Adam has some blue stickers and red stickers.
If 246 blue stickers are removed, 70% of the stickers will be red stickers.
If 79 blue stickers are added, 20% 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 |
3 u + 246 |
7 u |
28 u - 79 |
7 u |
Change |
- 246 |
No change |
+ 79 |
No change |
After |
3x1 = 3 u |
7x1 = 7 u |
4x7 = 28 u |
1x7 = 7 u |
(a)
70% =
70100 =
71020% =
20100 =
15Scenario 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 -
15 =
45 The number of red stickers remains unchanged in both scenarios.
LCM of 7 and 1 = 7
28 u - 79 = 3 u + 246
28 u - 3 u = 246 + 79
25 u = 325
1 u = 325 ÷ 25 = 13
Number of blue stickers
= 3 u + 246
= 3 x 13 + 246
= 39 + 246
= 285
(b)
Number of red stickers
= 7 u
= 7 x 13
= 91
Answer(s): (a) 285; (b) 91