IAS 2019 — CSAT Maths

Correct Option (2)

Let the original proper fraction be represented as $\frac{a}{b}$ba​, where $a$a and $b$b are positive integers such that $a < b$a<b. This condition signifies that the fraction's value is less than 1.

Let the positive quantity by which both the numerator and denominator are increased be $k$k, where $k > 0$k>0. The resulting fraction is $\frac{a+k}{b+k}$b+ka+k​.

To compare the original fraction $\frac{a}{b}$ba​ with the new fraction $\frac{a+k}{b+k}$b+ka+k​, we can examine their difference:

$$\frac{a+k}{b+k} - \frac{a}{b} = \frac{b(a+k) - a(b+k)}{b(b+k)}$$b+ka+k​−ba​=b(b+k)b(a+k)−a(b+k)​

$$= \frac{ab + bk - ab - ak}{b(b+k)}$$=b(b+k)ab+bk−ab−ak​

$$= \frac{bk - ak}{b(b+k)}$$=b(b+k)bk−ak​

$$= \frac{k(b - a)}{b(b+k)}$$=b(b+k)k(b−a)​

Given that $\frac{a}{b}$ba​ is a proper fraction, $a < b$a<b, which implies that $(b - a)$(b−a) is a positive quantity. Additionally, $k > 0$k>0, $b > 0$b>0, and $(b+k) > 0$(b+k)>0. Therefore, the entire expression $\frac{k(b - a)}{b(b+k)}$b(b+k)k(b−a)​ is always positive.

Since the difference $\frac{a+k}{b+k} - \frac{a}{b} > 0$b+ka+k​−ba​>0, it follows that $\frac{a+k}{b+k} > \frac{a}{b}$b+ka+k​>ba​. Thus, the resulting fraction is always greater than the original proper fraction.

Incorrect Options:

• Option 1 (always less than the original fraction): This statement is incorrect. As demonstrated by the algebraic analysis, adding the same positive quantity to both the numerator and denominator of a proper fraction consistently increases its value, moving it closer to 1.

• Option 3 (always equal to the original fraction): This statement is incorrect. For the fractions to be equal, the difference $\frac{k(b - a)}{b(b+k)}$b(b+k)k(b−a)​ would need to be zero. This is not possible because $k > 0$k>0 and $(b - a) > 0$(b−a)>0 for a proper fraction.

• Option 4 (such that nothing can be claimed definitely): This statement is incorrect. A definite conclusion can be drawn based on the mathematical properties of proper fractions and the specified operation. The resulting fraction is consistently greater than the original.