Adam has some red stickers and green stickers.
If 20 red stickers are added, 70% of the stickers will be green stickers.
If 345 red stickers are added, 20% 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 - 20 |
7 u |
28 u - 345 |
7 u |
Change |
+ 20 |
No change |
+ 345 |
No change |
After |
3x1 = 3 u |
7x1 = 7 u |
4x7 = 28 u |
1x7 = 7 u |
(a)
70% =
70100 =
71020% =
20100 =
15 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 -
15 =
45 The number of green stickers remains unchanged in both scenarios.
LCM of 7 and 1 = 7
28 u - 345 = 3 u - 20
28 u - 3 u = 345 - 20
25 u = 325
1 u = 325 ÷ 25 = 13
Number of red stickers
= 3 u - 20
= 3 x 13 - 20
= 39 - 20
= 19
(b)
Number of green stickers
= 7 u
= 7 x 13
= 91
Answer(s): (a) 19; (b) 91