Question

in a fraction if one is added to both numerator and dinominator it becomes 1/2. if 1 is subtracted from both numerator and dinominator it become 1/3. find the fraction​

Answer 1

• $\LARGE\boxed{\sf{\purple{Fraction = \dfrac{3}{7}}}}$

Explanation :

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• In a fraction if 1 is added to both numerator and denominator it becomes 1/2. if 1 is subtracted from both numerator and denominator it become 1/3. Find the fraction.

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• Let the fraction be p/q

A/q,

$\qquad\leadsto\quad\sf \dfrac{p + 1}{q + 1} = \dfrac{1}{2} \qquad\qquad\bigg\lgroup \sf{eq^{n}} \:(1) \bigg\rgroup$

$\qquad\leadsto\quad\sf \dfrac{p - 1}{q - 1} = \dfrac{1}{3} \qquad\qquad\bigg\lgroup \sf{eq^{n}} \:(2) \bigg\rgroup$

★ Using substitution method :-

• Finding value of variable "q" from eqⁿ (1) :

$\qquad\rightarrow\quad\sf \dfrac{p + 1}{q + 1} = \dfrac{1}{2}$

• By cross multiplication :-

$\qquad\rightarrow\quad\sf 2(p + 1) = 1(q + 1)$

$\qquad\rightarrow\quad\sf 2p + 2 = q + 1$

$\qquad\rightarrow\quad\sf 2p + 2 - 1 = q$

$\qquad\rightarrow\quad\bf {q = \red{2p + 1}} \qquad\qquad\bigg\lgroup \sf{eq^{n}} \:(3) \bigg\rgroup$

★ Putting value of q in eqⁿ (2) :-

$\qquad\rightarrow\quad\sf \dfrac{p - 1}{(2p + 1) - 1} = \dfrac{1}{3}$

$\qquad\rightarrow\quad\sf \dfrac{p - 1}{2p + 1 - 1} = \dfrac{1}{3}$

$\qquad\rightarrow\quad\sf \dfrac{p - 1}{2p + \cancel{1} - \cancel{1}} = \dfrac{1}{3}$

$\qquad\rightarrow\quad\sf \dfrac{p - 1}{2p} = \dfrac{1}{3}$

• By cross multiplication :-

$\qquad\rightarrow\quad\sf 3(p - 1) = 1(2p)$

$\qquad\rightarrow\quad\sf 3p - 3 = 2p$

$\qquad\rightarrow\quad\sf 3p - 2p = 3$

$\qquad\rightarrow\quad{\boxed{\frak{\orange{p = 3}}}}\:\blue{\bigstar}$

★ Putting value of p in eqⁿ (3) :

$\qquad\rightarrow\quad\sf q = 2(3) + 1$

$\qquad\rightarrow\quad\sf q = 6 + 1$

$\qquad\rightarrow\quad{\boxed{\frak{\orange{q = 7}}}}\:\blue{\bigstar}$

$\quad\red{\therefore}\:{\underline{\sf{Hence,\:fraction = \bf{\frac{p}{q} = \frac{3}{7}}\:\sf{respectively.}}}}$

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ㅤㅤㅤ✰ Substitution method ✰

• It is method used to solve linear equations. In this method the value of one variable from one equation is substituted in other equation.

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Answer 2

Solution :

• In a fraction , if one is added to both the numerator and the denominator , it becomes 1/2 .

• If 1 is subtracted from both the numerator and denominator it becomes 1/3.

We have to find the required fraction .

Let us assume that the fraction is x/y .

First Case :

One is added to both the numerator and denominator and it becomes ½ .

> ( x+1)/(y+1) = ½ .

Second Case -

If one is subtracted from both the numerator and denominator , it becomes ⅓.

> ( x - 1)/( y - 1) = ⅓

Expanding the first one

(x+1)/(y+1) = ½.

> 2x + 2 = y + 1

> 2x - y = -1

Expanding the second one

> (x-1)/(y-1) = ⅓.

> 3x - 3 = y - 1

> 3x - y = 2

Subtracting this from the result we got in the first

(2x - y) - ( 3x - y) = -1-2

> 2x - y - 3x + y = -3

> -x = -3

> x = 3

> y = 7 .

Thus the fraction becomes 3/7. This is the required answer.

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