Adam has some blue stickers and red stickers.
If 23 blue stickers are added, 70% of the stickers will be red stickers.
If 223 blue stickers are added, 30% 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 |
9 u - 23 |
21 u |
49 u - 223 |
21 u |
Change |
+ 23 |
No change |
+ 223 |
No change |
After |
3x3 = 9 u |
7x3 = 21 u |
7x7 = 49 u |
3x7 = 21 u |
(a)
70% =
70100 =
71030% =
30100 =
310 Scenario 1 Fraction of the stickers that are blue in the end
= 1 -
710 =
310 Scenario 2
Fraction of the stickers that are blue in the end
= 1 -
310 =
710 The number of red stickers remains unchanged in both scenarios.
LCM of 7 and 3 = 21
49 u - 223 = 9 u - 23
49 u - 9 u = 223 - 23
40 u = 200
1 u = 200 ÷ 40 = 5
Number of blue stickers
= 9 u - 23
= 9 x 5 - 23
= 45 - 23
= 22
(b)
Number of red stickers
= 21 u
= 21 x 5
= 105
Answer(s): (a) 22; (b) 105