Question

The sum of numerator and denominator of a fraction is greater by 1 tan thrice the numerator if numerator is decreased by 1 then the fraction reduced to 1/3 . Find the fraction

Answer 1

GIVEN:

• The sum of numerator and denominator of a fraction is greater by 1 than thrice the numerator.
• Numerator is decreased by 1 then the fraction reduced to 1/3.

TO FIND:

• What is the fraction ?

SOLUTION:

Let the fraction be x/y

CASE:- 1)

• The sum of numerator and denominator of a fraction is greater by 1 than thrice the numerator.

According to question:-

$\sf{\longmapsto x + y = 3x +1 }$

$\sf{\longmapsto y = 3x -x+1}$

$\bf{\longmapsto y = 2x +1...1) }$

CASE:- 2)

• Numerator is decreased by 1 then the fraction reduced to 1/3.

According to question:-

$\sf{\longmapsto \dfrac{x-1}{y} = \dfrac{1}{3}}$

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

$\sf{\longmapsto 3x -3 = y }$

$\bf{\longmapsto 3x -y = 3....2) }$

Put the value of 'x' from equation 1 in equation 2

$\sf{\longmapsto 3x -(2x +1) = 3 }$

$\sf{\longmapsto 3x -2x -1 = 3 }$

$\sf{\longmapsto x = 3+1 }$

$\bf{\longmapsto x= 4 }$

Put the value of 'x' in equation 1)

$\sf{\longmapsto y = 2 \times 4 +1}$

$\sf{\longmapsto y = 8+1 }$

$\bf{\longmapsto y = 9 }$

• NUMERATOR = x = 4
• DENOMINATOR = y = 9

$\large{\boxed{\bf{\star \: FRACTION = \dfrac{x}{y} = \dfrac{4}{9} \: \star}}}$

❝ Hence, the fraction become is 4/9 ❞

______________________

Answer 2

UR QUESTION:-

$\huge{\mathbf {Q.}}$The sum of numerator and denominator of a fraction is greater by 1 tan thrice the numerator if numerator is decreased by 1 then the fraction reduced to 1\3. Find the fraction

UR ANSWER ✓

$\Large\underline\bold{GIVEN,}$

The sum of numerator and denominator of a fraction is greater by 1 than thrice the numerator. Numerator is decreased by 1 then the fraction reduced to 1\3

$\Large\underline\bold{TO\:FIND,}$

$\sf\dashrightarrow FRACTION$

$\Large\underline\bold{SOLUTION,}$

$\sf\dashrightarrow\:let\:the\:fraction\:be\: \dfrac{a}{b}$

$\Large\underline\bold{NOW,,}$

$\sf\underline\bold{taking\:2\:cases}$

$\large {\fbox{CASE:-1}}$

$\sf\bold{we\:know,}$

the sum of both numerator and denominator is greater by 1 and then thrice the numerator

$\Large\underline\bold{ATQ.........i.e,....\:according\:to\:the\:question}$

$\sf{\implies a+b=3a+1}$

$\sf{\implies b=3a-a+1}$

$\sf{\implies b=2a+1...........eq^1}$

$\large{\fbox{CASE:-2}}$

$\sf\therefore if \:numerator\:is\:decreased\:by\:1\:then\:the\:fraction\:reduced\:to\:\dfrac{1}{3}$

$\Large\underline\bold{ATQ.........i.e,.... according\:to\:the\:question}$

$\sf{\implies \dfrac{a-1}{b} = \dfrac{1}{3}}$

$\sf{\implies 3(a-1) =b}$

$\sf{\implies 3a-3=b}$

$\sf{\implies 3a-b=3...........eq^2}$

$\sf\therefore substitute\:value\:of\:b\:in\:eq^2$

$\sf{\implies 3a-(2a +1)=3}$

$\sf{\implies 3a-2a-1 =3}$

$\sf{\implies a=3+1}$

$\sf{\implies a=4}$

$\sf {\fbox{a=4}}$

$\sf\therefore substitute\:value\:of\:a\:in\:eq^1$

$\sf{\implies b=2a+1...........eq^1}$

$\sf{\implies b=2 \times 4+1}$

$\sf{\implies b=8+1}$

$\sf{\implies b=9}$

$\sf {\fbox{b=9}}$

$\large\therefore \dfrac{a}{b} = \dfrac{4}{9}$

$\sf\therefore \dfrac{numerator}{denominator} = \dfrac{4}{9}$