Question

If the numerator and denominator of a proper fraction are increased by same quantity, then the resulting fraction is-
A. Always greater than the original fraction.
B. Always less than the original fraction.
C. Always equal to the original fraction.
D. None of the above.

Answer 1

Hint: In this question it is given that if the numerator and denominator of a proper fraction are increased by the same quantity, then the resulting fraction is always greater or always less or equal to the original fraction.
So to solve this type of question we have to take any proper fraction and after that add any positive quality with numerator as well as denominator and after that subtract the original fraction from the new one,
(i) If the result is greater than zero, then new fraction is greater than the original fraction, i.e, option A will be our correct answer,
(ii) If the result is less than zero, then new fraction is less than the original fraction, i.e, option B will be our correct answer
(iii) And if the result is equal to zero, then both the fraction is equal, i.e, option C will be our correct answer.

Complete step-by-step answer:
So let us consider $$\dfrac{x}{y}$$ to be any proper fraction, where x and y are any Integer, and since this fraction is a proper fraction so $x < y$.
Now we are going to add any value a(positive quantity) with numerator x and denominator y, so the new fraction will be,
$$\dfrac{\left( x+a\right) }{\left( y+a\right) }$$.

Now subtracting $$\dfrac{x}{y}$$ from $$\dfrac{\left( x+a\right) }{\left( y+a\right) }$$ we get,

$$\dfrac{\left( x+a\right) }{\left( y+a\right) } -\dfrac{x}{y}$$
To subtract or add any two fraction we need to find the Least common multiple i.e, LCM of denominator and convert the denominator into LCM quantity,
So the $$LCM\left( y+a,y\right) =y\left( y+a\right) $$, so we have to make each denominator as (y+a)y.
Therefore, we can write $$\dfrac{\left( x+a\right) }{\left( y+a\right) } -\dfrac{x}{y}$$ as,
$$\dfrac{\left( x+a\right) y}{\left( y+a\right) y} -\dfrac{x\left( y+a\right) }{y\left( y+a\right) }$$
$$=\dfrac{\left\{ \left( x+a\right) y\right\} -\left\{ x\left( y+a\right) \right\} }{\left( y+a\right) y}$$ [ since denominator is same , so we can subtract the numerator ]
$$=\dfrac{\left( xy+ay\right) -\left( xy+ax\right) }{\left( y+a\right) y}$$
$$=\dfrac{xy+ay-xy-ax}{\left( y+a\right) y}$$ [by omitting the braces]
$$=\dfrac{ay-ax}{\left( y+a\right) y}$$ [ by cancelling the xy ]
$$=\dfrac{a\left( y-x\right) }{y\left( y+a\right) }$$ [ taking common a from both the term]
As we know that y, (y+a) and a are positive quantities , and since the value of x is less than y so we can easily say that (y-x) is always positive.
So ultimately we can say that, $$\dfrac{a\left( y-x\right) }{y\left( y+a\right) }$$ is always greater than zero, i.e, $$\dfrac{a\left( y-x\right) }{y\left( y+a\right) } >0$$
So the correct option is option A.

So you can check by taking any example,
Let $\dfrac{2}{3}$ be any fraction and we add 1 with numerator and denominator so we ge, $\dfrac{3}{4}$
Now by subscribing we get,
$$\dfrac{3}{4} -\dfrac{2}{3}$$
$$=\dfrac{3\times 3}{4\times 3} -\dfrac{2\times 4}{3\times 4}$$
$$=\dfrac{9}{12} -\dfrac{8}{12}$$
$=\dfrac{9-8}{12}$
$=\dfrac{1}{12} >0$
So in this way also we can prove that the resulting fraction is greater than the original one.

Note: In the solution process one can take the negative value of x and y but while taking the values you have to keep in mind that x is always less than y.