Understanding Successive Percentage Price Changes

This problem involves a commodity's price undergoing a series of percentage changes: an increase, followed by a decrease, another increase, and finally another decrease, all by the same percentage k%. We need to find the relationship between the initial price (p) and the final price (q) after these four changes.

Calculating Price After Each Change

Let the initial price be $p$.

When a price is increased by k%, the new price is the original price plus k% of the original price. Mathematically, this is $p + p \times \frac{k}{100} = p \left(1 + \frac{k}{100}\right)$.

When a price is decreased by k%, the new price is the original price minus k% of the original price. Mathematically, this is $p - p \times \frac{k}{100} = p \left(1 - \frac{k}{100}\right)$.

Let's track the price after each step:

1. After the first increase of k%: The price becomes $p_1 = p \left(1 + \frac{k}{100}\right)$.
2. After the first decrease of k%: The price $p_1$ is decreased by k%. The new price is $p_2 = p_1 \left(1 - \frac{k}{100}\right) = p \left(1 + \frac{k}{100}\right) \left(1 - \frac{k}{100}\right)$.
3. After the second increase of k%: The price $p_2$ is increased by k%. The new price is $p_3 = p_2 \left(1 + \frac{k}{100}\right) = p \left(1 + \frac{k}{100}\right) \left(1 - \frac{k}{100}\right) \left(1 + \frac{k}{100}\right)$.
4. After the second decrease of k%: The price $p_3$ is decreased by k%. The new price is $p_4 = p_3 \left(1 - \frac{k}{100}\right) = p \left(1 + \frac{k}{100}\right) \left(1 - \frac{k}{100}\right) \left(1 + \frac{k}{100}\right) \left(1 - \frac{k}{100}\right)$.

The final price $q$ is $p_4$. So, we have:

\(q = p \left(1 + \frac{k}{100}\right) \left(1 - \frac{k}{100}\right) \left(1 + \frac{k}{100}\right) \left(1 - \frac{k}{100}\right)\)

Simplifying the Relation Between p and q

We can group the terms:

\(q = p \left[\left(1 + \frac{k}{100}\right) \left(1 - \frac{k}{100}\right)\right] \left[\left(1 + \frac{k}{100}\right) \left(1 - \frac{k}{100}\right)\right]\)

Using the algebraic identity \((a+b)(a-b) = a^2 - b^2\), let $a=1$ and $b=\frac{k}{100}$.

\(1 + \frac{k}{100}\) and \(1 - \frac{k}{100}\) terms become \(1^2 - \left(\frac{k}{100}\right)^2 = 1 - \frac{k^2}{10000}\).

So, the equation simplifies to:

\(q = p \left[1 - \frac{k^2}{10000}\right] \left[1 - \frac{k^2}{10000}\right]\)

\(q = p \left[1 - \frac{k^2}{10000}\right]^2\)

To match the format of the options, let's find a common denominator inside the bracket:

\(1 - \frac{k^2}{10000} = \frac{10000}{10000} - \frac{k^2}{10000} = \frac{10000 - k^2}{10000}\)

Substitute this back into the equation for q:

\(q = p \left[\frac{10000 - k^2}{10000}\right]^2\)

Using the rule \(\left(\frac{a}{b}\right)^2 = \frac{a^2}{b^2}\):

\(q = p \frac{(10000 - k^2)^2}{10000^2}\)

We know that \(10000 = 10^4\), so \(10000^2 = (10^4)^2 = 10^8\).

So, the equation becomes:

\(q = p \frac{(10^4 - k^2)^2}{10^8}\)

To isolate p and q, we can multiply both sides by \(10^8\):

\(q \times 10^8 = p (10^4 - k^2)^2\)

Or, rearranging to match the options format:

\(p (10^4 - k^2)^2 = q \times 10^8\)

Let's compare this derived relationship with the given options to find the correct one.

Comparison with Options

We found the relation: \(p(10^4 - k^2)^2 = q \times 10^8\)

The derived relationship matches Option 1 exactly.

Revision Table: Key Concepts

Additional Information: Successive Percentage Change Formula

When an initial value is changed by +a% and then by -b%, the net percentage change is given by \(\left(a - b - \frac{ab}{100}\right)\)%. This problem involves k% increase and k% decrease happening twice.

For the first increase (+k%) and first decrease (-k%), the net change on p results in a price \(p \left(1 + \frac{k}{100}\right) \left(1 - \frac{k}{100}\right) = p \left(1 - \frac{k^2}{10000}\right)\). This is a net decrease because \(\frac{k^2}{10000}\) is positive (assuming \(k \neq 0\)). The net percentage change is \(-\frac{k^2}{100}\)%. The price becomes \(p \left(1 - \frac{k^2}{100} \times \frac{1}{100}\right)\) which is \(p \left(1 - \frac{k^2}{10000}\right)\).

The question involves this sequence of (+k%, -k%) applied twice. So, the price after the first pair of changes is \(p' = p \left(1 - \frac{k^2}{10000}\right)\). The second pair of changes (+k%, -k%) is applied to this new price \(p'\). The final price \(q\) will be \(q = p' \left(1 - \frac{k^2}{10000}\right)\).

Substituting \(p'\):

\(q = p \left(1 - \frac{k^2}{10000}\right) \left(1 - \frac{k^2}{10000}\right) = p \left(1 - \frac{k^2}{10000}\right)^2\)

This confirms the result obtained through step-by-step calculation and further simplification:

\(q = p \left(\frac{10000 - k^2}{10000}\right)^2 = p \frac{(10000 - k^2)^2}{10000^2} = p \frac{(10^4 - k^2)^2}{10^8}\)

\(q \times 10^8 = p (10^4 - k^2)^2\)

Or, \(p (10^4 - k^2)^2 = q \times 10^8\).

Understanding successive percentage changes is crucial for solving problems involving repeated increases and decreases applied to a value.