Adam has some red stickers and green stickers.
If 31 red stickers are removed, 80% of the stickers will be green stickers.
If 44 red stickers are added, 20% of the stickers will be green stickers.
- How many red stickers are there?
- How many green stickers are there?
|
Scenario 1 |
Scenario 2 |
|
Red |
Green |
Red |
Green |
Before |
1 u + 31 |
4 u |
16 u - 44 |
4 u |
Change |
- 31 |
No change |
+ 44 |
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 red in the end
= 1 -
45 =
15 Scenario 2
Fraction of the stickers that are red in the end
= 1 -
15 =
45 The number of green stickers remains unchanged in both scenarios.
LCM of 4 and 1 = 4
16 u - 44 = 1 u + 31
16 u - 1 u = 31 + 44
15 u = 75
1 u = 75 ÷ 15 = 5
Number of red stickers
= 1 u + 31
= 1 x 5 + 31
= 5 + 31
= 36
(b)
Number of green stickers
= 4 u
= 4 x 5
= 20
Answer(s): (a) 36; (b) 20