Adam has some green stickers and red stickers.
If 26 green stickers are removed, 80% of the stickers will be red stickers.
If 149 green stickers are added, 30% 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 |
3 u + 26 |
12 u |
28 u - 149 |
12 u |
Change |
- 26 |
No change |
+ 149 |
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 green in the end
= 1 -
45 =
15 Scenario 2
Fraction of the stickers that are green in the end
= 1 -
310 =
710 The number of red stickers remains unchanged in both scenarios.
LCM of 4 and 3 = 12
28 u - 149 = 3 u + 26
28 u - 3 u = 26 + 149
25 u = 175
1 u = 175 ÷ 25 = 7
Number of green stickers
= 3 u + 26
= 3 x 7 + 26
= 21 + 26
= 47
(b)
Number of red stickers
= 12 u
= 12 x 7
= 84
Answer(s): (a) 47; (b) 84