# Indices MCQ

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The correct option is indicated by (√) mark with explanation in the rightmost column

Indices MCQ

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

a. (x+y)/x

b. xy/(x+y)

c. y/(x+y)

d. (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 1/z= 1/x + 1/y, or 1/z = (x+y)/xy.

Or z=xy/(x+y). So, option (b) is correct.

2. (xb+c)bc (xc+a)ca (xa+b)ab is equal to:

(a)      0

(b)      1

(c)      ab

(d)      ac

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

3. The value of (xa/xb)a+b X (xb/xc)b+c X (xc/xa)c+a is

(a)    1

(b)    0

(c)    x

(d)    x a + b

(xa/xb)a+b X (xb/xc)b+c X (xc/xa)c+a =   (xa b)a+b  x  (xbc)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)      ½

(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 (x1/4 1) (x1/4  + 1) will be:

(a)    6

(b)    8

(c)    2

(d)    1.

(x1/4 – 1) (x1/4  + 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 (Xm+3n . x4m-9n) / x6m-6n

(a)    xm

(b)    x m

(c)    xn

(d)    1.

(xm+3n . x4m-9n) / x6m-6n = (xm+ 3 n + 4 m 9n) / x6m-6n  = x5m 6n / x6m 6n = x 5m 6n – (6m 6n)

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

9. Value of  {xa(b c)  . xc(a b)} / xb(a c) is

(a)    a

(b)    1

(c)    xa+b+c

(d)    0.

{xa(b c)  . xc(a b)}  / xb(a c) = 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 .  81 . x3 y 2 is

(a)      2xy

(b)     xy/2

(c)      2

(d)     1.

16x 3 y2 .  81 . x3 y 2 = 16 . 81 . x. x. 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 X  ((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 = 1/2x, 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 +  bc. 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]} = x2abc. So, option (d) is correct.

15. Simplify (2x +1 + 2x) / (2x +3 2x+1)

(a)     2

(b)     ½

(c)      1

(d)     0.

(2x +1 + 2x) / (2x +3 – 2x+1) = (2x .  21 + 2x) / (2x . 23 – 2x . 21) = 2x(2+1)/ 2x (23-2)

= 3 X 2x / 6 X 2x=3/6=1/2. So, option (b) is correct.

16. Simplify (x m+3n. x 4m – 9n) / (x 6m – 6n)

(a)    X m

(b)    X m

(c)    X n

(d)    X n

(x m+3n. x 4m – 9n) / (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)     1/3

(b)     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 =

Let 4x = 5y = 20z = k. Hence, 4 = k /x, 5 = k/y, 20 =  k/z

Or, 4 x 5 =  k/z, Or, k/x .  k/y =  k/z, Or,  k/x +k/y =  k/z. Or, k(1/x + 1/y) = k/z

Or, (1/x + 1/y) = 1/z or, (x+y)/xy = 1/z, or z= xy/(x+y). So, option (d) is correct

19. Find value of (3½ / 9)5/2 .  (9/ (3X 3 ½ ) 7/2 .  9

(a)    1

(b)    4

(c)    9

(d)    27

(3½ / 9)5/2 .  (9/ (3X 3 ½ ) 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-(a/2)=4, or 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