The number of chickens to the number of ducks on a farm was 6:5. After 63 ducks were sold, there were 3 times as many chickens as duks left.
a) How many chicken were there on the farm?
b) How many chickens and ducks were there altogetheron the farm in the end?
+118696 The number of chickens to the number of ducks on a farm was 6:5. After 63 ducks were sold, there were 3 times as many chickens as ducks left.
I found this question a bit confusing too 
We are told two things so we will need to set up 2 equations and solve them simultaneously.
65=CD→C=6D5(1)31=CD−63→C=3(D−63)(2)
so
\begin{array}{rll} \frac{6D}{5}&=&3(D-63)}\\\\ \frac{2D}{5}&=&1(D-63)}\\\\ 2D&=&5(D-63)}\\\\ 2D&=&5D-315\\\\ 315&=&3D\\\\ D&=&105 \qquad \mbox{There were originally 105 ducks}\\\\\\ C&=&6*105\div 5\\\\ C&=&126 \qquad \mbox{There were originally 126 chickens}\\\\\\ \end{array}
105+126−63=168
In the end there were 168 birds.
I did check that all these numbers make the original described situation true and you should do that too :)
+118696 The number of chickens to the number of ducks on a farm was 6:5. After 63 ducks were sold, there were 3 times as many chickens as ducks left.
I found this question a bit confusing too
We are told two things so we will need to set up 2 equations and solve them simultaneously.
65=CD→C=6D5(1)31=CD−63→C=3(D−63)(2)
so
\begin{array}{rll} \frac{6D}{5}&=&3(D-63)}\\\\ \frac{2D}{5}&=&1(D-63)}\\\\ 2D&=&5(D-63)}\\\\ 2D&=&5D-315\\\\ 315&=&3D\\\\ D&=&105 \qquad \mbox{There were originally 105 ducks}\\\\\\ C&=&6*105\div 5\\\\ C&=&126 \qquad \mbox{There were originally 126 chickens}\\\\\\ \end{array}
105+126−63=168
In the end there were 168 birds.
I did check that all these numbers make the original described situation true and you should do that too :)
