Adam has some red stickers and green stickers.
If 379 red stickers are removed, 90% of the stickers will be green stickers.
If 89 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 |
1 u + 379 |
9 u |
27 u - 89 |
9 u |
Change |
- 379 |
No change |
+ 89 |
No change |
After |
1x1 = 1 u |
9x1 = 9 u |
3x9 = 27 u |
1x9 = 9 u |
(a)
90% =
90100 =
91025% =
25100 =
14Scenario 1 Fraction of the stickers that are red in the end
= 1 -
910 =
110 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 9 and 1 = 9
27 u - 89 = 1 u + 379
27 u - 1 u = 379 + 89
26 u = 468
1 u = 468 ÷ 26 = 18
Number of red stickers
= 1 u + 379
= 1 x 18 + 379
= 18 + 379
= 397
(b)
Number of green stickers
= 9 u
= 9 x 18
= 162
Answer(s): (a) 397; (b) 162