Adam has some blue stickers and red stickers.
If 17 blue stickers are removed, 70% of the stickers will be red stickers.
If 91 blue stickers are added, 25% of the stickers will be red stickers.
- How many blue stickers are there?
- How many red stickers are there?
|
Scenario 1 |
Scenario 2 |
|
Blue |
Red |
Blue |
Red |
Before |
3 u + 17 |
7 u |
21 u - 91 |
7 u |
Change |
- 17 |
No change |
+ 91 |
No change |
After |
3x1 = 3 u |
7x1 = 7 u |
3x7 = 21 u |
1x7 = 7 u |
(a)
70% =
70100 =
71025% =
25100 =
14Scenario 1 Fraction of the stickers that are blue in the end
= 1 -
710 =
310 Scenario 2
Fraction of the stickers that are blue in the end
= 1 -
14 =
34 The number of red stickers remains unchanged in both scenarios.
LCM of 7 and 1 = 7
21 u - 91 = 3 u + 17
21 u - 3 u = 17 + 91
18 u = 108
1 u = 108 ÷ 18 = 6
Number of blue stickers
= 3 u + 17
= 3 x 6 + 17
= 18 + 17
= 35
(b)
Number of red stickers
= 7 u
= 7 x 6
= 42
Answer(s): (a) 35; (b) 42