Adam has some red stickers and green stickers.
If 46 red stickers are removed, 75% of the stickers will be green stickers.
If 422 green stickers are added, 30% of the stickers will be red stickers.
- How many red stickers are there?
- How many green stickers are there?
|
Scenario 1 |
Scenario 2 |
|
Red |
Green |
Red |
Green |
Before |
1 u + 46 |
3 u |
3 p |
7 p - 422 |
Change |
- 46 |
No change |
No change |
+ 422 |
After |
1 u |
3 u |
3 p |
7 p |
(a)
75% =
75100
=
34 30% =
30100 =
310 Scenario 1 Fraction of the stickers that are red in the end
= 1 -
34 =
14 Scenario 2
Fraction of the stickers that are red in the end
= 1 -
310 =
710 1 u + 46 = 3 p --- (1)
3 u = 7 p - 422 --- (2)
(1)
x 3 3 u + 138 = 3 p
3 u = 3 p - 138 --- (3)
(2) = (3)
7 p - 422 = 3 p - 138
7 p - 3 p = 422 - 138
4 p = 284
1 p = 284 ÷ 4 = 71
Number of red stickers
= 3 p
= 3 x 71
= 213
(b)
Number of green stickers
= 7 p - 422
= 7 x 71 - 422
= 497 - 422
= 75
Answer(s): (a) 213; (b) 75