Adam has some red stickers and blue stickers.
If 361 red stickers are removed, 80% of the stickers will be blue stickers.
If 130 red stickers are removed, 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 + 361 |
4 u |
12 u + 130 |
4 u |
Change |
- 361 |
No change |
- 130 |
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
1 u + 361 = 12 u + 130
12 u - 1 u = 361 - 130
11 u = 231
1 u = 231 ÷ 11 = 21
Number of red stickers
= 1 u + 361
= 1 x 21 + 361
= 21 + 361
= 382
(b)
Number of blue stickers
= 4 u
= 4 x 21
= 84
Answer(s): (a) 382; (b) 84