8th Class Mathematics Algebraic Expressions Question Bank Algebra

Match column I with column II and select the correct answer II using the code given below the columns:

Column - I Column- II

(A) \[xy=b,\] (p) \[\frac{1}{{{x}^{3}}}+\frac{1}{{{y}^{3}}}\]

(B) \[\frac{{{a}^{3}}-3ab}{{{b}^{3}}}\] (q) \[\frac{{{a}^{3}}-3a}{{{b}^{3}}}\]

(C) \[\frac{{{a}^{3}}-3}{h}\] (r) \[\frac{{{a}^{3}}-3}{{{b}^{2}}}\]

(D) \[x=\frac{1}{2}\] (s) \[\begin{align} & x+\overset{1}{\mathop{\_\_\_\_\_\_\_}}\, \\ & 1+\overset{1}{\mathop{\_\_\_\_}}\, \\ & 1+\frac{1}{x} \\ \end{align}\]

(E) \[\frac{5}{4}\] (t) \[\frac{4}{5}\]

(F) \[\frac{3}{4}\] (u) \[{{2}^{2x-y}}=32\]

A) \[{{2}^{x+y}}=16\]\[{{x}^{2}}+{{y}^{2}}\]

B) \[x-\frac{1}{x-2}=2-\frac{1}{x-2},\]\[(E)\to (t),\,(F)\to (p)\]

C) \[{{x}^{2}}\]\[{{x}^{a+b+c}}\]

D) \[{{x}^{abc}}\]\[{{x}^{0}}\]

Correct Answer: C

Solution :

\[\frac{7}{2}\]

Column - I	Column- II
(A) \[xy=b,\]	(p) \[\frac{1}{{{x}^{3}}}+\frac{1}{{{y}^{3}}}\]
(B) \[\frac{{{a}^{3}}-3ab}{{{b}^{3}}}\]	(q) \[\frac{{{a}^{3}}-3a}{{{b}^{3}}}\]
(C) \[\frac{{{a}^{3}}-3}{h}\]	(r) \[\frac{{{a}^{3}}-3}{{{b}^{2}}}\]
(D) \[x=\frac{1}{2}\]	(s) \[\begin{align} & x+\overset{1}{\mathop{\_\_\_\_\_\_\_}}\, \\ & 1+\overset{1}{\mathop{\_\_\_\_}}\, \\ & 1+\frac{1}{x} \\ \end{align}\]
(E) \[\frac{5}{4}\]	(t) \[\frac{4}{5}\]
(F) \[\frac{3}{4}\]	(u) \[{{2}^{2x-y}}=32\]