| 61. | The octal equivalent of hexadecimal (A.B)16 is: |
| a. | 47.21 |
| b. | 12.74 |
| c. | 12.71 |
| d. | 17.21 |
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Answer: (b).12.74
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| 62. | The dual of the switching function x+yz is: |
| a. | x+yz |
| b. | x’+y’z’ |
| c. | x(y+z) |
| d. | x’(y’+z’) |
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Answer: (d).x’(y’+z’)
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| 63. | The octal equivalent of the hexadecimal number FF is: |
| a. | 100 |
| b. | 150 |
| c. | 377 |
| d. | 737 |
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Answer: (c).377
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| 64. | The idempotent law in Boolean algebra says that: |
| a. | ~(~x)=x |
| b. | x+x=x |
| c. | x+xy=x |
| d. | x(x+y)=x |
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Answer: (b).x+x=x
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| 65. | Simplified form of Boolean expression xy+(~x)z+yz is: |
| a. | xy+(~x)z |
| b. | (~x)y+(~x)z |
| c. | (~x)y+xz |
| d. | xy+xz |
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Answer: (a).xy+(~x)z
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| 66. | In order to build a MOD-18 counter, the minimum number of flip flops needed is equal to: |
| a. | 18 |
| b. | 9 |
| c. | 5 |
| d. | 4 |
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Answer: (c).5
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| 67. | 2’s complement of -100 is: |
| a. | 00011100 |
| b. | 10011101 |
| c. | 10011100 |
| d. | 11100100 |
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Answer: (c).10011100
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| 68. | Which of the following expression remove hazard form: xy+zx’ ? |
| a. | xy+x’ |
| b. | xy+zx’ |
| c. | xy+zx’+yz |
| d. | xy+zx’+wz |
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Answer: (c).xy+zx’+yz
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| 69. | How many 1’s are present in the binary representation of 15x256+5x16+3: |
| a. | 8 |
| b. | 9 |
| c. | 10 |
| d. | 11 |
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Answer: (a).8
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| 70. | If A⊕B=C, then: |
| a. | A⊕C=B |
| b. | B⊕C=A |
| c. | A⊕B⊕C=0 |
| d. | All of the above |
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Answer: (d).All of the above
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