Adam has some red stickers and blue stickers.
If 21 red stickers are added, 70% of the stickers will be blue stickers.
If 261 red stickers are added, 30% of the stickers will be blue stickers.
- How many red stickers are there?
- How many blue stickers are there?
|
Scenario 1 |
Scenario 2 |
|
Red |
Blue |
Red |
Blue |
Before |
9 u - 21 |
21 u |
49 u - 261 |
21 u |
Change |
+ 21 |
No change |
+ 261 |
No change |
After |
3x3 = 9 u |
7x3 = 21 u |
7x7 = 49 u |
3x7 = 21 u |
(a)
70% =
70100 =
71030% =
30100 =
310 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 -
310 =
710 The number of blue stickers remains unchanged in both scenarios.
LCM of 7 and 3 = 21
49 u - 261 = 9 u - 21
49 u - 9 u = 261 - 21
40 u = 240
1 u = 240 ÷ 40 = 6
Number of red stickers
= 9 u - 21
= 9 x 6 - 21
= 54 - 21
= 33
(b)
Number of blue stickers
= 21 u
= 21 x 6
= 126
Answer(s): (a) 33; (b) 126