
John and Jivanti together have 45 marbles. Both of them lost 5 marbles each, and the product of the number of marbles they have now is 124. From the quadratic equation, find out how many marbles they had to start with if John had x marbles.
A. 36, 9
B. 20, 25
C. 30, 15
D. 27, 18
Answer
564k+ views
Hint: We here have been given that the total number of marbles John and Jivanti have together are 45 and the number of marbles John has x and we have been asked to find the number of balls both of them have. For this, we will first assume the number of balls Jivanti has to be y and then calculate y in terms of x. Hence, we will get the number of marbles for both of them to have in terms of x. Then we have been given that both of them lost 5 marbles each and the product of the marbles they have left now is 124. We will use this relationship and form a quadratic equation in terms of ‘x’. Then we will solve that equation using the middle-term split method and hence we will obtain two values of x. They will be the number of balls both of them have. Hence, we will get the required answer.
Complete step-by-step solution:
We here have been given that the total number of marbles John and Jivanti together have are 45. Here, we have also been given that the number of marbles John has is ‘x’.
Now, if we assume the number of marbles Jivanti has to be ‘y’, we can say that:
$x+y=45$
Hence, we get the number of marbles Jivanti has as:
$y=45-x$
Now, we have been given that both of them lost 5 marbles each.
Thus the number of marbles each of them will have are:
John:
Initially: $x$
Finally: $x-5$
Jivanti:
Initially: $45-x$
Finally: $45-x-5\Rightarrow 40-x$
Now, we have been given that the product of the marbles left with both of them now is 124. Thus, we can say that:
$\left( x-5 \right)\left( 40-x \right)=124$
Simplifying this equation, we have:
$\begin{align}
& \left( x-5 \right)\left( 40-x \right)=124 \\
& \Rightarrow 40x-200-{{x}^{2}}+5x=124 \\
& \Rightarrow 45x-{{x}^{2}}-200=124 \\
& \therefore {{x}^{2}}-45x+324=0 \\
\end{align}$
Thus, the required quadratic equation is:
${{x}^{2}}-45x+324=0$
Now, if we solve this equation, we will get the number of marbles John and Jivanti have.
Thus, solving this equation by middle term splitting method, we get:
$\begin{align}
& {{x}^{2}}-45x+324=0 \\
& \Rightarrow {{x}^{2}}-36x-9x+324=0 \\
& \Rightarrow x\left( x-36 \right)-9\left( x-36 \right)=0 \\
& \Rightarrow \left( x-36 \right)\left( x-9 \right)=0 \\
\end{align}$
Thus, we have the values of x as:
$\begin{align}
& x-36=0 \\
& \therefore x=36 \\
& x-9=0 \\
& \therefore x=9 \\
\end{align}$
Hence, the number of marbles John and Jivanti have are 36 and 9.
Thus, option (A) is the correct option.
Note: The quadratic equation formed can be solved by many methods. One of them is by solving it using the quadratic formula for any quadratic equation $a{{x}^{2}}+bx+c=0$ given as:
$x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$
Here, the equation is:
${{x}^{2}}-45x+324=0$
Hence, here we have:
$\begin{align}
& a=1 \\
& b=-45 \\
& c=324 \\
\end{align}$
Thus, putting these values in the abovementioned formula we get:
$\begin{align}
& x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a} \\
& \Rightarrow x=\dfrac{-\left( -45 \right)\pm \sqrt{{{\left( -45 \right)}^{2}}-4\left( 1 \right)\left( 324 \right)}}{2\left( 1 \right)} \\
& \Rightarrow x=\dfrac{45\pm \sqrt{2025-1296}}{2} \\
& \Rightarrow x=\dfrac{45\pm \sqrt{729}}{2} \\
& \Rightarrow x=\dfrac{45\pm 27}{2} \\
\end{align}$
Thus, we get x as:
$\begin{align}
& x=\dfrac{45+27}{2},\dfrac{45-27}{2} \\
& \Rightarrow x=\dfrac{72}{2},\dfrac{18}{2} \\
& \therefore x=36,9 \\
\end{align}$
Complete step-by-step solution:
We here have been given that the total number of marbles John and Jivanti together have are 45. Here, we have also been given that the number of marbles John has is ‘x’.
Now, if we assume the number of marbles Jivanti has to be ‘y’, we can say that:
$x+y=45$
Hence, we get the number of marbles Jivanti has as:
$y=45-x$
Now, we have been given that both of them lost 5 marbles each.
Thus the number of marbles each of them will have are:
John:
Initially: $x$
Finally: $x-5$
Jivanti:
Initially: $45-x$
Finally: $45-x-5\Rightarrow 40-x$
Now, we have been given that the product of the marbles left with both of them now is 124. Thus, we can say that:
$\left( x-5 \right)\left( 40-x \right)=124$
Simplifying this equation, we have:
$\begin{align}
& \left( x-5 \right)\left( 40-x \right)=124 \\
& \Rightarrow 40x-200-{{x}^{2}}+5x=124 \\
& \Rightarrow 45x-{{x}^{2}}-200=124 \\
& \therefore {{x}^{2}}-45x+324=0 \\
\end{align}$
Thus, the required quadratic equation is:
${{x}^{2}}-45x+324=0$
Now, if we solve this equation, we will get the number of marbles John and Jivanti have.
Thus, solving this equation by middle term splitting method, we get:
$\begin{align}
& {{x}^{2}}-45x+324=0 \\
& \Rightarrow {{x}^{2}}-36x-9x+324=0 \\
& \Rightarrow x\left( x-36 \right)-9\left( x-36 \right)=0 \\
& \Rightarrow \left( x-36 \right)\left( x-9 \right)=0 \\
\end{align}$
Thus, we have the values of x as:
$\begin{align}
& x-36=0 \\
& \therefore x=36 \\
& x-9=0 \\
& \therefore x=9 \\
\end{align}$
Hence, the number of marbles John and Jivanti have are 36 and 9.
Thus, option (A) is the correct option.
Note: The quadratic equation formed can be solved by many methods. One of them is by solving it using the quadratic formula for any quadratic equation $a{{x}^{2}}+bx+c=0$ given as:
$x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$
Here, the equation is:
${{x}^{2}}-45x+324=0$
Hence, here we have:
$\begin{align}
& a=1 \\
& b=-45 \\
& c=324 \\
\end{align}$
Thus, putting these values in the abovementioned formula we get:
$\begin{align}
& x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a} \\
& \Rightarrow x=\dfrac{-\left( -45 \right)\pm \sqrt{{{\left( -45 \right)}^{2}}-4\left( 1 \right)\left( 324 \right)}}{2\left( 1 \right)} \\
& \Rightarrow x=\dfrac{45\pm \sqrt{2025-1296}}{2} \\
& \Rightarrow x=\dfrac{45\pm \sqrt{729}}{2} \\
& \Rightarrow x=\dfrac{45\pm 27}{2} \\
\end{align}$
Thus, we get x as:
$\begin{align}
& x=\dfrac{45+27}{2},\dfrac{45-27}{2} \\
& \Rightarrow x=\dfrac{72}{2},\dfrac{18}{2} \\
& \therefore x=36,9 \\
\end{align}$
Recently Updated Pages
Why are manures considered better than fertilizers class 11 biology CBSE

Find the coordinates of the midpoint of the line segment class 11 maths CBSE

Distinguish between static friction limiting friction class 11 physics CBSE

The Chairman of the constituent Assembly was A Jawaharlal class 11 social science CBSE

The first National Commission on Labour NCL submitted class 11 social science CBSE

Number of all subshell of n + l 7 is A 4 B 5 C 6 D class 11 chemistry CBSE

Trending doubts
10 examples of friction in our daily life

One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

Difference Between Prokaryotic Cells and Eukaryotic Cells

1 Quintal is equal to a 110 kg b 10 kg c 100kg d 1000 class 11 physics CBSE

State the laws of reflection of light

Explain zero factorial class 11 maths CBSE

