Question

A batsman scores ${\text{80}}$ runs in his sixth innings and thus increases his average by ${\text{5}}$. What is his average after six innings?
$\left( {\text{A}} \right){\text{ 50}}$
$\left( {\text{B}} \right){\text{ 55}}$
$\left( {\text{C}} \right){\text{ 60}}$
$\left( {\text{D}} \right){\text{ 65}}$

Answer 1

Here given that sixth innings score only.
We have to find the average value of after six innings.
First of all we have to find the average of five innings then add it by$5$.
Finally we get the required answer.

Formula used: Average ${\text{ = }}\dfrac{{{\text{sum of all data values}}}}{{{\text{number of values}}}}$

Complete step-by-step solution:
Let the first, second, third, fourth and fifth innings be ${{\text{I}}_{\text{1}}}{\text{, }}{{\text{I}}_{\text{2}}}{\text{, }}{{\text{I}}_{\text{3}}}{\text{, }}{{\text{I}}_{\text{4}}}{\text{ and }}{{\text{I}}_{\text{5}}}$ respectively.
First we have to use the average formula for the first five innings.
Average ${\text{ = }}\dfrac{{{\text{sum of all values}}}}{{{\text{number of values}}}}$
Here the average of the first five innings is not given so take it as by ${\text{X}}$.
That is, average of five innings {\text{ = }}\dfrac{{{\text{ }}{{\text{I}}_{\text{1}}}{\text{ + }}{{\text{I}}_{\text{2}}}{\text{ + }}{{\text{I}}_{\text{3}}}{\text{ + }}{{\text{I}}_{\text{4}}}{\text{ + }}{{\text{I}}_{\text{5}}}{\text{ }}}}{{\text{5}}}$$$${\text{ = X}}
$\Rightarrow {\text{ }}{{\text{I}}_{\text{1}}}{\text{ + }}{{\text{I}}_{\text{2}}}{\text{ + }}{{\text{I}}_{\text{3}}}{\text{ + }}{{\text{I}}_{\text{4}}}{\text{ + }}{{\text{I}}_{{\text{5 }}}}{\text{ = 5X}}$
Let the sixth innings be ${{\text{I}}_{\text{6}}}$
Given that batsman scores ${\text{80}}$ runs in his sixth innings
That is ${{\text{I}}_{\text{6}}}$ ${\text{ = 80}}$
Also given that average increases by ${\text{5}}$.
That is we add it to the average by ${\text{5}}$
Thus the new average becomes,
New average = Average after six innings {\text{ = }}\dfrac{{{\text{ }}{{\text{I}}_{\text{1}}}{\text{ + }}{{\text{I}}_{\text{2}}}{\text{ + }}{{\text{I}}_{\text{3}}}{\text{ + }}{{\text{I}}_{\text{4}}}{\text{ + }}{{\text{I}}_{\text{5}}}{\text{ + }}{{\text{I}}_{\text{6}}}{\text{ }}}}{{\text{6}}}$$${\text{ = X + 5}}$ Do cross multiply we get, \Rightarrow {\text{ }}{{\text{I}}{\text{1}}}{\text{ + }}{{\text{I}}{\text{2}}}{\text{ + }}{{\text{I}}{\text{3}}}{\text{ + }}{{\text{I}}{\text{4}}}{\text{ + }}{{\text{I}}{\text{5}}}{\text{ + }}{{\text{I}}{\text{6}}}{\text{ = 6}}\left( {{\text{X + 5}}} \right)$On simplifying we get,$ \Rightarrow {\text{ }}{{\text{I}}{\text{1}}}{\text{ + }}{{\text{I}}{\text{2}}}{\text{ + }}{{\text{I}}{\text{3}}}{\text{ + }}{{\text{I}}{\text{4}}}{\text{ + }}{{\text{I}}{\text{5}}}{\text{ + }}{{\text{I}}{\text{6}}}{\text{ = 6X + 30}} Since ${{\text{I}}_{\text{1}}}{\text{ + }}{{\text{I}}_{\text{2}}}{\text{ + }}{{\text{I}}_{\text{3}}}{\text{ + }}{{\text{I}}_{\text{4}}}{\text{ + }}{{\text{I}}_{{\text{5 }}}}{\text{ = 5X}}$ \Rightarrow {\text{ 5X + }}{{\text{I}}{\text{6}}}{\text{ = 6X + 30}}$On subtracting${\text{5X}}$on both sides and we get$ \Rightarrow {\text{ }}{{\text{I}}{\text{6}}}{\text{ = 6X - 5X + 30}}$On subtracting we get,${{\text{I}}{\text{6}}}{\text{ = X + 30}}$Given that${{\text{I}}{\text{6}}}{\text{ = 80}}. Substitute this value in the above equation we get, $ \Rightarrow {\text{ 80 = X + 30}}$ On subtracting 30on both sides, we get $ \Rightarrow {\text{ X = 80 - 30}}$ On subtracting we get, $ \Rightarrow {\text{ X = 50}}$ Now we got the average of five innings worth. But we have to find the average of six innings. That is we find the value of{\text{X + 5}}$.$ \Rightarrow $Average of six innings${\text{ = X + 5}}$${\text{ = 50 + 5}}$On adding we get, Average of after six innings${\text{ = 55}}$$.

Hence the correct option is $\left( {\text{B}} \right)$

Note: An average is also known as arithmetic mean.
Arithmetic mean is the central tendency of the given set of data observations.
We note that the average of the given data is less than the greatest observation and greater than the smallest observation of the given data.
The average of $4,8,10,14$ is $9$.
That is Average ${\text{ = }}\dfrac{{{\text{4 + 8 + 10 + 14}}}}{{\text{4}}}$
Adding we get,
$\Rightarrow {\text{ }}\dfrac{{36}}{4}$
Dividing we get,
$\Rightarrow {\text{ 9}}$.
Here, the average is $9$ which is less than the greatest observation $\left( {{\text{14}}} \right)$ and greater than the smallest observation $\left( {\text{4}} \right)$.