Adam has some red stickers and green stickers.
If 203 red stickers are removed, 90% of the stickers will be green stickers.
If 47 red stickers are removed, 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 + 203 |
9 u |
27 u + 47 |
9 u |
Change |
- 203 |
No change |
- 47 |
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
1 u + 203 = 27 u + 47
27 u - 1 u = 203 - 47
26 u = 156
1 u = 156 ÷ 26 = 6
Number of red stickers
= 1 u + 203
= 1 x 6 + 203
= 6 + 203
= 209
(b)
Number of green stickers
= 9 u
= 9 x 6
= 54
Answer(s): (a) 209; (b) 54