Question

The average of 13 results is 60. If the average of the first 7 results is 59 and that of the last seven results is 61, what will be the seventh result?
(a) 90
(b) 50
(c) 75
(d) 60
(e) None of these

Answer 1

Apply the formula to calculate mean of 13 students given as: - Mean =$\dfrac{{{x}_{1}}+{{x}_{2}}+{{x}_{3}}+....+{{x}_{13}}}{13}$, where ${{x}_{1}},{{x}_{2}},{{x}_{3}},....,{{x}_{13}}$ are the marks of 13 students. Substitute the given value of mean and find the sum of marks of these 13 students. Now, apply the same mean formula to calculate the sum of first 7 results and last 7 results given as: - $\left( {{x}_{1}}+{{x}_{2}}+....+{{x}_{7}} \right)$ and $\left( {{x}_{7}}+{{x}_{8}}+....+{{x}_{13}} \right)$ respectively. Take the sum of first 7 and last 7 results and subtract the sum of 13 results from it, to get the answer.

Complete step-by-step solution
Here, we have been provided with the information that an average of 13 results is 60. So, let us assume these results as ${{x}_{1}},{{x}_{2}},{{x}_{3}},....,{{x}_{13}}$.
Now, we know that average or mean of ‘n’ observations is given as: - Mean = $\dfrac{\left( \sum\limits_{i=1}^{n}{{{x}_{i}}} \right)}{n}$. Here, $\sum\limits_{i=1}^{n}{{{x}_{i}}}$ is the summation of all the observations and ‘n’ is the number of observations.
So, applying the above formula for n = 13, we get,

& \Rightarrow 60=\dfrac{\sum\limits_{i=1}^{13}{{{x}_{i}}}}{13} \\\ & \Rightarrow \sum\limits_{i=1}^{n=13}{}=13\times 60 \\\ \end{aligned}$$ $$\Rightarrow \left( {{x}_{1}}+{{x}_{2}}+{{x}_{3}}+....+{{x}_{13}} \right)=13\times 60$$ - (1) Now, it is given that the average of the first 7 results and the last 7 results are 59 and 61 respectively. So, we have, 1\. For first seven results: - The first seven results are $${{x}_{1}},{{x}_{2}},{{x}_{3}},....,{{x}_{7}}$$. So, applying the formula to calculate mean for n = 1 to 7, we get, $$\Rightarrow $$Mean = $$\dfrac{\sum\limits_{i=1}^{7}{{{x}_{i}}}}{7}$$ $$\Rightarrow 59=\dfrac{{{x}_{1}}+{{x}_{2}}+{{x}_{3}}+....+{{x}_{7}}}{7}$$ $$\Rightarrow \left( {{x}_{1}}+{{x}_{2}}+{{x}_{3}}+....+{{x}_{7}} \right)=59\times 7$$ - (2) 2\. For last seven results: - The last seven results are $${{x}_{7}},{{x}_{8}},....,{{x}_{13}}$$. So, applying the formula to calculate mean for n = 7 to 13, we get, $$\Rightarrow $$Mean = $$\dfrac{\sum\limits_{i=7}^{13}{{{x}_{i}}}}{7}$$ $$\Rightarrow 61=\dfrac{{{x}_{7}}+{{x}_{8}}+....+{{x}_{13}}}{7}$$ $$\Rightarrow \left( {{x}_{7}}+{{x}_{8}}+....+{{x}_{13}} \right)=61\times 7$$ - (3) Adding equations (2) and (3), we get, $$\Rightarrow \left( {{x}_{1}}+{{x}_{2}}+....+{{x}_{7}} \right)+\left( {{x}_{7}}+{{x}_{8}}+....+{{x}_{13}} \right)=\left( 59\times 7 \right)+\left( 61\times 7 \right)$$ Clearly, we can see that he result $${{x}_{7}}$$ is common in both the groups on the left hand side. Therefore, separating one $${{x}_{7}}$$, we get, $$\begin{aligned} & \Rightarrow \left( {{x}_{1}}+{{x}_{2}}+{{x}_{3}}+....+{{x}_{13}} \right)+{{x}_{7}}=\left( 59+61 \right)\times 7 \\\ & \Rightarrow \left( {{x}_{1}}+{{x}_{2}}+{{x}_{3}}+....+{{x}_{13}} \right)+{{x}_{7}}=7\times 120 \\\ \end{aligned}$$ Ow, subtracting equation (1) from the above expression, we get, $$\begin{aligned} & \Rightarrow \left( {{x}_{1}}+{{x}_{2}}+{{x}_{3}}+....+{{x}_{13}} \right)+{{x}_{7}}-\left( {{x}_{1}}+{{x}_{2}}+....+{{x}_{13}} \right)=120\times 7-13\times 60 \\\ & \Rightarrow {{x}_{7}}=60\times 14-16\times 60 \\\ & \Rightarrow {{x}_{7}}=60\left( 14-13 \right) \\\ & \Rightarrow {{x}_{7}}=60 \\\ \end{aligned}$$ **Hence, option (d) is the correct answer.** **Note:** One may note that it is important to find the sum of the given results because we cannot find the answer by directly adding or subtracting the given values of the mean. You may see that from n = 7 to 13 there are 7 results, i.e. $${{x}_{7}},{{x}_{8}},{{x}_{9}},....,{{x}_{13}}$$ and that is why we have substituted n = 7 in the formula to calculate mean.