Question : If the average of $s$ numbers is $r^4$ and the average of $r$ numbers is $s^4$, then find the average of all $r+s$ numbers.
Option 1: $r s\left(r^2+s^2-r s\right)$
Option 2: $rs$
Option 3: ${r}^2+s^2$
Option 4: $rs\left(r^2+s^2\right)$
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Correct Answer: $r s\left(r^2+s^2-r s\right)$
Solution : Given, the average of $s$ numbers is $r^4$ and the average of $r$ numbers is $s^4$ Total sum of $s$ numbers = $s\times r^4=sr^4$ Total sum of $r$ numbers = $r\times s^4= rs^4$ Now, the average of all $r+s$ numbers = $\frac{(rs^4+sr^4)}{ (r+s)}$ = $\frac{rs(s^3+r^3)}{ (r+s)}$ = $\frac{rs(s+r)(s^2-sr+r^2)}{ (r+s)}$ = $rs(r^2+s^2-rs)$ Hence, the correct answer is $rs(r^2+s^2-rs)$.
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Question : Let $p, q, r$ and $s$ be positive natural numbers having three exact factors including 1 and the number itself. If $q>p$ and both are two-digit numbers, and $r>s$ and both are one-digit numbers, then the value of the expression $\frac{p-q-1}{r-s}$ is:
Question : If $\frac{1}{x^2+a^2}=x^2-a^2$, then the value of $x$ is:
Question : If $x_{1}x_{2}x_{3}=4(4+x_{1}+x_{2}+x_{3})$, then what is the value of $\left [ \frac{1}{(2+x_{1})} \right ]+\left [ \frac{1}{(2+x_{2})} \right ]+\left [ \frac{1}{(2+x_{3})} \right ]?$
Question : If $\left(y+\frac{1}{y}\right)=4$, then find the value of $\left(y^6+\frac{1}{y^6}\right)$.
Question : Find the sum of $\left (1-\frac{1}{n+1} \right) + \left (1-\frac{2}{n+1} \right) + \left (1- \frac{3}{n+1} \right)+.....\left ( 1- \frac{n}{n+1} \right)$.
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