Adam has some red stickers and green stickers.
If 111 red stickers are removed, 80% of the stickers will be green stickers.
If 56 red stickers are removed, 25% 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 + 111 |
4 u |
12 u + 56 |
4 u |
Change |
- 111 |
No change |
- 56 |
No change |
After |
1x1 = 1 u |
4x1 = 4 u |
3x4 = 12 u |
1x4 = 4 u |
(a)
80% =
80100 =
4525% =
25100 =
14Scenario 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 -
14 =
34 The number of green stickers remains unchanged in both scenarios.
LCM of 4 and 1 = 4
1 u + 111 = 12 u + 56
12 u - 1 u = 111 - 56
11 u = 55
1 u = 55 ÷ 11 = 5
Number of red stickers
= 1 u + 111
= 1 x 5 + 111
= 5 + 111
= 116
(b)
Number of green stickers
= 4 u
= 4 x 5
= 20
Answer(s): (a) 116; (b) 20