Question

The value of ${{(1.02)}^{4}}+{{(0.98)}^{4}}$ correct to three decimal places is
A-$2.004$
B- $2.005$
C- $2.006$
D-$1.995$

Answer 1

Hint: These types of questions are always calculative without a calculator. It takes a really long time to get an exact result but if we know some formulas then we can easily convert these into much simpler form. Like we can write them in the form of summation of two numbers or in the form of subtraction of two numbers and then we can use the formula of the whole square of sum of two numbers or difference of two numbers.

Complete step-by-step answer:
 $ {{(1.02)}^{4}}+{{(0.98)}^{4}} $
We can write it as
 $ {{(1+0.02)}^{4}}+{{(1-0.02)}^{4}} $
Its implies
 $ {{({{(1+0.02)}^{2}})}^{2}}+{{({{(1-0.02)}^{2}})}^{2}} $
Now we will use formula of
 $ \begin{align}
  & {{(a+b)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab \\
 & {{(a-b)}^{2}}={{a}^{2}}+{{b}^{2}}-2ab \\
\end{align} $
Applying this in above expression
 $ \begin{align}
  & {{(1+0.02)}^{2}}={{1}^{2}}+{{(0.02)}^{2}}+2\times 1\times 0.02 \\
 & {{(1-0.02)}^{2}}={{1}^{2}}+{{(0.02)}^{2}}-2\times 1\times 0.02 \\
\end{align} $
On solving it will be
 $ \begin{align}
  & {{(1+0.02)}^{2}}=1+(0.0004)+0.04 \\
 & {{(1-0.02)}^{2}}=1+(0.0004)-0.04 \\
\end{align} $
 $ \begin{align}
  & {{(1+0.02)}^{2}}=(1.0004)+0.04 \\
 & {{(1-0.02)}^{2}}=(1.0004)-0.04 \\
\end{align} $
Again it will be
 $ {{(1.0004+0.04)}^{2}}+{{(1.0004-0.04)}^{2}} $
Again using same formula
 $ {{(1.0004)}^{2}}+{{(0.04)}^{2}}+2\times 1.0004\times 0.04+{{(1.0004)}^{2}}+{{(0.04)}^{2}}-2\times 1.0004\times 0.04 $
On simplifying
 $ 1.00080016+0.0016+1.00080016+0.0016 $
 $ 2\times 1.00080016+2\times 0.0016 $
 $ 2.00160032+0.0032 $
 $ 2.00480032 $
This is the final answer. Therefore option A is the correct answer.

Note:- As we can see, the solution is too calculative so we have to be careful with the signs and calculation and we should have a good knowledge of how and when we should break the given number in two parts so that it fulfills or needs to apply in formula.