There were 72 silver, green and blue pens in a box. 12 blue pens were removed from the box and the ratio of the number of silver pens to the number of green pens to the number of blue pens became 3 : 4 : 5. Then some black pens were added and the ratio of the number of blue pens to the number of black pens was 5 : 6.

1. What was the ratio of the number of silver pens to the number of green pens to the number of blue pens at first?
2. How many pens were in the box in the end?

	Silver	Green	Blue	Black	Total
Before	3 u	4 u	5 u + 12	0	72
Change 1	No change	No change	- 12		- 12
After 1	3 u	4 u	5 u		60
Change 2	No change	No change	No change	+ 6 u
After 2	3 u	4 u	5 u	6 u

(a)
Total number of pens at first
= 3 u + 4 u + 5 u + 12
= 12 u + 12

12 u + 12 = 72
12 u = 72 - 12
12 u = 60

1 u = 60 ÷ 12 = 5

Number of silver pens at first
= 3 u
= 3 x 5
= 15

Number of green pens at first
= 4 u
= 4 x 5
= 20

Number of blue pens at first
= 5 u + 12
= 5 x 5 + 12
= 37 

At first
Silver pens : Green pens : Blue pens
15 : 20 : 37

(b)
Total number of pens in the box in the end
= 3 u + 4 u + 5 u + 6 u
= 18 u
= 90

Answer(s): (a) 15 : 20 : 37; (b) 90