The denominator of a fraction is 1 less than twice its numerator . If

1.	The denominator of a fraction is 1 less than twice its numerator . If 1 is added to both the numerator and the denominator, the fraction becomes 2:3. Find the original fraction.
Answer» Answer: Let the numerator of the fraction be ‘x’ and its denominator be ‘y’. Then, the required fraction is x/y. According to the FIRST condition, the denominator is 1 less than twice its numerator. ∴ y = 2x – 1 ∴ 2x – y = 1 …(i) According to the second condition, if 1 is added to the numerator and the denominator, the ratio of numerator to denominator is 3 : 5. ∴ (x + 1)/(y + 1) = 3/5 ∴ y + 1 = 5 ∴ 5(x + 1) = 3(y + 1) ∴ 5X + 5 = 3y + 3 ∴ 5x – 3y = 3 – 5 ∴ 5x – 3y = -2 ……(ii) Multiplying equation(i) by 3, 6x – 3y = 3 …(iii) Subtracting equation (ii) from (iii), SUBSTITUTING x = 5 in equation (i), ∴ 2x – y = 1 ∴ 2(5) – y = 1 ∴ 10 – y = 1 ∴ y= 10 – 1 = 9 ∴ The required fraction is 5/9. Step-by-step explanation: HOPE IT HELPS..

The denominator of a fraction is 1 less than twice its numerator . If 1 is added to both the numerator and the denominator, the fraction becomes 2:3. Find the original fraction.

Answer»

Answer:

Let the numerator of the fraction be ‘x’ and its denominator be ‘y’.

Then, the required fraction is x/y. According to the FIRST condition, the denominator is 1 less than twice its numerator.

∴ y = 2x – 1

∴ 2x – y = 1 …(i)

According to the second condition, if 1 is added to the numerator and the denominator, the ratio of numerator to denominator is 3 : 5.

∴ (x + 1)/(y + 1) = 3/5

∴ y + 1 = 5

∴ 5(x + 1) = 3(y + 1)

∴ 5X + 5 = 3y + 3

∴ 5x – 3y = 3 – 5

∴ 5x – 3y = -2 ……(ii)

Multiplying equation(i) by 3,

6x – 3y = 3 …(iii)

Subtracting equation (ii) from (iii),

SUBSTITUTING x = 5 in equation (i),

∴ 2x – y = 1

∴ 2(5) – y = 1

∴ 10 – y = 1

∴ y= 10 – 1 = 9

∴ The required fraction is 5/9.

Step-by-step explanation:

HOPE IT HELPS..