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01. If p = 99, then the value of p(p² + 3p + 3) will be?

1999999
21000000
31000001
4999998

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Answer:- 1

Explanation:-

Solution:

\begin{aligned} p (p^{2} + 3 p + 3) \\ = p^{3} + 3 p^{2} + 3 p + 1 - 1 \end{aligned}

$\eqalign{ & p\left( {{p^2} + 3p + 3} \right) \cr & = {p^3} + 3{p^2} + 3p + 1 - 1 \cr}$
(On adding and subtracting one both side)

\begin{aligned} = {(p + 1)}^{3} - 1 \\ = {(99 + 1)}^{3} - 1 \\ = 100^{3} - 1 \\ = 1000000 - 3 \\ = 99999 \end{aligned}

$\eqalign{ & = {\left( {p + 1} \right)^3} - 1 \cr & = {\left( {99 + 1} \right)^3} - 1 \cr & = {100^3} - 1 \cr & = 1000000 - 3 \cr & = 99999 \cr}$

If for a non - zero x, 3² + 5x + 3 = 0, then the value of $x^{3} + \frac{1}{x^{3}}$ ${x^3} + \frac{1}{{{x^3}}}$ is?

1 $\frac{10}{27}$ $\frac{{10}}{{27}}$
2 $- (\frac{10}{27})$ $- \left( {\frac{{10}}{{27}}} \right)$
3 $\frac{2}{3}$ $\frac{2}{3}$
4 $- (\frac{2}{3})$ $- \left( {\frac{2}{3}} \right)$

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Answer:- 1

Explanation:-

Solution:

\begin{aligned} 3 x^{2} + 5 x + 3 = 0 \\ 3 x^{2} + 3 = - 5 x \\ Divide by 3 x both sides \\ x + \frac{1}{x} = \frac{- 5}{3} \\ Then, x^{3} + \frac{1}{x^{3}} \\ = {(\frac{- 5}{3})}^{3} - 3 \times (\frac{- 5}{3}) \\ = \frac{- 125}{27} + 5 \\ = \frac{10}{27} \end{aligned}

$\eqalign{ & 3{x^2} + 5x + 3 = 0 \cr & 3{x^2} + 3 = - 5x \cr & {\text{Divide by }}3x{\text{ both sides}} \cr & x + \frac{1}{x} = \frac{{ - 5}}{3} \cr & {\text{Then, }}{x^3} + \frac{1}{{{x^3}}} \cr & = {\left( {\frac{{ - 5}}{3}} \right)^3} - 3 \times \left( {\frac{{ - 5}}{3}} \right) \cr & = \frac{{ - 125}}{{27}} + 5 \cr & = \frac{{10}}{{27}} \cr}$

If $\frac{a}{b} = \frac{1}{2},$ $\frac{a}{b} = \frac{1}{2},$ find the value of the expression $\frac{(2 a - 5 b)}{(5 a + 3 b)}$ $\frac{{\left( {2a - 5b} \right)}}{{\left( {5a + 3b} \right)}}$ = ?

1-32
211
3 $\frac{- 8}{11}$ $\frac{{ - 8}}{{11}}$
417

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Answer:- 1

Explanation:-

Solution:

\begin{aligned} \frac{a}{b} = \frac{1}{2} \\ Let a = x, b = 2 x \\ Then, \frac{(2 a - 5 b)}{(5 a + 3 b)} \\ = \frac{2 x - 10 x}{5 x + 6 x} \\ = \frac{- 8 x}{11 x} \\ = \frac{- 8}{11} \end{aligned}

$\eqalign{ & \frac{a}{b} = \frac{1}{2} \cr & {\text{Let }}a = x,{\text{ }}b = 2x \cr & {\text{Then,}}\frac{{\left( {2a - 5b} \right)}}{{\left( {5a + 3b} \right)}} \cr & = \frac{{2x - 10x}}{{5x + 6x}} \cr & {\text{ = }}\frac{{ - 8x}}{{11x}} \cr & = \frac{{ - 8}}{{11}} \cr}$

If $a + b = 2 c,$ $a + b = 2c,$ find $\frac{a}{a - c}$ $\frac{a}{{a - c}}$ + $\frac{c}{b - c}$ $\frac{c}{{b - c}}$ = ?

127
23
354
49

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Answer:- 1

Explanation:-

Solution:

\begin{aligned} a + b = 2 c \\ Taking a = 2, b = 4 and c = 3 \\ So, 2 + 4 = 2 \times 3 \\ 6 = 6 \\ Now, \frac{a}{a - c} + \frac{c}{b - c} \\ = \frac{2}{2 - 3} + \frac{3}{4 - 3} \\ = \frac{2}{- 1} + \frac{3}{1} \\ = 1 \end{aligned}

$\eqalign{ & a + b = 2c \cr & {\text{Taking }}a = 2,{\text{ }}b = 4{\text{ and }}c = 3 \cr & {\text{So,}}2 + 4 = 2 \times 3 \cr & \boxed{6 = 6} \cr & {\text{Now,}}\frac{a}{{a - c}} + {\text{ }}\frac{c}{{b - c}} \cr & = \frac{2}{{2 - 3}} + {\text{ }}\frac{3}{{4 - 3}} \cr & = \frac{2}{{ - 1}} + \frac{3}{1} \cr & = 1 \cr}$

05. If a² = b + c, b² = a + c, c² = b + a, then what will be the value of $\frac{1}{a + 1}$ $\frac{1}{{a + 1}}$ + $\frac{1}{b + 1}$ $\frac{1}{{b + 1}}$ + $\frac{1}{c + 1}$ $\frac{1}{{c + 1}}$ ?

1-1
22
31
40

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Answer:- 1

Explanation:-

Solution:

\begin{aligned} a^{2} = b + c, b^{2} = a + c, c^{2} = b + a \\ Taking a = 2, b = 2 and c = 2 \\ So, {(2)}^{2} = 2 + 2 \\ 4 = 4 \\ Now, \frac{1}{a + 1} + \frac{1}{b + 1} + \frac{1}{c + 1} \\ Put a = 2, b = 2 and c = 2 \\ = \frac{1}{2 + 1} + \frac{1}{2 + 1} + \frac{1}{2 + 1} \\ = \frac{1}{3} + \frac{1}{3} + \frac{1}{3} \\ = 1 \end{aligned}

$\eqalign{ & {a^2} = b + c,{\text{ }}{b^2} = a + c,{\text{ }}{c^2} = b + a \cr & {\text{Taking }}a = 2,{\text{ }}b = 2{\text{ and }}c = 2 \cr & {\text{So,}}{\left( 2 \right)^2} = 2 + 2 \cr & \boxed{4 = 4} \cr & {\text{Now,}}\frac{1}{{a + 1}} + {\text{ }}\frac{1}{{b + 1}} + {\text{ }}\frac{1}{{c + 1}} \cr & {\text{Put }}a = 2,{\text{ }}b = 2{\text{ and }}c = 2 \cr & = \frac{1}{{2 + 1}} + {\text{ }}\frac{1}{{2 + 1}} + {\text{ }}\frac{1}{{2 + 1}} \cr & = \frac{1}{3} + \frac{1}{3} + \frac{1}{3} \cr & = 1 \cr}$

If $x = \sqrt{a \sqrt[3]{a b \sqrt{a \sqrt[3]{a b}}}} . . . . \propto$ $x = \sqrt {a\root 3 \of {ab\sqrt {a\root 3 \of {ab} } } } .... \propto$ then the value of x is?

1 $\sqrt[5]{a^{2} b}$ $\root 5 \of {{a^2}b}$
2 $\sqrt[5]{a^{4} b^{4}}$ $\root 5 \of {{a^4}{b^4}}$
3 $\sqrt[6]{a^{5} b}$ $\root 6 \of {{a^5}b}$
4 $\sqrt[5]{a^{4} b}$ $\root 5 \of {{a^4}b}$

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Answer:- 1

Explanation:-

Solution:

\begin{aligned} x = \sqrt{a \sqrt[3]{a b \sqrt{a \sqrt[3]{a b}}}} . . . . . . . \propto \\ Square both sides \\ x^{2} = a \sqrt[3]{a b x} (∴ \sqrt{a \sqrt[3]{a b}} . . . . \propto) \\ Again cube both sides \\ x^{6} = a^{3} a b x \\ x^{5} = a^{4} b \\ x = \sqrt[5]{a^{4} b} \end{aligned}

$\eqalign{ & x = \sqrt {a\root 3 \of {ab\sqrt {a\root 3 \of {ab} } } } ....... \propto \cr & {\text{Square both sides}} \cr & {x^2} = a\root 3 \of {abx} {\text{ }}\left( {\therefore \sqrt {a\root 3 \of {ab} } .... \propto } \right) \cr & {\text{Again cube both sides}} \cr & {x^6} = {a^3}abx \cr & {x^5} = {a^4}b \cr & x = \root 5 \of {{a^4}b} \cr}$

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

127
224
319
418

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Answer:- 1

Explanation:-

Solution:

\begin{aligned} Given , a + \frac{1}{a} = 3 \\ Cube both sides \\ a^{3} + \frac{1}{a^{3}} + 3 \times a \times \frac{1}{a} (a + \frac{1}{a}) = {(3)}^{3} \\ \Rightarrow a^{3} + \frac{1}{a^{3}} + 3 \times 3 = 27 \\ \Rightarrow a^{3} + \frac{1}{a^{3}} = 27 - 9 \\ \Rightarrow a^{3} + \frac{1}{a^{3}} = 18 \end{aligned}

$\eqalign{ & {\text{Given , }}a + \frac{1}{a} = 3 \cr & {\text{Cube both sides}} \cr & {a^3} + \frac{1}{{{a^3}}} + 3 \times a \times \frac{1}{a}\left( {a + \frac{1}{a}} \right) = {\left( 3 \right)^3} \cr & \Rightarrow {a^3} + \frac{1}{{{a^3}}} + 3 \times 3 = 27 \cr & \Rightarrow {a^3} + \frac{1}{{{a^3}}} = 27 - 9 \cr & \Rightarrow {a^3} + \frac{1}{{{a^3}}} = 18 \cr}$

If $(\sqrt{a} + \sqrt{b})$ $\left( {\sqrt a + \sqrt b } \right)$ = 15 and $(\sqrt{a} - \sqrt{b})$ $\left( {\sqrt a - \sqrt b } \right)$ = 3, then the value of $\frac{\sqrt{a b}}{4}$ $\frac{{\sqrt {ab} }}{4}$ is?

16
27
3 $\frac{27}{2}$ $\frac{{27}}{2}$
45

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Answer:- 1

Explanation:-

Solution:

\begin{aligned} (\sqrt{a} + \sqrt{b}) = 15 \\ Square both sides \\ \Rightarrow a + b + 2 \sqrt{a b} = 225 \\ \Rightarrow a + b = 225 - 2 \sqrt{a b} . . . . . (i) \\ (\sqrt{a} - \sqrt{b}) = 3 \\ Square both sides \\ \Rightarrow a + b - 2 \sqrt{a b} = 9 \\ \Rightarrow a + b = 9 + 2 \sqrt{a b} . . . . . (i i) \\ From (i) and (ii) \\ \Rightarrow 225 - 2 \sqrt{a b} = 9 + 2 \sqrt{a b} \\ \Rightarrow 216 = 4 \sqrt{a b} \\ \Rightarrow 54 = \sqrt{a b} \\ Divided by 4 on both sides \\ \Rightarrow \frac{\sqrt{a b}}{4} = \frac{54}{4} \\ \Rightarrow \frac{\sqrt{a b}}{4} = \frac{27}{2} \end{aligned}

$\eqalign{ & \left( {\sqrt a + \sqrt b } \right) = 15{\text{ }} \cr & {\text{Square both sides}} \cr & \Rightarrow a + b + 2\sqrt {ab} = 225 \cr & \Rightarrow a + b = 225 - 2\sqrt {ab} \,.....(i) \cr & {\text{ }}\left( {\sqrt a - \sqrt b } \right) = 3 \cr & {\text{Square both sides}} \cr & \Rightarrow a + b - 2\sqrt {ab} = 9 \cr & \Rightarrow a + b = 9 + 2\sqrt {ab} \,.....(ii) \cr & {\text{From (i) and (ii)}} \cr & \Rightarrow 225 - 2\sqrt {ab} = 9 + 2\sqrt {ab} \cr & \Rightarrow 216 = 4\sqrt {ab} \cr & \Rightarrow 54 = \sqrt {ab} \cr & {\text{Divided by 4 on both sides}} \cr & \Rightarrow \frac{{\sqrt {ab} }}{4} = \frac{{54}}{4} \cr & \Rightarrow \frac{{\sqrt {ab} }}{4} = \frac{{27}}{2} \cr}$

If a + b + c = m and $\frac{1}{a}$ $\frac{1}{a}$ + $\frac{1}{b}$ $\frac{1}{b}$ + $\frac{1}{c},$ $\frac{1}{c}{\text{,}}$ then average of a², b², c² is?

1 $m^{2}$ ${m^2}$
2 $\frac{m^{2}}{3}$ $\frac{{{m^2}}}{3}$
3 $\frac{m^{2}}{9}$ $\frac{{{m^2}}}{9}$
4 $\frac{m^{2}}{27}$ $\frac{{{m^2}}}{{27}}$

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Answer:- 1

Explanation:-

Solution:

\begin{aligned} a + b + c = m \\ \Rightarrow \frac{1}{a} + \frac{1}{b} + \frac{1}{c} = 0 \\ \Rightarrow \frac{a b + b c + c a}{a b c} = 0 \\ \Rightarrow a b + b c + c a = 0 \\ {(a + b + c)}^{2} = a^{2} + b^{2} + c^{2} + 2 (a b + b c + c a) \\ \Rightarrow m^{2} = a^{2} + b^{2} + c^{2} \\ \Rightarrow \frac{m^{2}}{3} = \frac{a^{2} + b^{2} + c^{2}}{3} \end{aligned}

$\eqalign{ & a + b + c = m \cr & \Rightarrow \frac{1}{a} + \frac{1}{b} + \frac{1}{c} = 0 \cr & \Rightarrow \frac{{ab + bc + ca}}{{abc}} = 0 \cr & \Rightarrow ab + bc + ca = 0 \cr & {\left( {a + b + c} \right)^2} = {\text{ }}{a^2} + {b^2} + {c^2} + 2\left( {ab + bc + ca} \right) \cr & \Rightarrow {m^2} = {\text{ }}{a^2} + {b^2} + {c^2} \cr & \Rightarrow \frac{{{m^2}}}{3} = {\text{ }}\frac{{{a^2} + {b^2} + {c^2}}}{3} \cr}$

10. If a + b + c = 26 and ab + bc + ca = 109, find the value of a² + b² + c² = ?

1458
2472
3452
4476

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Answer:- 1

Explanation:-

Solution:

\begin{aligned} {(a + b + c)}^{2} = a^{2} + b^{2} + c^{2} + 2 (a b + b c + c a) \\ \Rightarrow {(26)}^{2} = a^{2} + b^{2} + c^{2} + 2 (109) \\ \Rightarrow a^{2} + b^{2} + c^{2} = 676 - 218 \\ \Rightarrow a^{2} + b^{2} + c^{2} = 458 \end{aligned}

$\eqalign{ & {\left( {a + b + c} \right)^2} = {\text{ }}{a^2} + {b^2} + {c^2} + 2\left( {ab + bc + ca} \right) \cr & \Rightarrow {\left( {26} \right)^2} = {\text{ }}{a^2} + {b^2} + {c^2} + 2\left( {109} \right) \cr & \Rightarrow {a^2} + {b^2} + {c^2} = 676 - 218 \cr & \Rightarrow {a^2} + {b^2} + {c^2} = 458 \cr}$

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algebra - 04

01. If p = 99, then the value of p(p² + 3p + 3) will be?

If for a non - zero x, 3² + 5x + 3 = 0, then the value of $x^{3} + \frac{1}{x^{3}}$ ${x^3} + \frac{1}{{{x^3}}}$ is?

If $\frac{a}{b} = \frac{1}{2},$ $\frac{a}{b} = \frac{1}{2},$ find the value of the expression $\frac{(2 a - 5 b)}{(5 a + 3 b)}$ $\frac{{\left( {2a - 5b} \right)}}{{\left( {5a + 3b} \right)}}$ = ?

If $a + b = 2 c,$ $a + b = 2c,$ find $\frac{a}{a - c}$ $\frac{a}{{a - c}}$ + $\frac{c}{b - c}$ $\frac{c}{{b - c}}$ = ?

05. If a² = b + c, b² = a + c, c² = b + a, then what will be the value of $\frac{1}{a + 1}$ $\frac{1}{{a + 1}}$ + $\frac{1}{b + 1}$ $\frac{1}{{b + 1}}$ + $\frac{1}{c + 1}$ $\frac{1}{{c + 1}}$ ?

If $x = \sqrt{a \sqrt[3]{a b \sqrt{a \sqrt[3]{a b}}}} . . . . \propto$ $x = \sqrt {a\root 3 \of {ab\sqrt {a\root 3 \of {ab} } } } .... \propto$ then the value of x is?

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

If $(\sqrt{a} + \sqrt{b})$ $\left( {\sqrt a + \sqrt b } \right)$ = 15 and $(\sqrt{a} - \sqrt{b})$ $\left( {\sqrt a - \sqrt b } \right)$ = 3, then the value of $\frac{\sqrt{a b}}{4}$ $\frac{{\sqrt {ab} }}{4}$ is?

If a + b + c = m and $\frac{1}{a}$ $\frac{1}{a}$ + $\frac{1}{b}$ $\frac{1}{b}$ + $\frac{1}{c},$ $\frac{1}{c}{\text{,}}$ then average of a², b², c² is?

10. If a + b + c = 26 and ab + bc + ca = 109, find the value of a² + b² + c² = ?

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algebra - 04

01. If p = 99, then the value of p(p2 + 3p + 3) will be?

02. If for a non - zero x, 32 + 5x + 3 = 0, then the value of x3+1x3x3+1x3{x^3} + \frac{1}{{{x^3}}} is?

03. If ab=12,ab=12,\frac{a}{b} = \frac{1}{2}, find the value of the expression (2a−5b)(5a+3b)(2a−5b)(5a+3b)\frac{{\left( {2a - 5b} \right)}}{{\left( {5a + 3b} \right)}} = ?

04. If a+b=2c,a+b=2c,a + b = 2c, find aa−caa−c\frac{a}{{a - c}} + cb−ccb−c\frac{c}{{b - c}} = ?

05. If a2 = b + c, b2 = a + c, c2 = b + a, then what will be the value of 1a+11a+1\frac{1}{{a + 1}} + 1b+11b+1\frac{1}{{b + 1}} + 1c+11c+1\frac{1}{{c + 1}} ?

06. If x=aabaab−−√3−−−−−√−−−−−−−−√3−−−−−−−−−−−√....∝x=aabaab33....∝x = \sqrt {a\root 3 \of {ab\sqrt {a\root 3 \of {ab} } } } .... \propto then the value of x is?

07. If a+1a=3,a+1a=3,a + \frac{1}{a} = 3, then the value of a3+1a3a3+1a3{a^3} + \frac{1}{{{a^3}}} is?

08. If (a−−√+b√)(a+b)\left( {\sqrt a + \sqrt b } \right) = 15 and (a−−√−b√)(a−b)\left( {\sqrt a - \sqrt b } \right) = 3, then the value of ab−−√4ab4\frac{{\sqrt {ab} }}{4} is?

09. If a + b + c = m and 1a1a\frac{1}{a} + 1b1b\frac{1}{b} + 1c,1c,\frac{1}{c}{\text{,}} then average of a2, b2, c2 is?

10. If a + b + c = 26 and ab + bc + ca = 109, find the value of a2 + b2 + c2 = ?

Subject Wise

Exam Wise

NCERT 6 to 12

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Practice Batch

01. If p = 99, then the value of p(p² + 3p + 3) will be?

If for a non - zero x, 3² + 5x + 3 = 0, then the value of $x^{3} + \frac{1}{x^{3}}$ ${x^3} + \frac{1}{{{x^3}}}$ is?

If $\frac{a}{b} = \frac{1}{2},$ $\frac{a}{b} = \frac{1}{2},$ find the value of the expression $\frac{(2 a - 5 b)}{(5 a + 3 b)}$ $\frac{{\left( {2a - 5b} \right)}}{{\left( {5a + 3b} \right)}}$ = ?

If $a + b = 2 c,$ $a + b = 2c,$ find $\frac{a}{a - c}$ $\frac{a}{{a - c}}$ + $\frac{c}{b - c}$ $\frac{c}{{b - c}}$ = ?

05. If a² = b + c, b² = a + c, c² = b + a, then what will be the value of $\frac{1}{a + 1}$ $\frac{1}{{a + 1}}$ + $\frac{1}{b + 1}$ $\frac{1}{{b + 1}}$ + $\frac{1}{c + 1}$ $\frac{1}{{c + 1}}$ ?

If $x = \sqrt{a \sqrt[3]{a b \sqrt{a \sqrt[3]{a b}}}} . . . . \propto$ $x = \sqrt {a\root 3 \of {ab\sqrt {a\root 3 \of {ab} } } } .... \propto$ then the value of x is?

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

If $(\sqrt{a} + \sqrt{b})$ $\left( {\sqrt a + \sqrt b } \right)$ = 15 and $(\sqrt{a} - \sqrt{b})$ $\left( {\sqrt a - \sqrt b } \right)$ = 3, then the value of $\frac{\sqrt{a b}}{4}$ $\frac{{\sqrt {ab} }}{4}$ is?

If a + b + c = m and $\frac{1}{a}$ $\frac{1}{a}$ + $\frac{1}{b}$ $\frac{1}{b}$ + $\frac{1}{c},$ $\frac{1}{c}{\text{,}}$ then average of a², b², c² is?

10. If a + b + c = 26 and ab + bc + ca = 109, find the value of a² + b² + c² = ?

NCERT
6 to 12