Adam has some blue stickers and green stickers.
If 363 blue stickers are removed, 80% of the stickers will be green stickers.
If 88 blue stickers are removed, 30% of the stickers will be green stickers.
- How many blue stickers are there?
- How many green stickers are there?
|
Scenario 1 |
Scenario 2 |
|
Blue |
Green |
Blue |
Green |
Before |
3 u + 363 |
12 u |
28 u + 88 |
12 u |
Change |
- 363 |
No change |
- 88 |
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 green stickers remains unchanged in both scenarios.
LCM of 4 and 3 = 12
3 u + 363 = 28 u + 88
28 u - 3 u = 363 - 88
25 u = 275
1 u = 275 ÷ 25 = 11
Number of blue stickers
= 3 u + 363
= 3 x 11 + 363
= 33 + 363
= 396
(b)
Number of green stickers
= 12 u
= 12 x 11
= 132
Answer(s): (a) 396; (b) 132