Adam has some green stickers and red stickers.
If 272 green stickers are removed, 70% of the stickers will be red stickers.
If 122 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 |
3 u + 272 |
7 u |
28 u + 122 |
7 u |
Change |
- 272 |
No change |
- 122 |
No change |
After |
3x1 = 3 u |
7x1 = 7 u |
4x7 = 28 u |
1x7 = 7 u |
(a)
70% =
70100 =
71020% =
20100 =
15Scenario 1 Fraction of the stickers that are green in the end
= 1 -
710 =
310 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 7 and 1 = 7
3 u + 272 = 28 u + 122
28 u - 3 u = 272 - 122
25 u = 150
1 u = 150 ÷ 25 = 6
Number of green stickers
= 3 u + 272
= 3 x 6 + 272
= 18 + 272
= 290
(b)
Number of red stickers
= 7 u
= 7 x 6
= 42
Answer(s): (a) 290; (b) 42