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) 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
Question. 1. Simplify : (i) (27)2/3 (ii) (1024)-4/5 (iii)( 8 / 125 )-4/3
Answer .
(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
Question. 2. Evaluate: (i) (.00032)3/5 (ii)l (256)0.16 x (16)0.18.
Answer .
(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.
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 -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 2x - 1 + 2x + 1 = 1280, then find the value of x.
Answer .
2x - 1 + 2X+ 1 = 1280 ó 2x-1 (1 +22) = 1280
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.
Question. 5. Find the value of [ 5 ( 81/3 + 271/3)3]1/ 4
Answer .
[ 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 ( 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.
Question. 6. Find the Value of {(16)3/2 + (16)-3/2}
Answer .
[(16)3/2 +(16)-3/2 = (42)3/2 +(42)-3/2 = 4(2 * 3/2) + 4{ 2* (-3/2)}
[(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.
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 ó 3y = 3 ó Y = 1.
(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.
Question. 8. Find the value of (243)n/5 ´ 32n + 19n ´ 3n -1 .
Answer .
(243)n/5 x32n+l = 3 (5 * n/5) ´ 32n+l _ = 3n ´32n+1
(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)
Question. 9. Find the value Of (21/4-1)(23/4+21/2+21/4+1)
Answer .
Putting 21/4 = x, we get :
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.
Question. 10. Find the value of 62/3 ´ 3√67
3√66
Answer .
62/3 ´ 3√67 = 62/3 ´ (67)1/3 = 62/3 ´ 6(7 * 1/3) = 62/3 ´ 6(7/3)
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.
Question. 11. If x= ya, y=zb and z=xc,then find the value of abc.
Answer .
z1= xc =(ya)c [since x= ya]
=y(ac) = (zb)ac [since y=zb]
=zb(ac)= zabc
\ abc = 1.
= 24
Question. 12. Simplify [(xa / xb)^(a2+b2+ab)] * [(xb / xc )^ b2+c2+bc)] * [(xc/xa)^(c2+a2+ca)]
Answer .
Given Expression
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.
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:
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.
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:
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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Sunday 25 December 2016
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