Question

the sum of the numerator and denominator of a fraction is greator by 1 than thrice the numerator if the numerator is decreased by 1 then the fraction reduce to 1/3 find the fraction​

Answer 1

Answer:

$\frac{4}{9}$

Step-by-step explanation:

Let the fraction be $\frac{x}{y}$

Sum of numerator and denominator = x + y

⇒ x + y = 3x + 1

⇒ y = 1 + 2x          . . . (1)

If the numerator is decreased by 1, the fraction becomes $\frac{x - 1}{y}$

⇒ $\frac{x - 1}{y}$ = $\frac{1}{3}$

⇒ 3x - 3 = y          . . . (2)

Equating (1) and (2)

3x - 3 = 1 + 2x

⇒ x = 4

and y = 1 + 2 * (4) = 9          from (1)

So, the fraction $\frac{x}{y}$ = $\frac{4}{9}$

Answer 2

Given :-

The sum of numerator and denominator of a fraction is greater by 1 than thrice the numerator is decreased by 1 then the fraction reduced to 1/3.

To Find :-

What is the fraction.

Solution :-

Let, the fraction be x/y

According to the question,

➣ The sum of numerator and denominator of a fraction is greater by 1 than thrice the numerator.

⇒ x + y = 3x + 1

⇒ y = 3x - x + 1

⇒ y = 2x + 1 .......... equation no (1)

➣ Denominator is decreased by 1 than the fraction reduced by 1/3.

⇒ x - 1/y = 1/3

⇒ 3(x - 1) = y

⇒ 3x - 3 = y

⇒ 3x - y = 3 ......... equation no (2)

➣ Now, putting the value of x from equation no (1) in equation no (2) we get,

⇒ 3x - y = 3

⇒ 3x - (2x + 1) = 3

⇒ 3x - 2x - 1 = 3

⇒ x = 3 + 1

➠ x = 4

➣ Again, putting the value of x in the equation no (1) we get,

⇒ y = 2x + 1

⇒ y = 2(4) + 1

⇒ y = 8 + 1

➥ y = 9

Hence, the required fraction will be,

↦ x/y

➽ 4/9

∴ The fraction will be 4/9 .