$5 \frac{1}{3}$ kgm of $15.00 candies #### Explanation: Define $\textcolor{w h i t e}{\text{XX}} m =$weight of$10.50 candies (in kgm)
$\textcolor{w h i t e}{\text{XX}} n =$weight of $15.00 candies (in kgm) We are told [1]$\textcolor{w h i t e}{\text{XX}} m + n = 12$and $\textcolor{w h i t e}{\text{XX}} 10.50 m + 15.00 n = 12.50 \cdot 12$which simplifies to: [2]$\textcolor{w h i t e}{\text{XX}} 7 m + 10 n = 100$Multiplying [1] by 7 [3]$\textcolor{w h i t e}{\text{XX}} 7 m + 7 n = 84$Subtracting [3] from [2] [4]$\textcolor{w h i t e}{\text{XX}} 3 n = 16$Dividing by 3 [5]$\textcolor{w h i t e}{\text{XX}} n = \frac{16}{3} = 5 \frac{1}{3}$Substituting $5 \frac{1}{3}$for $n$in [1] (and simplifying) [6]$\textcolor{w h i t e}{\text{XX}} m = 6 \frac{2}{3}\$