Question

Evaluate the expression $mn\left( {{m}^{2}}+{{n}^{2}} \right)$ when m = 2 and n = 0.1
[a] 0.800
[b] 0.801
[c] 0.802
[c] 0.803

Answer 1

Hint: Evaluate each of the terms ${{m}^{2}},{{n}^{2}}$ when m = 2 and n = 0.1 and hence find the value of ${{m}^{2}}+{{n}^{2}}$. Determine the value of $mn$ when m = 2 and n = 0.1 and hence determine the value of $mn\left( {{m}^{2}}+{{n}^{2}} \right)$ when m = 2 and n = 0.1

Complete step by step answer:
Before solving the question, we need to understand how to evaluate an algebraic expression at particular values.
Suppose we have the expression ${{x}^{2}}$ and we want to evaluate it at 2. We simply replace x by 2 and evaluate the resulting expression. Hence ${{x}^{2}}$ evaluates to ${{2}^{2}}=4$ at x = 2.
When two variables are written together with no sign in between them then during evaluation we insert a multiplication sign in between them. Suppose we have to evaluate ${{x}^{2}}y$ when x = 2 and y = 3. We simply replace x by 2 and y by 3 and insert the multiplication sign between them. Hence ${{x}^{2}}y$ evaluates to ${{2}^{2}}\times 3=12$ when x = 2 and y = 3.
Hence we solve the question as follows:
When m = 2 then ${{m}^{2}}={{2}^{2}}=4$
When n = 0.1, we have ${{n}^{2}}={{0.1}^{2}}=0.01$
Hence, we have when m = 2 and n = 0.1, then ${{m}^{2}}+{{n}^{2}}=4+0.01=4.01$
When m =2 and n = 0.1, then $mn=2\times 0.1=0.2$
Hence, we have when m = 2 and n= 0.1, then $mn\left( {{m}^{2}}+{{n}^{2}} \right)=0.2\times 4.01=0.802$
Hence, when m = 2 and n = 0.1, then $mn\left( {{m}^{2}}+{{n}^{2}} \right)=0.802$

So, the correct answer is “Option C”.

Note: [1] Alternative solution:
We know that $mn\left( {{m}^{2}}+{{n}^{2}} \right)=mn\times {{m}^{2}}+mn\times {{n}^{2}}={{m}^{3}}n+m{{n}^{3}}$
Hence, when m = 2 and n = 0.1, we have ${{m}^{3}}n={{2}^{3}}\times 0.1=0.8$ and $m{{n}^{3}}=2\times {{0.1}^{3}}=0.002$
Hence, we have $mn\left( {{m}^{2}}+{{n}^{2}} \right)=m{{n}^{3}}+{{m}^{3}}n=0.8+0.002=0.802$, which is the same as obtained above.
Hence option [c] is correct.