Adam has some red stickers and green stickers.
If 249 red stickers are removed, 70% of the stickers will be green stickers.
If 111 red stickers are added, 25% of the stickers will be green stickers.
- How many red stickers are there?
- How many green stickers are there?
|
Scenario 1 |
Scenario 2 |
|
Red |
Green |
Red |
Green |
Before |
3 u + 249 |
7 u |
21 u - 111 |
7 u |
Change |
- 249 |
No change |
+ 111 |
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 red in the end
= 1 -
710 =
310 Scenario 2
Fraction of the stickers that are red in the end
= 1 -
14 =
34 The number of green stickers remains unchanged in both scenarios.
LCM of 7 and 1 = 7
21 u - 111 = 3 u + 249
21 u - 3 u = 249 + 111
18 u = 360
1 u = 360 ÷ 18 = 20
Number of red stickers
= 3 u + 249
= 3 x 20 + 249
= 60 + 249
= 309
(b)
Number of green stickers
= 7 u
= 7 x 20
= 140
Answer(s): (a) 309; (b) 140