Question

A fraction becomes 2/3 if 1 is added to both to the numerrator and denominator .if 3 is added to both to the numerator and denominator it becomes 4/5 then the fraction is

Answer 1

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$\huge{ \underline{ \sf{ \purple{Answer...}}}}$

♣️ GiveN:

• When 1 is added to numerator and denominator, fraction becomes 2/3
• When 3 is added to numerator and denominator, fraction becomes 4/5

♣️ To FinD:

• Find the fraction.

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$\huge{ \underline{ \sf{ \purple{ Explanation \: of \: Q.}}}}$

The above question is based on the usage of two variables and for that two different conditions are given, for which we can frame two simultaneous equations and solve for the two different variables. Here, the variables are the numerator and denominator. The linear equation in two variable can be solve by three methods, Substitution, Elimination and Cross multiplication method. I will be using Elimination method, You can also use method of your choice.

So, let's solve by framing and solving equations,

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$\huge{ \underline{ \purple{ \sf{ Solution...}}}}$

Let the num. be x and deno. be y,

$\large{ \sf{ \dashrightarrow{then \: fraction = \Large{\frac{x}{y}}}}}$

Given, When 1 is added to both numerator and denominator, the fraction becomes 2/3

So, the equation would be,

$\Large{ \sf{ \dashrightarrow \: \frac{x + 1}{y + 1} = \frac{2}{3} }}$

Simplifying this equation,

$\large{ \sf{ \dashrightarrow \: 3(x + 1) = 2(y + 1)}} \\ \\ \large{ \sf{ \dashrightarrow \: 3x + 3 = 2y + 2}} \\ \\ \large{ \sf{ \dashrightarrow \: 3x - 2y = 2 - 3}} \\ \\ \large{ \sf{ \dashrightarrow \: 3x - 2y = - 1........(1)}}$

Now, also given that when numerator and denominator are both added with 3, the fraction becomes 4/5

So equation would be,

$\Large{ \sf{ \dashrightarrow \: \frac{x + 3}{y + 3} = \frac{4}{5} }}$

Simplifying the equation,

$\large{ \sf{ \dashrightarrow \: 5(x + 3) = 4(y + 3)}} \\ \\ \large{ \sf{ \dashrightarrow \: 5x + 15 = 4y + 12}} \\ \\ \large{ \sf{ \dashrightarrow \: 5x - 4y = 12 - 15}} \\ \\ \large{ \sf{ \dashrightarrow \: 5x - 4y = - 3........(2)}}$

Multiplying eq.(1) with 2,

$\large{ \sf{ \dashrightarrow \: 2(3x - 2y) = 2 \times - 1}} \\ \\ \large{ \sf{ \dashrightarrow \: 6x - 4y = - 2}}$

Subtracting eq.(2) from eq.(1),

$\large{ \sf{ \dashrightarrow \: 6x - 4y - (5x - 4y) = - 2 - ( - 3)}} \\ \\ \large{ \sf{ \dashrightarrow \: 6x - \cancel{4y} - 5x + \cancel{4y} = - 2 + 3}} \\ \\ \large{ \dashrightarrow \: \boxed{ \green{ \sf{x = 1}}}}$

Putting value of x in eq.(1),

$\large{ \sf{ \dashrightarrow \: 3(1) - 2y = - 1}} \\ \\ \large{ \sf{ \dashrightarrow \: 3 - 2y = - 1}} \\ \\ \large{ \sf{ \dashrightarrow \: - 2y = - 1 - 3}} \\ \\ \large{ \sf{ \dashrightarrow \: y = \cancel{\frac{ - 4}{ - 2}}}} \\ \\ \large{ \sf{ \dashrightarrow \: \boxed{ \green{ \sf{y = 2}}}}} \\ \\ \large{ \therefore{ \sf{ \underline{ \red{then \: the \: fraction( \frac{x}{y} ) = \frac{1}{2}}}}}}$

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Answer 2

ANSWER :–

$\\ \longrightarrow{ \red{ \bold{ Fraction \: \: = \: \dfrac{1}{2} }}} \:$

EXPLANATION :–

GIVEN :–

• When 1 is added to both to the numerator and denominator then fraction become ⅔.

• When 3 is added to both to the numerator and denominator then fraction becomes ⅘.

TO FIND :–

• Fraction = ?

SOLUTION :–

• Let the fraction be $\: { \bold{ \dfrac{x}{y} }} \: .$

☞ According to the first condition –

$\\ \implies \: { \bold{ \dfrac{x + 1}{y + 1} = \dfrac{2}{3} }} \: \:$

$\\ \implies \: { \bold{3(x + 1) = 2(y + 1) }} \: \:$

$\\ \implies \: { \bold{3x + 3 = 2y + 2 }} \: \:$

$\\ \implies \: { \bold{3x - 2y = 2 - 3 }} \: \:$

$\\ \implies \: { \bold{3x - 2y = - 1 }} \: \:$

$\\ \implies \: { \bold{x = \dfrac{1}{3} (2y - 1) \: \: \: \: - - - - eq.(1)}} \: \:$

☞ According to the second condition –

$\\ \implies \: { \bold{ \dfrac{x + 3}{y + 3} = \dfrac{4}{5} }} \: \:$

$\\ \implies \: { \bold{5(x + 3) = 4(y + 3) }} \: \:$

$\\ \implies \: { \bold{5x + 15 = 4y + 12 }} \: \:$

$\\ \implies \: { \bold{5x - 4y =12 - 15}} \: \:$

$\\ \implies \: { \bold{5x - 4y = - 3}} \: \:$

$\\ \implies \: { \bold{5[ \dfrac{1}{3}(2y - 1) ] - 4y = - 3 \: \: \: \: [using \: \: eq .(1)] }} \: \:$

$\\ \implies \: { \bold{\dfrac{5}{3}(2y - 1) - 4y = - 3 }} \: \:$

$\\ \implies \: { \bold{\dfrac{10y}{3} - \dfrac{5}{3} - 4y = - 3 }} \: \:$

$\\ \implies \: { \bold{\dfrac{10y - 12y}{3} = \dfrac{5}{3} - 3 }} \: \:$

$\\ \implies \: { \bold{\dfrac{- 2y}{3} = - \dfrac{4}{3} }} \: \:$

$\\ \implies \large { \red{ \boxed{ \bold{y = 2 }}}} \: \:$

• Now put the value of 'y' in eq.(1) –

$\\ \implies \: { \bold{x = \dfrac{1}{3} (2 \times 2 - 1) }} \: \:$

$\\ \implies \: { \bold{x = \dfrac{1}{3} (4 - 1) }} \: \:$

$\\ \implies \large { \red{ \boxed{ \bold{x = 1 }}}} \: \:$

Hence , The fraction be $\: { \bold{ \dfrac{1}{2} }} \: .$

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