Question

Lcm of 144 and 95 is​

Answer 1

Given :

The numerator of a fraction is 4 less than the denominator .

If 1 is subtracted from numerator and 3 is added to the denominator the fraction becomes 1/5 .

To Find :

Original Fraction .

Solution :

$\longmapsto\tt{Let\:Denominator\:be=x}$

As Given that The numerator of a fraction is 4 less than the denominator . So ,

$\longmapsto\tt{Numerator=x-4}$

Now ,

If 1 is subtracted from numerator and 3 is added to the denominator the fraction becomes 1/5 .

$\longmapsto\tt{Numerator=x-4-1=x-5}$

$\longmapsto\tt{Denominator=x+3}$

A.T.Q :

$\longmapsto\tt{\dfrac{x-5}{x+3}=\dfrac{1}{5}}$

$\longmapsto\tt{5(x-5)=1(x+3)}$

$\longmapsto\tt{5x-25=x+3}$

$\longmapsto\tt{5x-x=3+25}$

$\longmapsto\tt{4x=28}$

$\longmapsto\tt{x=\cancel\dfrac{28}{4}}$

$\longmapsto\tt\bf{x=7}$

Value of x is 7 .

Therefore :

$\longmapsto\tt{Numerator=7-4}$

$\longmapsto\tt\bf{3}$

$\longmapsto\tt{Denominator=x}$

$\longmapsto\tt\bf{7}$

So , The Fraction is 3/7 .

Answer 2

Given :

The numerator of a fraction is 4 less than the denominator .

If 1 is subtracted from numerator and 3 is added to the denominator the fraction becomes 1/5 .

To Find :

Original Fraction .

Solution :

$\longmapsto\tt{Let\:Denominator\:be=x}$

As Given that The numerator of a fraction is 4 less than the denominator . So ,

$\longmapsto\tt{Numerator=x-4}$

Now ,

If 1 is subtracted from numerator and 3 is added to the denominator the fraction becomes 1/5 .

$\longmapsto\tt{Numerator=x-4-1=x-5}$

$\longmapsto\tt{Denominator=x+3}$

A.T.Q :

$\longmapsto\tt{\dfrac{x-5}{x+3}=\dfrac{1}{5}}$

$\longmapsto\tt{5(x-5)=1(x+3)}$

$\longmapsto\tt{5x-25=x+3}$

$\longmapsto\tt{5x-x=3+25}$

$\longmapsto\tt{4x=28}$

$\longmapsto\tt{x=\cancel\dfrac{28}{4}}$

$\longmapsto\tt\bf{x=7}$

Value of x is 7 .

Therefore :

$\longmapsto\tt{Numerator=7-4}$

$\longmapsto\tt\bf{3}$

$\longmapsto\tt{Denominator=x}$

$\longmapsto\tt\bf{7}$

So , The Fraction is 3/7 .

Answer 3

Given :

The numerator of a fraction is 4 less than the denominator .

If 1 is subtracted from numerator and 3 is added to the denominator the fraction becomes 1/5 .

To Find :

Original Fraction .

Solution :

$\longmapsto\tt{Let\:Denominator\:be=x}$

As Given that The numerator of a fraction is 4 less than the denominator . So ,

$\longmapsto\tt{Numerator=x-4}$

Now ,

If 1 is subtracted from numerator and 3 is added to the denominator the fraction becomes 1/5 .

$\longmapsto\tt{Numerator=x-4-1=x-5}$

$\longmapsto\tt{Denominator=x+3}$

A.T.Q :

$\longmapsto\tt{\dfrac{x-5}{x+3}=\dfrac{1}{5}}$

$\longmapsto\tt{5(x-5)=1(x+3)}$

$\longmapsto\tt{5x-25=x+3}$

$\longmapsto\tt{5x-x=3+25}$

$\longmapsto\tt{4x=28}$

$\longmapsto\tt{x=\cancel\dfrac{28}{4}}$

$\longmapsto\tt\bf{x=7}$

Value of x is 7 .

Therefore :

$\longmapsto\tt{Numerator=7-4}$

$\longmapsto\tt\bf{3}$

$\longmapsto\tt{Denominator=x}$

$\longmapsto\tt\bf{7}$

So , The Fraction is 3/7 .

Answer 4

Given :

The numerator of a fraction is 4 less than the denominator .

If 1 is subtracted from numerator and 3 is added to the denominator the fraction becomes 1/5 .

To Find :

Original Fraction .

Solution :

$\longmapsto\tt{Let\:Denominator\:be=x}$

As Given that The numerator of a fraction is 4 less than the denominator . So ,

$\longmapsto\tt{Numerator=x-4}$

Now ,

If 1 is subtracted from numerator and 3 is added to the denominator the fraction becomes 1/5 .

$\longmapsto\tt{Numerator=x-4-1=x-5}$

$\longmapsto\tt{Denominator=x+3}$

A.T.Q :

$\longmapsto\tt{\dfrac{x-5}{x+3}=\dfrac{1}{5}}$

$\longmapsto\tt{5(x-5)=1(x+3)}$

$\longmapsto\tt{5x-25=x+3}$

$\longmapsto\tt{5x-x=3+25}$

$\longmapsto\tt{4x=28}$

$\longmapsto\tt{x=\cancel\dfrac{28}{4}}$

$\longmapsto\tt\bf{x=7}$

Value of x is 7 .

Therefore :

$\longmapsto\tt{Numerator=7-4}$

$\longmapsto\tt\bf{3}$

$\longmapsto\tt{Denominator=x}$

$\longmapsto\tt\bf{7}$

So , The Fraction is 3/7 .

Answer 5

Given :

The numerator of a fraction is 4 less than the denominator .

If 1 is subtracted from numerator and 3 is added to the denominator the fraction becomes 1/5 .

To Find :

Original Fraction .

Solution :

$\longmapsto\tt{Let\:Denominator\:be=x}$

As Given that The numerator of a fraction is 4 less than the denominator . So ,

$\longmapsto\tt{Numerator=x-4}$

Now ,

If 1 is subtracted from numerator and 3 is added to the denominator the fraction becomes 1/5 .

$\longmapsto\tt{Numerator=x-4-1=x-5}$

$\longmapsto\tt{Denominator=x+3}$

A.T.Q :

$\longmapsto\tt{\dfrac{x-5}{x+3}=\dfrac{1}{5}}$

$\longmapsto\tt{5(x-5)=1(x+3)}$

$\longmapsto\tt{5x-25=x+3}$

$\longmapsto\tt{5x-x=3+25}$

$\longmapsto\tt{4x=28}$

$\longmapsto\tt{x=\cancel\dfrac{28}{4}}$

$\longmapsto\tt\bf{x=7}$

Value of x is 7 .

Therefore :

$\longmapsto\tt{Numerator=7-4}$

$\longmapsto\tt\bf{3}$

$\longmapsto\tt{Denominator=x}$

$\longmapsto\tt\bf{7}$

So , The Fraction is 3/7 .

Answer 6

Given :

The numerator of a fraction is 4 less than the denominator .

If 1 is subtracted from numerator and 3 is added to the denominator the fraction becomes 1/5 .

To Find :

Original Fraction .

Solution :

$\longmapsto\tt{Let\:Denominator\:be=x}$

As Given that The numerator of a fraction is 4 less than the denominator . So ,

$\longmapsto\tt{Numerator=x-4}$

Now ,

If 1 is subtracted from numerator and 3 is added to the denominator the fraction becomes 1/5 .

$\longmapsto\tt{Numerator=x-4-1=x-5}$

$\longmapsto\tt{Denominator=x+3}$

A.T.Q :

$\longmapsto\tt{\dfrac{x-5}{x+3}=\dfrac{1}{5}}$

$\longmapsto\tt{5(x-5)=1(x+3)}$

$\longmapsto\tt{5x-25=x+3}$

$\longmapsto\tt{5x-x=3+25}$

$\longmapsto\tt{4x=28}$

$\longmapsto\tt{x=\cancel\dfrac{28}{4}}$

$\longmapsto\tt\bf{x=7}$

Value of x is 7 .

Therefore :

$\longmapsto\tt{Numerator=7-4}$

$\longmapsto\tt\bf{3}$

$\longmapsto\tt{Denominator=x}$

$\longmapsto\tt\bf{7}$

So , The Fraction is 3/7 .

Answer 7

Given :

The numerator of a fraction is 4 less than the denominator .

If 1 is subtracted from numerator and 3 is added to the denominator the fraction becomes 1/5 .

To Find :

Original Fraction .

Solution :

$\longmapsto\tt{Let\:Denominator\:be=x}$

As Given that The numerator of a fraction is 4 less than the denominator . So ,

$\longmapsto\tt{Numerator=x-4}$

Now ,

If 1 is subtracted from numerator and 3 is added to the denominator the fraction becomes 1/5 .

$\longmapsto\tt{Numerator=x-4-1=x-5}$

$\longmapsto\tt{Denominator=x+3}$

A.T.Q :

$\longmapsto\tt{\dfrac{x-5}{x+3}=\dfrac{1}{5}}$

$\longmapsto\tt{5(x-5)=1(x+3)}$

$\longmapsto\tt{5x-25=x+3}$

$\longmapsto\tt{5x-x=3+25}$

$\longmapsto\tt{4x=28}$

$\longmapsto\tt{x=\cancel\dfrac{28}{4}}$

$\longmapsto\tt\bf{x=7}$

Value of x is 7 .

Therefore :

$\longmapsto\tt{Numerator=7-4}$

$\longmapsto\tt\bf{3}$

$\longmapsto\tt{Denominator=x}$

$\longmapsto\tt\bf{7}$

So , The Fraction is 3/7 .