Answer

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**Hint:**Assume the marks of Udaya as a variable. Then by the information given you form equations and solve them to get marks of both.

**Complete step-by-step answer:**

Let marks of Udaya be $x$

This means marks of Ramya is $2x - 9$ where $2x$ is twice of Udaya’s marks and $9$ is subtracted as she got $9$ less marks than her twice

Product of their marks is $ = 195$

Marks of Ramya$ \times $Marks of Udaya $ = 195$

$

\Rightarrow (2x - 9) \times x = 195 \\

\Rightarrow 2{x^2} - 9x = 195 \\

\Rightarrow 2{x^2} - 9x - 195 = 0 \\

$

This is a quadratic equation of the form $a{x^2} + bx + c = 0$

$x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$

In this equation $a = 2\,,\,b = - 9\,,\,c = - 195$. So,

\[

x = \dfrac{{ - ( - 9) \pm \sqrt {{{( - 9)}^2} - 4 \times (2) \times ( - 195)} }}{{2 \times 2}} \\

x = \dfrac{{9 \pm \sqrt {81 + 1560} }}{4} \\

x = \dfrac{{9 \pm \sqrt {1641} }}{4} \\

x = \dfrac{{9 \pm 40.5}}{4} \\

x = \dfrac{{9 + 40.5}}{4},\dfrac{{9 - 40.5}}{4} \\

x = 12.375, - 7.875 \\

\]

\[ - 7.875\] won’t be considered as marks are not considered to be negative until given in the question.

Hence, is \[x = 12.375\]

$ \Rightarrow 2x - 9 = 15.75$

Marks of Udaya are $12.375$

Marks of Ramya are $15.75$

Therefore, Ramya got more marks i.e. $15.75$

**Note:**As this was a question of marks, we ignored the negative solution obtained by solving the quadratic equation. But in other cases like that of temperature, we do need to consider the negative solution as well.

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