Practice MCQ Questions and Answer on Algebra

511.

If a³ - b³ - c³ - 3abc = 0, then -

• (A) a = b = c
• (B) a + b + c = 0
• (C) a + c = b
• (D) a = b + c

Show Answer

 $$\eqalign{ & {a^3} - {b^3} - {c^3} - 3abc = 0 \cr & \therefore a - b - c = 0 \cr & \Rightarrow a = b + c \cr} $$

512.

The simplified value of following is: $$\left(\frac{3}{15}{a}^5{b}^6{c}^3\times \frac{5}{9}a{b}^5{c}^4\right)$$     $$\text{Math input error}$$ $$\frac{10}{27}{a}^{2}b{c}^3$$

• (A) $$\frac{3}{10}a{b}^4{c}^3$$
• (B) $$\frac{9}{10}{a}^{2}b{c}^4$$
• (C) $$\frac{3}{10}{a}^4{b}^{10}{c}^4$$
• (D) $$\frac{1}{10}{a}^4{b}^4{c}^{10}$$

Show Answer

 $$\eqalign{ & \left( {\frac{3}{{15}}{a^5}{b^6}{c^3} \times \frac{5}{9}a{b^5}{c^4}} \right) \div \frac{{10}}{{27}}{a^2}b{c^3} \cr & \Rightarrow \frac{1}{9}{a^6}{b^{11}}{c^7} \div \frac{{10}}{{27}}{a^2}b{c^3} \cr & \Rightarrow \frac{{\frac{1}{9}{a^6}{b^{11}}{c^7}}}{{\frac{{10}}{{27}}{a^2}b{c^3}}} \cr & \Rightarrow \frac{3}{{10}}{a^4}{b^{10}}{c^4} \cr} $$

513.

If $$x+\frac{9}{x}=6\text{,}$$   then $$\left(x^2+\frac{9}{x^2}\right)$$   is equal to?

• (A) 8
• (B) 9
• (C) 10
• (D) 12

Show Answer

 $$\eqalign{ & x + \frac{9}{x} = 6 \cr & {\text{Take value of x}} \cr & {\text{Let }}x = 3 \cr & 3 + \frac{9}{3} = 6{\text{ }}\left( {{\text{Proved}}} \right) \cr & {\text{So, }}x = 3 \cr & \therefore {x^2} + \frac{9}{{{x^2}}} \cr & = 9 + \frac{9}{9} \cr & = 10 \cr & \cr & {\bf{Alternate:}} \cr & x + \frac{9}{x} = 6 \cr & {\text{On squaring,}} \cr & \Rightarrow {\left( {x + \frac{9}{x}} \right)^2} = 36 \cr & \Rightarrow {x^2} + \frac{{81}}{{{x^2}}} + 2 \times x \times \frac{9}{x} = 36 \cr & \Rightarrow {x^2} + \frac{{81}}{{{x^2}}} - 18 = 0 \cr & \Rightarrow {\left( {x - \frac{9}{x}} \right)^2} = 0 \cr & \Rightarrow x = \frac{9}{x} \cr & \Rightarrow {x^2} = 9 \cr & {\text{Hence, }}\left( {{x^2} + \frac{9}{{{x^2}}}} \right) \cr & = \left( {9 + \frac{9}{9}} \right) \cr & = 10 \cr} $$

514.

If x = 2 - p, then x³ + 6xp + p³ is equal to:

• (A) 12
• (B) 6
• (C) 8
• (D) 4

Show Answer

 x = 2 - p x + p = 2 Cube both side (x + p)3 = 23 x3 + p3 + 3xp(x + p) = 8 x3 + p3 + 2xp × 2 = 8 x3 + 6xp + p3 = 8

515.

If x = 222, y = 223, z = 225, then the value of x³ + y³ + z³ - 3xyz is?

• (A) 4590
• (B) 4690
• (C) 4950
• (D) 4960

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 $${x^3} + {y^3} + {z^3} - 3xyz$$ $$ = \frac{1}{2}\left( {x + y + z} \right)$$  $$\left[ {{{\left( {x - y} \right)}^2} + {{\left( {y - z} \right)}^2} + {{\left( {z - x} \right)}^2}} \right]$$ $$ = \frac{1}{2}\left( {222 + 223 + 225} \right)$$    $$\left[ {{{\left( {222 - 223} \right)}^2} + {{\left( {223 - 225} \right)}^2} + {{\left( {225 - 222} \right)}^2}} \right]$$ $$\eqalign{ & = \frac{1}{2}\left( {670} \right)\left( {1 + 4 + 9} \right) \cr & = \frac{1}{2} \times 670 \times 14 \cr & = 4690 \cr} $$

516.

a = $$\sqrt{6}$$ - $$\sqrt{5}$$ , b = $$\sqrt{5}$$ - 2, c = 2 - $$\sqrt{3}$$  then point out the correct alternative among the four alternatives given below ?

• (A) b a c
• (B) a c b
• (C) b c a
• (D) a b c

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 $$\eqalign{ & a = \sqrt 6 - \sqrt 5 \cr & b = \sqrt 5 - 2 \cr & c = 2 - \sqrt 3 \cr & {\text{Rationalize all the terms}} \cr & a = \frac{{\sqrt 6 - \sqrt 5 }}{{\sqrt 6 + \sqrt 5 }} \times \sqrt 6 + \sqrt 5 \cr & \,\,\,\,\, = \frac{1}{{\sqrt 6 + \sqrt 5 }} \cr & b = \frac{{\sqrt 5 - 2}}{{\sqrt 5 + 2}} \times \sqrt 5 + 2 \cr & \,\,\,\,\,\, = \frac{1}{{\sqrt 5 + 2}} \cr & \,\,\,\,\,\, = \frac{1}{{\sqrt 5 + \sqrt 4 }} \cr & c = \frac{{2 - \sqrt 3 }}{{2 + \sqrt 3 }} \times 2 + \sqrt 3 \cr & \,\,\,\,\,\, = \frac{1}{{2 + \sqrt 3 }} \cr & \,\,\,\,\,\, = \frac{1}{{\sqrt 4 + \sqrt 3 }} \cr & {\text{Now, }} \cr & a = \frac{1}{{\sqrt 6 + \sqrt 5 }} \cr & b = \frac{1}{{\sqrt 5 + \sqrt 4 }} \cr & c = \frac{1}{{\sqrt 4 + \sqrt 3 }} \cr} $$ Now the term whose denominator is largest is the smallest term. So, a b c

517.

If a + b = 1 and a³ + b³ + 3ab = k, then the value of k is?

• (A) 1
• (B) 3
• (C) 5
• (D) 7

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 $$\eqalign{ & a + b = 1 \cr & {\text{By cubing,}} \cr & {a^3} + {b^3} + 3ab\left( {a + b} \right) = {1^3} \cr & \Rightarrow {a^3} + {b^3} + 3ab = 1\left[ {\because a + b = 1} \right] \cr & \Rightarrow {a^3} + {b^3} + 3ab = k \cr & {\text{From above both equations,}} \cr & k = 1 \cr} $$

518.

If $$\sqrt{0.04\times 0.4\times a}$$    = 0.004 × 0.4 × $$\sqrt{b},$$  then the value of $$\frac{a}{b}\text{is?}$$

• (A) $$\text{16}\times \text{1}{\text{0}}^{-3}$$
• (B) $$\text{16}\times \text{1}{\text{0}}^{-4}$$
• (C) $$\text{16}\times \text{1}{\text{0}}^{-5}$$
• (D) $$\text{16}\times \text{1}{\text{0}}^{-6}$$

Show Answer

 $$\eqalign{ & {\text{ }}\sqrt {0.04 \times 0.4 \times a} = 0.004 \times 0.4 \times \sqrt b \cr & \Rightarrow \sqrt {\frac{4}{{100}} \times \frac{4}{{10}} \times a} = \frac{4}{{1000}} \times \frac{4}{{10}} \times \sqrt b \cr & \Rightarrow \frac{4}{{10}}\sqrt {\frac{a}{{10}}} = \frac{4}{{10}} \times \frac{4}{{1000}} \times \sqrt b \cr & \Rightarrow \frac{a}{{10}} = \frac{{16}}{{1000000}} \times b \cr & \Rightarrow \frac{a}{b} = \frac{{16 \times 10}}{{1000000}} \cr & \Rightarrow \frac{a}{b} = \frac{{16}}{{100000}} \cr & \Rightarrow \frac{a}{b} = 16 \times {10^{ - 5}} \cr} $$

519.

If $${\text{7}}^x\text{ = }\frac{1}{343}\text{,}$$   then the value of x is?

• (A) 3
• (B) -3
• (C) $$\frac{1}{3}$$
• (D) $$\frac{1}{7}$$

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 $$\eqalign{ & {{\text{7}}^x}{\text{ = }}\frac{1}{{343}} \cr & \Rightarrow {{\text{7}}^x}{\text{ = }}\frac{1}{{{7^3}}} \cr & \Rightarrow {{\text{7}}^x}{\text{ = }}{{\text{7}}^{ - 3}} \cr & \Rightarrow x = - 3 \cr} $$ (If bases are equal then their power are also equal)

520.

If a³ + 3a² + 9a = 1, then what is the value of $$a^3+\frac{3}{a}?$$

• (A) 31
• (B) 26
• (C) 28
• (D) 24

Show Answer

 $$\eqalign{ & {a^3} + 3{a^2} + 9a = 1.......\left( {\text{i}} \right) \cr & {a^3} + \frac{3}{a} = ? \cr & {\text{Multiply by '}}a{\text{' and '3' in equation}}\left( {\text{i}} \right) \cr & \left( {{a^3} + 3{a^2} + 9a = 1} \right) \times a \cr & \left( {{a^3} + 3{a^2} + 9a = 1} \right) \times 3 \cr & {a^4} + 3{a^3} + 9{a^2} = 1 \times a \cr & \underline {3{a^3} + 9{a^2} + 27a = 1 \times 3} \to \left( {{\text{Subtracting}}} \right) \cr & {a^4} - 27a = a - 3 \cr & {a^4} + 3 = 28a \cr & {\text{On dividing by }}a \cr & {a^3} + \frac{3}{a} = 28 \cr} $$