Question

If ${\alpha }_1,{\alpha }_2,{\alpha }_3,......,{\alpha }_n$ are n-values of a variable $\alpha$ such that ${\sum }_{i=1}^n{u}_i=155$ and ${\sum }_{i=1}^n{v}_i=225$, where $u_i=2{\alpha }_i-3$ and $v_i=3{\alpha }_i–5$, for i = 1, 2, 3, ..., n, then the value of $\left(\frac{{\sum }_{i=1}^n{u}_i+{\sum }_{i=1}^n{v}_i}{n}\right)$ is

• A. $25.\overline{3}$
• B. $35.\overline{3}$
• C. $27.\overline{9}$
• D. 25

Answer 1

The correct option is A $25.\overline{3}$

Given: ${\sum }_{i=1}^n{u}_i=155$ and ${\sum }_{i=1}^n{v}_i=225$
where $u_i=2{\alpha }_i-3,v_i=3{\alpha }_i–5$; i = 1, 2, ..., n,
Now, ${\sum }_{i=1}^n{u}_i+{\sum }_{i=1}^n(2{\alpha }_i-3)$
$\Rightarrow 155=2{\sum }_{i=1}^n{\alpha }_i-3n$
$\Rightarrow 155=2({\alpha }_1+{\alpha }_2+....+{\alpha }_n)-3n$
$\Rightarrow {\alpha }_1+{\alpha }_2+....+{\alpha }_n=\frac{155+3n}{2}$............(i)
Similarly, ${\sum }_{i=1}^n{v}_i={\sum }_{i=1}^n(3{\alpha }_i-5)$
$\Rightarrow 225=3{\sum }_{i=1}^n{\alpha }_i-5n$
$\Rightarrow 225=3({\alpha }_1+{\alpha }_2+....+{\alpha }_n)-5n$
$\Rightarrow {\alpha }_1+{\alpha }_2+....+{\alpha }_n=\frac{225+5n}{3}$............(ii)
Equating (i) and (ii), we get
$\frac{155+3n}{2}=\frac{225+5n}{3}$
$\Rightarrow 465+9n=450+10n$
$\Rightarrow 15=n$
Putting all the values in $\left(\frac{{\sum }_{i=1}^n{u}_i+{\sum }_{i=1}^n{v}_i}{n}\right)$, we get
$\frac{{\sum }_{i=1}^n{u}_i+{\sum }_{i=1}^n{v}_i}{n}=\frac{155+225}{15}$
$=\frac{380}{15}$
$=\frac{76}{3}$
= 25.33....
$=25.\overline{3}$
Hence, the correct answer is option (a).