Adam has some red stickers and blue stickers.
If 4 red stickers are removed, 80% of the stickers will be blue stickers.
If 26 red stickers are added, 40% of the stickers will be blue stickers.
- How many red stickers are there?
- How many blue stickers are there?
|
Scenario 1 |
Scenario 2 |
|
Red |
Blue |
Red |
Blue |
Before |
1 u + 4 |
4 u |
6 u - 26 |
4 u |
Change |
- 4 |
No change |
+ 26 |
No change |
After |
1x1 = 1 u |
4x1 = 4 u |
3x2 = 6 u |
2x2 = 4 u |
(a)
80% =
80100 =
4540% =
40100 =
25Scenario 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 -
25 =
35 The number of blue stickers remains unchanged in both scenarios.
LCM of 4 and 2 = 4
6 u - 26 = 1 u + 4
6 u - 1 u = 4 + 26
5 u = 30
1 u = 30 ÷ 5 = 6
Number of red stickers
= 1 u + 4
= 1 x 6 + 4
= 6 + 4
= 10
(b)
Number of blue stickers
= 4 u
= 4 x 6
= 24
Answer(s): (a) 10; (b) 24