
80% =
80100 =
4520% =
20100 =
15 Scenario 1 Fraction of the stickers that are green in the end
= 1 -
45 =
15 Scenario 2
Fraction of the stickers that are green in the end
= 1 -
15 =
45 The number of blue stickers remains unchanged in both scenarios.
LCM of 4 and 1 = 4
16 u - 382 = 1 u - 22
16 u - 1 u = 382 - 22
15 u = 360
1 u = 360 ÷ 15 = 24
Number of green stickers
= 1 u - 22
= 1 x 24 - 22
= 24 - 22
= 2
(b)
Number of blue stickers
= 4 u
= 4 x 24
= 96
Answer(s): (a) 2; (b) 96