Mathematics

Mathematical Indices MCQ | Multiple Choice Questions

Posted by Asok K Nadhani on

Last Updated on: 18th December 2024, 03:17 pm

Mathematical Indices MCQ

Competitive Exams, Entrance Exams are conducted on MCQ.  Click Here to understand the Forms, Structure, Rules of MCQ, techniques of understanding, analysing and selection correct answer of MCQ,

Play the Video explaining some interesting aspects of selection of correct answer of MCQ.

Complete resources on Mathematics

Complete resources on Indices

1. If 3x = 4y = 12z, then z is equal to:

a. \displaystyle \frac{{(x+y)}}{x}

b. \displaystyle \frac{{xy}}{{(x+y)}}

c. \displaystyle \frac{y}{{(x+y)}}

d. \displaystyle \frac{{(x+y)}}{y}

Let 3x = 4y = 12z = k. So, 3=k1/x, 4=k1/y, 12=k1/z.

So, k1/x X k1/y = 3X 4=12. Or k1/x + 1/y = 12. k1/z =12 = k (1/x + 1/y). So \displaystyle \frac{1}{z}= \displaystyle \frac{1}{x} + \displaystyle \frac{1}{y}, or \displaystyle \frac{1}{z} = \displaystyle \frac{{(x+y)}}{{xy}}.

Or z=\displaystyle \frac{{xy}}{{(x+y)}}.

So, option (b) is correct.

2. (xb+c)b–c (xc+a)c–a (xa+b)a–b is equal to:

(a)      0

(b)      1

(c)      ab

(d)      ac

(xb+c)b–c  (xc+a)c–a  (xa+b)a–b = x b²– c²  .  x c²– a²  .  x a²– b² =  x0 =1. So, option (b) is correct.

3. The value of \displaystyle {{\left( {\frac{{{{x}^{a}}}}{{{{x}^{b}}}}} \right)}^{{a+b}}}\times {{\left( {\frac{{{{x}^{b}}}}{{{{x}^{c}}}}} \right)}^{{b+c}}}\times {{\left( {\frac{{{{x}^{c}}}}{{{{x}^{a}}}}} \right)}^{{c+a}}} is

(a)    1

(b)    0

(c)    x

(d)    x a + b

\displaystyle {{\left( {\frac{{{{x}^{a}}}}{{{{x}^{b}}}}} \right)}^{{a+b}}}\times {{\left( {\frac{{{{x}^{b}}}}{{{{x}^{c}}}}} \right)}^{{b+c}}}\times {{\left( {\frac{{{{x}^{c}}}}{{{{x}^{a}}}}} \right)}^{{c+a}}} =   (xa- b)a+b  x  (xb-c)b+c  x  (xc- a)c+a

= (x a²– b² ) x     (x b²– c² ) x   (x c²– a²)=x0=1. So, option (a) is correct.

4. If ax ¸ a2 = 1, then value of x is:

(a)     \displaystyle \frac{1}{2}

(b)     2

(c)      1

(d)     0.

ax ¸ a2 = 1 = a0. So, ax– 2 = a0, or, x – 2 = 0    or, x = 2. So, option (b) is correct.

5. If x = 81, then value of (\displaystyle {{x}^{{\frac{1}{4}}}} – 1) (\displaystyle {{x}^{{\frac{1}{4}}}}  + 1) will be:

(a)    6

(b)    8

(c)    2

(d)    1.

(\displaystyle {{x}^{{\frac{1}{4}}}} – 1) (\displaystyle {{x}^{{\frac{1}{4}}}}  + 1) = {(81)1/4 – 1} X {(81)1/4 + 1}= {(34)1/4 – 1} X {(34)1/4 + 1} = (3 – 1) X ( 3 + 1) = 2 x 4 = 8. So, option (b) is correct.

6. If 3x– 1 = 81, then x will be:

(a)    8

(b)    5

(c)    3

(d)    2.

3x– 1 = 81, So, 3x– 1 = 81 = 34. So, x – 1 = 4, Hence x = 4 + 1 = 5. So, option (b) is correct.

7. If 42x – 22x = 12 then x is:

(a)    6

(b)    1

(c)    5

(d)    3.

42x – 22x = 12, Or, (4x)2 – (22)x  = 12, Or, (4x)2 – (4)x  = 12, Or, (4x)2 – 4x – 12 = 0

Or, (4x)2 + 3X4x – 4X 4x – 12 = 0, Or, 4x (4x + 3) – 4 (4x + 3) = 0, Or, (4x + 3) (4x – 4) = 0

Or, 4x – 4 = 0, Or, 4x = 4 = 41   So, x = 1. So, option (b) is correct.

8. Find value of \displaystyle \left( {\frac{{{{X}^{{m+3n}}}.{{x}^{{4m-9n}}}}}{{{{x}^{{6m-6n}}}}}} \right)

(a)    xm

(b)    x– m

(c)    xn

(d)    1.

\displaystyle \left( {\frac{{{{X}^{{m+3n}}}.{{x}^{{4m-9n}}}}}{{{{x}^{{6m-6n}}}}}} \right)= \displaystyle \left( {\frac{{{{x}^{{m+3n+4m-9n}}}}}{{{{x}^{{6m-6n}}}}}} \right)  = \displaystyle \left( {\frac{{{{x}^{{5m-6n}}}}}{{{{x}^{{6m-6n}}}}}} \right) = x 5m – 6n – (6m – 6n)

= x 5m – 6n – 6m + 6n = x-m.  So, option (b) is correct.

9. Value of  \displaystyle \left( {\frac{{{{x}^{{a(b-c)}}}.{{x}^{{c(a-b)}}}}}{{{{x}^{{b(a-c)}}}}}} \right) is

(a)    a

(b)    1

(c)    xa+b+c

(d)    0.

\displaystyle \left( {\frac{{{{x}^{{a(b-c)}}}.{{x}^{{c(a-b)}}}}}{{{{x}^{{b(a-c)}}}}}} \right) = x (ab-ac)+(ac-bc) – (ab-ac) = x ab-ac+ac-bc – ab+bc =x0=1. So, option (b) is correct.

10. Value of 16x – 3 y2 .  8–1 . x3 y –2 is

(a)      2xy

(b)     \displaystyle \frac{{xy}}{2}

(c)      2

(d)     1.

16x– 3 y2 .  8–1 . x3 y –2 = 16 . 8–1 . x– 3  . x3  . y2 . y – 2 = 16/8 . x0. y0 = 2 X 1 X 1=2. So, option (c) is correct.

11. Find the simplest value of 4 x 8 –2/3

a. 1

b. -1

c. ¼

d. ½

4 x 8 -2/3 = 22 \displaystyle \times  ((2)3)-2/3  = 22 X 2(-2) = 22-2= 20=1. So, option (a) is correct.

12. If 3x = 2 – x, Find the value of x.

(a)    1

(b)    -1

(c)    0

(d)    6

3x = 2 – x, or, 3x = \displaystyle \frac{1}{{{{2}^{x}}}}, Or, 3x . 2x = 1  Or,  (3 x 2)x = 1  Or,  6x = 1 = 60  Or, x = 0. So, option (c) is correct.

13. If x = 8, y =27, find the value of (x4/3  + y 2/3)1/2.

(a)    1

(b)    2

(c)    4

(d)    5.

Putting the value of x & y, we get (x4/3  + y 2/3)1/2 = (84/3  + 27 2/3)1/2 = {(23) 4/3 + (33) 2/3}1/2 = (24+32)1/2 = (16+9)1/2 =(25)1/2 = 5.

So, option (d) is correct.

14. Simplify [(xa +  b–c. xa – b + c) b] c.

(a)    1

(b)    2

(c)    0

(d)    x2abc.

[(xa +  b  –  c . xa – b + c) b] c = [(xa +  b –  c +  a – b + c) b] c = {(x2a) b]} c  = x2abc. So, option (d) is correct.

15. Simplify \displaystyle \left( {\frac{{{{2}^{{x+1}}}+{{2}^{x}}}}{{{{2}^{{x+3}}}-{{2}^{{x+1}}}}}} \right)

(a)     2

(b)     \displaystyle \frac{1}{2}

(c)      1

(d)     0.

\displaystyle \left( {\frac{{{{2}^{{x+1}}}+{{2}^{x}}}}{{{{2}^{{x+3}}}-{{2}^{{x+1}}}}}} \right) = \displaystyle \left( {\frac{{{{2}^{x}}{{{.2}}^{1}}+{{2}^{x}}}}{{{{2}^{x}}{{{.2}}^{3}}-{{2}^{x}}{{{.2}}^{1}}}}} \right)= \displaystyle {\frac{{{{2}^{x}}(2+1)}}{{{{2}^{x}}({{2}^{3}}-2)}}}

= \displaystyle \frac{{3\times {{2}^{x}}}}{{6\times {{2}^{x}}}}=\frac{3}{6}=\frac{1}{2}.

So, option (b) is correct.

16. Simplify \displaystyle \frac{{({{x}^{{m+3n}}}.{{x}^{{4m-5n}}})}}{{({{x}^{{6m-6n}}})}}

(a)    X m

(b)    X –m

(c)    X n

(d)    X –n

\displaystyle \frac{{({{x}^{{m+3n}}}.{{x}^{{4m-5n}}})}}{{({{x}^{{6m-6n}}})}} = x (m+3n) + (4m-9n) – (6m-6n) = x(-m) .

So, option (b) is correct.

17. Simplify 1/(1+za–b + z a–c) + 1/(1+zb–c + z b–a )  + (1/1+zc–a + z c–b)   

 (a)     \displaystyle \frac{1}{3}

(b)     \displaystyle \frac{1}{2}

(c)      1

(d)     0

  1/(1+za–b + z a–c) + 1/(1+zb–c + z b–a ) + (1/1+zc–a + z c–b)

= [z– a / {z– a(1+za–b + z a–c)}] + [z-b / {z-b (1+zb–c + z b–a )}] + [z-c / z-c {(1 + z c – a + z c – b )}]

= [z– a / (z– a + z– b + z– c )] +  [z– b / (z– b + z– c + z– a )] +  [z– c /( z– c + z– a + z– b)]

 = (z– a + z– b + z– c) / (z– a + z– b + z– c) =1. So, option (c) is correct

18. If 4x = 5y = 20z then z =

(a) 20xy

(b) \displaystyle \frac{{x+y}}{{xy}}

(c) \displaystyle \frac{4}{{xy}}

(d) \displaystyle \frac{{xy}}{{x+y}}

Let 4x = 5y = 20z = k. Hence, 4 = \displaystyle \frac{k}{x}, 5 = \displaystyle \frac{k}{y}, 20 =  \displaystyle \frac{k}{z}

Or, 4 x 5 =  k/z, Or, \displaystyle \frac{k}{x} .  \displaystyle \frac{k}{y} =  \displaystyle \frac{k}{z}, Or,  \displaystyle \frac{k}{x} +\displaystyle \frac{k}{y} =  \displaystyle \frac{k}{z}. Or, k(\displaystyle \frac{1}{x}+ \displaystyle \frac{1}{y}) = \displaystyle \frac{k}{z}

Or, (\displaystyle \frac{1}{x} + \displaystyle \frac{1}{y}) = \displaystyle \frac{1}{z} or, (x+y)/xy = \displaystyle \frac{1}{z}, or z= \displaystyle \frac{{xy}}{{x+y}}.

So, option (d) is correct

19. Find value of \displaystyle {{\left( {\frac{{{{3}^{{\frac{1}{2}}}}}}{9}} \right)}^{{\frac{5}{2}}}} .  \displaystyle \frac{9}{{{{{\left( {3\times {{3}^{{\frac{1}{2}}}}} \right)}}^{{\frac{7}{2}}}}}} .  9   

(a)    1

(b)    4

(c)    9

(d)    27

\displaystyle {{\left( {\frac{{{{3}^{{\frac{1}{2}}}}}}{9}} \right)}^{{\frac{5}{2}}}} .  \displaystyle \frac{9}{{{{{\left( {3\times {{3}^{{\frac{1}{2}}}}} \right)}}^{{\frac{7}{2}}}}}} .  9  = {3 ½ . 3 – 2} 5/2 . {32. 3 – 3/2} 7/2 . {32}

= (3– 2) 5/2. (3 ½ )7/2. 32 = 3 – 15/4. 3 7/4. 32 = 3 – 15/4 + 7/4 +2 = 3 0/4 = 30 =1. So, option (a) is correct

20. If 2x – 2 x–1 = 4, then the value of xx is

(a)    8

(b)    0

(c)    125

(d)    27

Let 2x = a. So, 2x – 2x -1  = 4, a-(\displaystyle \frac{a}{2})=4, or \displaystyle \frac{a}{2}=4. Or a=8. Now 2x =8 = 23. So, x=3. So, xx=3 3 = 27. So, option (d) is correct

21. If x = y a, y = z b and z = x c then abc =

(a)    2

(b)    1

(c)    0

(d)    5

x = y a = (z b)a = {(x c)b}a = xabc. So, x= xabc. So, abc=1. So, option (b) is correct

22. If ax = b, by = c and cz = a and a\displaystyle \ne 0, then which one is correct:
(a) xyz = 0
(b) xyz = 1
(c) xy = 1 + z
(d) xz = 2y

ax  = b, by  = c, cz  = a,  a\displaystyle \ne 0. Now, ax = b,  (ax) y = by = c  [ as by = c].  

Again, cz  = a, So, c xyz  = c1 or, xyz = 1  

So, option (b) is correct

23. If \displaystyle {{\left[ {\frac{2}{3}} \right]}^{x}}{{\left[ {\frac{3}{2}} \right]}^{{2x}}}=\left[ {\frac{{27}}{8}} \right], then x is equal to:
(a) 1
(b) 0
(c) 3
(d) 5

\displaystyle {{\left[ {\frac{2}{3}} \right]}^{x}}{{\left[ {\frac{3}{2}} \right]}^{{2x}}}=\left[ {\frac{{27}}{8}} \right]

Or, \displaystyle \frac{{{{2}^{x}}}}{{{{3}^{x}}}}.\frac{{{{3}^{{2x}}}}}{{{{2}^{{2x}}}}}=\frac{{{{3}^{3}}}}{{{{2}^{3}}}}

Or, \displaystyle \frac{{{{2}^{{x+3}}}}}{{{{2}^{{2x}}}}}=\frac{{{{3}^{{x+3}}}}}{{{{3}^{{2x}}}}}

Or, \displaystyle {{2}^{{x+3-2x}}}={{3}^{{x+3-2x}}}

Or, \displaystyle {{2}^{{3-x}}}={{3}^{{3-x}}}

Or, \displaystyle \frac{{{{2}^{{3-x}}}}}{{{{3}^{{3-x}}}}}=1

Or, \displaystyle {{\left( {\frac{2}{3}} \right)}^{{3-x}}}=1={{\left( {\frac{2}{3}} \right)}^{0}}

Or, 3 – x = 0 or, x = 3

So, option (c) is correct

24. The value of \displaystyle {{\frac{{243}}{{\left[ {32} \right]}}}^{{-4/5}}} is:

(a) \displaystyle \frac{2}{9}

(b) \displaystyle \frac{{16}}{{81}}

(c) \displaystyle \frac{1}{2}

(d) \displaystyle \frac{4}{5}

Image 1

So, option (b) is correct

25. If \displaystyle {{\left( {{{4}^{{\sqrt{x}}}}} \right)}^{{\sqrt{x}}}} = 256 then x is:
(a) 3
(b) 2
(c) 6
(d) 1
\displaystyle {{\left( {{{4}^{{\sqrt{x}}}}} \right)}^{{\sqrt{x}}}}= 256=(4)2
Or, 4x = 42
Or, x = 2
So, option (b) is correct

26. Value of \displaystyle 4\sqrt{{{{{\left( {16} \right)}}^{{-3}}}}} is:

(a) \displaystyle \frac{1}{5}

(b) \displaystyle \frac{1}{8}

(c) 8

(d) \displaystyle \frac{1}{6}

\displaystyle 4\sqrt{{{{{\left( {16} \right)}}^{{-3}}}}}

= {(16)-3}1/4
= [{(2)4}-3] 1/4
= (2-12) 1/4
= 2-3
=\displaystyle \frac{1}{{{{2}^{3}}}}=\frac{1}{8}
So, option (b) is correct

27. If \displaystyle \sqrt{{5+3\sqrt{x}}}=3 then  x is:

(a) \displaystyle \frac{{16}}{9}

(b) \displaystyle \frac{9}{{16}}

(c) \displaystyle \frac{1}{4}

(d) 3
\displaystyle \sqrt{{5+3\sqrt{x}}}=3

Or, \displaystyle {{\left( {\sqrt{{5+3\sqrt{x}}}} \right)}^{2}} = (3)2

Or, \displaystyle 5+3\sqrt{x}=9

Or, \displaystyle 3\sqrt{x}=9-5=4

Or, \displaystyle \sqrt{x}=\frac{4}{3}\text{ or, x=}{{\left[ {\frac{4}{3}} \right]}^{2}}=\frac{{16}}{9}
So, option (a) is correct

28 If  \displaystyle {{x}^{{\sqrt{x}}}}={{\left( {\sqrt{x}} \right)}^{x}} then value of x=
(a) 1
(b) 4
(c) 2
(d) 3

\displaystyle {{x}^{{\sqrt{x}}}}={{\left( {\sqrt{x}} \right)}^{x}}

Or, \displaystyle {{x}^{{{{x}^{{\frac{1}{2}}}}}}}={{\left( {{{x}^{{\frac{1}{2}}}}} \right)}^{x}}

Or, \displaystyle {{x}^{{{{x}^{{\frac{1}{2}}}}}}}={{x}^{{\frac{x}{2}}}}

Or, \displaystyle {{x}^{{\frac{1}{2}}}}=\frac{x}{2}

Or, \displaystyle {{\left( {{{x}^{{\frac{1}{2}}}}} \right)}^{2}}={{\left( {\frac{x}{2}} \right)}^{2}}

Or, \displaystyle x=\frac{{{{x}^{2}}}}{4}\text{ Or, }{{x}^{2}}=4x\text{ Or, }{{x}^{2}}-4x=0

Or, Or, x(x – 4) = 0 Or, x = 0 Or, x = 4
So, option (b) is correct

29. The value of   \displaystyle \sqrt[3]{{{{x}^{{15}}}}}\text{ }\sqrt[4]{{{{x}^{{12}}}}}  is
(a) x15
(b) x5
(c) x10
(d) x8
\displaystyle \sqrt[3]{{{{x}^{{15}}}}}\text{ }\sqrt[4]{{{{x}^{{12}}}}}

\displaystyle ={{\left( {{{x}^{{15}}}} \right)}^{{\frac{1}{3}}}}\times {{\left( {{{x}^{{12}}}} \right)}^{{\frac{1}{4}}}}
= x5 \displaystyle \times x3 = x5+3 = x8
So, option (d) is correct

30. If a3 – b3 = (a – b) (a2 – ab – b2), then the simplified from of \displaystyle {{\left[ {\frac{{{{x}^{l}}}}{{{{x}^{m}}}}} \right]}^{{{{l}^{2}}+lm+{{m}^{2}}}}}\times {{\left[ {\frac{{{{x}^{m}}}}{{{{x}^{n}}}}} \right]}^{{{{m}^{2}}+mn+{{n}^{2}}}}}\times {{\left[ {\frac{{{{x}^{n}}}}{{{{x}^{l}}}}} \right]}^{{{{n}^{2}}+\ln +{{l}^{2}}}}}
(a) 0
(b) 1
(c) x3
(d) 2

\displaystyle {{\left[ {\frac{{{{x}^{l}}}}{{{{x}^{m}}}}} \right]}^{{{{l}^{2}}+lm+{{m}^{2}}}}}\times {{\left[ {\frac{{{{x}^{m}}}}{{{{x}^{n}}}}} \right]}^{{{{m}^{2}}+mn+{{n}^{2}}}}}\times {{\left[ {\frac{{{{x}^{n}}}}{{{{x}^{l}}}}} \right]}^{{{{n}^{2}}+\ln +{{l}^{2}}}}}
= {xl – m} l²+ l m + m² x {xm – n} m²+ mn + n²x {xn – l}n² + l n + l²
= x l3 – m3. xm3 – n3 . xn3 – l3
= x l3 – m3 + m3 – n3 + n3 – l3
= x 0 = 1
So, option (b) is correct

31. (17)3.5 x (17)x = 178, value of x is :

  1. 2.29
  2. 2.75
  3. 4.25
  4. 4.5

Let (17)3.5 x (17)x = 178.
Then, (17)3.5 + x = 178.
So, 3.5 + x = 8, or x = (8 – 3.5), or  x = 4.5
So, option (d) is correct

    32. Given that 100.48 = x, 100.70 = y and xz = y2, then the value of z is close to:

    1. 1.45
    2. 1.88
    3. 2.9
    4. 3.7

    xz = y2 , so,   10(0.48z) = 10(2 x 0.70) = 101.40

    or, 0.48z = 1.40, or z=\displaystyle \frac{{140}}{{48}}=\frac{{35}}{{12}}=2.9 appx

    So, option (c) is correct

    33. If 3(x – y) = 27 and 3(x + y) = 243, then x is equal to:

    1. 0
    2. 2
    3. 4
    4. 6

    3x – y = 27 = 33 , or,  x – y = 3 ….(i)

    3x + y = 243 = 35 , or  x + y = 5 ….(ii)

    Adding both sides of the 2 equations, we get 2x=8, or x=4,

    So, option (c) is correct

    34. What is value of (0.000001)⅓ ?
    a. 0.000001
    b. 0.001
    c. 0.01
    d. 0.1
    Cube root of 1 is 1. Now we have to deduce the number of decimal places. There are 6 decimal places in the given number 0.000001. So, in cube root of 1, there would be 6/2=2 decimal places. So, 11/3 will have 2 decimal places, i.e .01.
    So, option (c) is correct

    35. What will be value of 46.7856 if, (684)2= 467856?
    a. 6.84
    b. 0.0684
    c. 0.684
    d. 0.000684
    If a number has 4 decimals, its square root will have \displaystyle \frac{4}{2}=2 decimals
    As (684)2= 467856, so, √(467856) = 684, so √(46.7856) will be 6.84.
    So, option (a) is correct

    36.  If value of √28 is approximately 5.2915, then value of \displaystyle \frac{7}{4} is approximately
    a. 1.2000
    b. 0.5687
    c. 1.3228
    d. 1.4652
    28= 5.2915. As it has 4 decimal places, √28 = .052915
    √(\displaystyle \frac{7}{4}) = √{(7 X 4)/(4 X4)} = \displaystyle \frac{{\sqrt{{28}}}}{{\sqrt{{16}}}}= \displaystyle \frac{{5.2915}}{4}=1.3228
    So, option (c) is correct

    37. Which of the following is the biggest ?

    1. 41/3,
    2. 61/4,
    3. 151/6
    4. 2451/12

    To compare, we must bring all the numbers to the same power. The LCM of 3, 4.6 and 12 is 12. So, convert all the numbers to the power \displaystyle \frac{1}{{12}}

    41/3 = 4 (4/12) =(44)x1/12 = (256)1/12
    61/4=63/12 = (63)1/12 = (216)1/12
    151/6=152/12 = (152) 1 /12 = (225)1/12
    So, option (a) is correct

    38. Find the value of  \displaystyle {{\left( {\frac{{1024}}{{742}}} \right)}^{{-4/5}}}

    (a) \displaystyle \frac{{81}}{{16}}

    (b) \displaystyle \frac{{81}}{{256}}

    (c) \displaystyle \frac{4}{9}

    (d) None of the above

    Let us simplify the numerator and denominator for easy understanding
    1024=210, So, 1024 -4/5  = 2(10 X -4/5) = 2-8 
    243=25, 243 -4/5= 35X-4/5 = 3-4
    So, \displaystyle {{\left( {\frac{{1024}}{{742}}} \right)}^{{-4/5}}} = \displaystyle \frac{{{{2}^{{-8}}}}}{{{{3}^{{-4}}}}} = \displaystyle \frac{{{{3}^{4}}}}{{{{2}^{8}}}}  = \displaystyle \frac{{81}}{{256}}
    So, option (b) is correct

    39. If m and n are whole numbers such that mn= 121, the value of (m − 1)n + 1 is:

    1. 1
    2. 11
    3. 121
    4. 1000

    We know that 112 = 121.
    Putting m = 11 and n = 2, we get:
    (m − 1)n + 1 = (11 − 1)(2 + 1) = 103 = 1000.
    So, option (d) is correct