Question

solve the problem given​

Answer 1

$\sf{\bold{\green{\underline{\underline{Given}}}}}$

• Numerator is 6 less than the denominator
• If numerator is decreased by 4 and denominator is decreased by 1 the fraction becomes 5 / 8

______________________

$\sf{\bold{\green{\underline{\underline{To\:Find}}}}}$

• Original fraction = ??

______________________

$\sf{\bold{\green{\underline{\underline{Solution}}}}}$

• Numerator is 6 less than denominator

⠀⠀⠀⠀

$\sf : \implies\:{\bold{ Required \: fraction = \dfrac{ x - 6 }{x}}}$

⠀⠀⠀⠀

• If numerator is decreased by 4 and denominator is decreased by 1 the fraction becomes 5 / 8

⠀⠀⠀⠀

$\sf : \implies\:{\bold{ \dfrac{ x - 6 - 4 }{x - 1 }= \dfrac{5}{8}}}$

⠀⠀⠀⠀

$\sf : \implies\:{\bold{ \dfrac{ x - 10 }{x - 1 }= \dfrac{5}{8}}}$

⠀⠀⠀

$\sf : \implies\:{\bold{ 8 ( x - 10 ) = 5 ( x - 1 )}}$

⠀⠀⠀⠀

$\sf : \implies\:{\bold{ 8x - 80 = 5x - 5}}$

⠀⠀⠀⠀

$\sf : \implies\:{\bold{ 8x - 5x = 80 - 5}}$

⠀⠀⠀⠀

$\sf : \implies\:{\bold{ 3x = 75 }}$

⠀⠀⠀⠀

$\sf : \implies\:{\bold{ x = \dfrac{75}{3}}}$

⠀⠀⠀⠀

$\sf : \implies\:{\bold{ x = 25 }}$

⠀⠀⠀⠀

• Required fraction

$\sf : \implies\:{\bold{ Required \: fraction = \dfrac{ x - 6 }{x}}}$

⠀⠀⠀⠀

$\sf : \implies\:{\bold{ Required \: fraction = \dfrac{ 25- 6 }{25}}}$

⠀⠀⠀⠀

$\sf : \implies\:{\bold{ Required \: fraction = \dfrac{ 19 }{25}}}$

⠀⠀⠀⠀

⠀⠀⠀⠀

___________________

$\sf{\bold{\green{\underline{\underline{Answer}}}}}$

• Therefore ; ${\bold{ Required \: fraction = \dfrac{ x - 6 }{x}}}$

Answer 2

Given:

• Numerator of a fraction is 6 less than its denominator.
• If numerator is decreased by 4 and the denominator is decreased by 1 the fraction becomes 5/8

Find:

• What is Fraction

Solution:

Let, the numerator be x

and denominator be y

$\mathbb{ACCORDING \: TO \: QUESTION}$

Case 1:

$\sf \to \dfrac{x}{y - 6}$

$\sf \to x = y- 6.....(1)$

Case 2:

$\sf \to \dfrac{x - 4}{y - 1} = \dfrac{5}{8}$

$\sf \to 8(x - 4) = 5(y - 1)$

$\sf \to 8x - 32 = 5y - 5$

$\sf \to 8x - 5y = - 5 + 32$

$\sf \to 8x - 5y = 27$

Now, use eq(1)

$\sf \to 8x - 5y = 27$

$\sf \to 8(y - 6) - 5y = 27$

$\sf \to 8y - 48 - 5y = 27$

$\sf \to 8y - 5y = 27 + 48$

$\sf \to 3y = 75$

$\sf \to y = \cancel{\dfrac{75}{3}} = 25$

$\sf \to y = 25$

Hence, y = 25

_______________

Now, use the value of y in eq(1) we, get

$\sf \to x = y- 6$

$\sf \to x = 25- 6$

$\sf \to x = 19$

Hence, x = 19

________________

So, fraction will be

$\sf \leadsto \dfrac{x}{y} = \dfrac{19}{25}$

________________

Verification

$\sf \to \dfrac{x - 4}{y - 1} = \dfrac{5}{8}$

where,

• x = 19
• y = 25

So,

$\sf \implies \dfrac{x - 4}{y - 1} = \dfrac{5}{8}$

$\sf \implies \dfrac{19- 4}{25 - 1} = \dfrac{5}{8}$

$\sf \implies \dfrac{15}{24} = \dfrac{5}{8}$

$\sf \implies \cancel{ \dfrac{15}{24}} = \dfrac{5}{8}$

$\sf \implies \dfrac{5}{8} = \dfrac{5}{8}$

Here,

L.H.S = R.H.S

Hence, VERIFIED