Adam has some red stickers and blue stickers.
If 171 red stickers are removed, 90% of the stickers will be blue stickers.
If 29 red stickers are added, 30% of the stickers will be blue stickers.
- How many red stickers are there?
- How many blue stickers are there?
|
Scenario 1 |
Scenario 2 |
|
Red |
Blue |
Red |
Blue |
Before |
1 u + 171 |
9 u |
21 u - 29 |
9 u |
Change |
- 171 |
No change |
+ 29 |
No change |
After |
1x1 = 1 u |
9x1 = 9 u |
7x3 = 21 u |
3x3 = 9 u |
(a)
90% =
90100 =
91030% =
30100 =
310Scenario 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 -
310 =
710 The number of blue stickers remains unchanged in both scenarios.
LCM of 9 and 3 = 9
21 u - 29 = 1 u + 171
21 u - 1 u = 171 + 29
20 u = 200
1 u = 200 ÷ 20 = 10
Number of red stickers
= 1 u + 171
= 1 x 10 + 171
= 10 + 171
= 181
(b)
Number of blue stickers
= 9 u
= 9 x 10
= 90
Answer(s): (a) 181; (b) 90