Adam has some blue stickers and red stickers.
If 25 blue stickers are added, 70% of the stickers will be red stickers.
If 235 blue stickers are added, 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 - 25 |
14 u |
21 u - 235 |
14 u |
Change |
+ 25 |
No change |
+ 235 |
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 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
21 u - 235 = 6 u - 25
21 u - 6 u = 235 - 25
15 u = 210
1 u = 210 ÷ 15 = 14
Number of blue stickers
= 6 u - 25
= 6 x 14 - 25
= 84 - 25
= 59
(b)
Number of red stickers
= 14 u
= 14 x 14
= 196
Answer(s): (a) 59; (b) 196