Pierre had 36 coins. ³₄ were blue and the rest were black and silver. The ratio of the number of black coins to the number of silver coins was 1 : 2. Pierre decided to give away some blue coins to decrease the number of blue coins to half of the total number of coins.

1. How many more blue coins than black coins did he have at first?
2. How many blue coins did he have to give away?

Blue coins	Black coins	Silver coins	Total
3x3	1x3
	1x1	2x1
9 u	1 u	2 u	36

	Blue coins	Black coins	Silver coins
Before	9 u	1 u	2 u
Change	- ?
After	3 u	3 u
Comparing blue, black and silver coins in the end	1	1

(a)
The total number of black coins and silver coins is repeated. Make the total number of black coins and silver coins the same.  LCM of 1 and 3 is 3.

Total number of coins at first
= 9 u + 1 u + 2 u
= 12 u

12 u = 36

1 u = 36 ÷ 12 = 3

Number of more blue coins than black coins at first
= 9 u - 1 u
= 8 u
= 8 x 3
= 24

(b)
Number of more blue coins that Pierre had to give away
= 9 u - 3 u
= 6 u
= 6 x 3
= 18

Answer(s): (a) 24; (b) 18