Adam has some green stickers and red stickers.
If 293 green stickers are removed, 70% of the stickers will be red stickers.
If 27 green stickers are added, 30% 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 |
9 u + 293 |
21 u |
49 u - 27 |
21 u |
Change |
- 293 |
No change |
+ 27 |
No change |
After |
3x3 = 9 u |
7x3 = 21 u |
7x7 = 49 u |
3x7 = 21 u |
(a)
70% =
70100 =
71030% =
30100 =
310Scenario 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 -
310 =
710 The number of red stickers remains unchanged in both scenarios.
LCM of 7 and 3 = 21
49 u - 27 = 9 u + 293
49 u - 9 u = 293 + 27
40 u = 320
1 u = 320 ÷ 40 = 8
Number of green stickers
= 9 u + 293
= 9 x 8 + 293
= 72 + 293
= 365
(b)
Number of red stickers
= 21 u
= 21 x 8
= 168
Answer(s): (a) 365; (b) 168