Adam has some red stickers and green stickers.
If 289 red stickers are added, 40% of the stickers will be green stickers.
If 204 green stickers are added, 25% 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 |
3 u - 289 |
2 u |
1 p |
3 p - 204 |
Change |
+ 289 |
No change |
No change |
+ 204 |
After |
3 u |
2 u |
1 p |
3 p |
(a)
40% =
40100 =
25 25% =
25100 =
14 Scenario 1 Fraction of the stickers that are red
= 1 -
25 =
35 Number of red stickers at first = 3 u - 289
Number of green stickers at first = 2 u
Scenario 2 Fraction of the stickers that are green
= 1 -
14=
34 Number of red stickers at first = 1 p
Number of green stickers at first = 3 p - 204
3 u - 289 = 1 p --- (1)
2 u = 3 p - 204
2 u + 204 = 3 p --- (2)
(1)
x 3 9 u - 867 = 3 p --- (3)
(3) = (2)
9 u - 867 = 2 u + 204
9 u - 2 u = 867 + 204
7 u = 1071
1 u = 1071 ÷ 7 = 153
Number of red stickers
= 3 u - 289
= 3 x 153 - 289
= 459 - 289
= 170
(b)
Number of green stickers
= 2 u
= 2 x 153
= 306
Answer(s): (a) 170; (b) 306