21. State ' \( T \) ' for true or ' F ' for false. I. If \( \frac{p}{q} \) is a rational number, then \( p \neq 0 \). II. If \( \frac{b}{a} \) is multiplicative inverse of \( \frac{a}{b} \), then \( a \neq 0 \). III. For any rational number \( x, x+(-1)=-x \). IV. \( \frac{x+y}{2} \) is a rational number which lies between \( x \) and \( y \). V. If \( \frac{p}{q}=\frac{3}{5} \) and \( \frac{x}{y}=\frac{2}{3} \), then \( \frac{p y+q x}{3 q y}=\frac{19}{45} \) Codes \( \begin{array}{ll}\text { I II III IV V } & \text { (a) F T F T T' }\end{array} \) \( \begin{array}{ll}\text { (c) F F F F F } & \text { (b) T F T F F F }\end{array} \)

Let's evaluate each statement one by one: I. False: A rational number can be \( \frac{p}{q} \) even if \( p = 0 \), as \( \frac{0}{q} \) (where \( q \neq 0 \)) is still a valid rational number. II. True: The multiplicative inverse of \( \frac{a}{b} \) is \( \frac{b}{a} \), which is defined only when \( a \neq 0 \) to avoid division by zero. III. False: For any rational number \( x \), \( x + (-1) \) is actually \( x - 1 \), which is not necessarily equal to \( -x \). IV. True: The average \( \frac{x+y}{2} \) is indeed a rational number that will lie between \( x \) and \( y \) if \( x \) and \( y \) are both rational. V. False: Given the values \( p, q, x, y \), the calculated expression does not equate to \( \frac{19}{45} \). So the answers turn out to be: I - F, II - T, III - F, IV - T, V - F. Thus, the correct coding for your option is: (b) T F T F F.