Adam has 22 stickers.
After
23 of the green stickers and
27 of the blue stickers are given away,
there is an equal number of green and blue stickers left.
- How many green stickers are there in the end?
- How many more green stickers are there than blue stickers at first?
- How many blue stickers are given away?
|
Green |
Blue |
Total |
Before |
3 x 5 = 15 u |
7 x 1 = 7 u |
22 |
Change |
- 2 x 5 = - 10 u |
- 2 x 1 = - 2 u |
|
After |
1 x 5 = 5 u |
5 x 1 = 5 u |
|
(a)
Fraction of the green stickers left
= 1 -
23 =
13 Fraction of the blue stickers left
= 1 -
27 =
57 Number of green stickers and blue stickers left is the same.
Make the number of green stickers and blue stickers left the same using the LCM of 1 and 5.
LCM of 1 and 5 = 5
Total number of stickers
= 15 u + 7 u
= 22 u
22 u = 22
1 u = 22 ÷ 22 = 1
Number of green stickers in the end
= 5 u
= 5 x 1
= 5
(b)
Number of more green stickers than blue stickers at first
= 15 u - 7 u
= 8 u
= 8 x 1
= 8
(c)
Number of blue stickers given away
= 7 u - 5 u
= 2 u
= 2 x 1
= 2
Answer(s): (a) 5; (b) 8; (c) 2