Adam has some red stickers and blue stickers.
If 237 red stickers are removed, 80% of the stickers will be blue stickers.
If 72 red stickers are removed, 20% 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 |
1 u + 237 |
4 u |
16 u + 72 |
4 u |
Change |
- 237 |
No change |
- 72 |
No change |
After |
1x1 = 1 u |
4x1 = 4 u |
4x4 = 16 u |
1x4 = 4 u |
(a)
80% =
80100 =
4520% =
20100 =
15Scenario 1 Fraction of the stickers that are red in the end
= 1 -
45 =
15 Scenario 2
Fraction of the stickers that are red in the end
= 1 -
15 =
45 The number of blue stickers remains unchanged in both scenarios.
LCM of 4 and 1 = 4
1 u + 237 = 16 u + 72
16 u - 1 u = 237 - 72
15 u = 165
1 u = 165 ÷ 15 = 11
Number of red stickers
= 1 u + 237
= 1 x 11 + 237
= 11 + 237
= 248
(b)
Number of blue stickers
= 4 u
= 4 x 11
= 44
Answer(s): (a) 248; (b) 44