Question

a fraction becomes 1/3 if 1 is subtracted from both its numerator and dinominator if 1 is added to both numerator and dinominator, it becomes 1/2 find the fraction​

Answer 1

Answer:

3/7

Step-by-step explanation:

Let the required fraction be 'x/y'  (x is numerator and y is denominator)

  When 1 is subtracted from x and y:

⇒ (x - 1)/(y - 1) = 1/3

⇒ 3(x - 1) = y - 1

⇒ 3x - 3 = y - 1

⇒ 3x - 2 = y         ...(1)

    When 1 is added to x and y:

⇒ (x + 1)/(y + 1) = 1/2

⇒ 2(x + 1) = (y + 1)

⇒ 2x + 2 = y + 1

⇒ 2x + 1 = y

⇒ 2x + 1 = 3x - 2        [from (1)]

⇒ 1 + 2 = 3x - 2x

⇒ 3 = x

              Therefore, y = 3(3) - 2 = 7

Hence the required fraction is x/y = 3/7

Answer 2

Answer:

• $\bold{\dfrac{3}{7} }$ is the required fraction.

Step-by-step explanation:

Given :

A fraction becomes $\bold{\dfrac{1}{3} }$ if 1 is subtracted from both its numerator and denominator.

If 1 is added to both numerator and denominator, it becomes $\bold{\dfrac{1}{2} }$.

To find :

What is the fraction ?

Solution :

• Let the fraction be $\bold{\dfrac{x}{y} }$.

1 is subtracted from both its numerator and denominator. It becomes  $\bold{\dfrac{1}{3} }$ .

• $\bold{\dfrac{x-1}{y-1}=\dfrac{1}{3} }$

$\bold{3(x-1)=y-1}$

$\bold{3x-3=y-1}$

$\bold{3x-y=-1+3}$

$\bold{3x-y=2}$

• $\bold{3x-2=y}$ - (1)

1 is added to both its numerator and denominator. It becomes $\bold{\dfrac{1}{2} }$.

• $\bold{\dfrac{x+1}{y+1}=\dfrac{1}{2} }$

$\bold{2(x+1)=y+1}$

$\bold{2x+2=y+1}$

$\bold{2x+2-1=y}$

• $\bold{2x+1=y}$

Now, substitute (1) in y place.

$\bold{2x+1=3x-2}$

$\bold{2x-3x=-2-1}$

$\bold{-x=-3}$

• $\bold{x=3}$

Now, substitute (x) value in (1).

$\bold{3x-2=y}$

$\bold{3(3)-2=y}$

$\bold{9-2=y}$

• $\bold{y=7}$

∴ The fraction is $\bold{\dfrac{x}{y}=\dfrac{3}{7} }$