Question

in a class test, the sum of kamal's marks in mathematics and english is 40. had he got 3 marks more in mathematics and 4 marks less in english, the product of his marks would have been 360. Find his marks in the two subjects.​

Answer 1

$\huge\mathfrak{Answer:}$

Given:

• It is given that sum of marks obtained by Kamal in Mathematics and English is 40.
• He got 3 marks more in Mathematics and 4 marks less in English.
• The product his marks is 360.

To Find:

• We need to find his marks in both subjects.

Solution:

Let the marks obtained in Maths be x, then the marks obtained in English is (40 - x).

Now, according to the question, we have

$\longrightarrow\sf{ (x + 3)(40 - x - 4) = 360}$

$\longrightarrow\sf{ (x + 3)(36 - x) = 360}$

$\longrightarrow\sf{ {- x}^{2} + 36x + 108 - 3x = 360}$

$\longrightarrow\sf{ {- x}^{2} + 33x - 252 = 0}$

$\longrightarrow\sf{ {x}^{2} - 21x - 12x + 252 = 0}$

$\longrightarrow\sf{ x(x - 21) - 12(x - 21) = 0}$

$\longrightarrow\sf{ (x - 21)(x - 12)}$

Either (x - 21) = 0 or (x - 12) = 0.

When (x - 21) = 0

$\mapsto\sf{ x = 21}$

When (x - 12) = 0

$\mapsto\sf{ x = 12}$

If x = 21, then marks obtained in Mathematics is 21 and marks obtained in English is (40 -21) = 19.

If x = 12, then marks obtained in Mathematics is 12 and marks obtained in English is (40 - 12) = 28.

Answer 2

$\huge\mathfrak\blue{Answer:}$

Given:

• Given that the sum of Kamal Marks in Mathematics and English is 40
• Had he got 3 marks more in Mathematics and 4 marks less in English . The product would be 360

To Find:

• We have to find Kamal's marks in Mathematics and English

Solution:

Let Marks in English = y

Marks in Mathematics = x

We have been given that

$\hookrightarrow \boxed{\sf{\pink{ Sum \: of \: Marks = 40 }}}$

$\hookrightarrow \sf{ \: x + y = 40 }$

$\hookrightarrow \sf{ \: y = ( 40 - x ) }$ ---------- ( 1 )

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$\underline{\large\mathfrak\orange{According \: to \: the \: Question}}$

Had he got 3 marks more in Mathematics and 4 marks less in English the product would be 360

$\implies \sf{( x + 3 ) ( y - 4 ) = 36}$

Using Equation ( 1 )

$\implies \sf{ ( x + 3 ) ( 40 - x - 4 ) = 360}$

$\implies \sf{( x + 3 )(36 - x) = 360}$

$\implies \sf{36x - x^2 + 108 - 3x = 360}$

$\implies \sf{33x - x^2 + 108 = 360 }$

$\implies \sf{ x^2 - 33x + 252 = 0 }$

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Solving the Quadratic Equation using middle term spitting method

$\implies \sf{ x^2 - 33x + 252 = 0 }$

$\implies \sf{ x^2 - 12x - 21x + 252 = 0 }$

$\implies \sf{ x( x - 12 ) - 21( x - 12 ) = 0 }$

$\implies \sf{( x - 12 ) ( x - 21 ) = 0 }$

Either

$\implies \sf{ x - 12 = 0 }$

$\implies \boxed{\sf{ x = 12 }}$

Or

$\implies \sf{ x - 21 = 0 }$

$\implies \boxed{\sf{ x = 21 }}$

_______________________________

$\underline{\large\mathfrak\orange{Finding \: the \: Values}}$

When x = 12

$\implies \sf{ y = 40 - 12}$

$\implies \sf{ y = 28}$

When x = 21

$\implies \sf{ y = 40 - 21 }$

$\implies \sf{ y = 19 }$

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$\huge\underline{\sf{\red{A}\orange{n}\green{s}\pink{w}\blue{e}\purple{r}}}$

$\large\boxed{\sf{\purple{Marks \: in \: Mathematics = 12 }}}$

$\large\boxed{\sf{\purple{Marks \: in \: English = 28 }}}$

$\large\sf{\green{O}\orange{R}}$

$\large\boxed{\sf{\red{Marks \: in \: Mathematics = 21 }}}$

$\large\boxed{\sf{\red{Marks \ : in \: English = 19 }}}$

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