The price (p) 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?
p(10⁴ - k²)² = q × 10⁸
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:
- After the first increase of k%: The price becomes $p_1 = p \left(1 + \frac{k}{100}\right)$.
- After the first decrease of k%: The price $p_1$ is decreased by k%. The new price is $p_2 = p_1 \left(1 - \frac{k}{100}\right) = p \left(1 + \frac{k}{100}\right) \left(1 - \frac{k}{100}\right)$.
- After the second increase of k%: The price $p_2$ is increased by k%. The new price is $p_3 = p_2 \left(1 + \frac{k}{100}\right) = p \left(1 + \frac{k}{100}\right) \left(1 - \frac{k}{100}\right) \left(1 + \frac{k}{100}\right)$.
- 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: \(p(10^4 - k^2)^2 = q \times 10^8\)
| Option | Relation | Matches Derived Relation? |
|---|---|---|
| 1 | \(p(10^4 - k^2)^2 = q \times 10^8\) | Yes |
| 2 | \(p(10^4 - k^2)^2 = q \times 10^4\) | No |
| 3 | \(p(10^4 - k^2) = q \times 10^4\) | No |
| 4 | \(p(10^4 - k^2) = q \times 10^8\) | No |
The derived relationship matches Option 1 exactly.
Revision Table: Key Concepts
| Concept | Explanation | Mathematical Representation |
|---|---|---|
| Percentage Increase | Adding a percentage of the original value. | Initial Value \(\times (1 + \frac{\text{k}}{100})\) |
| Percentage Decrease | Subtracting a percentage of the original value. | Initial Value \(\times (1 - \frac{\text{k}}{100})\) |
| Successive Changes | Applying multiple percentage changes one after another on the new value each time. | Multiply the original value by the change factors sequentially. |
| Difference of Squares | A fundamental algebraic identity useful for simplification. | \((a+b)(a-b) = a^2 - b^2\) |
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.
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