Ken had 70 pens.
47 were green and the rest were red and black. The ratio of the number of red pens to the number of black pens was 1 : 5. Ken decided to give away some green pens to decrease the number of green pens to half of the total number of pens.
- How many more green pens than red pens did he have at first?
- How many green pens did he have to give away?
| Green pens |
Red pens |
Black pens |
Total |
| 4x2 |
3x2 |
|
| |
1x1 |
5x1 |
|
| 8 u |
1 u |
5 u |
70 |
| |
Green pens |
Red pens |
Black pens |
| Before |
8 u |
1 u |
5 u |
| Change |
- ? |
|
|
| After |
6 u |
6 u |
Comparing green, red
and black pens
in the end |
1 |
1 |
(a)
The total number of red pens and black pens is repeated. Make the total number of red pens and black pens the same. LCM of 3 and 6 is 6.
Total number of pens at first
= 8 u + 1 u + 5 u
= 14 u
14 u = 70
1 u = 70 ÷ 14 = 5
Number of more green pens than red pens at first
= 8 u - 1 u
= 7 u
= 7 x 5
= 35
(b)
Number of more green pens that Ken had to give away
= 8 u - 6 u
= 2 u
= 2 x 5
= 10
Answer(s): (a) 35; (b) 10
