What will be the ones digit in the squares of the following numbers?

(a)

36

(b)

252

(c)

543

\(\\ \sqrt { 48 } \) is approximately equal to

(a)

5

(b)

6

(c)

7

(d)

8

\(\sqrt { 128 } -\sqrt { 98 } +\sqrt { 18 } =\) 

(a)

\(\sqrt { 2 } \)

(b)

\(\sqrt { 8 } \)

(c)

\(\sqrt { 48 } \)

(d)

\(\sqrt { 32 } \)

\(\sqrt { 22+\sqrt { 7 } +\sqrt { 4 } } \) =

(a)

\(\sqrt { 25 } \)

(b)

\(\sqrt { 33 } \)

(c)

\(\sqrt { 31 } \)

(d)

\(\sqrt { 29 } \)

The number of digits in the square root of 123454321 is

(a)

4

(b)

5

(c)

6

(d)

7

By what number should (-4)^-1 be multiplied so that the product becomes 10^-1?

(a)

\(\cfrac { 2 }{ 3 } \)

(b)

\(\cfrac { -2 }{ 5 } \)

(c)

\(\cfrac { 5 }{ 2 } \)

(d)

\(\cfrac { -5 }{ 2 } \)

0.0000000002020 in scientific form is

(a)

2.02 \(\times\) 10⁹

(b)

2.02 \(\times\) 10⁻⁹

(c)

2.02 \(\times\) 10⁻⁸

(d)

2.02 \(\times\) 10⁻¹⁰

(-2)^-3 x (- 2)^-2 is

(a)

\(\cfrac { -1 }{ 32 } \)

(b)

\(\cfrac { 1 }{ 32 } \)

(c)

32

(d)

-32

Which is not correct?

(a)

\(\left( \cfrac { -1 }{ 4 } \right) ={ 4 }^{ -2 }\)

(b)

\(\left( \cfrac { -1 }{ 4 } \right) =\left( \cfrac { 1 }{ 2 } \right) ^{ 4 }\)

(c)

\(\left( \cfrac { -1 }{ 4 } \right) ^{ 2 }={ 16 }^{ -1 }\)

(d)

\(-\left( \cfrac { 1 }{ 4 } \right) ^{ 2 }={ 16 }^{ -1 }\)

If \(\left( \cfrac { p }{ q } \right) ^{ 1-3x }=\left( \cfrac { q }{ p } \right) ^{ \frac { 1 }{ 2 } }\), then x is __________

(a)

4⁻¹

(b)

3⁻¹

(c)

2⁻¹

(d)

1^-1