Marion had 105 pens.
37 were red and the rest were yellow and gold. The ratio of the number of yellow pens to the number of gold pens was 1 : 2. Marion decided to buy some more red pens to increase the number of red pens to half of the total number of pens.
- How many less yellow pens than red pens did he have at first?
- How many more red pens did he have to buy?
| Red pens |
Yellow pens |
Gold pens |
Total |
| 3x3 |
4x3 |
|
| |
1x4 |
2x4 |
|
| 9 u |
4 u |
8 u |
105 |
| |
Red pens |
Yellow pens |
Gold pens |
| Before |
9 u |
4 u |
8 u |
| Change |
+ ? |
|
|
| After |
12 u |
12 u |
Comparing red, yellow
and gold pens
in the end |
1 |
1 |
(a)
The total number of yellow pens and gold pens is repeated. Make the total number of yellow pens and gold pens the same. LCM of 4 and 3 is 12.
Total number of pens at first
= 9 u + 4 u + 8 u
= 21 u
21 u = 105
1 u = 105 ÷ 21 = 5
Number of less yellow pens than red pens at first
= 9 u - 4 u
= 5 u
= 5 x 5
= 25
(b)
Number of more red pens that Marion had to buy
= 12 u - 9 u
= 3 u
= 3 x 5
= 15
Answer(s): (a) 25; (b) 15
