Question

The numerator of a fraction is 3 less that the denominator are increased by 2the new fraction become 6/7.find the original number

Answer 1

Correct Question:

• The numerator of a fraction is 3 less than the denominator. If both the numerator and denominator are increased by 2, the new fraction becomes 6/7. Find the original fraction.

To Find:

• The original fraction.

Let us assume:

• The denominator of a fraction be x.

The numerator of a fraction is 3 less than the denominator.

• Numerator = (x - 3)

Both the numerator and denominator are increased by 2.

• New numerator = (x - 3) + 2 = (x - 1)
• New numerator = (x + 2)
• New fraction = 6/7

We know that:

• Fraction = Numerator / Denominator

Finding the original fraction:

According to the question.

↣ (x - 1)/(x + 2) = 6/7

Cross multiplication.

↣ 7(x - 1) = 6(x + 2)

↣ 7x - 7 = 6x + 12

↣ 7x - 6x = 12 + 7

↣ x = 19

We get:

• Denominator = x = 19
• Numerator = (x - 3) = (19 - 3) = 16
• Fraction = 16/19

Hence,

• The original fraction is 16/19.

Answer 2

Answer:

Appropriate Question :-

• The numerator of a fraction is 3 less than the denominator. If the both numerator and denominator are increased by 2, then the fraction becomes 6/7. Find the original fraction.

Given :-

• The numerator of a fraction is 3 less than the denominator.
• The both numerator and denominator are increased by 2.
• The new fraction become 6/7.

To Find :-

• What is the original fraction.

Solution :-

Let,

$\mapsto \bf{Denominator =\: x}$

$\mapsto \bf{Numerator =\: x - 3}$

Hence, the required original fraction will be :

$\leadsto \sf \dfrac{Numerator}{Denominator}$

$\leadsto \sf\bold{\green{\dfrac{x - 3}{x}}}$

$\purple{\bigstar}\: \: \bf{According\: to\: the\: question\: :-}$

$\implies \sf \dfrac{Numerator + 2}{Denominator + 2} =\: New\: fraction$

$\implies \sf \dfrac{x - 3 + 2}{x + 2} =\: \dfrac{6}{7}$

$\implies \sf \dfrac{x - 1}{x + 2} =\: \dfrac{6}{7}$

$\purple{\bigstar}\: \: \bf{By\: doing\: cross\: multiplication\: we\: get\: :-}$

$\implies \sf 7(x - 1) =\: 6(x + 2)$

$\implies \sf 7x - 7 =\: 6x + 12$

$\implies \sf 7x - 6x =\: 12 + 7$

$\implies \sf\bold{\green{x =\: 19}}$

Hence, the required original fraction will be :

$\longrightarrow \sf Original\: Fraction =\: \dfrac{x - 3}{x}$

$\longrightarrow \sf Original\: Fraction =\: \dfrac{19 - 3}{19}$

$\longrightarrow \sf\bold{\red{Original\: Fraction =\: \dfrac{16}{19}}}$

${\small{\bold{\underline{\therefore\: The\: original\: fraction\: is\: \dfrac{16}{19}\: .}}}}$

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VERIFICATION :-

$\rightarrow \sf \dfrac{x - 3 + 2}{x + 2} =\: \dfrac{6}{7}$

By putting x = 19 we get,

$\rightarrow \sf \dfrac{19 - 3 + 2}{19 + 2} =\: \dfrac{6}{7}$

$\rightarrow \sf \dfrac{19 - 1}{21} =\: \dfrac{6}{7}$

$\rightarrow \sf \dfrac{\cancel{18}}{\cancel{21}} =\: \dfrac{6}{7}$

$\rightarrow \sf\bold{\dfrac{6}{7} =\: \dfrac{6}{7}}$

Hence, Verified.