Question

78. When 2 is added to both the numerator and
denominator of a possible fraction, the fraction
thus obtained is 3/5. If the denominator is 6 more
than the numerator, what is the original value of
the denominator ?
​

Answer 1

To solve such problems, we need to remember some points

• An equation is in the form of ax + by + c = 0, where a, b, c are real numbers & a ≠ 0 & b ≠ 0 is known as linear equation in two variables

• Take equations be a₁x + b₁y + c₁ = 0 and a₂x + b₂ + c₂ = 0

To find : What is the original value of the denominator ?

Given, When 2 is added to both the numerator and denominator of a possible fraction, the fraction thus obtained is 3/5. If the denominator is 6 more than the numerator.

• First condition

When 2 is added to both the numerator and denominator of a possible fraction, the fraction thus obtained is 3/5.

Consider numerator be x and denominator be y

$\implies { \pmb{ \tt{\dfrac{Numerator}{Denominator} = \dfrac{3}{5}}}} \\ \\ \implies \sf \dfrac{x + 2}{y + 2} = \dfrac{3}{5} \\ \\ \qquad\underline{ \pmb{ \sf{Cross \: multiplication}}} \\ \\ \implies \sf 5(x + 2) = 3(y + 2) \\ \\ \implies \sf 5x + 10 = 3y + 6 \\ \\ \implies \sf 5x - 3y = 6 - 10 \\ \\ \implies \sf 5x - 3y = - 4 \qquad \bf\lgroup{equation \: 1} \rgroup$

$\:$ ━━━━━━━━━━━━━━━━━━━━━━━━

• Second condition

If the denominator is 6 more than the numerator.

$\implies \sf y = x + 6$

$\implies \sf x - y = - 6 \quad\bf \lgroup{equation \: 2} \rgroup$

$\:$ ━━━━━━━━━━━━━━━━━━━━━━━━

Solve for the value x & y by adding & subtracting method. Multiply equation (1) by 1 & multiply equation (2) by 5. So the equations become

• 5x - 3y = - 4
• 5x - 5y = - 30

Subtract both the equations

$\implies \sf 5x - 3y - (5x - 5y) = - 4 - ( - 30) \\ \\ \implies \sf \cancel{5x }- 3y - { \cancel{5x}} + 5y = - 4 + 30 \\ \\ \implies \sf - 3y + 5y = 26 \\ \\ \implies \sf 2y = 26 \\ \\ \implies \sf y = \dfrac{ \cancel{26} \: ^{13}}{ \cancel{2}} \\ \\ \therefore \sf \: y = 13$

Final answer

• Original value of denominator = y = 13

$\:$ ━━━━━━━━━━━━━━━━━━━━━━━━

Answer 2

${\bf{Given\::}}$

• When 2 is added to both the numerator and denominator of a fraction, then the fraction is ³/₅.

• The denominator is 6 more than the numerator.

$\\ {\bf{To\: Find\::}}$

• The original value of denominator.

$\\ {\bf{Solution\::}}$

Let,

• Numerator be x.

• Denominator be y.

➣ When 2 is added to both the numerator and denominator, then

⇢ Numerator = (x + 2)

⇢ Denominator = (y + 2)

According to the question,

☛ When 2 is added to both the numerator and denominator of a fraction, then the fraction is ³/₅.

➛ $\tt{\dfrac{x\:+\:2}{y\:+\:2}\:=\:\dfrac{3}{5}\:}$

➛ $\tt{5\times(x\:+\:2)\:=\:3\times(y\:+\:2)\:}$

➛ $\tt{5x\:+\:10\:=\:3y\:+\:6\:}$

➛ $\tt{5x\:-\:3y\:=\:6\:-\:10\:}$

➛ $\bf{5x\:-\:3y\:=\:-4\:}---(1)$

Given that,

☛ The denominator is 6 more than the numerator.

➛ y = x + 6

➛ x = y - 6 ---(2)

Now,

➣ Putting the value of y in equation (1), we get

➠ [5 × (y - 6)] - 3y = - 4

➠ 5y - 30 - 3y = - 4

➠ 5y - 3y = - 4 + 30

➠ 2y = 26

➠ y = $\tt{\dfrac{26}{2}}$

➠ y = 13

∴ The original value of denominator is 13.