Adam has some blue stickers and red stickers.
If 50 blue stickers are added, 70% of the stickers will be red stickers.
If 290 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 - 50 |
14 u |
21 u - 290 |
14 u |
Change |
+ 50 |
No change |
+ 290 |
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 - 290 = 6 u - 50
21 u - 6 u = 290 - 50
15 u = 240
1 u = 240 ÷ 15 = 16
Number of blue stickers
= 6 u - 50
= 6 x 16 - 50
= 96 - 50
= 46
(b)
Number of red stickers
= 14 u
= 14 x 16
= 224
Answer(s): (a) 46; (b) 224