Adam has some green stickers and blue stickers.
If 21 green stickers are removed, 80% of the stickers will be blue stickers.
If 174 green stickers are added, 20% of the stickers will be blue stickers.
- How many green stickers are there?
- How many blue stickers are there?
|
Scenario 1 |
Scenario 2 |
|
Green |
Blue |
Green |
Blue |
Before |
1 u + 21 |
4 u |
16 u - 174 |
4 u |
Change |
- 21 |
No change |
+ 174 |
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 green in the end
= 1 -
45 =
15 Scenario 2
Fraction of the stickers that are green in the end
= 1 -
15 =
45 The number of blue stickers remains unchanged in both scenarios.
LCM of 4 and 1 = 4
16 u - 174 = 1 u + 21
16 u - 1 u = 21 + 174
15 u = 195
1 u = 195 ÷ 15 = 13
Number of green stickers
= 1 u + 21
= 1 x 13 + 21
= 13 + 21
= 34
(b)
Number of blue stickers
= 4 u
= 4 x 13
= 52
Answer(s): (a) 34; (b) 52