Question

the denominator of a fraction is 12 more than the numerator. If 3 is added to the denominator and 5 is added to the numerator, then the fraction is 3/4. Find the original fraction.

Answer 1

Answer:-

Let the fraction be x/y.

Given:

The denominator is 12 more than the numerator.

⟶ Denominator = Numerator + 12

⟶ y = x + 12 -- equation (1)

And,

If 3 is added to the denominator and 5 is added to the numerator , the fraction becomes 3/4.

$\longrightarrow \sf \dfrac{x + 5}{y + 3} = \dfrac{3}{4}$

Substitute the value of y from equation (1).

$\: \longrightarrow \sf \dfrac{x + 5}{x + 12 + 3} = \dfrac{3}{4} \\ \\ \longrightarrow \sf 4(x + 5) = 3(x + 15) \\ \\ \longrightarrow \sf 4x + 20 = 3x + 45 \\ \\ \longrightarrow \sf 4x - 3x = 45 - 20 \\ \\ \longrightarrow \boxed{ \sf x = 25}$

Substitute the value of x in equation (1)

⟶ y = x + 12

⟶ y = 25 + 12

⟶ y = 37

∴ The required fraction x/y is 25/37.

Answer 2

Answer :-

$\: \: \underline{\boxed{\bf{\blue{\mapsto \: \: Firstly \: let's \: understand \: the \: concept \: used \: :-}}}}$

Here the concept of Linear Equations in Two Variables has been used. According to this, we can find the required two unknown values. We can make the value of one, dependent on other so that we can find them both . Now we are going to take the numerator and denominator as unknown variable, and by the help of constants we are going to find them.

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★ Question :-

The denominator of a fraction is 12 more than the numerator. If 3 is added to the denominator and 5 is added to the numerator, then the fraction is 3/4. Find the original fraction.

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★ Solution :-

Given,

» The denominator of fraction = 12 + The numerator fraction

» (3 + Denominator) and (5 + Numerator) = ¾

$\: \: \bf{\green{\longrightarrow \: \: \: Let \: the \: numerator \: of \: the \: fraction \: be \: \underline{\red{x}}}}$

$\: \: \bf{\green{\longrightarrow \: \: \: Let \: the \: denominator \: of \: the \: fraction \: be \: \underline{\red{y}}}}$

$\: \: \bf{\green{\longrightarrow \: \: \: By \: using \: the \: variables \: the \: fraction \: will \: be \: \red{\dfrac{x}{y}}}}$

Then, according to the question :-

~ Case I :

⌬ y = 12 + x ... (i)

~ Case II :-

$\: \: \bf{\Longrightarrow \: \: \dfrac{x \: + \: 5}{y \: + \: 3} \: = \: \dfrac{3}{4}}$

By cross multiplication, we get,.

⌬ 4(x + 5) = 3(y + 3)

⌬ 4x + 20 = 3y + 9

⌬ 4x - 3y = 9 - 20

⌬ 4x - 3y = - 11 ... (ii)

From equation (i) and (ii), we get,

⌬ 4x - 3(12 + x) = -11

⌬ 4x - 36 - 3x = -11

⌬ 4x - 3x = -11 + 36

$\: \: \huge{\bf{\orange{:\Longrightarrow \: \: x \: = \: 25}}}$

Now, using equation (i) and the value of x, we get,

⌬ y = 12 + x

⌬ y = 12 + 25

$\: \: \huge{\bf{\orange{:\Longrightarrow \: \: y \: = \: 37}}}$

Now by applying these values, in fraction, we get,

$\: \: \underline{\boxed{\purple{\rm{Hence, \: the \: required \: fraction \: is \: \red{\dfrac{25}{37}}}}}}$

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$\: \: \underline{\boxed{\bf{\pink{\mapsto \: \: Confused? \: Don't \: worry \: let's \: verify \: it \: :}}}}$

For verification, we need to simply apply the values we got, into the equations we formed. Then,

~ Case I :-

⌬ y = 12 + x

⌬ 37 = 12 + 25

⌬ 37 = 37

Clearly, LHS = RHS

~ Case II :-

$\: \: \bf{\orange{\Longrightarrow \: \: \dfrac{x \: + \: 5}{y \: + \: 3} \: = \: \dfrac{3}{4}}}$

$\: \: \bf{\orange{\Longrightarrow \: \: \dfrac{25 \: + \: 5}{37 \: + \: 3} \: = \: \dfrac{3}{4}}}$

$\: \: \bf{\orange{\Longrightarrow \: \: \dfrac{30}{40} \: = \: \dfrac{3}{4} = \: \dfrac{3}{4}}}$

Clearly, LHS = RHS

Here both the conditions satisfy, so our answer is correct.

Hence, Verified.

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$\: \: \underline{\boxed{\red{\rm{\mapsto \: \: \: Let's \: know \: more \: :-}}}}$

• Linear Equations are the equations formed using constant and variable terms of single degrees.

• Polynomials are the equations formed using constant and variable terms but can be of many degrees.