Adam has some green stickers and blue stickers.
If 111 green stickers are removed, 70% of the stickers will be blue stickers.
If 21 green stickers are removed, 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 + 111 |
7 u |
21 u + 21 |
7 u |
Change |
- 111 |
No change |
- 21 |
No change |
After |
3x1 = 3 u |
7x1 = 7 u |
3x7 = 21 u |
1x7 = 7 u |
(a)
70% =
70100 =
71025% =
25100 =
14Scenario 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
3 u + 111 = 21 u + 21
21 u - 3 u = 111 - 21
18 u = 90
1 u = 90 ÷ 18 = 5
Number of green stickers
= 3 u + 111
= 3 x 5 + 111
= 15 + 111
= 126
(b)
Number of blue stickers
= 7 u
= 7 x 5
= 35
Answer(s): (a) 126; (b) 35