Adam has some blue stickers and red stickers.
If 298 blue stickers are removed, 70% of the stickers will be red stickers.
If 28 blue stickers are removed, 40% 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 |
6 u + 298 |
14 u |
21 u + 28 |
14 u |
Change |
- 298 |
No change |
- 28 |
No change |
After |
3x2 = 6 u |
7x2 = 14 u |
3x7 = 21 u |
2x7 = 14 u |
(a)
70% =
70100 =
71040% =
40100 =
25Scenario 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 -
25 =
35 The number of red stickers remains unchanged in both scenarios.
LCM of 7 and 2 = 14
6 u + 298 = 21 u + 28
21 u - 6 u = 298 - 28
15 u = 270
1 u = 270 ÷ 15 = 18
Number of blue stickers
= 6 u + 298
= 6 x 18 + 298
= 108 + 298
= 406
(b)
Number of red stickers
= 14 u
= 14 x 18
= 252
Answer(s): (a) 406; (b) 252