Question

The denominator of a rational number is greater than its numerator by 5.If the numerator is decreased by 20and the denominator is increased by 3 ,the number obtained is 1/3 . find the rational number,

Answer 1

Answer:

Given :-

The denominator of a rational number is greater than its numerator by 5.If the numerator is decreased by 20 and the denominator is increased by 3 ,the number obtained is 1/3

To Find :-

The Fraction

Solution :-

Let us assume the numerator be x and denominator be x + 5

Fraction = x/x + 5

When numerator decreased by 20 and denominator increased by 3

$\sf \: \dfrac{x - 20}{x + 5 + 3} = \dfrac{1}{3}$

$\sf \dfrac{x - 20}{x + 8} = \dfrac{1}{3}$

By Cross Multiplication

3(x - 20) = 1(x + 8)

3x - 60 = x + 8

3x + x = 60 + 8

4x = 68

x = 68/4

x = 17

Hence :-

$\large \sf \: Fraction \: = \dfrac{17 - 20}{17 + 5 + 3} = \dfrac{ - 3}{25}$

Answer 2

☆ Given :

• The denominator of a rational number is greater than its numerator by 5.
• If numerator is decreased by 20 and denominator is increased by 3 ,we got a number that is 1/3.

☆ To find :

• The rational number.

☆ Solution :

Let the numerator be x

Then the denominator will be x + 5

Now,

• If the numerator decreased by 20

→ Then new numerator = x - 20

• If the denominator increased by 3

→ Then new denominator = x + 5 + 3 = x + 8

According to Question ,

$\dashrightarrow\sf\dfrac{x-20}{x+8}=\dfrac{1}{3}$

$\dashrightarrow\sf{3(x-20)=1(x+8)}$

$\dashrightarrow\sf{3x-60=x+8}$

$\dashrightarrow\sf{3x-x=8+60}$

$\dashrightarrow\sf{2x=68}$

$\dashrightarrow\sf{x=}\cancel\dfrac{68}{2}^{34}$

$\dashrightarrow\sf{x=34}$

Therefore,

• Numerator = x = 34
• Denominator = (x+5) = 34 + 5 = 39

${\boxed{\sf{Required \ rational \ no. = {\dfrac{34}{39}}}}}$

Verification :

As it is given that when numerator is decreased by 20 and denominator is increased by 3 then the result is 1/3

Thus,

$\dashrightarrow\sf\dfrac{x-20}{x+8}=\dfrac{1}{3}$

Putting the value of x.

$\to\sf\dfrac{34-20}{34+8}=\dfrac{1}{3}$

$\to\sf\dfrac{\cancel{14}^1}{\cancel{42}^{3}}=\dfrac{1}{3}$

$\to\sf\dfrac{1}{3}=\dfrac{1}{3}$

Thus, Verified.