Adam has some green stickers and red stickers.
If 362 green stickers are removed, 80% of the stickers will be red stickers.
If 107 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 + 362 |
4 u |
16 u + 107 |
4 u |
Change |
- 362 |
No change |
- 107 |
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 + 362 = 16 u + 107
16 u - 1 u = 362 - 107
15 u = 255
1 u = 255 ÷ 15 = 17
Number of green stickers
= 1 u + 362
= 1 x 17 + 362
= 17 + 362
= 379
(b)
Number of red stickers
= 4 u
= 4 x 17
= 68
Answer(s): (a) 379; (b) 68