Re: In year 1980, the price of an article was $X. The price was increased
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25 May 2025, 04:00
Let's express the prices at different years based on the initial price \$X.
1. Price in $\(1980\left(\mathbf{P}_{\mathrm{o}}\right)\)$ : $\(\$ \mathrm{X}\)$.
2. Price in $\(1982\left(\mathbf{P}_{\mathbf{1}}\right)\)$ : Increased by $\(12 \%\)$ from 1980.
- Increase $\(=12 \%\)$ of $\(X=0.12 * X\)$.
- $\(P_1=X+0.12 X=1.12 X\)$.
3. Price in $\(1984\left(\mathbf{P}_{\mathbf{2}}\right)\)$ : Increased by $\(24 \%\)$ from 1980.
- Increase $\(=24 \%\)$ of $\(X=0.24 * X\)$.
- $\(\mathrm{P}_2=\mathrm{X}+0.24 \mathrm{X}=1.24 \mathrm{X}\)$.
Now, we need to find the percentage change in price from 1982 to 1984, which is Quantity A.
Calculating Quantity A: Percentage Change from 1982 to 1984
The percentage change from 1982 to 1984 is calculated as:
$$
\(\text { Percentage Change }=\left(\frac{\text { New Price - Old Price }}{\text { Old Price }}\right) \times 100 \%\)
$$
Here:
- Old Price $\((1982)=P_1=1.12 X\)$.
- New Price $\((1984)=P_2=1.24 X\)$.
Plugging in the values:
$$
\(\text { Quantity } \mathrm{A}=\left(\frac{1.24 X-1.12 X}{1.12 X}\right) \times 100 \%=\left(\frac{0.12 X}{1.12 X}\right) \times 100 \%\)
$$
The X's cancel out:
$$
\(\text { Quantity } \mathrm{A}=\left(\frac{0.12}{1.12}\right) \times 100 \% \approx 0.1071 \times 100 \% \approx 10.71 \%\)
$$
So, Quantity $\([m]\mathrm{A} \approx 10.71 \%\)[/m]$.
Comparing Quantity A and Quantity B
- Quantity A: \(~10.71\%\).
- Quantity B: $\(12 \%\)$.
Clearly, $\(10.71 \%\)$ is less than $\(12 \%\)$.
Therefore, Quantity B is greater than Quantity A.