Adam has some red stickers and blue stickers.
If 386 red stickers are removed, 80% of the stickers will be blue stickers.
If 136 red stickers are removed, 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 |
3 u + 386 |
12 u |
28 u + 136 |
12 u |
Change |
- 386 |
No change |
- 136 |
No change |
After |
1x3 = 3 u |
4x3 = 12 u |
7x4 = 28 u |
3x4 = 12 u |
(a)
80% =
80100 =
4530% =
30100 =
310Scenario 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 -
310 =
710 The number of blue stickers remains unchanged in both scenarios.
LCM of 4 and 3 = 12
3 u + 386 = 28 u + 136
28 u - 3 u = 386 - 136
25 u = 250
1 u = 250 ÷ 25 = 10
Number of red stickers
= 3 u + 386
= 3 x 10 + 386
= 30 + 386
= 416
(b)
Number of blue stickers
= 12 u
= 12 x 10
= 120
Answer(s): (a) 416; (b) 120