Adam has some green stickers and red stickers.
If 383 green stickers are added, 40% of the stickers will be red stickers.
If 288 red stickers are added, 20% of the stickers will be green stickers.
- How many green stickers are there?
- How many red stickers are there?
|
Scenario 1 |
Scenario 2 |
|
Green |
Red |
Green |
Red |
Before |
3 u - 383 |
2 u |
1 p |
4 p - 288 |
Change |
+ 383 |
No change |
No change |
+ 288 |
After |
3 u |
2 u |
1 p |
4 p |
(a)
40% =
40100 =
25 20% =
20100 =
15 Scenario 1 Fraction of the stickers that are green
= 1 -
25 =
35 Number of green stickers at first = 3 u - 383
Number of red stickers at first = 2 u
Scenario 2 Fraction of the stickers that are red
= 1 -
15=
45 Number of green stickers at first = 1 p
Number of red stickers at first = 4 p - 288
3 u - 383 = 1 p --- (1)
2 u = 4 p - 288
2 u + 288 = 4 p --- (2)
(1)
x 4 12 u - 1532 = 4 p --- (3)
(3) = (2)
12 u - 1532 = 2 u + 288
12 u - 2 u = 1532 + 288
10 u = 1820
1 u = 1820 ÷ 10 = 182
Number of green stickers
= 3 u - 383
= 3 x 182 - 383
= 546 - 383
= 163
(b)
Number of red stickers
= 2 u
= 2 x 182
= 364
Answer(s): (a) 163; (b) 364