Adam has some blue stickers and red stickers.
If 32 blue stickers are added, 70% of the stickers will be red stickers.
If 392 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 - 32 |
7 u |
21 u - 392 |
7 u |
Change |
+ 32 |
No change |
+ 392 |
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 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 - 392 = 3 u - 32
21 u - 3 u = 392 - 32
18 u = 360
1 u = 360 ÷ 18 = 20
Number of blue stickers
= 3 u - 32
= 3 x 20 - 32
= 60 - 32
= 28
(b)
Number of red stickers
= 7 u
= 7 x 20
= 140
Answer(s): (a) 28; (b) 140