Adam has some blue stickers and red stickers.
If 201 blue stickers are removed, 80% of the stickers will be red stickers.
If 69 blue stickers are added, 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 + 201 |
4 u |
16 u - 69 |
4 u |
Change |
- 201 |
No change |
+ 69 |
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
16 u - 69 = 1 u + 201
16 u - 1 u = 201 + 69
15 u = 270
1 u = 270 ÷ 15 = 18
Number of blue stickers
= 1 u + 201
= 1 x 18 + 201
= 18 + 201
= 219
(b)
Number of red stickers
= 4 u
= 4 x 18
= 72
Answer(s): (a) 219; (b) 72