Adam has some green stickers and red stickers.
If 167 green stickers are removed, 90% of the stickers will be red stickers.
If 148 green stickers are added, 20% 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 |
1 u + 167 |
9 u |
36 u - 148 |
9 u |
Change |
- 167 |
No change |
+ 148 |
No change |
After |
1x1 = 1 u |
9x1 = 9 u |
4x9 = 36 u |
1x9 = 9 u |
(a)
90% =
90100 =
91020% =
20100 =
15Scenario 1 Fraction of the stickers that are green in the end
= 1 -
910 =
110 Scenario 2
Fraction of the stickers that are green in the end
= 1 -
15 =
45 The number of red stickers remains unchanged in both scenarios.
LCM of 9 and 1 = 9
36 u - 148 = 1 u + 167
36 u - 1 u = 167 + 148
35 u = 315
1 u = 315 ÷ 35 = 9
Number of green stickers
= 1 u + 167
= 1 x 9 + 167
= 9 + 167
= 176
(b)
Number of red stickers
= 9 u
= 9 x 9
= 81
Answer(s): (a) 176; (b) 81