Question

the denominator of a fraction exceeds its numerator by 3 if one is added to both numerator and denominator the difference between the new and the original fractions 1/24 find the original fraction

Answer 1

let the numerator be 'n' and the denominator be 'd'

the fraction is given by 'n/d'

given that, the denominator of a fraction exceeds its numerator by 3 i.e,

d= n+3

now the fraction becomes 'n/(n+3)'

now, if one is added to both numerator and denominator the difference between the new and the original fractions = 1/24

therefore,

(n+1/n+4) - (n/n+3) = (1/24)

we get the quadratic equation as:

n^2 +7n -60 = 0

after solving, we get n=5 and n= -12

so we have d= n+3

so, d=8 or d= -9

therefore the possible fractions are 5/8 or 12/9

if we take 5/8, the conditions will not be satisfied.

so, the fraction will be 12/9 = (4/3)

Answer 2

Answer:

4/3

Step-by-step explanation:

let the numerator be 'n' and the denominator be 'd'

the fraction is given by 'n/d'

given that, the denominator of a fraction exceeds its numerator by 3 i.e,

d= n+3

now the fraction becomes 'n/(n+3)'

now, if one is added to both numerator and denominator the difference between the new and the original fractions = 1/24

therefore,

(n+1/n+4) - (n/n+3) = (1/24)

we get the quadratic equation as:

n^2 +7n -60 = 0

after solving, we get n=5 and n= -12

so we have d= n+3

so, d=8 or d= -9

therefore the possible fractions are 5/8 or 12/9

if we take 5/8, the conditions will not be satisfied.

so, the fraction will be 12/9 = (4/3)