Adam has some blue stickers and red stickers.
If 30 blue stickers are removed, 80% of the stickers will be red stickers.
If 507 red stickers are added, 30% of the stickers will be blue stickers.
- How many blue stickers are there?
- How many red stickers are there?
|
Scenario 1 |
Scenario 2 |
|
Blue |
Red |
Blue |
Red |
Before |
1 u + 30 |
4 u |
3 p |
7 p - 507 |
Change |
- 30 |
No change |
No change |
+ 507 |
After |
1 u |
4 u |
3 p |
7 p |
(a)
80% =
80100
=
45 30% =
30100 =
310 Scenario 1 Fraction of the stickers that are blue in the end
= 1 -
45 =
15 Scenario 2
Fraction of the stickers that are blue in the end
= 1 -
310 =
710 1 u + 30 = 3 p --- (1)
4 u = 7 p - 507 --- (2)
(1)
x 4 4 u + 120 = 4 p
4 u = 4 p - 120 --- (3)
(2) = (3)
7 p - 507 = 4 p - 120
7 p - 4 p = 507 - 120
3 p = 387
1 p = 387 ÷ 3 = 129
Number of blue stickers
= 3 p
= 3 x 129
= 387
(b)
Number of red stickers
= 7 p - 507
= 7 x 129 - 507
= 903 - 507
= 396
Answer(s): (a) 387; (b) 396