Adam has some red stickers and blue stickers.
If 57 red stickers are removed, 80% of the stickers will be blue stickers.
If 459 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 + 57 |
4 u |
3 p |
7 p - 459 |
Change |
- 57 |
No change |
No change |
+ 459 |
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 + 57 = 3 p --- (1)
4 u = 7 p - 459 --- (2)
(1)
x 4 4 u + 228 = 4 p
4 u = 4 p - 228 --- (3)
(2) = (3)
7 p - 459 = 4 p - 228
7 p - 4 p = 459 - 228
3 p = 231
1 p = 231 ÷ 3 = 77
Number of red stickers
= 3 p
= 3 x 77
= 231
(b)
Number of blue stickers
= 7 p - 459
= 7 x 77 - 459
= 539 - 459
= 80
Answer(s): (a) 231; (b) 80