### APTITUDE- Problems on SURDS AND INDICES

SURDS AND INDICES

Simple problems:

1. Laws of Indices:

(i) am * an = a(m+n)
(ii) am / an = a(m-n)
(iii) (am)n = a(m*n)
(iv) (ab)n = an * bn
(v) (a/b)n = an / bn
(vi) a0 = 1

2.Surds :Let ‘a’ be a rational number & ‘n’ be a positive integer such that a1/n = nth root a is irrational.Then nth root a is called ‘a’ surd of ‘n’.

Problems:-
(1)
(i) (27)2/3 = (33)2/3 = 32 = 9.
(ii) (1024)-4/5 = (45)-4/5 = (4)-4= 1/(4)4 = 1/256.
(iii)(8/125)-4/3 =((2/5)3)-4/3 = (2/5)-4 = (5/2)4 = 625/16

(2) If 2(x-1)+ 2(x+1) = 1280 then find the value of x .
Solution: 2x/2+2x.2 = 1280
2x(1+22) = 2*1280
2x = 2560/5
2x = 512 => 2x = 29
x = 9

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

Solution: [5[(23)1/3+(33)1/3]3]1/4
[5[2+3]3]1/4
1/4 => 5.

(4) If (1/5)3y= 0.008 then find the value of (0.25)y
Solution: (1/5)3y = 0.008
(1/5)3y =[0.2]3
(1/5)3y =(1/5)3
3y= 3 => y=1.
(0.25)y = (0.25)1 => 0.25 = 25/100 = 1/4

(5) Find the value of (243)n/5 * 32n+1 / 9n * 3 n-1

Solution: (35)n/5 * 32n +1 / (32)n * 3n-1
33n+1 / 33n-1 3
33n+1 * 3-3n+1 => 32 =>9.

(6) Find the value of (21/4-1)( 23/4 +21/2+21/4+1)
Solution: Let us say 21/4 = x
(x-1)(x3+x2+x+1)
(x-1)(x2(x+1)+(x+1))
(x-1) (x2+1) (x+1) [(x-1)(x+1) = (x2-1)]
(x2+1) (x2-1) => (x4-1)
((21/4))4 – 1) = > (2-1) = > 1.

(7) If x= ya , y = zb , z = xc then find the value of abc.
Solution: z= xc
z= (ya)c [ x= ya ]
z= (y)ac
z= (zb)ac [y= zb]
z= zabc
abc = 1

(8)Simplify (xa/xb)a2+ab+b2*(xb/xc)b2+bc+c2*(xc/xa)c2+ca+a2
Solution:[xa-b]a2+ab+b2 * [xb-c]b2+bc+c2 * [xc-a]c2+ca+a2
[ (a-b)(a2+ab+b2) = a3-b3]
from the above formula
=> xa3-b3 xb3-c3 xc3-a3
=> xa3-b3+b3-c3+c3-a3
=> x0 = 1

(9) (1000)7 /1018 = ?
(a) 10 (b) 100 (c ) 1000 (d) 10000
Solution: (1000)7 / 1018
(103)7 / (10)18 = > (10)21 / (10)18
=> (10)21-18 => (10)3 => 1000
Ans :( c )

(10) The value of (8-25-8-26) is
(a) 7* 8-25 (b) 7*8-26 (c ) 8* 8-26 (d) None
Solution: ( 8-25 – 8-26 )
=> 8-26 (8-1 )
=> 7* 8-26
Ans: (b)

(11) 1 / (1+ an-m ) +1/ (1+am-n) = ?
(a) 0 (b) 1/2 (c ) 1 (d) an+m

Solution: 1/ (1+ an/am) + 1/ ( 1+ am/an)
=> am / (am+ an ) + an /(am +an )
=> (am +an ) /(am + an)
=> 1
Ans: ( c)

(12) 1/(1+xb-a+xc-a)+1/(1+xa-b+xc-b)+1/(1+xb-c+xa-c)=?
(a) 0 (b) 1 ( c ) xa-b-c (d) None of the above
Solution: 1/ (1+xb/xa+xc/xa) + 1/(1+xa/xb +xc/xb) +
1/(1+xb/xc +xa/xc)
=> xa /(xa +xb+xc) + xb/(xa +xb+xc) +xc/(xa +xb+xc)
=>(xa +xb+xc) /(xa +xb+xc)
=>1
Ans: (b)

(13) If x=3+2 âˆš2 then the value of (âˆšx â€“ 1/ âˆšx)
is [ âˆš=root]

(a) 1 (b) 2 (c ) 2âˆš2 ( d) 3âˆš3
Solution: (âˆšx-1/âˆšx)2 = x+ 1/x-2
=> 3+2âˆš2 + (1/3+2âˆš2 )-2
=> 3+2âˆš2 + 3-2âˆš2 -2
=> 6-2 = 4
(âˆšx-1/âˆšx)2 = 4
=>(âˆšx-1/âˆšx)2 = 22
(âˆšx-1/âˆšx) = 2.
Ans : (b)

(14) (xb/xc)b+c-a (xc/xa)c+a-b (xa/xb)b+a-c = ? (a) xabc (b) 1 ( c) xab+bc+ca (d) xa+b+c

Solution: [xb-c]b+c-a [xc-a]c+a-b [xa-b]a+b-c

=>x(b-c)(b+c-a) x(c-a)(c+a-b) x(a-b)(a+b-c)
=>x(b2-c2-ab-ac) x(c2-a2-bc-ab) x(a2-b2-ac-bc)
=>x(b2-c2-ab-ac+c2-a2-bc-ab+a2-b2-ac-bc)
=> x0
=>1
Ans: (b)

(15) If 3x-y = 27 and 3x+y = 243 then x is equal to
(a) 0 (b) 2 (c ) 4 (d) 6

Solution: 3x-y = 27 => 3x-y = 33
x-y= 3
3x+y = 243 => 3x+y = 35
x+y = 5
From above two equations x = 4 , y=1
Ans: (c )

(16) If ax = by = cz and b2 = ac then â€˜yâ€™equals
(a)xz/x+z (b)xz/2(x-z) (c)xz/2(z-x) (d)2xz/x+z

Solution: Let us say ax = by = cz = k
ax =k => [ax]1/x = k1/x
=> a = k1/x
Simillarly b = k1/y
c = k1/z
b2 = ac
[k1/y]2=k1/xk1/z
=>k2/y = k1/x+1/z
=> 2/y = 1/x+1/y
=>y= 2xz/x+z
Ans: (d)

(17) ax = b,by = c ,cz = a then the value of xyz is is
(a) 0 (b) 1 (c ) 1/abc (d) abc

Solution: ax = b
(cz)x = b [cz = a]
by)xz = b [by = c]
=>xyz =1
Ans: (b)

(18) If 2x = 4y =8z and (1/2x +1/4y +1/6z) =24/7 then the value of ‘z’ is
(a) 7/16 (b) 7 / 32 (c ) 7/48 (d) 7/64

Solution: 2x = 4y=8z
2x = 22y = 23z
x= 2y = 3z
Multiply above equation with â€˜ 2â€™
2x = 4y= 6z