Question

A fraction is such that if numerator is multiply by 3 and the denominator is reduce by 2 we get 3/5 but if the numerator is increased by 4 and the denominator is doubled we get 5/14. Find the fraction. ​

Answer 1

Given :

• If numerator is multiply by 3 and the denominator is reduce by 2 we get 3/5.
• If the numerator is increased by 4 and the denominator is doubled we get 5/14.

To Find :

• The original fraction.

Solution :

Let the numerator of the fraction be x.

Let the denominator of the fraction be y.

Fraction → $\mathtt{\dfrac{x}{y}}$

Case 1 :

Numerator → 3 × x = 3x

Denominator → ( y - 2)

★ Fraction → $\mathtt{\dfrac{3x}{(y-2) }}$

Equation :

➣ $\mathtt{\dfrac{3x}{(y-2)}\:=\:{\dfrac{3}{5}}}$

➣ $\mathtt{5(3x)=3(y-2)}$

➣ $\mathtt{15x=3y-6}$

$\mathtt{15x-3y=-6}$ ___(1)

Case 2 :

Numerator → x + 4

Denominator → 2 × y = 2y

★ Fraction → $\mathtt{\dfrac{(x+4)}{(2y) }}$

Equation :

➣ $\mathtt{\dfrac{(x+4)}{2y}\:=\:{\dfrac{5}{14}}}$

➣ $\mathtt{14(x+4)=5(2y)}$

➣ $\mathtt{14x+56=10y}$

$\mathtt{14x-10y=-56}$ ___(2)

★ Divide equation (2) by 2,

$\mathtt{7x-5y=-28}$ ___(3)

★ Multiply equation (1) by 7,

$\mathtt{105x-21y=-42}$ ___(4)

★ Multiply, equation (3) by 15,

$\mathtt{105x-75y=-420}$ ___(5)

★ Solve equation (4) and (5) to find value of x and y.

★ Subtract equation (4) from (5),

$\mathtt{105x-21y-(105x-75y=-42-(-420)}$

➣ $\mathtt{105x-21y-105x+75y=-42+420}$

➣ $\mathtt{-21y+75y=378}$

➣ $\mathtt{54y=378}$

➣ $\mathtt{y\:=\:{\cancel{\dfrac{378}{54}}}}$

➣ $\mathtt{y=7}$

★ Substitute, y = 7 in equation (1),

➣ $\mathtt{15x-3y=-6}$

➣ $\mathtt{15x-3(7)=-6}$

➣ $\mathtt{15x-21=-6}$

➣ $\mathtt{15x=-6+21}$

➣ $\mathtt{15x=15}$

➣ $\mathtt{x={\cancel{\dfrac{15}{15}}}}$

➣ $\mathtt{x=1}$

Fraction :

$\large{\boxed{\sf{\red{Numerator\:=\:x\:=\:1}}}}$

$\large{\boxed{\sf{\red{Denominator\:=\:y\:=\:7}}}}$

$\large{\boxed{\sf{\purple{Fraction\:=\:{\dfrac{x}{y}\:=\:\dfrac{1}{7}}}}}}$

Answer 2

$\huge \fcolorbox{red}{pink}{Solution :)}$

Let ,

The numerator and denominator of the fraction be x and y

Condition 1 : If numerator is multiply by 3 and the denominator is reduced by 2 , we get 3/5

$\sf \hookrightarrow \frac{3x}{y - 2} = \frac{3}{5} \\ \\ \sf \hookrightarrow 15x = 3y - 6 \\ \\ \sf \hookrightarrow15x - 3y = - 6\\ \\ \sf \hookrightarrow 5x - y = - 2 \: - - - \: (i)$

Condition 2 : If numerator is increased by 4 and the denominator is doubled , we get 5/14

$\sf \hookrightarrow \frac{x + 4}{2y} = \frac{5}{14} \\ \\ \sf \hookrightarrow 14x + 56 = 10y \\ \\ \sf \hookrightarrow 14x - 10y = - 56 \\ \\ \sf \hookrightarrow 7x - 5y = - 28 \: - - - \: (ii)$

Multiply eq (i) by 7 and eq (ii) by 5 , we get

$\star \sf \: \: 35x - 7y = - 14 \: - - - \: (iii) \\ \\ \sf \star \: \: 35x - 25y = - 140 \: - - - (iv)$

Subtract eq (iii) from eq (iv) , we get

$\sf \hookrightarrow 35x - 25y - (35x - 7y) = - 140 - ( - 14) \\ \\ \sf \hookrightarrow - 25y + 7y = - 140 + 14 \\ \\ \sf \hookrightarrow - 18y = - 126 \\ \\\sf \hookrightarrow y = \frac{ - 126}{ - 18} \\ \\\sf \hookrightarrow y = 7$

Put the value of y = 7 in eq (iii) , we get

$\sf \hookrightarrow 5x - 7 = -2 \\ \\ \sf \hookrightarrow </p><p>5x = -2 + 7 \\ \\ \sf \hookrightarrow </p><p>5x = 5 \\ \\ \sf \hookrightarrow </p><p>x = 1$

Hence , the required fraction is 1/7

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