Question

Match column I with column II and select the correct answer by using the given codes.
$\begin{array}{cccc}& \text{Column I}& & \text{Column II}\\ \text{(A)}& {\left(\frac{4^{10}}{4^{15}}\right)}^{1/5}& \text{(P)}& \frac{1}{16}\\ \text{(B)}& {\left(\frac{7^{20}}{7^{30}}\right)}^{1/5}& \text{(Q)}& \frac{1}{49}\\ \text{(C)}& {\left(\frac{{49}^{49}}{{49}^{51}}\right)}^{1/4}& \text{(R)}& \frac{1}{4}\\ \text{(D)}& {\left(\frac{{16}^{27}}{{16}^{31}}\right)}^{1/4}& \text{(S)}& \frac{1}{7}\end{array}$

• A. $A\to R;B\to S;C\to Q;D\to P$
• B. $A\to R;B\to S;C\to P;D\to Q$
• C. $A\to R;B\to Q;C\to S;D\to P$
• D. $A\to R;B\to P;C\to Q;D\to S$

Answer 1

The correct option is C $A\to R;B\to Q;C\to S;D\to P$

${\left(\frac{4^{10}}{4^{15}}\right)}^{1/5}=(4^{10-15}{)}^{1/5}=(4^{-5}{)}^{1/5}=(4{)}^{-5\times \frac{1}{5}}=4^{-1}=\frac{1}{4}$

${\left(\frac{7^{20}}{7^{30}}\right)}^{1/5}=(7^{20-30}{)}^{1/5}=(7^{-10}{)}^{1/5}=(7{)}^{-10\times \frac{1}{5}}=7^{-2}=\frac{1}{7^2}=\frac{1}{49}$

${\left(\frac{{49}^{49}}{{49}^{51}}\right)}^{\frac{1}{4}}=({49}^{49-51}{)}^{1/4}=({49}^{-2}{)}^{\frac{1}{4}}=(49{)}^{-2\times \frac{1}{4}}=(7^2{)}^{-\frac{1}{2}}=7^{2\times (-\frac{1}{2})}=7^{-1}=\frac{1}{7}$

${\left(\frac{{16}^{27}}{{16}^{31}}\right)}^{\frac{1}{4}}=({16}^{27-31}{)}^{\frac{1}{4}}=({16}^{-4}{)}^{\frac{1}{4}}=(16{)}^{-4\times \frac{1}{4}}=(16{)}^{-1}=\frac{1}{16}$