Question

the sum of numerator and denominator of a fraction is greater by 1 than thrice the numerator . if fraction is decreased by 1 then fraction reduced to 1/3 find the fraction​

Answer 1

Answer

let the fraction be x/y

according to data given

x+y= 3x+1 ---------equ (1)

greater by 1 than thrice numerator means (numerator×3) +1

come to equ (1)

y= 3x-x+1

y=2x+1 ------- equ(2)

now its given that if 1 is subtracted from x/y then it becomes 1/3

lets write this as an equation

$\frac{x}{y} -1= \frac{1}{3}$   in LHS take y as LCM

$\frac{x-y}{y} =\frac{1}{3}$     cross multiply

3x-3y= y

3x=4y

so x/y = 4/3

so the fraction is 4/3

hope it helps...

Answer 2

Given:-

• The sum of numerator and denominator of a fraction is greater by 1 than thrice the numerator.
• Numerator and denominator by 1 then fraction reduced to 1/3.

To find:-

• Find the fraction..?

Solutions:-

• Let the fraction be x/y

The sum of numerator and denominator of a fraction is greater by 1 than thrice the numerator.

=> x + y = 3x + 1

=> y = 3x - x + 1

=> y = 2x + 1 ........(i).

Numerator and denominator by 1 then fraction reduced to 1/3.

=> x - 1/y = 1/3

=> 3(x - 1) = y

=> 3x - 3 = y

=> 3x - y = 3 ......(ii).

putting the value of x from Eq (i)in Eq (ii).

=> 3x - y = 3

=> 3x - 2x - 1 = 3

=> x - 1 = 3

=> x = 3 + 1

=> x = 4

Putting the value of x in Eq (i).

=> y = 2x + 1

=> y = 2(4) + 1

=> y = 8 + 1

=> y = 9

So,

• Numerator = x = 4
• Denominator = y = 9
• Fraction = x/y = 4/9

Hence, the fraction become id 4/9.