Adam has some green stickers and red stickers.
If 209 green stickers are removed, 80% of the stickers will be red stickers.
If 29 green stickers are removed, 20% of the stickers will be red stickers.
- How many green stickers are there?
- How many red stickers are there?
|
Scenario 1 |
Scenario 2 |
|
Green |
Red |
Green |
Red |
Before |
1 u + 209 |
4 u |
16 u + 29 |
4 u |
Change |
- 209 |
No change |
- 29 |
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 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 red stickers remains unchanged in both scenarios.
LCM of 4 and 1 = 4
1 u + 209 = 16 u + 29
16 u - 1 u = 209 - 29
15 u = 180
1 u = 180 ÷ 15 = 12
Number of green stickers
= 1 u + 209
= 1 x 12 + 209
= 12 + 209
= 221
(b)
Number of red stickers
= 4 u
= 4 x 12
= 48
Answer(s): (a) 221; (b) 48