Question

The sum of numerator and denominator of a fraction is 4 more than twice the numerator. If the numerator and denominator are decreased by 1, the numerator becomes half the denominator. Find the fraction.​

Answer 1

Answer:

5/9

Explanation:

let  the fraction be x/y, so x is numerator and y is denominator

given:

sum of numerator and denominator = 4 more than twice the numerator

x+y = 2x+4

y = x+4       --------1

the numerator and denominator are decreased by 1,  the numerator becomes half the denominator

x-1 = (y-1)/2

2x-2 = y-1

y = 2x-1  -----------2

equate 1 and 2 equations,

2x-1= x+4

2x-x=4+1

X= 5

sustitute X=5 in 1,

y =5+4 = 9

so, the fraction is 5/9

Answer 2

$\large \underline{\pmb{\mathfrak{\red{Given....}}}}$

Given that, the sum of a numerator and denominator of a fraction is 4 more than twice the numerator. If the numerator and denominator are decreased by 1, the numerator becomes half the denominator.

$\large \underline{\pmb{\mathfrak{\red{To~ find....}}}}$

We have to find the fraction.

$\large \underline{\pmb{\mathfrak{\red{Solution....}}}}$

Let the fraction be $\sf \dfrac{a}{b}$ with a as numerator and b as denominator.

The sum of numerator and denominator is 4 more than twice the numerator.

$\: \: \: \sf : \implies \: a + b = 2a + 4$

$\: \: \: \sf : \implies \: a + b - 2a - 4 = 0$

$\: \: \: \red{ \sf : \implies \: - a + b - 4 = 0 \: \: \: \: ....(eq. \: 1)}$

The numerator becomes half the denominator when both are decreased by 1.

$\: \: \: \sf : \implies \: (a - 1) = \dfrac{1}{2} (b - 1)$

$\: \: \: \sf : \implies \: \dfrac{(a - 1)}{(b - 1)} = \dfrac{1}{2}$

Cross multiplying them :

$\: \: \: \sf : \implies \: 2(a - 1) = (b - 1)$

$\: \: \: \sf : \implies \: 2a - 2 - b + 1 = 0$

$\: \: \: \red{\sf : \implies \: 2a - b - 1 = 0 \: \: \: \: \: ....(eq. \: 2)}$

Solving (eq. 1) and (eq. 2) for getting a and b using substituting method :

$\: \: \: \sf : \implies \: - a + b = 4$

$\: \: \: \red{ \sf: \implies \: b = 4 + a \: \: \: \: ....(eq .\: 3)}$

Substituting (eq. 3) in (eq. 2) :

$\: \: \: \sf : \implies \: 2a - (4 + a) - 1 = 0$

$\: \: \: \sf : \implies \: 2a - 4 - a - 1 = 0$

$\: \: \: \sf : \implies \: a - 5 = 0$

$\: \: \: \red{ \sf : \implies \: a = 5}$

Substituting a value in (eq. 1) to get b :

$\: \: \: \sf : \implies \: - a + b - 4 = 0$

$\: \: \: \sf : \implies \: - 5 + b - 4 = 0$

$\: \: \: \sf: \implies \: b - 9 = 0$

$\: \: \: \red{\sf : \implies \: b = 9}$

_________________________

Therefore,the fraction is $\bf \dfrac{5}{9}.$