Adam has some red stickers and blue stickers.
If 321 red stickers are removed, 70% of the stickers will be blue stickers.
If 196 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 |
3 u + 321 |
7 u |
28 u + 196 |
7 u |
Change |
- 321 |
No change |
- 196 |
No change |
After |
3x1 = 3 u |
7x1 = 7 u |
4x7 = 28 u |
1x7 = 7 u |
(a)
70% =
70100 =
71020% =
20100 =
15Scenario 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 -
15 =
45 The number of blue stickers remains unchanged in both scenarios.
LCM of 7 and 1 = 7
3 u + 321 = 28 u + 196
28 u - 3 u = 321 - 196
25 u = 125
1 u = 125 ÷ 25 = 5
Number of red stickers
= 3 u + 321
= 3 x 5 + 321
= 15 + 321
= 336
(b)
Number of blue stickers
= 7 u
= 7 x 5
= 35
Answer(s): (a) 336; (b) 35