A packet contained gold, white and blue pens.
35 of the pens were gold and
79 of the remainder were white pens.The rest were blue pens.
- Find the ratio of the number of blue pens to the number of gold pens.
- There were 143 fewer white pens than gold pens. How many pens were there in the packet?
| Gold pens |
White pens |
Blue pens |
| 3x9 |
2x9 |
| |
7x2 |
2x2 |
| 27 u |
14 u |
4 u |
(a)
Fraction of gold pens =
35 Fraction of pens that are not gold = 1 -
35 =
25Remaining fraction of white pens =
79Remaining fraction of blue pens = 1 -
79 =
29The total number of white pens and blue pens is the repeated identity.
Make the repeated identity the same by using LCM.
LCM of 2 and 9 = 18
Blue pens : Gold pens
4 : 27
(b)
Difference in the number of gold pens and white pens
= 27 u - 14 u
= 13 u
13 u = 143
1 u = 143 ÷ 13 = 11
Total number of pens
= 27 u + 14 u + 4 u
= 45 u
= 45 x 11
= 495
Answer(s): (a) 4 : 27; (b) 495
