Question

a fraction before 1/ 3 when 1 is substance for the numerator and its become one1/4 when 8 is added to it denominator find the faction​

Answer 1

Given :-

• A fraction becomes 1/3 when 1 is subtracted for the numerator and its become one 1/4 when 8 is added to its denominator

To find :-

• Required fraction

Solution :-

Let the required fraction be x/y

• A fraction becomes ⅓ when 1 is subtracted for the numerator

→ x - 1/y = ⅓

→ 3(x - 1) = y

→ 3x - 3 = y

→ 3x - y = 3 ------(i)

• A fraction becomes one 1/4 when 8 is added to its denominator

→ x/y + 8 = ¼

→ 4x = y + 8

→ 4x - y = 8 -----(ii)

Subtract both the equations

→ (3x - y) - (4x - y) = 3 - 8

→ 3x - y - 4x + y = - 5

→ 3x - 4x = - 5

→ - x = - 5

→ x = 5

Put the value of x in equation (i)

→ 3x - y = 3

→ 3 × 5 - y = 3

→ 15 - y = 3

→ y = 15 - 3

→ y = 12

Hence,

• Required fraction

• Numerator/Denominator = x/y = 5/12

Answer 2

AnswEr :-

• The fraction is 5/12.

Given :-

• A fraction becomes 1/3 when 1 is subtracted from the numerator.

• It becomes 1/4 when 8 is added to the denominator.

To Find :-

• The Fraction.

SoluTion :-

Let,

• Numerator = x

• Denominator = y

Fraction = x/y

According to question :-

• x - 1/y = 1/3

→ 3x - y = 3 ...i)

• x/y + 8 = 1/4

→ 4x - y = 8 ...ii)

Subtracting equation i) from equation ii) , We get

• x = 5 ...iii)

Putting this value in equation i) , We get

• 15 - y = 3

→ y = 12

Hence, the fraction is 5/12.

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