Here is in this article we are sharing Important quiz on surds and indices topic of Quantitative Aptitude subject. It is recommended for SSC, IBPS, SBI, Clerk and PO online exams. Top 14 example of surds indices aptitude questions are listed below. Generally this part question asked in most of all competitive exams, And also repeat many question. So interested candidates can study these question and revise the practice.

**SURDS AND INDICES**__I IMPORTANT FACTS AND FORMULAE__

__I__

**1. LAWS OF INDICES:**

(i)

*a*x^{m}*a*=^{n}*a*^{m }^{+ n}
(ii)

*a*^{m}_{}/ a^{n}= a^{m-n}
(iii)

*(a*^{m})^{n}= a^{mn}
(iv)

*(ab)*^{n}= a^{n}b^{n}
(v)

*( a/ b )*^{n}= ( a^{n}/ b^{n})
(vi)

*a*^{0}= 1**2.**

**SURDS**:

*Let a be a rational number and*n

*be a positive integer such that a*

^{1/n}=^{n}sqrt(a)*is irrational. Then*sqrt(a) is

^{n}*called a surd of order*n.

**3.**

**LAWS OF SURDS:**

(i)

^{n}√a = a^{1/2}
(ii)

^{n }√ab =^{n }√a *^{n }√b
(iii)

^{n }√a/b =^{n }√a /^{n }√b
(iv) (

^{n}√a)^{n}= a
(v)

^{m}√(^{n}√(a)) =^{mn}√(a)
(vi) (

^{n}√a)^{m}=^{n}√a^{m}

__I SOLVED EXAMPLES__**Question. 1. Simplify : (i) (27)**

^{2/3}(ii) (1024)^{-4/5}(iii)( 8 / 125 )^{-4/3}**Answer .**

(i) (27)

^{2/3}= (3

^{3})

^{2/3}= 3

^{( 3 * ( 2/ 3))}= 3

^{2}= 9

(ii) (1024)

^{-4/5}= (4^{5})^{-4/5 }= 4^{{ 5 * ( (-4) / 5 )}}= 4^{-4}= 1 / 4^{4}= 1 / 256
(iii) ( 8 / 125 )

^{-4/3}= {(2/5)^{3}}^{-4/3}= (2/5)^{{ 3 * ( -4/3)}}= ( 2 / 5 )^{-4}= ( 5 / 2 )^{4 }= 5^{4 }/ 2^{4}= 625 / 16

**Question**. 2. Evaluate: (i)*(.00032)*(256)^{3/5}(ii)l^{0.16 }x (16)^{0.18}.**Answer .**

(i) (0.00032)

^{3/5}= ( 32 / 100000 )

^{3/5}. = (2

^{5}/ 10

^{5})

^{3/5 }= {( 2 / 10 )

^{5}}

^{3/5}= ( 1 / 5 )

^{(5 * 3 / 5)}= (1/5)

^{3}= 1 / 125

(ii) (256)

^{0. 16}* (16)^{0. 18}= {(16)^{2}}^{0. 16}* (16)^{0. 18}= (16)^{(2 * 0. 16)}* (16)^{0. 18}
=(16)

^{0.32}* (16)^{0.18}= (16)^{(0.32+0.18)}= (16)^{0.5}= (16)^{1/2}= 4.

**Question**. 3.*What is the quotient when*(x^{-1}- 1)*is divided by*(x - 1) ?**Answer .**

__x____=__

^{-1}-1__(1/x)-1__= _

__1 -x__

__* 1__=

__-1__

*x*- 1

*x*- 1

*x (x*- 1)

*x*

*Hence, the required quotient is -1/x*

**Question. 4. If 2**

^{x - 1}+ 2^{x + 1}=*1280, then find the value*of x.**Answer .**

2

^{x - 1}+

*2*

^{X+ }^{1}= 1280 Ã³ 2

^{x-1}(1 +2

^{2}) = 1280

Ã³ 2

^{x-1}= 1280 / 5 = 256 = 2^{8}Ã³ x -1 = 8 Ã³ x = 9.
Hence, x = 9.

**Question. 5.**

*Find*the value of [ 5 ( 8^{1/3}+ 27^{1/3})^{3}]^{1/ 4}**Answer .**

[ 5 ( 8

^{1/3}+ 27

^{1/3})

^{3}]

^{1/ 4 }= [ 5 { (2

^{3})

^{1/3}+ (3

^{3})

^{1/3}}

^{3}]

^{1/ 4}= [ 5 { (2

^{3 * 1/3})

^{1/3}+ (3

^{3 *1/3 })

^{1/3}}

^{3}]

^{1/ 4}

= {5(2+3)

^{3}}^{1/4}= (5 * 5^{3})^{1/ 4}=5^{(4 * 1/ 4) }= 5^{1}= 5.**Question. 6. Find the Value of {(16)**

^{3/2}+ (16)^{-3/2}}**Answer .**

[(16)

^{3/2}+(16)

^{-3/2}= (4

^{2})

^{3/2}+(4

^{2})

^{-3/2}= 4

^{(2 * 3/2)}+ 4

^{{ 2* (-3/2)}}

= 4

^{3}+ 4^{-3}= 4^{3}+ (1/4^{3}) = ( 64 + ( 1/64)) = 4097/64.**Question. 7.**

*If (1/5)*=^{3y}*0.008, then find the value of(0.25)*^{y}.**Answer .**

(1/5)

^{3y}= 0.008 = 8/1000 = 1/125 = (1/5)

^{3}Ã³

*3y*= 3 Ã³

*Y*= 1.

\

*(0.25)*= (0.25)^{y}^{1}= 0.25.**Question. 8.**

*Find the value*of__(243)__^{n/5 }__´__*3*^{2n }^{+ 1}*9*´^{n }*3*^{n }^{-1}.**Answer .**

__(243)__=

^{n/5}*x3*^{2n+l }__3__

^{(5 * n/5)}*_ =*

__´ 3__^{2n+l}

__3__^{n }

__´3__^{2n+1}*(3*´

^{2})^{n}*3*

^{n }^{- 1}

*3*´

^{2n}*3*

^{n }^{- 1}

*32n*

*´*

*3*

^{n-l }*=*

__3__^{n }__=__

^{+ (2n + 1)}__3__=

^{(3n+1)}*3*= 3

^{(3n+l)-(3n-l)}^{2}= 9.

3

^{2n+n-1}3^{(3n-1)}**Question. 9.**

*Find the value Of (2*^{1/4}-1)(2^{3/4}+2^{1/2}+2^{1/4}+1)**Answer .**

Putting 2

^{1/4}= x, we get :

(2

^{1/4}-1)*(2*^{3/4}+2^{1/2}+2^{1/4}+1)=(x-1)(x^{3}+x^{2}+x+1) , where x = 2^{1/4}*=(x-1)[x*

^{2}(x+1)+(x+1)]*=(x-1)(x+1)(x*

^{2}+1) = (x^{2}-1)(x^{2}+1)*=(x*

^{4}-1) = [(2^{1/4})^{4}-1] = [2^{(1/4}*^{4)}–1] = (2-1) = 1.

**Question. 10. Find the value of**

__6__^{2/3}__´__^{3}√6^{7}

^{3}√6^{6}**Answer .**

__6__

^{2/3}__´__=

^{3}√6^{7}__6__

^{2/3 }__´ (6__=

^{7})^{1/3}__6__

^{2/3}__´ 6__=

^{(7 * 1/3)}__6__

^{2/3 }__´ 6__

^{(7/3)}

^{3}√6^{6}(6^{6})^{1/3}6^{(6 * 1/3)}6^{2}*=6*

^{2/3}*´ 6*

^{((7/3)-2)}= 6^{2/3}*´ 6*

^{1/3}= 6^{1 }= 6.

*Question. 11. If x= y*^{a}, y=z^{b}and z=x^{c},then find the value of abc.**Answer .**

*z*

^{1}= x^{c}=(y^{a})^{c}[since x= y^{a}]*=y*

^{(ac)}= (z^{b})^{ac}[since y=z^{b}]*=z*

^{b(ac)}= z^{abc}*\ abc = 1.*

^{ }
= 24

**Question. 12. Simplify [(x**

^{a}/ x^{b})^(a^{2}+b^{2}+ab)] * [(x^{b}/ x^{c})^ b^{2}+c^{2}+bc)] * [(x^{c}/x^{a})^(c^{2}+a^{2}+ca)]**Answer .**

Given Expression

= [

*{x*^{(o }^{- b)}*}^(a*+^{2}*b*+^{2}*ob)].['(x*^{(b }^{- c)}*}^ (b*+ c^{2}^{2}+*bc)].['(x*^{(c }^{- a)}*}^(c*+ a^{2}^{2}+*ca])*
= [

*x*^{(a }^{- b)(a2 + b2 + ab)}.*x*^{(b }^{- c) (b2 +c2+ bc)}*.x*^{(c}^{- a) (c2 + a2 + ca)}*]*
= [

*x^(a*=^{3}-b^{3})].[x^(b^{3}-e^{3})].[x^(c^{3}-a^{3})]*x^(a*=^{3}-b^{3}+b^{3}-c^{3}+c^{3}-a^{3})*x*= 1.^{0}**Question. 13.**

*Which is larger √2*or^{3}√3 ?**Answer .**

Given surds are of order 2 and 3. Their L.C.M. is 6. Changing each to a surd of order 6, we get:

√2 = 2

^{1/2}= 2^{((1/2)*(3/2))}=2^{3/6 }= 8^{1/6}=^{6}√8^{3}√3= 3

^{1/3}= 3

^{((1/3)*(2/2))}= 3

^{2/6}= (3

^{2})

^{1/6}= (9)

^{1/6}=

^{ 6}√9.

Clearly,

^{6}√9 >^{6}√8 and hence^{3}√3 > √2.**Question. 14.**

*Find the largest from among*4√6,*√2 and*^{3}√4.**Answer .**

Given surds are of order 4, 2 and 3 respectively. Their L.C,M, is 12, Changing each to a surd of order 12, we get:

^{4}√6 = 6

^{1/4}= 6

^{((1/4)*(3/3))}= 6

^{3/12}= (6

^{3})

^{1/12 }= (216)

^{1/12.}

√2 = 2

^{1/2}= 2^{((1/2)*(6/6))}= 2^{6/12}= (2^{6})^{1/12 }= (64)^{1/12}.^{3}√4 = 4

^{1/3}= 4

^{((1/3)*(4/4))}= 4

^{4/12 }= (4

^{4})

^{1/12}= (256)

^{1/12}.

Clearly, (256)

^{1/12}> (216)^{1/12 }> (64)^{1/12}
Largest one is (256)

^{1/12}. i.e.^{3}√4 .^{ }

.

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