Adam has some red stickers and blue stickers.
If 219 red stickers are removed, 70% of the stickers will be blue stickers.
If 111 red stickers are removed, 25% 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 |
3 u + 219 |
7 u |
21 u + 111 |
7 u |
Change |
- 219 |
No change |
- 111 |
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 red in the end
= 1 -
710 =
310 Scenario 2
Fraction of the stickers that are red in the end
= 1 -
14 =
34 The number of blue stickers remains unchanged in both scenarios.
LCM of 7 and 1 = 7
3 u + 219 = 21 u + 111
21 u - 3 u = 219 - 111
18 u = 108
1 u = 108 ÷ 18 = 6
Number of red stickers
= 3 u + 219
= 3 x 6 + 219
= 18 + 219
= 237
(b)
Number of blue stickers
= 7 u
= 7 x 6
= 42
Answer(s): (a) 237; (b) 42