The average of sixteen numbers is 48. The average of the first six of these numbers is 45 and that of the last seven numbers is 53. The seventh and the eighth numbers are, respectively, 3 and 7 greater than the ninth number. What is the average of the ninth and seventh numbers?

Solution(By Examveda Team)

Sum of 16 numbers = 16 × 48 = 768
Sum of first 6 numbers = 6 × 45 = 270
Sum of last 7 numbers = 7 × 53 = 371
\[\begin{array}{*{20}{c}} {{7^{{\text{th}}}}}&{{8^{{\text{th}}}}}&{\,\,\,{9^{{\text{th}}}}} \\ {\left( {x + 3} \right)}&{\left( {x + 7} \right)}&{\left( x \right)} \end{array}\]
$$\eqalign{ & \Rightarrow 3x = 127 - 10 \cr & \Rightarrow x = 39 \cr & \therefore {\text{Average of}}\,{{\text{9}}^{{\text{th}}}}\,{\text{and}}\,{{\text{7}}^{{\text{th}}}}\,{\text{number}} \cr & = \frac{{x + \left( {x + 3} \right)}}{2} \cr & = \frac{{2x + 3}}{2} \cr & = x + 10.5 \cr & = 39 + 1.5 \cr & = 40.5 \cr} $$