Question

the sum of numerator and denominator of a fraction is greater by 1 than thrice the numerator if the numerator is decreased by 1, then the fraction reduced to 1/5 find the fraction​

Answer 1

Step-by-step explanation:

let, the numerator of the fraction be x.

the denominator of the fraction is y.

required fraction=x

y

a/c to given condition 1,

3x=x+y+1

3x-x-y=1

2x-y=1........................1

if numerator and denominator is reduced by 1,

new numerator=(x-1)

denominator=(y-1)

a/c to given condition 2,

(x-1) = 1

(y-1) 5

by cross multiplication,

5×(x-1)=(y-1)×1

5x-5=y-1

5x-y=-1+5

5x-y=4................…………2

by solving the problem,

we get, x=1 and y=1

therefore, required fraction= x = 1

y 1

Answer 2

Answer:

Given :-

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

To Find :-

What is the fraction.

Solution :-

Let, the fraction be x/y

According to the question,

➣ 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 .......... equation no (1)

➣ Denominator is decreased by 1 than the fraction reduced by 1/3.

⇒ x - 1/y = 1/3

⇒ 3(x - 1) = y

⇒ 3x - 3 = y

⇒ 3x - y = 3 ......... equation no (2)

➣ Now, putting the value of x from equation no (1) in equation no (2) we get,

⇒ 3x - y = 3

⇒ 3x - (2x + 1) = 3

⇒ 3x - 2x - 1 = 3

⇒ x = 3 + 1

➠ x = 4

➣ Again, putting the value of x in the equation no (1) we get,

⇒ y = 2x + 1

⇒ y = 2(4) + 1

⇒ y = 8 + 1

➥ y = 9

Hence, the required fraction will be,

↦ x/y

➽ 4/9

∴ The fraction will be 4/9 .