Adam has some red stickers and blue stickers.
If 81 red stickers are added, 70% of the stickers will be blue stickers.
If 396 red stickers are added, 40% 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 |
6 u - 81 |
14 u |
21 u - 396 |
14 u |
Change |
+ 81 |
No change |
+ 396 |
No change |
After |
3x2 = 6 u |
7x2 = 14 u |
3x7 = 21 u |
2x7 = 14 u |
(a)
70% =
70100 =
71040% =
40100 =
25 Scenario 1 Fraction of the stickers that are red in the end
= 1 -
710 =
310 Scenario 2
Fraction of the stickers that are red in the end
= 1 -
25 =
35 The number of blue stickers remains unchanged in both scenarios.
LCM of 7 and 2 = 14
21 u - 396 = 6 u - 81
21 u - 6 u = 396 - 81
15 u = 315
1 u = 315 ÷ 15 = 21
Number of red stickers
= 6 u - 81
= 6 x 21 - 81
= 126 - 81
= 45
(b)
Number of blue stickers
= 14 u
= 14 x 21
= 294
Answer(s): (a) 45; (b) 294