Adam has some red stickers and blue stickers.
If 31 red stickers are removed, 80% of the stickers will be blue stickers.
If 433 blue stickers are added, 30% of the stickers will be red stickers.
- How many red stickers are there?
- How many blue stickers are there?
|
Scenario 1 |
Scenario 2 |
|
Red |
Blue |
Red |
Blue |
Before |
1 u + 31 |
4 u |
3 p |
7 p - 433 |
Change |
- 31 |
No change |
No change |
+ 433 |
After |
1 u |
4 u |
3 p |
7 p |
(a)
80% =
80100
=
45 30% =
30100 =
310 Scenario 1 Fraction of the stickers that are red in the end
= 1 -
45 =
15 Scenario 2
Fraction of the stickers that are red in the end
= 1 -
310 =
710 1 u + 31 = 3 p --- (1)
4 u = 7 p - 433 --- (2)
(1)
x 4 4 u + 124 = 4 p
4 u = 4 p - 124 --- (3)
(2) = (3)
7 p - 433 = 4 p - 124
7 p - 4 p = 433 - 124
3 p = 309
1 p = 309 ÷ 3 = 103
Number of red stickers
= 3 p
= 3 x 103
= 309
(b)
Number of blue stickers
= 7 p - 433
= 7 x 103 - 433
= 721 - 433
= 288
Answer(s): (a) 309; (b) 288