Question

Represent the following situations in the form of quadratic equations :
(i) The area of a rectangular plot is $528{{m}^{2}}$. The length of the plot (in meters) is one more than twice its breadth. We need to find the length and breadth of the plot.
(ii) The product of two consecutive positive integers is 306. We need to find the integers.
(iii) Rohan's mother is 26 years older than him. The product of their ages (in years) 3 years from now will be 360. We would like to find Rohan’s present age.
(iv) A train travels a distance of 480 km at a uniform speed. If the speed had been 8 km/h less, then it would have taken 2 hours more to cover the same distance. We need to find the speed of the train

Answer 1

Hint: We have to form quadratic equations from the given situations. Consider (i). Let us consider x to be the breadth. Hence, we can represent length as $\text{Length}=1+2x$ . We have, area of rectangular plot $=528{{m}^{2}}$ . Hence, $528{{m}^{2}}=\left( 1+2x \right)\times x$ . Solving this gives the required answer. Consider (ii). Let us consider x to be one integer and x+1 to be the other. Hence, we get $x\times \left( x+1 \right)=306$ .On simplifying this, we will get the required quadratic equation. Now, consider (iii). Let us consider the age of Rohan to be x. Hence, the age of Rohan’s mother $=26+x$ . Rohan’s age 3 years from now $=x+3$ and age of Rohan’s mother 3 years from now $=\left( 26+x \right)+3$ . Hence, $\left( x+3 \right)\left( 26+x+3 \right)=360$ . On simplifying this, we will get the answer. Let’s consider (iv). let us consider x to be the speed. Using $\text{Speed}=\dfrac{\text{Distance}}{\text{Time}}$ we will get $\text{Time}=\dfrac{480}{x}$ . It is given that if the speed had been 8 km/h less, then it would have taken 2 hours more to cover the same distance. Thus, $\text{Speed}=x-8$ and $\text{Time}=\dfrac{480}{x}+3$ . Substituting these in the formula for speed and simplifying, the required result can be obtained.

Complete step-by-step solution:
We have to form quadratic equations from the given situations.
(i) It is given that the length of the plot is one more than twice its breadth. Let us consider x to be the breadth. Hence, we can represent length as
$\text{Length}=1+2x$
We have, area of rectangular plot $=528{{m}^{2}}$.
We know that the area of a rectangle =$\text{Length}\times \text{Breadth}$ . Hence,
$528{{m}^{2}}=\left( 1+2x \right)\times x$
Let us expand this as
$x+2{{x}^{2}}=528$
When we rearrange the terms, we will get
$2{{x}^{2}}+x=528$
Let us take the constant from RHS to LHS. We will get
$2{{x}^{2}}+x-528=0$
We know that the above equation is of the form $a{{x}^{2}}+bx+c=0$ .
Hence, the quadratic equation is $2{{x}^{2}}+x-528=0$ .
(ii) We have the product of two consecutive positive integers is 306.
We know that consecutive integers mean the next number, that is, if the number is 2, its consecutive number is 3. Hence, let us consider x to be one integer and x+1 to be the other.
Now, we have to multiply these to get the result 306.
$\Rightarrow x\times \left( x+1 \right)=306$
Let us simplify the LHS. We will get
${{x}^{2}}+x=306$
On taking the constant to the LHS, we will get
${{x}^{2}}+x-306=0$
We know that the above equation is of the form $a{{x}^{2}}+bx+c=0$ .
Hence, the quadratic equation is ${{x}^{2}}+x-306=0$ .
(iii) It is given that Rohan's mother is 26 years older than him. Let us consider the age of Rohan to be x. Hence,
Age of Rohan’s mother $=26+x$
It is given that the product of their ages 3 years from now will be 360.
So let us find the Rohan’s age 3 years from.
Rohan’s age 3 years from now $=x+3...(i)$
Also let's find the age of Rohan’s mother 3 years from now.
Age of Rohan’s mother 3 years from now $=\left( 26+x \right)+3...(ii)$
Now, let’s take the product of (i) and (ii) that results in 360.
$\Rightarrow \left( x+3 \right)\left( 26+x+3 \right)=360$
Let us simplify the equation. We will get
$\left( x+3 \right)\left( x+29 \right)=360$
On multiplying the terms in LHS, we will get
${{x}^{2}}+29x+3x+87=360$
Let’s simplify this. We will get
${{x}^{2}}+32x+87=360$
On taking 360 to RHS, we will get
$\begin{align}
  & {{x}^{2}}+32x+87-360=0 \\
 & \Rightarrow {{x}^{2}}+32x-273=0 \\
\end{align}$
We know that the above equation is of the form $a{{x}^{2}}+bx+c=0$ .
Hence, the quadratic equation is ${{x}^{2}}+32x-273=0$ .
(iv) Since we have found the speed of the train, let us consider x to be the speed.
It is given that the train travels a distance of 480 km at a uniform speed.
We know that $\text{Speed}=\dfrac{\text{Distance}}{\text{Time}}...(i)$
Let us substitute the values. We will get
$x=\dfrac{480}{\text{Time}}$
From this, we can find time as
$\text{Time}=\dfrac{480}{x}$
It is given that if the speed had been 8 km/h less, then it would have taken 2 hours more to cover the same distance.
We can write speed and time as follows.
$\text{Speed}=x-8$
$\text{Time}=\dfrac{480}{x}+3$
We know that the distance is the same. Now, let us substitute these values in the equation (i).
$x-8=\dfrac{480}{\dfrac{480}{x}+3}$
We can write this as
$\left( x-8 \right)\left( \dfrac{480}{x}+3 \right)=480$
Now, let's simplify this.
$\left( x-8 \right)\left( \dfrac{480+3x}{x} \right)=480$
Let us take x from the denominator of LHS to RHS.
$\left( x-8 \right)\left( 3x+480 \right)=480x$
$\Rightarrow 3{{x}^{2}}+480x-24x-3840=480x$
On solving this, we will get
$3{{x}^{2}}+456x-3840=480x$
Let’s take the term in RHS to LHS, we will get
$\begin{align}
  & 3{{x}^{2}}+456x-3840-480x=0 \\
 & \Rightarrow 3{{x}^{2}}-24x-3840=0 \\
\end{align}$
We can take 3 common from LHS. We will get
$3\left( {{x}^{2}}-8x-1280 \right)=0$
Taking 3 to RHS, we get
${{x}^{2}}-8x-1280=0$
We know that the above equation is of the form $a{{x}^{2}}+bx+c=0$. Hence, the quadratic equation is ${{x}^{2}}-8x-1280=0$.

Note: You have to consider the parameter which we have to find as x. The given conditions should be carefully read. To solve the last question, we have to know the formula of speed. You may make mistakes when writing the speed formula as $\text{Speed}=\dfrac{\text{Time}}{\text{Distance}}$ . Do not forget to take the common factors out, if any, and simplify further in the equation obtained at the end.