Question

if 1 is subtracted from the numerator and 2 is added to the denominator of a fraction the value obtained is 1/2. if 3 is added to its numerator and 2 is subtracted from its denominator the resulting value is 5/4. find the fraction.​

Answer 1

Given,

• If 1 is subtracted from the numerator and 2 is added to the denominator of a fraction the value obtained is 1/2.
• If 3 is added to its numerator and 2 is subtracted from its denominator the resulting value is 5/4.

To Find,

• The Fraction

Solution :

⇒Suppose the Numerator be a

And,Suppose the Denominator be b

Therefore,

$\boxed{\sf{The \ Fraction = \dfrac{a}{b}}}$

$\mapsto \underline{\sf{\pink{According \ to \ the \ First \ Condition :}}}$

• If 1 is subtracted from the numerator and 2 is added to the denominator of a fraction the value obtained is 1/2.

$\implies \sf{\dfrac{a - 1}{b + 2} = \dfrac{1}{2}}$

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

$\implies \sf{2a - 2 = b + 2}$

$\implies \sf{2a - b = 2 + 2}$

$\implies \sf{2a - b = 4}$

$\implies \boxed{\sf{b = 2a - 4}}$

$\mapsto \underline{\sf{\pink{According \ to \ the \ Second \ Condition :}}}$

• If 3 is added to its numerator and 2 is subtracted from its denominator the resulting value is 5/4.

$\implies \sf{\dfrac{a + 3}{b - 2} = \dfrac{5}{4}}$

$\implies \sf{4(a + 3) = 5(b - 2)}$

$\implies \sf{4a + 12 = 5b - 10}$

$\implies \sf{4a - 5b = - 10 - 12}$

$\implies \sf{4a - 5b = -22}$

(Now put the Value of b From the First Condition)

$\implies \sf{4a - 5(2a - 4) = -22}$

$\implies \sf{4a - 10a + 20 = -22}$

$\implies \sf{-6a = -22 -20}$

$\implies \sf{-6a = -42}$

$\implies \sf{a = \dfrac{42}{6}}$

$\implies \boxed{\sf{a = 7}}$

Now Put the Value of a in First Condition :-

$\implies \sf{b = 2a - 4}$

$\implies \sf{b = 2(7) - 4}$

$\implies \sf{b = 14 - 4}$

$\implies \boxed{\sf{b = 10}}$

Therefore,

$\boxed{\bold{\purple{The \ Fraction = \dfrac{a}{b} = \dfrac{7}{10}}}}$

$\rule{200}2$

Answer 2

Answer:

Let the Numerator be N and Denominator be D of the Fraction respectively.

$\underline{\bigstar\:\textsf{According to 1st Statement :}}$

$:\implies\sf \dfrac{Numerator-1}{Denominator+2}=\dfrac{1}{2}$

$:\implies\sf\dfrac{N-1}{D+2}=\dfrac{1}{2}$

$:\implies$ 2(N – 1) = 1(D + 2)

$:\implies$ 2N – 2 = D + 2

$:\implies$ 2N – 2 – 2 = D

$:\implies$ D = 2N – 4 ⠀⠀⠀⠀— eq. ( I )

$\rule{90}{1}$

$\underline{\bigstar\:\textsf{According to 2nd Statement :}}$

$:\implies\sf \dfrac{Numerator+3}{Denominator-2}=\dfrac{5}{4}$

$:\implies\sf\dfrac{N+3}{D-2}=\dfrac{5}{4}$

$:\implies$ 4(N + 3) = 5(D – 2)

$:\implies$ 4N + 12 = 5D – 10

• Putting the value of D from eq. ( I )

$:\implies$ 4N + 12 = 5(2N – 4) – 10

$:\implies$ 4N + 12 = 10N – 20 – 10

$:\implies$ 12 + 20 + 10 = 10N – 4N

$:\implies$ 42 = 6N

$:\implies\sf\dfrac{42}{6}=N$

$:\implies$ N = 7

$\rule{90}{1}$

$\underline{\bigstar\:\textsf{Putting value of N in eq. ( I ) :}}$

$\dashrightarrow\:$ D = 2N – 4

$\dashrightarrow\:$ D = 2(7) – 4

$\dashrightarrow\:$ D = 14 – 4

$\dashrightarrow\:$ D = 10

$\rule{170}{2}$

$\bf{\dag}\:\underline{\boxed{\sf Fraction=\dfrac{Numerator}{Denominator}= \dfrac{N}{D} = \dfrac{\textsf{\textbf{7}}}{\textsf{\textbf{10}}}}}$