Question

In a class test, the sum of Gagan’s marks in mathematics and English is $45$. If he had $1$ more mark in Mathematics and $1$ less in English, the product of marks would have been $500$. Find the original marks obtained by Gagan in Mathematics and English separately.

Answer 1

Hint: Take $x$ be the original marks in Mathematics and $y$ be the original marks in English. Then using the given condition find two linear equations of two variables. This solves the system of linear equations of two variables and finds the value of $x$ and $y$.

Complete step-by-step answer:
We have to find the original marks obtained by Gagan in Mathematics and English.
Let us consider $x$ be the original marks in Mathematics and $y$ be the original marks in English obtained by Gagan.
The sum of Gagan’s marks in mathematics and English is $45$. Therefore $x+y=45$.
Now if he had $1$ more marks in mathematics then the new mark in mathematics is $x+1$.
If he had $1$ less mark in English then new mark in English $y-1$.
By given condition If he had $1$ more mark in Mathematics and $1$ less in English, the product of marks would have been $500$.
Hence $\left( x+1 \right)\left( y-1 \right)=500$
Therefore two linear equations are
$x+y=45...\left( i \right)$ and
$\left( x+1 \right)\left( y-1 \right)=500...\left( ii \right)$
We will solve this system of linear equations by eliminating $y$.
From the equation $\left( i \right)$ we get $y=45-x...\left( iii \right)$.
Substitute the value of $y$ from $\left( iii \right)$ in equation $\left( ii \right)$ we get $\left( x+1 \right)\left( 45-x-1 \right)=500$.
Simplifying further we get $\left( x+1 \right)\left( 44-x \right)=500$.
Multiplying the left hand side we get $44x+44-{{x}^{2}}-x=500$
Taking all the left hand side terms to the right hand side we get $500-44x-44+{{x}^{2}}+x=0$.
Taking the same power of $x$ together we get ${{x}^{2}}+x-44x+500-44=0$.
Simplifying the equation we have ${{x}^{2}}-43x+456=0$.
Splitting the middle term we get ${{x}^{2}}-24x-19x+456=0$.
Further calculating we get $x\left( x-24 \right)-19\left( x-24 \right)=0$.
Hence we get $\left( x-24 \right)\left( x-19 \right)=0$
Since \[a\] and \[b\] are two non-zero numbers such that $ab=0$ then either $a=0$ or $b=0$.
Using this result we get $x-24=0$ or $x-19=0$.
If $x-24=0$ then $x=24$.
Substituting $x=24$ in the equation $\left( iii \right)$ we get $y=45-24$. So $y=21$.
And if $x-19=0$ then $x=19$.
Substituting $x=19$ in the equation $\left( iii \right)$ we get $y=45-19$. So $y=26$.
Therefore when $x=24$ and $y=21$ we get original marks in Mathematics is $24$ and original marks in English is $21$.
When $x=19$and $y=26$ original marks in Mathematics is $19$ and original marks in English is $26$.
This is the required solution.

Note: We can solve this problem by taking one variable also. By assuming original marks in Mathematics is $x$ we can take the original marks in English as $45-x$ as the sum of original marks in Mathematics and English is $45$. While solving this problem we will get two values of $x$ and so students must take care while finding the corresponding value of $y$.