Adam has some blue stickers and green stickers.
If 385 blue stickers are added, 40% of the stickers will be green stickers.
If 392 green stickers are added, 25% of the stickers will be blue stickers.
- How many blue stickers are there?
- How many green stickers are there?
|
Scenario 1 |
Scenario 2 |
|
Blue |
Green |
Blue |
Green |
Before |
3 u - 385 |
2 u |
1 p |
3 p - 392 |
Change |
+ 385 |
No change |
No change |
+ 392 |
After |
3 u |
2 u |
1 p |
3 p |
(a)
40% =
40100 =
25 25% =
25100 =
14 Scenario 1 Fraction of the stickers that are blue
= 1 -
25 =
35 Number of blue stickers at first = 3 u - 385
Number of green stickers at first = 2 u
Scenario 2 Fraction of the stickers that are green
= 1 -
14=
34 Number of blue stickers at first = 1 p
Number of green stickers at first = 3 p - 392
3 u - 385 = 1 p --- (1)
2 u = 3 p - 392
2 u + 392 = 3 p --- (2)
(1)
x 3 9 u - 1155 = 3 p --- (3)
(3) = (2)
9 u - 1155 = 2 u + 392
9 u - 2 u = 1155 + 392
7 u = 1547
1 u = 1547 ÷ 7 = 221
Number of blue stickers
= 3 u - 385
= 3 x 221 - 385
= 663 - 385
= 278
(b)
Number of green stickers
= 2 u
= 2 x 221
= 442
Answer(s): (a) 278; (b) 442