Adam has some red stickers and blue stickers.
If 92 red stickers are removed, 80% of the stickers will be blue stickers.
If 73 red stickers are added, 25% 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 + 92 |
4 u |
12 u - 73 |
4 u |
Change |
- 92 |
No change |
+ 73 |
No change |
After |
1x1 = 1 u |
4x1 = 4 u |
3x4 = 12 u |
1x4 = 4 u |
(a)
80% =
80100 =
4525% =
25100 =
14Scenario 1 Fraction of the stickers that are red in the end
= 1 -
45 =
15 Scenario 2
Fraction of the stickers that are red in the end
= 1 -
14 =
34 The number of blue stickers remains unchanged in both scenarios.
LCM of 4 and 1 = 4
12 u - 73 = 1 u + 92
12 u - 1 u = 92 + 73
11 u = 165
1 u = 165 ÷ 11 = 15
Number of red stickers
= 1 u + 92
= 1 x 15 + 92
= 15 + 92
= 107
(b)
Number of blue stickers
= 4 u
= 4 x 15
= 60
Answer(s): (a) 107; (b) 60