Solve: (27-5×3)÷(6×2) + (15×4)÷3 | Order of Operations Challenge

According to the order of arithmetic operations, first we place the multiplication exercises within parentheses:

$\frac{27-(5\cdot3)}{(6\cdot2)}+\frac{(15\cdot4)}{3}=$

We then solve the exercises within parentheses:

$5\times3=15$

$6\times2=12$

$15\times4=60$

Now we obtain the exercise:

$\frac{27-15}{12}+\frac{60}{3}=$

We solve the numerator of the fraction:

$27-15=12$

We obtain:

$\frac{12}{12}+\frac{60}{3}=$

We solve the fractions:

$\frac{12}{12}=1$

$60:3=20$

Finally we obtain the exercise:

$1+20=21$