Adam has some red stickers and green stickers.
If 157 red stickers are removed, 80% of the stickers will be green stickers.
If 32 red stickers are removed, 30% 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 |
3 u + 157 |
12 u |
28 u + 32 |
12 u |
Change |
- 157 |
No change |
- 32 |
No change |
After |
1x3 = 3 u |
4x3 = 12 u |
7x4 = 28 u |
3x4 = 12 u |
(a)
80% =
80100 =
4530% =
30100 =
310Scenario 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 -
310 =
710 The number of green stickers remains unchanged in both scenarios.
LCM of 4 and 3 = 12
3 u + 157 = 28 u + 32
28 u - 3 u = 157 - 32
25 u = 125
1 u = 125 ÷ 25 = 5
Number of red stickers
= 3 u + 157
= 3 x 5 + 157
= 15 + 157
= 172
(b)
Number of green stickers
= 12 u
= 12 x 5
= 60
Answer(s): (a) 172; (b) 60