the sum of numerator and denominator of a fraction is greater than thrice the numerator if numerator is decreased by 1 then the fraction is 1/3. Find the fraction?
Answers
Correct Question -
The sum of numerator and denominator of a fraction is greater than thrice the numerator by 12.
If numerator is decreased by 1 then the fraction is 1/3.
Find the fraction.
Solution -
Let us denote the numerator as n and the denominator as d .
Now ,
We have the first condition as -
The sum of numerator and denominator of a fraction is greater than thrice the numerator by 12.
So,
n + d = 3n + 12
=> d = 2n + 12 ....... ( 1 )
Now , we have the following second condition -
If numerator is decreased by 1 then the fraction is 1/3.
=> n + 1 / d = 1 / 3
=> 3n + 3 = d ............. ( 2 )
Equation 1 = Equation 2
=> 2n + 12 = 3n + 3
=> n = 9
=> d = 3 × 9 + 3
=> d = 30
Required Fraction
=> ( 9 / 30 )
=> 3 / 10 .... ...... [ Answer ]
Answer:
Given :-
The sum of numerator and denominator of a fraction is greater by 1 than thrice the numerator is decreased by 1 then the fraction reduced to 1/3.
To Find :-
What is the fraction.
Solution :-
Let, the fraction be x/y
According to the question,
➣ The sum of numerator and denominator of a fraction is greater by 1 than thrice the numerator.
⇒ x + y = 3x + 1
⇒ y = 3x - x + 1
⇒ y = 2x + 1 .......... equation no (1)
➣ Denominator is decreased by 1 than the fraction reduced by 1/3.
⇒ x - 1/y = 1/3
⇒ 3(x - 1) = y
⇒ 3x - 3 = y
⇒ 3x - y = 3 ......... equation no (2)
➣ Now, putting the value of x from equation no (1) in equation no (2) we get,
⇒ 3x - y = 3
⇒ 3x - (2x + 1) = 3
⇒ 3x - 2x - 1 = 3
⇒ x = 3 + 1
➠ x = 4
➣ Again, putting the value of x in the equation no (1) we get,
⇒ y = 2x + 1
⇒ y = 2(4) + 1
⇒ y = 8 + 1
➥ y = 9
Hence, the required fraction will be,
↦ x/y
➽ 4/9
∴ The fraction will be 4/9 .