Adam has some blue stickers and red stickers.
If 337 blue stickers are removed, 80% of the stickers will be red stickers.
If 62 blue stickers are removed, 25% 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 |
1 u + 337 |
4 u |
12 u + 62 |
4 u |
Change |
- 337 |
No change |
- 62 |
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 blue in the end
= 1 -
45 =
15 Scenario 2
Fraction of the stickers that are blue in the end
= 1 -
14 =
34 The number of red stickers remains unchanged in both scenarios.
LCM of 4 and 1 = 4
1 u + 337 = 12 u + 62
12 u - 1 u = 337 - 62
11 u = 275
1 u = 275 ÷ 11 = 25
Number of blue stickers
= 1 u + 337
= 1 x 25 + 337
= 25 + 337
= 362
(b)
Number of red stickers
= 4 u
= 4 x 25
= 100
Answer(s): (a) 362; (b) 100