Question

the numeration of a fraction is one more than its demominator if its reciprocal is subtracted from it the differnce is 11 upon 30 find the fraction

Answer 1

Answer:

Given :-

• The numerator of a fraction is one more than its denominator, if it's reciprocal is subtracted from it and the difference is 11/30.

To Find :-

• What is the fraction.

Solution :-

Let, the denominator be x

And, the numerator will be x + 1

Then, the fraction is $\sf \dfrac{x + 1}{x}$

According to the question,

⇒ $\sf \dfrac{x + 1}{x} -\: \dfrac{x}{x + 1} =\: \dfrac{11}{30}$

⇒ $\sf \dfrac{\cancel{{x}^{2}} + 2x + 1 - \cancel{{x}^{2}}}{{x}^{2} + x} =\: \dfrac{11}{30}$

⇒ $\sf \dfrac{2x + 1}{{x}^{2} + x} =\: \dfrac{11}{30}$

$\mapsto$ By doing cross multiplication we get,

⇒ $\sf 30(2x + 1) =\: 11({x}^{2} + x)$

⇒ $\sf 60x + 30 =\: 11{x}^{2} + 11x$

⇒ $\sf 11{x}^{2} + 11x - 60x - 30 =\: 0$

⇒ $\sf 11{x}^{2} - 49x - 30 =\: 0$

$\mapsto$ By splitting the middle term we get,

⇒ $\sf 11{x}^{2} - (55 - 6)x - 30 =\: 0$

⇒ $\sf 11{x}^{2} - 55x + 6x - 30 =\: 0$

⇒ $\sf 11x(x - 5) + 6(x - 5) =\: 0$

⇒ $\sf (11x + 6)(x - 5) =\: 0$

⇒ $\sf 11x + 6 = 0$

⇒ $\sf x =\: \dfrac{- 6}{11}$

➠ $\sf\bold{\green{x =\: \dfrac{- 6}{11}}}$

Either,

⇒ $\sf x - 5 =\: 0$

⇒ $\sf x =\: 5$

➠ $\sf\bold{\green{x =\: 5}}$

We can't take x as negative.

So x = 5

Hence, the required fraction are,

↦ $\sf \dfrac{x + 1}{x}$

↦ $\sf \dfrac{5 + 1}{5}$

➦ $\sf\bold{\purple{\dfrac{6}{5}}}$

${\underline{\boxed{\small{\bf{\therefore The\: required\: fraction\: is\: \dfrac{6}{5}.}}}}}$

VERIFICATION :-

↦ $\sf \dfrac{x + 1}{x} -\: \dfrac{x}{x + 1} =\: \dfrac{11}{30}$

By putting x = 5 we get,

↦ $\sf \dfrac{5 + 1}{5} -\: \dfrac{5}{5 + 1} =\: \dfrac{11}{30}$

↦ $\sf \dfrac{6}{5} -\: \dfrac{5}{6} =\: \dfrac{11}{30}$

↦ $\sf \dfrac{36 - 25}{30} =\: \dfrac{11}{30}$

↦ $\sf\bold{\dfrac{11}{30} =\: \dfrac{11}{30}}$

➦ LHS = RHS

Hence, Verified ✔