Question

Divide 10 into two parts in such a way that when a certain one of these parts is divided by the other and the fraction is multiplied by its numerator, the result gives 9.

Answer 1

Answer:

The fraction is $\frac{x}{10-x}\times x$ where the numeric value of x = 6 and 10-x =4.

Step-by-step explanation:

Given:

A number 10 we have to divide it into two parts and then when one part is divided by other with multiplication of its numerator the number obtained is 9.

Lets the number in numerator be $x$

Then

The number in denominator is $10-x$

According to the question:

⇒ $\frac{x}{10-x}\times x =9$

⇒ $\frac{x^2}{10-x} =9$

⇒ $x^2=9(10-x)$  

⇒ $x^2+9x-90=0$

⇒ $x^2+15x-6x-90=0$ solving by middle term splitting.

⇒ $x(x+15)-6(x+15)=0$

⇒ $(x-6)(x+15)=0$

⇒ $x=6,-15$

Discarding the negative values.

⇒ $x=6$

Check:

So the fraction is $\frac{x}{10-x} \times x =9$

then,

⇒ $\frac{6}{10-6} \times 6 =9$

⇒ $\frac{6\times 6}{4} =9$

⇒ $\frac{36}{4}=9$

⇒ $9=9$

Left terms= Right terms

The fraction is $\frac{x}{10-x}\times x$ where the number 'x' = 6 and 10-x = 4.