Adam has some red stickers and green stickers.
If 167 red stickers are removed, 70% of the stickers will be green stickers.
If 58 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 + 167 |
7 u |
28 u - 58 |
7 u |
Change |
- 167 |
No change |
+ 58 |
No change |
After |
3x1 = 3 u |
7x1 = 7 u |
4x7 = 28 u |
1x7 = 7 u |
(a)
70% =
70100 =
71020% =
20100 =
15Scenario 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 - 58 = 3 u + 167
28 u - 3 u = 167 + 58
25 u = 225
1 u = 225 ÷ 25 = 9
Number of red stickers
= 3 u + 167
= 3 x 9 + 167
= 27 + 167
= 194
(b)
Number of green stickers
= 7 u
= 7 x 9
= 63
Answer(s): (a) 194; (b) 63