Adam has some red stickers and blue stickers.
If 52 red stickers are removed, 90% of the stickers will be blue stickers.
If 123 red stickers are added, 20% 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 + 52 |
9 u |
36 u - 123 |
9 u |
Change |
- 52 |
No change |
+ 123 |
No change |
After |
1x1 = 1 u |
9x1 = 9 u |
4x9 = 36 u |
1x9 = 9 u |
(a)
90% =
90100 =
91020% =
20100 =
15Scenario 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 -
15 =
45 The number of blue stickers remains unchanged in both scenarios.
LCM of 9 and 1 = 9
36 u - 123 = 1 u + 52
36 u - 1 u = 52 + 123
35 u = 175
1 u = 175 ÷ 35 = 5
Number of red stickers
= 1 u + 52
= 1 x 5 + 52
= 5 + 52
= 57
(b)
Number of blue stickers
= 9 u
= 9 x 5
= 45
Answer(s): (a) 57; (b) 45