Adam has some green stickers and red stickers.
If 333 green stickers are removed, 90% of the stickers will be red stickers.
If 52 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 + 333 |
9 u |
36 u - 52 |
9 u |
Change |
- 333 |
No change |
+ 52 |
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 - 52 = 1 u + 333
36 u - 1 u = 333 + 52
35 u = 385
1 u = 385 ÷ 35 = 11
Number of green stickers
= 1 u + 333
= 1 x 11 + 333
= 11 + 333
= 344
(b)
Number of red stickers
= 9 u
= 9 x 11
= 99
Answer(s): (a) 344; (b) 99