Adam has some red stickers and green stickers.
If 51 red stickers are added, 70% of the stickers will be green stickers.
If 393 red stickers are added, 25% of the stickers will be green stickers.
- How many red stickers are there?
- How many green stickers are there?
|
Scenario 1 |
Scenario 2 |
|
Red |
Green |
Red |
Green |
Before |
3 u - 51 |
7 u |
21 u - 393 |
7 u |
Change |
+ 51 |
No change |
+ 393 |
No change |
After |
3x1 = 3 u |
7x1 = 7 u |
3x7 = 21 u |
1x7 = 7 u |
(a)
70% =
70100 =
71025% =
25100 =
14 Scenario 1 Fraction of the stickers that are red in the end
= 1 -
710 =
310 Scenario 2
Fraction of the stickers that are red in the end
= 1 -
14 =
34 The number of green stickers remains unchanged in both scenarios.
LCM of 7 and 1 = 7
21 u - 393 = 3 u - 51
21 u - 3 u = 393 - 51
18 u = 342
1 u = 342 ÷ 18 = 19
Number of red stickers
= 3 u - 51
= 3 x 19 - 51
= 57 - 51
= 6
(b)
Number of green stickers
= 7 u
= 7 x 19
= 133
Answer(s): (a) 6; (b) 133