The price $\left(p\right)$ of a commodity is first increased by $k\%$; then decreased by $k\%$; again increased by $k\%$; and again decreased by $k\%$. If the new price is $q$, then what is the relation between $p$ and $q$?

The correct answer is Option 1:
$p{({10}^4–k^2)}^2=q\times {10}^8$

Step-by-step Explanation:

1. Understanding Percentage Increase and Decrease Multipliers:
When an initial price is increased by $k\%$, the original price is multiplied by the factor:
$(1+\frac{k}{100})$

When the price is decreased by $k\%$, the current price is multiplied by the factor:
$(1-\frac{k}{100})$

2. Sequential Changes Applied to the Initial Price $p$:
The price $p$ undergoes four consecutive changes:
1. Increased by $k\%$
2. Decreased by $k\%$
3. Increased by $k\%$
4. Decreased by $k\%$

Therefore, the new price $q$ can be expressed as:
$q=p\times (1+\frac{k}{100})\times (1-\frac{k}{100})\times (1+\frac{k}{100})\times (1-\frac{k}{100})$

3. Combining Terms:
We can group the paired terms using the algebraic identity $(a+b)(a-b)=a^2-b^2$:
$(1+\frac{k}{100})(1-\frac{k}{100})=(1-\frac{k^2}{{100}^2})=(1-\frac{k^2}{{10}^4})$

Since this paired operation happens twice, the equation for $q$ simplifies to:
$q=p{(1-\frac{k^2}{{10}^4})}^2$

4. Simplifying the Fractional Expression:
Taking the common denominator ${10}^4$ inside the parentheses:
$q=p{\left(\frac{{10}^4–k^2}{{10}^4}\right)}^2$

Squaring the numerator and denominator separately:
$q=p\times \frac{{({10}^4–k^2)}^2}{{\left({10}^4\right)}^2}$

Since ${\left({10}^4\right)}^2={10}^8$, we have:
$q=\frac{p{({10}^4–k^2)}^2}{{10}^8}$

Multiplying both sides by ${10}^8$ gives the final relation:
$p{({10}^4–k^2)}^2=q\times {10}^8$