Adam has some red stickers and green stickers.
If 31 red stickers are added, 70% of the stickers will be green stickers.
If 356 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 - 31 |
7 u |
28 u - 356 |
7 u |
Change |
+ 31 |
No change |
+ 356 |
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 - 356 = 3 u - 31
28 u - 3 u = 356 - 31
25 u = 325
1 u = 325 ÷ 25 = 13
Number of red stickers
= 3 u - 31
= 3 x 13 - 31
= 39 - 31
= 8
(b)
Number of green stickers
= 7 u
= 7 x 13
= 91
Answer(s): (a) 8; (b) 91