## Saturday, 21 May 2011

### AGARVAL APPTITUDE - SURDS AND INDICES

SURDS AND INDICES

IMPORTANT FACTS AND FORMULAE

1. LAWS OF INDICES:

i. am x an = am + n
ii. am / an = am-n
iii. (am)n = amn
iv. (ab)n = anbn
v. ( a/ b )n = ( an / bn )
vi. a0 = 1

2. 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.

3. 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
I SOLVED EXAMPLES

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  2x-1 (1 +22) = 1280
 2x-1 = 1280 / 5 = 256 = 28  x -1 = 8  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  3y = 3  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 _ = 3n 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 .