Question

7. A fraction becomes 1/3 when 1 is subtracted from the numerator and it becomes 1/4 when 8 is added to its denominator. The fraction obtained is:

Answer 1

Fraction – 5⁄12

Question states that a fraction becomes ⅓ when 1 is subtracted from the numerator of the fraction. Similarly, when 8 is added to the denominator of the fraction the fraction becomes ¼. & we've to calculate the fraction.

Let's say, that the Numerator be x and Denominator be y of the fraction.

• C A S E : I

• A fraction becomes ⅓ when 1 is subtracted from the numerator of the fraction.

↠⠀⠀ (x – 1)/y = ⅓

↠⠀⠀ 3(x – 1) = y

↠⠀⠀ 3x – 3 = y

↠⠀⠀ 3x – y = 3 ⠀⠀⠀⠀⠀ ⠀⠀⠀⠀ ⠀ —eqₙ ( i )

• C A S E : I I

• A fraction becomes ¼ when 8 is added to the denominator of the fraction.

↠⠀⠀x/(y + 8) = ¼

↠⠀⠀4x = y + 8

↠⠀⠀4x – y = 8⠀⠀ ⠀⠀ ⠀⠀⠀⠀⠀ ⠀ —eqₙ ( ii )

⠀⠀⠀━━━━━━━━━━━━━━━━━━━━━━━━━━━━

• E L I M I N A T I O N⠀M E T H O D :

↠⠀⠀ 3x – y = 3

↠⠀⠀ 4x – y = 8

↠⠀⠀ x = 5

• Putting the value of x in eqₙ ( i ) :

↠⠀⠀ 3x – y = 3

↠⠀⠀ 3(5) – y = 3

↠⠀⠀ 15 – y = 3

↠⠀⠀ – y = 3 – 15

↠⠀⠀ – y = – 12

↠⠀⠀ y = 12

$\rule{230px}{.2ex}$

❝ Therefore, the required fraction obtained is 5⁄12 ❞

Answer 2

Given :-

A fraction becomes 1/3 when 1 is subtracted from the numerator and it becomes 1/4 when 8 is added to its denominator.

To Find :-

The Fraction.

Solution :-

Let us assume that , the fraction is

$\quad \qquad { \bigstar { \underline { \boxed { \tt { \dfrac{a}{b} } } } } { \bigstar } }$

According to Question ;

$\quad \qquad { \tt { \dfrac{ a - 1 }{b} = \dfrac{1}{3} } }$

Performing Cross - Multiplication

$\quad { : \longmapsto { \tt { 3 ( a - 1 ) = b × 1 } } }$

$\quad { : \longmapsto { \tt { 3a - 3 = b \quad -----(i) } } }$

Also ;

$\quad \qquad { \tt { \dfrac{a}{b + 8} = \dfrac{1}{4} } }$

Performing Cross - Multiplication

$\quad { : \longmapsto { \tt { 4 × a = 1 × ( b + 8 ) } } }$

$\quad { : \longmapsto { \tt { 4a = b + 8 } } }$

$\quad { : \longmapsto { \tt { 4a - 8 = b \quad ------(ii) } } }$

As the RHS of both eq.(i) and (ii) . So , we can equate their LHS ;

$\quad \qquad { \tt { 4a - 8 = 3a - 3 } }$

$\quad { : \longmapsto { \tt { 4a - 3a = - 3 + 8 } } }$

$\quad { : \longmapsto { \tt { a = 5 } } }$

Now By ( i ) ;

$\quad \qquad { \tt { b = 3a - 3 } }$

$\quad { : \longmapsto { \tt { b = 3 × 5 - 3 } } }$

$\quad { : \longmapsto { \tt { b = 15 - 3 } } }$

$\quad { : \longmapsto { \tt { b = 12 } } }$

Now , Just Put the value of "a" and "b" in our assumption we get The Fraction ;

$\quad \qquad { \bigstar { \underline { \boxed { \tt { \dfrac{a}{b} = \dfrac{5}{12} } } } } { \bigstar } }$

❝ Henceforth , The Required Fraction is 5/12 ❞