Mathematics Question Answers On Surds And Indices Free PDF Download For Competitive Exams

Mathematics Question Answers On Surds And Indices Free PDF Download For Competitive Exams

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                      Mathematics Question Answers On Surds

Mathematics Question Answers On Surds And Indices

I IMPORTANT FACTS AND FORMULAE I

  1. LAWS OF INDICES:

 

  • am x an = am + n
  • am­ / an = am-n
  • (am)n = amn
  • (ab)n = anbn
  • ( a/ b )n = ( an / bn )
  • a0 = 1

 

 

  1. SURDS: Let a be a rational number and n be a positive integer such that a1/n = nsqrt(a)

   is irrational. Then nsqrt(a)  is called a surd of order n.

 

  1. LAWS OF SURDS:

 

(i)  n√a = a1/2

(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√am

Mathematics Question Answers On Surds And Indices Free PDF Download.jpg

Mathematics Question Answers On Surds And Indices Free PDF                                                            Download

I –SOLVED EXAMPLES Of Mathematics Question Answers On Surds

 

 

Ex. 1. Simplify : (i) (27)2/3       (ii) (1024)-4/5    (iii)( 8 / 125 )-4/3

 

Sol .     (i) (27)2/3 = (33)2/3 = 3( 3 * ( 2/ 3)) = 32 = 9

(ii) (1024)-4/5 = (45)-4/5 = 4 { 5 * ( (-4) / 5 )} = 4-4 = 1 / 44 = 1 / 256

(iii) ( 8 / 125 )-4/3 = {(2/5)3}-4/3 = (2/5){ 3 * ( -4/3)} = ( 2 / 5 )-4 = ( 5 / 2 )4  = 54 / 24 = 625 / 16

 

 

Ex. 2. Evaluate: (i) (.00032)3/5            (ii)l (256)0.16 x (16)0.18.

 

Sol.      (i) (0.00032)3/5 = ( 32 / 100000 )3/5. = (25 / 105)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.

196

 

Ex. 3. What is the quotient when (x-1 – 1) is divided by (x – 1) ?

 

Sol.     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

 

Ex. 4. If 2x – 1 + 2x + 1 = 1280, then find the value of  x.

Sol.      2x – 1 + 2X+ 1 = 1280  ó 2x-1 (1 +22) = 1280

ó 2x-1  =   1280 / 5 = 256 =  28   ó x -1 = 8 ó x  =  9.

 

 

Hence, x = 9.

Ex. 5. Find the value of [ 5 ( 81/3 + 271/3)3]1/ 4

 

Sol.      [ 5 ( 81/3 + 271/3)3]1/ 4  =  [ 5 { (23)1/3 + (33)1/3}3]1/ 4 =   [ 5 { (23 * 1/3)1/3 + (33 *1/3 )1/3}3]1/ 4

= {5(2+3)3}1/4 = (5 * 53)1/ 4 =5(4 * 1/ 4)  = 51 = 5.

 

 

Ex. 6. Find the Value of {(16)3/2 + (16)-3/2}

 

Sol.     [(16)3/2 +(16)-3/2 = (42)3/2 +(42)-3/2 = 4(2 * 3/2) + 4{ 2* (-3/2)}

= 43 + 4-3 = 43 + (1/43) = ( 64 + ( 1/64)) = 4097/64.

 

Ex. 7. If (1/5)3y = 0.008, then find the value of(0.25)y.

 

Sol. (1/5)3y = 0.008 =  8/1000 =  1/125 = (1/5)3 ó 3y = 3 ó Y = 1.

\ (0.25)y = (0.25)1 = 0.25.

 

Ex. 8. Find the value of     (243)n/5 ´ 32n + 1

                                             9n ´ 3n -1 .

 

Sol. (243)n/5 x32n+l   =   3 (5 * n/5) ´ 32n+l _ = 3´32n+1

       (32)n ´ 3n – 1          32n ´ 3n – 1                 32n ´ 3n-l

 

                                    = 3n + (2n + 1)   =   3(3n+1)   =  3(3n+l)-(3n-l) = 32 = 9.

32n+n-1           3(3n-1)

 

Ex. 9. Find the value Of (21/4-1)(23/4+21/2+21/4+1)

 

Sol.

Putting 21/4 = x, we get :

 

(21/4-1) (23/4+21/2+21/4+1)=(x-1)(x3+x2+x+1) , where x = 21/4

                                                     =(x-1)[x2(x+1)+(x+1)]

                                                     =(x-1)(x+1)(x2+1) = (x2-1)(x2+1)

                                                     =(x4-1) = [(21/4)4-1] = [2(1/4*4) –1] = (2-1) = 1.

           

Ex. 10. Find the value of   62/3 ´  3√67

                                                3√66

 

Sol.       62/3 ´  3√67   =  62/3  ´ (67)1/3   =  62/3  ´  6(7 * 1/3)  =   62/3  ´  6(7/3)

                  3√66                 (66)1/3                  6(6 * 1/3)                    62

 

                                    =62/3 ´ 6((7/3)-2) = 62/3 ´ 61/3  = 61 = 6.

Ex. 11. If x= ya, y=zb and z=xc,then find the value of abc.

 

Sol.    z1= xc =(ya)c       [since x= ya]

             =y(ac) = (zb)ac   [since y=zb]

             =zb(ac)= zabc

\           abc = 1.          

 = 24

 

Ex. 12. Simplify [(xa / xb)^(a2+b2+ab)] * [(xb / xc )^ b2+c2+bc)] * [(xc/xa)^(c2+a2+ca)]

Sol.

 Given Expression

= [{x(o b)}^(a2 + b2 + ob)].[‘(x(b – c)}^ (b2 + c2 + bc)].[‘(x(c a)}^(c2 + a2 + ca])

= [x(a b)(a2 + b2 + ab) . x(b – c) (b2 +c2+ bc).x(c– a) (c2 + a2 + ca)]

= [x^(a3-b3)].[x^(b3-e3)].[x^(c3-a3)] = x^(a3-b3+b3-c3+c3-a3) = x0 = 1.

 

Ex. 13. Which is larger √2 or 3√3 ?

 

Sol. 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 = 21/2 = 2((1/2)*(3/2)) =23/6 =  81/6 = 6√8

3√3= 31/3 = 3((1/3)*(2/2)) = 32/6 = (32)1/6 = (9)1/6 = 6√9.

Clearly, 6√9 > 6√8 and hence 3√3  > √2.

 

Ex. 14. Find the largest from among 4√6, √2 and 3√4.

Sol. 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 = 61/4 = 6((1/4)*(3/3)) = 63/12 = (63)1/12  = (216)1/12.

√2 = 21/2 = 2((1/2)*(6/6)) = 26/12 = (26)1/12  = (64)1/12.

3√4 = 41/3 = 4((1/3)*(4/4))  =  44/12  = (44)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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