Adam has some green stickers and blue stickers.
If 233 green stickers are added, 40% of the stickers will be blue stickers.
If 148 blue stickers are added, 20% of the stickers will be green stickers.
- How many green stickers are there?
- How many blue stickers are there?
|
Scenario 1 |
Scenario 2 |
|
Green |
Blue |
Green |
Blue |
Before |
3 u - 233 |
2 u |
1 p |
4 p - 148 |
Change |
+ 233 |
No change |
No change |
+ 148 |
After |
3 u |
2 u |
1 p |
4 p |
(a)
40% =
40100 =
25 20% =
20100 =
15 Scenario 1 Fraction of the stickers that are green
= 1 -
25 =
35 Number of green stickers at first = 3 u - 233
Number of blue stickers at first = 2 u
Scenario 2 Fraction of the stickers that are blue
= 1 -
15=
45 Number of green stickers at first = 1 p
Number of blue stickers at first = 4 p - 148
3 u - 233 = 1 p --- (1)
2 u = 4 p - 148
2 u + 148 = 4 p --- (2)
(1)
x 4 12 u - 932 = 4 p --- (3)
(3) = (2)
12 u - 932 = 2 u + 148
12 u - 2 u = 932 + 148
10 u = 1080
1 u = 1080 ÷ 10 = 108
Number of green stickers
= 3 u - 233
= 3 x 108 - 233
= 324 - 233
= 91
(b)
Number of blue stickers
= 2 u
= 2 x 108
= 216
Answer(s): (a) 91; (b) 216