Adam has some blue stickers and red stickers.
If 211 blue stickers are removed, 80% of the stickers will be red stickers.
If 121 blue stickers are removed, 20% 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 + 211 |
4 u |
16 u + 121 |
4 u |
Change |
- 211 |
No change |
- 121 |
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 blue in the end
= 1 -
45 =
15 Scenario 2
Fraction of the stickers that are blue in the end
= 1 -
15 =
45 The number of red stickers remains unchanged in both scenarios.
LCM of 4 and 1 = 4
1 u + 211 = 16 u + 121
16 u - 1 u = 211 - 121
15 u = 90
1 u = 90 ÷ 15 = 6
Number of blue stickers
= 1 u + 211
= 1 x 6 + 211
= 6 + 211
= 217
(b)
Number of red stickers
= 4 u
= 4 x 6
= 24
Answer(s): (a) 217; (b) 24