Adam has some red stickers and blue stickers.
If 259 red stickers are removed, 70% of the stickers will be blue stickers.
If 101 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 + 259 |
14 u |
21 u - 101 |
14 u |
Change |
- 259 |
No change |
+ 101 |
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 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 - 101 = 6 u + 259
21 u - 6 u = 259 + 101
15 u = 360
1 u = 360 ÷ 15 = 24
Number of red stickers
= 6 u + 259
= 6 x 24 + 259
= 144 + 259
= 403
(b)
Number of blue stickers
= 14 u
= 14 x 24
= 336
Answer(s): (a) 403; (b) 336