Adam has some blue stickers and red stickers.
If 371 blue stickers are removed, 90% of the stickers will be red stickers.
If 111 blue stickers are removed, 30% 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 |
1 u + 371 |
9 u |
21 u + 111 |
9 u |
Change |
- 371 |
No change |
- 111 |
No change |
After |
1x1 = 1 u |
9x1 = 9 u |
7x3 = 21 u |
3x3 = 9 u |
(a)
90% =
90100 =
91030% =
30100 =
310Scenario 1 Fraction of the stickers that are blue in the end
= 1 -
910 =
110 Scenario 2
Fraction of the stickers that are blue in the end
= 1 -
310 =
710 The number of red stickers remains unchanged in both scenarios.
LCM of 9 and 3 = 9
1 u + 371 = 21 u + 111
21 u - 1 u = 371 - 111
20 u = 260
1 u = 260 ÷ 20 = 13
Number of blue stickers
= 1 u + 371
= 1 x 13 + 371
= 13 + 371
= 384
(b)
Number of red stickers
= 9 u
= 9 x 13
= 117
Answer(s): (a) 384; (b) 117