Adam has some red stickers and blue stickers.
If 28 red stickers are removed, 80% of the stickers will be blue stickers.
If 182 red stickers are added, 20% 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 + 28 |
4 u |
16 u - 182 |
4 u |
Change |
- 28 |
No change |
+ 182 |
No change |
After |
1x1 = 1 u |
4x1 = 4 u |
4x4 = 16 u |
1x4 = 4 u |
(a)
80% =
80100 =
4520% =
20100 =
15Scenario 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 -
15 =
45 The number of blue stickers remains unchanged in both scenarios.
LCM of 4 and 1 = 4
16 u - 182 = 1 u + 28
16 u - 1 u = 28 + 182
15 u = 210
1 u = 210 ÷ 15 = 14
Number of red stickers
= 1 u + 28
= 1 x 14 + 28
= 14 + 28
= 42
(b)
Number of blue stickers
= 4 u
= 4 x 14
= 56
Answer(s): (a) 42; (b) 56