Adam has some green stickers and blue stickers.
If 21 green stickers are added, 70% of the stickers will be blue stickers.
If 291 green stickers are added, 25% 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 |
3 u - 21 |
7 u |
21 u - 291 |
7 u |
Change |
+ 21 |
No change |
+ 291 |
No change |
After |
3x1 = 3 u |
7x1 = 7 u |
3x7 = 21 u |
1x7 = 7 u |
(a)
70% =
70100 =
71025% =
25100 =
14 Scenario 1 Fraction of the stickers that are green in the end
= 1 -
710 =
310 Scenario 2
Fraction of the stickers that are green in the end
= 1 -
14 =
34 The number of blue stickers remains unchanged in both scenarios.
LCM of 7 and 1 = 7
21 u - 291 = 3 u - 21
21 u - 3 u = 291 - 21
18 u = 270
1 u = 270 ÷ 18 = 15
Number of green stickers
= 3 u - 21
= 3 x 15 - 21
= 45 - 21
= 24
(b)
Number of blue stickers
= 7 u
= 7 x 15
= 105
Answer(s): (a) 24; (b) 105