## 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​

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​

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1 month 2021-08-12T10:07:20+00:00 2 Answers 0 views 0

• 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.

### ToFind:–

• 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

## _____________________

2. ### 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