Mary had some black coins and green coins in 2 bags. In Bag B, the ratio of the number of black coins to green coins was 10 : 7. In Bag C, the number of black coins was 4 times the number of green coins. Mary transferred ⁴₇ of the green coins from Bag B to Bag C. The number of coins in Bag B became 182 and the ratio of the number of black coins to green coins in Bag C became 8 : 9.

1. How many green coins were transferred from Bag B to Bag C?
2. What was the number of coins in Bag C after the transfer?

	Bag B		Bag C
	Black	Green	Black	Green
Comparing black coins and green coins at first	10 u	7 u	4x2 = 8 p	1x2 = 2 p
Before		7 u
Change		- 4 u		+ 4 u
After		3 u
Comparing black coins and green coins in the end	10 u	3 u	8 p	9 p

(a)
Total number of coins in the end for Bag B
= 10 u + 3 u
= 13 u

13 u = 182

1 u = 182 ÷ 13 = 14

Number of green coins that were transferred from Bag B to Bag C
= 4 u
= 4 x 14
= 56

(b)
The number of black coins in Bag C remains unchanged.  Make the number of black coins in Bag C the same.  LCM of 4 and 8 is 8. 

Increase in the number of green coins in Bag C
= 9 p - 2 p
= 7 p

7 p = 4 u
7 p = 56

1 p = 56 ÷ 7 = 8

Number of coins in Bag C in the end
= 8 p + 9 p
= 17 p
= 17 x 8
= 136

Answer(s): (a) 56; (b) 136