Answer
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Hint: Let the marks obtained by Shefali in English and Mathematics be x and y respectively. For two equations with the data given in the question, solve the equation to obtain the value of x and y.
Complete step-by-step answer:
Let the marks obtained in English $ = x$
Let the marks in Maths $ = y$
According to the question, the sum of marks in both subjects is $30$.
Therefore,
$
x + y = 30 \\
y = 30 - x \\
$ (1)
If she had got two more marks in maths, her marks would have increased to $y + 2$ and if she had got three marks less in English, her marks would have been reduced to $x - 3$.
According to question, product of the marks would have been $210$, i.e.,
$
\left( {x - 3} \right)\left( {y + 2} \right) = 210 \\
xy + 2x - 3y - 6 = 210 \\
$ (2)
Substituting the value of y from equation (1) in equation (2), we get,
\[
x\left( {30 - x} \right) + 2x - 3\left( {30 - x} \right) - 6 = 210 \\
30x - {x^2} + 2x - 90 + 3x - 6 = 210 \\
- {x^2} + 35x - 96 - 210 = 0 \\
{x^2} - 35x + 306 = 0 \\
\]
By splitting the middle term,
$
{x^2} - 17x - 18x + 306 = 0 \\
x\left( {x - 17} \right) - 18\left( {x - 17} \right) = 0 \\
\left( {x - 17} \right)\left( {x - 18} \right) = 0 \\
x = 17,18 \\
$
That is,
When
$
x = 17 \\
y = 30 - x = 30 - 17 = 13 \\
$
And when,
$
x = 18 \\
y = 30 - x = 30 - 18 = 12 \\
$
Hence, if her marks in English is $17$, then in Maths she got $13$. If her mark in English is $18$, then in Maths she got $12$.
Note: This problem could also have been solved using one variable by letting the marks in English be x and in maths $\left( {30 - x} \right)$. Similar to this question, you would have got a quadratic equation in x. Here too, x would have had two roots.
Complete step-by-step answer:
Let the marks obtained in English $ = x$
Let the marks in Maths $ = y$
According to the question, the sum of marks in both subjects is $30$.
Therefore,
$
x + y = 30 \\
y = 30 - x \\
$ (1)
If she had got two more marks in maths, her marks would have increased to $y + 2$ and if she had got three marks less in English, her marks would have been reduced to $x - 3$.
According to question, product of the marks would have been $210$, i.e.,
$
\left( {x - 3} \right)\left( {y + 2} \right) = 210 \\
xy + 2x - 3y - 6 = 210 \\
$ (2)
Substituting the value of y from equation (1) in equation (2), we get,
\[
x\left( {30 - x} \right) + 2x - 3\left( {30 - x} \right) - 6 = 210 \\
30x - {x^2} + 2x - 90 + 3x - 6 = 210 \\
- {x^2} + 35x - 96 - 210 = 0 \\
{x^2} - 35x + 306 = 0 \\
\]
By splitting the middle term,
$
{x^2} - 17x - 18x + 306 = 0 \\
x\left( {x - 17} \right) - 18\left( {x - 17} \right) = 0 \\
\left( {x - 17} \right)\left( {x - 18} \right) = 0 \\
x = 17,18 \\
$
That is,
When
$
x = 17 \\
y = 30 - x = 30 - 17 = 13 \\
$
And when,
$
x = 18 \\
y = 30 - x = 30 - 18 = 12 \\
$
Hence, if her marks in English is $17$, then in Maths she got $13$. If her mark in English is $18$, then in Maths she got $12$.
Note: This problem could also have been solved using one variable by letting the marks in English be x and in maths $\left( {30 - x} \right)$. Similar to this question, you would have got a quadratic equation in x. Here too, x would have had two roots.
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