Answer
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Hint: Assume the price of the first article to be x and second article to be y. Write equations based on the data given in the question. Solve the pair of linear equations in two variables to find the value of variables x and y and thus get the price of the second article.
Complete step-by-step answer:
We have two articles whose prices are related by data given in the question. We have to find the price of the second article. We will solve this question by writing linear equations, relating the prices of both the articles.
Let us assume that the price of the first article is \[Rs.x\] and the price of the second article is \[Rs.y\].
We know that the prices of both the articles are in the ratio \[3:4\]. Thus, we have \[\dfrac{x}{y}=\dfrac{3}{4}\].
We can write this equation as \[x=\dfrac{3}{4}y.....\left( 1 \right)\].
We increased the price of the second article by \[10%\].
We know that \[a%\] of b has the value \[\dfrac{ab}{100}\].
Thus, the value of \[10%\] of \[Rs.x\]\[=\dfrac{10x}{100}=\dfrac{Rs.x}{10}\].
So, the new price of the first article \[=Rs.\left( x+\dfrac{x}{10} \right)=Rs.\dfrac{11x}{10}\].
We increased the price of the second article by \[Rs.4\]. Thus, the new price of the second article is \[Rs.(y+4)\].
We know that the ratio of new prices is still the same, i.e., \[3:4\].
Thus, we have \[\dfrac{\dfrac{11x}{10}}{y+4}=\dfrac{3}{4}.....\left( 2 \right)\].
Substituting equation (2) in equation (1), we have \[\dfrac{\dfrac{11\left( \dfrac{3}{4}y \right)}{10}}{y+4}=\dfrac{3}{4}\].
Simplifying the above equation, we have \[\dfrac{11\left( \dfrac{3}{4}y \right)}{10}=\dfrac{3}{4}\left( y+4 \right)\].
\[\begin{align}
& \Rightarrow \dfrac{11y}{10}=y+4 \\
& \Rightarrow \dfrac{11y}{10}-y=4 \\
& \Rightarrow \dfrac{y}{10}=4 \\
& \Rightarrow y=4\times 10=40 \\
\end{align}\]
Hence, the price of the second article is \[Rs.y=Rs.40\].
Note: We can also solve this question by forming linear equations in one variable. Let us assume that the price of the first article is \[Rs.x\]. Write the price of the second article in terms of the price of the first article. Form another equation based on the given data and solve them to get the value of both the articles.
Complete step-by-step answer:
We have two articles whose prices are related by data given in the question. We have to find the price of the second article. We will solve this question by writing linear equations, relating the prices of both the articles.
Let us assume that the price of the first article is \[Rs.x\] and the price of the second article is \[Rs.y\].
We know that the prices of both the articles are in the ratio \[3:4\]. Thus, we have \[\dfrac{x}{y}=\dfrac{3}{4}\].
We can write this equation as \[x=\dfrac{3}{4}y.....\left( 1 \right)\].
We increased the price of the second article by \[10%\].
We know that \[a%\] of b has the value \[\dfrac{ab}{100}\].
Thus, the value of \[10%\] of \[Rs.x\]\[=\dfrac{10x}{100}=\dfrac{Rs.x}{10}\].
So, the new price of the first article \[=Rs.\left( x+\dfrac{x}{10} \right)=Rs.\dfrac{11x}{10}\].
We increased the price of the second article by \[Rs.4\]. Thus, the new price of the second article is \[Rs.(y+4)\].
We know that the ratio of new prices is still the same, i.e., \[3:4\].
Thus, we have \[\dfrac{\dfrac{11x}{10}}{y+4}=\dfrac{3}{4}.....\left( 2 \right)\].
Substituting equation (2) in equation (1), we have \[\dfrac{\dfrac{11\left( \dfrac{3}{4}y \right)}{10}}{y+4}=\dfrac{3}{4}\].
Simplifying the above equation, we have \[\dfrac{11\left( \dfrac{3}{4}y \right)}{10}=\dfrac{3}{4}\left( y+4 \right)\].
\[\begin{align}
& \Rightarrow \dfrac{11y}{10}=y+4 \\
& \Rightarrow \dfrac{11y}{10}-y=4 \\
& \Rightarrow \dfrac{y}{10}=4 \\
& \Rightarrow y=4\times 10=40 \\
\end{align}\]
Hence, the price of the second article is \[Rs.y=Rs.40\].
Note: We can also solve this question by forming linear equations in one variable. Let us assume that the price of the first article is \[Rs.x\]. Write the price of the second article in terms of the price of the first article. Form another equation based on the given data and solve them to get the value of both the articles.
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