Adam has some green stickers and red stickers.
If 251 green stickers are removed, 70% of the stickers will be red stickers.
If 199 green stickers are added, 25% of the stickers will be red stickers.
- How many green stickers are there?
- How many red stickers are there?
|
Scenario 1 |
Scenario 2 |
|
Green |
Red |
Green |
Red |
Before |
3 u + 251 |
7 u |
21 u - 199 |
7 u |
Change |
- 251 |
No change |
+ 199 |
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 red stickers remains unchanged in both scenarios.
LCM of 7 and 1 = 7
21 u - 199 = 3 u + 251
21 u - 3 u = 251 + 199
18 u = 450
1 u = 450 ÷ 18 = 25
Number of green stickers
= 3 u + 251
= 3 x 25 + 251
= 75 + 251
= 326
(b)
Number of red stickers
= 7 u
= 7 x 25
= 175
Answer(s): (a) 326; (b) 175