Adam has some red stickers and green stickers.
If 27 red stickers are removed, 75% of the stickers will be green stickers.
If 507 green stickers are added, 10% 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 + 27 |
3 u |
1 p |
9 p - 507 |
Change |
- 27 |
No change |
No change |
+ 507 |
After |
1 u |
3 u |
1 p |
9 p |
(a)
75% =
75100
=
34 10% =
10100 =
110 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 -
110 =
910 1 u + 27 = 1 p --- (1)
3 u = 9 p - 507 --- (2)
(1)
x 3 3 u + 81 = 3 p
3 u = 3 p - 81 --- (3)
(2) = (3)
9 p - 507 = 3 p - 81
9 p - 3 p = 507 - 81
6 p = 426
1 p = 426 ÷ 6 = 71
Number of red stickers
= 1 p
= 1 x 71
= 71
(b)
Number of green stickers
= 9 p - 507
= 9 x 71 - 507
= 639 - 507
= 132
Answer(s): (a) 71; (b) 132