Adam has some blue stickers and red stickers.
If 392 blue stickers are removed, 80% of the stickers will be red stickers.
If 158 blue stickers are added, 30% 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 + 392 |
12 u |
28 u - 158 |
12 u |
Change |
- 392 |
No change |
+ 158 |
No change |
After |
1x3 = 3 u |
4x3 = 12 u |
7x4 = 28 u |
3x4 = 12 u |
(a)
80% =
80100 =
4530% =
30100 =
310Scenario 1 Fraction of the stickers that are blue in the end
= 1 -
45 =
15 Scenario 2
Fraction of the stickers that are blue in the end
= 1 -
310 =
710 The number of red stickers remains unchanged in both scenarios.
LCM of 4 and 3 = 12
28 u - 158 = 3 u + 392
28 u - 3 u = 392 + 158
25 u = 550
1 u = 550 ÷ 25 = 22
Number of blue stickers
= 3 u + 392
= 3 x 22 + 392
= 66 + 392
= 458
(b)
Number of red stickers
= 12 u
= 12 x 22
= 264
Answer(s): (a) 458; (b) 264