Adam has some blue stickers and red stickers.
If 217 blue stickers are removed, 70% of the stickers will be red stickers.
If 158 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 + 217 |
14 u |
21 u - 158 |
14 u |
Change |
- 217 |
No change |
+ 158 |
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
21 u - 158 = 6 u + 217
21 u - 6 u = 217 + 158
15 u = 375
1 u = 375 ÷ 15 = 25
Number of blue stickers
= 6 u + 217
= 6 x 25 + 217
= 150 + 217
= 367
(b)
Number of red stickers
= 14 u
= 14 x 25
= 350
Answer(s): (a) 367; (b) 350