Question

please solve the question 18​

Answer 1

$\blue\bigstar$Answer:

• Original Fraction = $\dfrac{3}{11}$

$\pink\bigstar$Given:

• The denominator of a fraction is greater than its numerator by 8. If one is added to the numerator and 4 is subtracted from the denominator, the fraction becomes $\dfrac{4}{7}$

$\red\bigstar$To find:

• The original Fraction.

$\green\bigstar$ Solution:

Let the numerator be x.

$\hookrightarrow$ Numerator = x

Now, the denominator is greater than its numerator by 8.

$\hookrightarrow$ Denominator = x + 8

$\therefore$ the fraction = $\dfrac{x}{x\: + \: 8\:}$

Now, if one is added to the numerator,

$\hookrightarrow$ Numerator = x + 1

And 4 is subtracted from the denominator,

$\hookrightarrow$ Denominator = x + 8 - 4

$\implies$ Denominator = x + 4

So now the fraction becomes = $\dfrac{x\: + \: 1\:}{x \: + \: 4\:}$

But it is given that when 1 is added to the Numerator and 4 is subtracted from the denominator, the fraction becomes $\dfrac{4}{7}$

A/q $\sf\dfrac{x \: + \: 1 \:}{x \: + \: 4}$ = $\sf\dfrac{4}{7}$

$\sf\implies \: 7(x \: + \: 1) \: = \: 4(x \: + \: 4)$

( By Cross Multiplication )

$\sf\implies \: 7(x) \: + \: 7(1) \: = \: 4(x) \: + \: 4(4)$

$\sf\implies \: 7x\: + \: 7 \: = \: 4x \: + \: 16$

$\sf\implies \: 7x \: - \: 4x \: = \: 16 - 7$

$\sf\implies \: 3x \: = \: 9$

$\sf\implies \: x \: = \: \frac{9}{3}$

$\sf \implies \boxed{x \: = 3} \:$

$\therefore$ $\boxed{x \: = \:3}$

So,by putting the value of x ,we get the Original Fraction as,

$\sf \hookrightarrow \: \frac{x}{x \: + \: 8}$

$\sf \hookrightarrow \: \frac{3}{3 \: + \: 8}$

$\sf\hookrightarrow \: \frac{3}{11}$

$\therefore$ Original Fraction = $\dfrac{3}{11}$

$\blue\bigstar$ Verification:

When 1 is added to the Numerator and 4 is subtracted from the denominator in the original Fraction, the fraction should become $\dfrac{4}{7}$

So,

$\hookrightarrow$ $\frac{3 \: + \:1 }{11 \: - \: 4} = \frac{4}{7}$

$\because$LHS = RHS

Hence Verified too.

$\pink\bigstar$ Concepts Used:

• Assumption of numbers in variables
• Equating two expressions
• Substituting the Values
• Cross Multiplication

$\red\bigstar$Extra - Information:

$\hookrightarrow$ The linear equations in one variable is an equation which is expressed in the form of ax + b = 0, where a and b are two integers, and x is a variable and has only one solution.

$\hookrightarrow$ The standard form of linear equations in one variable is represented as:

• ax + b = 0

Where,

• ‘a’ and ‘b’ are real numbers,

• Both ‘a’ and ‘b’ are not equal to zero.

$\hookrightarrow$ How to solve them?

• Step 1: Using LCM, clear the fractions if any.

• Step 2: Simplify both sides of the equation.

• Step 3: Equate the variable.

• Step 4: Verify your answer.