Xylia had some grey pens and black pens in 2 bags. In Bag E, the ratio of the number of grey pens to black pens was 9 : 7. In Bag F, the number of grey pens was 2 times the number of black pens. Xylia transferred ⁴₇ of the black pens from Bag E to Bag F. The number of pens in Bag E became 108 and the ratio of the number of grey pens to black pens in Bag F became 4 : 5.

1. How many black pens were transferred from Bag E to Bag F?
2. What was the number of pens in Bag F after the transfer?

	Bag E		Bag F
	Grey	Black	Grey	Black
Comparing grey pens and black pens at first	9 u	7 u	2x2 = 4 p	1x2 = 2 p
Before		7 u
Change		- 4 u		+ 4 u
After		3 u
Comparing grey pens and black pens in the end	9 u	3 u	4 p	5 p

(a)
Total number of pens in the end for Bag E
= 9 u + 3 u
= 12 u

12 u = 108

1 u = 108 ÷ 12 = 9

Number of black pens that were transferred from Bag E to Bag F
= 4 u
= 4 x 9
= 36

(b)
The number of grey pens in Bag F remains unchanged.  Make the number of grey pens in Bag F the same.  LCM of 2 and 4 is 4. 

Increase in the number of black pens in Bag F
= 5 p - 2 p
= 3 p

3 p = 4 u
3 p = 36

1 p = 36 ÷ 3 = 12

Number of pens in Bag F in the end
= 4 p + 5 p
= 9 p
= 9 x 12
= 108

Answer(s): (a) 36; (b) 108