Question

the denominator of a rational number is greater than numerator by 8 . if the numerator is increased by17 and denominator is decreased by 1 , the number obtained is 9/4.find the rational number.​

Answer 1

Answer:

Let the numerator and denominator of the required rational number be x and y respectively.

Rational Number = x/y.

Given, y=x+8.

Also , x+17 = 9

____ _

y-1 4

4 (x+17) = 9(y-1)

4x+68 = 9 (x+8-1)

4x+68= 9 (x+7)

4x+68= 9x+63.

68-63=9x-4x.

5=5x.

x=1

y=x+8

y=9.

Required Rational number = x/y=1/9.

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Answer 2

ANSWER:

• The rational number = 1/9

GIVEN:

• The denominator of a rational number is greater than numerator by 8.

• If the numerator is increased by 17 and denominator is decreased by 1 , the number obtained is 9/4.

TO FIND:

• The rational number.

EXPLANATION:

Let the numerator be x and the denominator be y.

y = x + 8

$\sf \dfrac{x + 17}{y - 1} = \dfrac{9}{4}$

$\sf 4x + 68 = 9y - 9$

$\sf 4x - 9y = - 77$

Substitute y = x + 8

$\sf 4x - 9(x + 8) = - 77$

$\sf 4x - 9x - 72 = - 77$

$\sf - 5x= - 5$

$\sf x = 1$

$\sf We\ know \ that \ y = x + 8$

$\sf y = 1 + 8$

$\sf y = 9$

HENCE THE RATIONAL NUMBER IS 1 / 9.

VERIFICATION:

y = x + 8

Substitute y = 9 and x = 1

9 = 8 + 1

9 = 9

$\sf \dfrac{x + 17}{y - 1} = \dfrac{9}{4}$

Substitute y = 9 and x = 1

$\sf \dfrac{1+ 17}{9 - 1} = \dfrac{9}{4}$

$\sf \dfrac{18}{8} = \dfrac{9}{4}$

$\sf \dfrac{9}{4} = \dfrac{9}{4}$

HENCE VERIFIED.