Adam has some red stickers and blue stickers.
If 35 red stickers are added, 70% of the stickers will be blue stickers.
If 335 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 - 35 |
14 u |
21 u - 335 |
14 u |
Change |
+ 35 |
No change |
+ 335 |
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 - 335 = 6 u - 35
21 u - 6 u = 335 - 35
15 u = 300
1 u = 300 ÷ 15 = 20
Number of red stickers
= 6 u - 35
= 6 x 20 - 35
= 120 - 35
= 85
(b)
Number of blue stickers
= 14 u
= 14 x 20
= 280
Answer(s): (a) 85; (b) 280