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.
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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:
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