Question

A fraction becomes 1/2 when 1 is subtracted from its numerator and 1 is added to its denominator.also it becomes 1/3 when 6 is subtracted from its numerator and 1 from the denomination.find the original fraction.

Answer 1

Answer:Let the original fraction be x / y

Then, (x+1) / (y+1) = 2/3 (given)

So, 3x +3 = 2y +2

Or, 3x- 2y = 2–3 = -1

Or, 3x - 2y = -1. Eq.1

Also,

(x-1) / (y-1) = 1/2 (given)

So, 2x-2 = y-1

Or, 2x - y = -1 + 2 = 1

Or, 2x- y = 1 Eq. 2

Multiplying EQ. 2 by 2, we have:

4x - 2y = 2 Eq. 3

Subtracting EQ. 1 from EQ. 3, we get:

x=3

Substituting the value of x in EQ.1, we get:

9- 2y = -1

Or, 2y = 9+1 = 10,

Or, y = 5.

So, the original fraction ( x / y ) is 3 / 5

Check:

Adding 1 to the numerator as well as denominator of original fraction 3/ 5, we get : (3+1) / (5+1)= 4 / 6 = 2/3 ✓

By subtracting 1 from the numerator as well as denominator of the original fraction 3 / 5, we get :

(3–1) / (5 - 1) = 2 / 4 = 1 / 2 ✓

Step-by-step explanation:

Answer 2

Answer:-

14/25

Explanation:-

Given :

• A fractions becomes 1/2 when 1 is  subtracted from it’s numerator and 1 is added to it’s denominator.
• It becomes 1/3 when 6 is subtracted from it’s numerator and 1 from the denominator.

To Find :

• The original fraction = ?

Solution :

Let x be the numerator and y be the denominator

Fraction formed :-

$\rightarrow\rm{\dfrac{x}{y}}$

ATQ,

1)   $\rm{\dfrac{x-1}{y-1}}=\rm{\dfrac{1}{2}}$

$\rightarrow \rm{(x-1)2=1(y+1)} \ \ \ \ \rm{[By\ cross\ multiplication]}$

$\rightarrow \rm{2x-1=y+1}$

$\rightarrow \rm{2x-y-3=0}.......(1)$

2)    $\rm{\dfrac{x-6 }{y-1}}=\rm{\dfrac{1}{3}}$

$\rightarrow \rm{3(x-6)=1(y-1)} \ \ \ \ \rm{[By\ cross\ multiplication]}$

$\rightarrow \rm{3x-18=y-1}$

$\rightarrow \rm{3x-18-y+1=0}........(1)$

Subtract equation (i) from (ii)

$\ \ \ \rm{3x - y - 17 = 0}\\ - \rm{2x - y - \ 3 \ = 0}$

$- - - - - - - - - \\ \ \ -\rm{ x\ +0 \ - 14 \ = 0} \\ - - - - - - - - -$

⇒ - x - 14 = 0

∴ x = 14

So, x = 14

Therefore substitute the value of x = 14 in equation (1)

2(14) - y - 3 = 0

⇒ 28 - 3 - y= 0

⇒ 25 - y = 0

⇒ 25 = y

∴ y = 25

x = 14

y = 25

Therefore,the original fraction is :-

$\rm{\dfrac{14}{25}}$