Adam has some red stickers and blue stickers.
If 41 red stickers are removed, 75% of the stickers will be blue stickers.
If 531 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 + 41 |
3 u |
3 p |
7 p - 531 |
Change |
- 41 |
No change |
No change |
+ 531 |
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 + 41 = 3 p --- (1)
3 u = 7 p - 531 --- (2)
(1)
x 3 3 u + 123 = 3 p
3 u = 3 p - 123 --- (3)
(2) = (3)
7 p - 531 = 3 p - 123
7 p - 3 p = 531 - 123
4 p = 408
1 p = 408 ÷ 4 = 102
Number of red stickers
= 3 p
= 3 x 102
= 306
(b)
Number of blue stickers
= 7 p - 531
= 7 x 102 - 531
= 714 - 531
= 183
Answer(s): (a) 306; (b) 183