Question

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 find the fraction​

Answer 1

S O L U T I O N :

Let the place of numerator be x & let the place of denominator be y respectively.

• The fraction become = x/y

According to the question :

1st case :

➝ x + y = 3x + 1

➝ x - 3x = 1 - y

➝ -2x = 1 - y............(1)

2nd case :

➝ x - 1/y = 1/3

➝ 3(x - 1) = y(1)

➝ 3x - 3 = y............(2)

∴Putting the value of y in equation (1), we get;

➝ -2x = 1 - (3x - 3)

➝ -2x = 1 - 3x + 3

➝ -2x = -3x + 4

➝ -2x + 3x = 4

➝ x = 4

∴ Putting the value of x in equation (1), we get;

➝ -2(4) = 1 - y

➝ -8 = 1 - y

➝ -8 - 1 = -y

➝ -9 = -y

➝ y = 9

Thus,

• Numerator, (N) = 4
• Denominator, (D) = 9

The fraction will be x/y = 4/9 .

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 .