Adam has some blue stickers and red stickers.
If 261 blue stickers are removed, 70% of the stickers will be red stickers.
If 86 blue stickers are removed, 20% of the stickers will be red stickers.
- How many blue stickers are there?
- How many red stickers are there?
|
Scenario 1 |
Scenario 2 |
|
Blue |
Red |
Blue |
Red |
Before |
3 u + 261 |
7 u |
28 u + 86 |
7 u |
Change |
- 261 |
No change |
- 86 |
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 blue in the end
= 1 -
710 =
310 Scenario 2
Fraction of the stickers that are blue in the end
= 1 -
15 =
45 The number of red stickers remains unchanged in both scenarios.
LCM of 7 and 1 = 7
3 u + 261 = 28 u + 86
28 u - 3 u = 261 - 86
25 u = 175
1 u = 175 ÷ 25 = 7
Number of blue stickers
= 3 u + 261
= 3 x 7 + 261
= 21 + 261
= 282
(b)
Number of red stickers
= 7 u
= 7 x 7
= 49
Answer(s): (a) 282; (b) 49