Answer
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Hint: We can easily solve this type of problem by finding the expression of the amount the man actually paid with the help of $x$ . Then we equate that expression to $2730$ and get a quadratic equation of variable $x$ from there. Further solving the quadratic equation by split factor method, we will get the solution of the equation i.e., the value of $x$ .
Complete step by step solution:
According to the given problem the price of each plant was $x$ in rupees
If the supplier has reduced the price of each plant by one rupee then the price of each plants will be $x-1$ in rupees
Earlier the man spent Rs. 2800 on buying a number of plants priced at Rs. $x$ each.
Hence, the number of plants he bought earlier was $=\dfrac{2800}{x}$
Later, he got $10$ more additional plants which makes the total number of plants he got $=\dfrac{2800}{x}+10$
Now, the total amount of money (in rupees) the man gave to the supplier according to the above expressions is
\[=\text{present price of each plant}\times \text{total number of plants}\]
$=\left( x-1 \right)\left( \dfrac{2800}{x}+10 \right)$
We equate the above expression to the given present value of net amount, $2730$as shown below
$\left( x-1 \right)\left( \dfrac{2800}{x}+10 \right)=2730$
Further, simplifying as
\[\Rightarrow \dfrac{\left( 2800+10x \right)\left( x-1 \right)}{x}=2730\]
\[\Rightarrow \left( 2800+10x \right)\left( x-1 \right)=2730x\]
\[\Rightarrow 2800x-2800+10{{x}^{2}}-10x-2730x=0\]
\[\Rightarrow 10{{x}^{2}}+60x-2800=0\]
Dividing both the terms of the above equation by $10$ we get
\[\Rightarrow {{x}^{2}}+6x-280=0\]
Using the split factor method as shown below
\[\Rightarrow {{x}^{2}}+20x-14x-280=0\]
\[\Rightarrow x\left( x+20 \right)-14\left( x+20 \right)=0\]
\[\Rightarrow \left( x+20 \right)\left( x-14 \right)=0\]
Now,
$x+20=0$ and \[x-14=0\]
Simplifying we get $x=-20,14$
As, $x$ is the price of each plant it can’t be negative. So, we only consider the solution $x=14$
Therefore, we conclude that the value of $x$ is $14$.
Note: We must be careful while making the expressions of the amount the man actually paid with the help of $x$ as, an incorrect expression of the same can cause unavoidable mistakes. Also, it is not necessary to solve the quadratic equation by split factor method. It can also be solved using Sridhar Acharya’s formula or by completing a square.
Complete step by step solution:
According to the given problem the price of each plant was $x$ in rupees
If the supplier has reduced the price of each plant by one rupee then the price of each plants will be $x-1$ in rupees
Earlier the man spent Rs. 2800 on buying a number of plants priced at Rs. $x$ each.
Hence, the number of plants he bought earlier was $=\dfrac{2800}{x}$
Later, he got $10$ more additional plants which makes the total number of plants he got $=\dfrac{2800}{x}+10$
Now, the total amount of money (in rupees) the man gave to the supplier according to the above expressions is
\[=\text{present price of each plant}\times \text{total number of plants}\]
$=\left( x-1 \right)\left( \dfrac{2800}{x}+10 \right)$
We equate the above expression to the given present value of net amount, $2730$as shown below
$\left( x-1 \right)\left( \dfrac{2800}{x}+10 \right)=2730$
Further, simplifying as
\[\Rightarrow \dfrac{\left( 2800+10x \right)\left( x-1 \right)}{x}=2730\]
\[\Rightarrow \left( 2800+10x \right)\left( x-1 \right)=2730x\]
\[\Rightarrow 2800x-2800+10{{x}^{2}}-10x-2730x=0\]
\[\Rightarrow 10{{x}^{2}}+60x-2800=0\]
Dividing both the terms of the above equation by $10$ we get
\[\Rightarrow {{x}^{2}}+6x-280=0\]
Using the split factor method as shown below
\[\Rightarrow {{x}^{2}}+20x-14x-280=0\]
\[\Rightarrow x\left( x+20 \right)-14\left( x+20 \right)=0\]
\[\Rightarrow \left( x+20 \right)\left( x-14 \right)=0\]
Now,
$x+20=0$ and \[x-14=0\]
Simplifying we get $x=-20,14$
As, $x$ is the price of each plant it can’t be negative. So, we only consider the solution $x=14$
Therefore, we conclude that the value of $x$ is $14$.
Note: We must be careful while making the expressions of the amount the man actually paid with the help of $x$ as, an incorrect expression of the same can cause unavoidable mistakes. Also, it is not necessary to solve the quadratic equation by split factor method. It can also be solved using Sridhar Acharya’s formula or by completing a square.
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