Adam has some blue stickers and red stickers.
If 121 blue stickers are removed, 90% of the stickers will be red stickers.
If 35 blue stickers are added, 25% of the stickers will be red stickers.
- How many blue stickers are there?
- How many red stickers are there?
|
Scenario 1 |
Scenario 2 |
|
Blue |
Red |
Blue |
Red |
Before |
1 u + 121 |
9 u |
27 u - 35 |
9 u |
Change |
- 121 |
No change |
+ 35 |
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 blue in the end
= 1 -
910 =
110 Scenario 2
Fraction of the stickers that are blue in the end
= 1 -
14 =
34 The number of red stickers remains unchanged in both scenarios.
LCM of 9 and 1 = 9
27 u - 35 = 1 u + 121
27 u - 1 u = 121 + 35
26 u = 156
1 u = 156 ÷ 26 = 6
Number of blue stickers
= 1 u + 121
= 1 x 6 + 121
= 6 + 121
= 127
(b)
Number of red stickers
= 9 u
= 9 x 6
= 54
Answer(s): (a) 127; (b) 54