Adam has some red stickers and green stickers.
If 42 red stickers are removed, 75% of the stickers will be green stickers.
If 598 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 + 42 |
3 u |
3 p |
7 p - 598 |
Change |
- 42 |
No change |
No change |
+ 598 |
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 + 42 = 3 p --- (1)
3 u = 7 p - 598 --- (2)
(1)
x 3 3 u + 126 = 3 p
3 u = 3 p - 126 --- (3)
(2) = (3)
7 p - 598 = 3 p - 126
7 p - 3 p = 598 - 126
4 p = 472
1 p = 472 ÷ 4 = 118
Number of red stickers
= 3 p
= 3 x 118
= 354
(b)
Number of green stickers
= 7 p - 598
= 7 x 118 - 598
= 826 - 598
= 228
Answer(s): (a) 354; (b) 228