Question

Match column I with column II and select the correct answer by using the given codes.
$\begin{array}{cccc}& \text{Column I}& & \text{Column II}\\ & & & \\ \text{(A)}& \frac{9^6}{(27{)}^6}\times 3^3& \text{(P)}& 3\\ & & & \\ & & & \\ \text{(B)}& \frac{{16}^3}{8^3}\times 2^2& \text{(Q)}& 27\\ & & & \\ & & & \\ \text{(C)}& \frac{2^3\times (12{)}^4}{4^4\times 6^3}& \text{(R)}& 32\\ & & & \\ & & & \\ \text{(D)}& \frac{{12}^5}{4^5}\times \frac{1}{3^2}& \text{(S)}& \frac{1}{27}\\ & & & \end{array}$

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

Answer 1

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

$\bullet \frac{9^6}{(27{)}^6}\times 3^3={\left(\frac{9}{27}\right)}^6\times 3^3=\frac{1}{3^6}\times 3^3$

$=\frac{3^3}{3^6}$ $=3^{3-6}$ $=3^{-3}=\frac{1}{3^3}=\frac{1}{27}$

$\bullet \frac{{16}^3}{8^3}\times 2^2={\left(\frac{16}{8}\right)}^3\times 2^2=2^3\times 2^2$

$=2^{3+2}=2^5=32$

$\bullet \frac{2^3\times (12{)}^4}{4^4\times 6^3}=\frac{2^3\times (12{)}^4}{6^3\times 4^4}$

$={\left(\frac{2}{6}\right)}^3\times {\left(\frac{12}{4}\right)}^4$


$=\frac{1}{3^3}\times 3^4$ $=\frac{3^4}{3^3}$

$=3^{4-3}=3^1=3$

$\bullet \frac{{12}^5}{4^5}\times \frac{1}{3^2}={\left(\frac{12}{4}\right)}^5\times \frac{1}{3^2}$

$=3^5\times \frac{1}{3^2}$ $=\frac{3^5}{3^2}$

$=3^{5-2}=3^3=27$