Question

A fraction becomes 1, if we add 1 to the numerator and subtract 1 from the denominator.
It becomes 1/2 if we only add 1 to the denominator. What is the fraction? Guys pls send it fast

Answer 1

Answer:

Step-by-step explanation:

Given :-

If we add 1 to numerator, and subtract 1 from the denomination, a fraction reduce the 1.

It becomes ½ if we only add 1 to the denominator.

To Find :-

The Fraction

Solution :-

Let the fraction be x/y                  

According to the question,

⇒ x + 1/y - 1 = 1                  

⇒ x - y = - 2 ... (i)

⇒ x/y + 1 = 1/2                  

⇒ 2x - y = 1 ... (ii)             

Subtracting equation (i) from equation (ii), we get                  

⇒ x = 3 ... (iii)              

Putting this value in equation (i), we get                  

⇒ 3 - y = - 2                  

⇒ -y =  5                  

⇒ y = 5                  

Hence, the fraction is 3/5

Answer 2

GIVEN :-

• A fraction becomes 1, if we add 1 to the numerator and subtract 1 from the denominator.

• It becomes 1/2 if we only add 1 to the denominator.

TO FIND :-

The fraction.

PRE-REQUISITE :-

Let the numerator of the fraction be x and the denominator be y.

SOLUTION :-

Case-I

$\sf{\dfrac{x+1}{y-1}=1 }$

$\sf{\implies x+1=y-1}\\\\\sf{\implies x-y+2=0}\:\:\:\:\:\:\:\: \cdots(i)$

$\rule{160}{2}$

Case-II

$\sf{\dfrac{x}{y+1}=\dfrac{1}{2} }$

$\sf{\implies y+1=2x}\\\\\sf{\implies 2x-y-1=0}\:\:\:\:\:\:\:\: \cdots(ii)$

$\rule{160}{2}$

Subtracting eq.(i) from eq.(ii) :-

$\sf{\:\:2x-y-1=0}\\-\\\sf{\:\:x-y+2=0}\\\rule{60}{1}\\\sf{\:\:x-3=0}$

$\boxed{\sf{\implies x=3}}$

$\rule{160}{2}$

Putting the value of x in eq.(i) :-

$\sf{x-y+2=0}\\\\\sf{\implies 3+2=y}\\\\\boxed{\sf{\implies y=5}}$

Therefore, the fraction is 3/5.