Adam has some blue stickers and red stickers.
If 28 blue stickers are removed, 80% of the stickers will be red stickers.
If 42 blue stickers are added, 40% of the stickers will be red stickers.
- How many blue stickers are there?
- How many red stickers are there?
|
Scenario 1 |
Scenario 2 |
|
Blue |
Red |
Blue |
Red |
Before |
1 u + 28 |
4 u |
6 u - 42 |
4 u |
Change |
- 28 |
No change |
+ 42 |
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 blue in the end
= 1 -
45 =
15 Scenario 2
Fraction of the stickers that are blue in the end
= 1 -
25 =
35 The number of red stickers remains unchanged in both scenarios.
LCM of 4 and 2 = 4
6 u - 42 = 1 u + 28
6 u - 1 u = 28 + 42
5 u = 70
1 u = 70 ÷ 5 = 14
Number of blue stickers
= 1 u + 28
= 1 x 14 + 28
= 14 + 28
= 42
(b)
Number of red stickers
= 4 u
= 4 x 14
= 56
Answer(s): (a) 42; (b) 56