IMPORTANT COMPUTER APTITUDE PRACTICE QUESTIONS FOR SBI PO & CLERK MAINS

 

Hope all are ready for SBI Clerk & Po exams.Here is some of the important computer aptitude questions, you can practice very well for your mains examination and can score more marks in the upcoming exams.

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1) Convert the given Decimal number into Binary form -173

a) 00101111₂

b) 00101001₂

c) 10101111₂

d) 10101001₂

e) 10101101₂

2) Convert the given Decimal into octal form -130

a) 301₈

b) 303₈

c) 202₈

d) 104₈

e) 203₈

3) Convert the given base 10 into base 16 – 310

a) D2₁₆

b) C2₁₆

c) C3₁₆

d) A3₁₆

e) D3₁₆

4) Convert the given Decimal number into Binary number – 195

a) 00111110₂

b) 11111000₂

c) 11000111₂

d) 00111000₂

e) 00101000₂

5) Convert the binary number into octal number – 100100110

a) 544₈

b) 446₈

c) 564₈

d) 454₈

e) 554₈

6) Convert the given binary number into octal number – 111100110

a) 656₈

b) 786₈

c) 556₈

d) 746₈

e) 756₈

7) Convert the given base 10 to base 16 – 200

a) B8₁₆

b) D2₁₆

c) A3₁₆

d) C6₁₆

e) C8₁₆

8) Convert the given Binary number into Decimal number – 100011₂

a) 30

b) 31

c) 32

d) 34

e) 35

9) Convert the given binary number into Decimal number – 001110₂

a) 17

b) 12

c) 14

10) Convert the given Decimal into octal – 135

a) 216₈

b) 209₈

c) 205₈

d) 207₈

e) 206₈

11) Convert the given Binary number into octal number 111001001

a) 722₈

b) 711₈

c) 655₈

d) 611₈

e) 766₈

12) What will be the 1’s complement of decimal number 44

a) 111000

b) 100110

c) 010011

d) 110010

e) 001100

13) What will be the 1’s complement of decimal number 22 ?

a) 01000

b) 01101

c) 01010

d) 01011

e) 01001

14) What will be the 2’s complement of 100100₂

a) 11001

b) 01100

c) 10101

d) 11100

e) 11111

15) What will be the 2’s complement of 111011₂ ?

a) 000101

b) 110001

c) 110000

d) 100011

e) 110011

16) What will be the 2’s complement of 100011₂ ?

a) 111101

b) 010101

c) 011100

d) 111101

e) 011101

17) What will be the 1’s complement of decimal number 45 ?

a) 111000

b) 010010

c) 011110

d) 111000

e) 011101

18) Convert the given hexadecimal to decimal number (base 10) – BC2

a) 3040₁₀

b) 3010₁₀

c) 2018₁₀

d) 2010₁₀

e) 4010₁₀

19) Convert the given hexadecimal to decimal number (base 10) – AC2

a) 3754₁₀

b) 2753₁₀

c) 2654₁₀

d) 2744₁₀

e) 2754₁₀

20) Convert the given hexadecimal to decimal number (base 10) – CC2

a) 3241₁₀

b) 3311₁₀

c) 3266₁₀

d) 3261₁₀

e) 3211₁₀

 

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Answer Key with Explanation

1) e

173/2 – Remainder 1

86/2 – Remainder 0

43/2 – Remainder 1

21/2 – Remainder 1

10/2 – Remainder 0

5/2 – Remainder 1

2/2 – Remainder 0

1

173=> 10101101₂

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2) c

130/8 – Remainder 2

16/8 – Remainder 0

2

130 => 202₈

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3) a

310 / 16 -remainder 2

13 – D

(A = 10 , B = 11 , C = 12 , D = 13)

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4) c

11000111₂

195 /2 – Remainder 1

97/2 – Remainder 1

48/2 – Remainder 1

24/2 – Remainder 0

12/2 – Remainder 0

6/2 – Remainder 0

3/2 – Remainder 1

1

195 -> 11000111₂

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5) b

446₈

100100110 = > 100/100/110

4 / 4 / 6

100100110 => 446₈

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6) d

746₈

111100110 => 111/100/110

7 / 4 / 6

111100110 => 746₈

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7) e

C8₁₆

200 / 16 – Remainder 8

12 – C => C8₁₆

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8) d

34

100011₂ = 2⁵ + 0 + 0 + 0 + 1 + 1

32 + 0 + 0 + 0 + 2 = > 34

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9) c

001110₂

0 + 0 + 2³ + 2² + 2¹ + 0

0 + 0 + 8 + 4 + 2 + 0 => 14

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10) d

135/8 – Remainder 7

16/8 – Remainder 0

2

135 => 207₈

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11) b

111001001₂

111/001/001

7/ 1/ 1

111001001₂ => 711

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12) c

44/2 – Remainder – 0

22/2 – Remainder 0

11/2 – Remainder 0

5/2 – Remainder 1

2/2 – Remainder 0

1

44 => 010011

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13) e

22/2 – Remainder 0

11/2 – Remainder 1

5/2 – Remainder 1

2/2 – Remainder 0

1

22 => 01001

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14) d

100100₂ = 011011

(+1)

= 011100

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15) a

111011₂ = 000100

(+1)

= 000101

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16) c

100011₂ = 011100

(+1)

= 011101

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17) b

45/2 – Remainder 1

22/2 – Remainder 0

11/2 – Remainder 1

5/2 – Remainder 1

2/2 – Remainder 0

1

45 => 101101

=> 010010

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18) b

BC2 => 11 12 2

11 x 16² + 12 x 16¹ + 2 x 16⁰

2816 + 192 + 2

3010₁₀

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19) e

AC2 => 10 12 2

10 x 16² + 12 x 16¹ + 2 x 16⁰

2560 + 192 + 2

2754₁₀

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20) c

CC2 = 12 12 2

12x 16² + 12 x 16¹ + 2 x 16⁰

3072 + 192 + 2

3266₁₀

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