IBPS has introduced Computer Aptitude this year (clubbed with Reasoning Section) in the Mains Examination of Probationary Officer Recruitment. Aspirants preparing for SBI must be familiar with Computer Aptitude Section. You must prepare basics of Computer Knowledge for this section to score well. The only difference between Computer Aptitude and Computer Awareness/Knowledge is the practical approach of questions asked in Computer Aptitude. Practice with these 15 questions of Computer Aptitude for IBPS PO Mains 2017 Exam.

Q1. Which of the following is the correct binary form of 4A2.8D16 ?
(a) 010010100010.100011012
(b) 010110100010.111011012
(c) 011110100010.100011012
(d) 010010111110.100011012
(e) None of these

S1. Ans.(a)
Sol. The required conversion is as follows:

Therefore, 4A2.8D16 = 010010100010.100011012

Q2.  Which of the following symbol represents decisional or conditional statement in a flow chart?

S2. Ans.(c)
Sol. Diamond (rhombus) shaped figure represents decisional or conditional statement in a flow chart.

Q3. Which of the following is not a valid binary number?
(a) 00000
(b) 11111
(c) FFFFF
(d) 101010
(e) 010101

S3. Ans.(c)
Sol. FFFFF is not a valid binary number, there are only 2 digits in binary number system- 0 and 1.

Q4. If ∆ represents ‘1’ and ○ represents ‘0’. What will be the one’s complement of - ○∆∆○○∆?
(a) 011001
(b) 100110
(c) 101010
(d) 000000
(e) 111111

S4. Ans.(b)
Sol. ○∆∆○○∆ represents 011001. For finding one’s complement we change each 1 to 0 and each 0 to 1. Hence One’s complement of 011001 will be 100110.

Q5.  Following diagram depicts which logic gate?
(a) NOR gate
(b) NOT gate
(c) OR gate
(d) NAND gate
(e) None of the above

S5. Ans.(b)
Sol.

Q6. Which of the following is a correct set of hexadecimal digits?
(a) 102n
(b) a19fk
(c) ZOLO12
(d) FACE
(e) 0101xx

S6. Ans.(d)
Sol. hexadecimal digits are written using symbols from 0-9 or A-F

Q7. Which of the following is a Logical Conjunction?
(a) AND gate
(b) OR gate
(c) NOT gate
(d) both(a) and (b)
(e) All of the above

S7. Ans.(a)
Sol. When two statements are combined with an 'and,' then it is a conjunction. For conjunctions, both statements must be true for the compound (result) statement to be true. When your two statements are combined with an 'or,' it is a disjunction.

Q8. If Ó¢ represents ‘1’ and á¹ˆ represents ‘0’. A and B are two input ends of OR logic gate and C is the only output. What will be the output if the input is A= Ó¢ and B= Ó¢?
(a) á¹ˆ Ó¢
(b) á¹ˆ
(c) Ó¢
(d) Ó¢ á¹ˆ
(e) Ó¢ á¹ˆ Ó¢

S8. Ans.(c)
Sol.

á¹ˆ= 0 and  Ó¢=1
Given that A=1 and B=1 Following the truth table for OR gate

Directions (9-11) : Triangle represents ∆ (1) and circle represents ○ (0). If triangle appears in unit's place then its value is 1. If it appears in 10's place its value is doubled to 2 like that it continues. Using the given terminology answer the following questions. For example:

∆ = 1
∆○∆ = 4, 0, 1 = 4+0+1
∆○ = 2

Q9. How will you represent ‘87’ in this code language?
(a) ○∆∆∆○∆∆
(b) ∆○∆○∆∆∆
(c) ∆∆○∆∆∆∆
(d) ∆○○∆○○∆
(e) ∆∆○∆∆∆○

S9. Ans.(b)
Sol.

1010111 = ∆○∆○∆∆∆

Q10. What will be the code for ∆∆○○○∆○?
(a) 98
(b) 97
(c) 90
(d) 94
(e) 99

S10. Ans.(a)
Sol.

1100010 = 64+32+2=98

Q11. What will be the code for one’s complement of ∆∆○∆?
(a) 2
(b) 3
(c) 4
(d) 5
(e) 6

S11. Ans.(a)
Sol. 1’s complement of 1101 = 0010 = 2

Q12. Which of the following number systems is considered as a base-10 number system?
(a) Decimal number system
(b) Binary number system
(c) Octal number system
(e) None of these

S12. Ans.(a)
Sol. Decimal number system has unique 10 digits, that is, from 0 to 9, to represent a number and its base (or radix) is 10. The decimal number system is also called a base-10 system as it has 10 digits.

Q13. The hexadecimal number system represents symbols __________ and __________.
(a) 0-9, A-E
(b) 0-9, A-F
(c) 1-10, A-E
(d) 1-10, A-F
(e) None of these

S13. Ans.(b)
Sol. The hexadecimal number system has 16 possible digits represented by the symbols 0 to 9 and the alphabets A, B, C, D, E and F.

Q14. The conversion of the binary number 101001110112 into its decimal form is __________.
(a) 133910
(b) 134910
(c) 139310
(d) 193910
(e) None of these

S14. Ans.(a)
Sol. The required conversion is as follows:
101001110112 = 1 × 210 + 0 × 29 + 1 × 28 + 0 × 27 + 0 × 26 + 1 × 25 + 1 × 24 + 1 × 23 + 0 × 22 +1 × 21 + 1 × 20
= 1024 + 0 + 256 + 0 + 0 + 32 + 16  +8 + 0 + 2 + 1
= 133910

Q15. __________ are the gates that can be used to implement any type of Boolean logic.
(a) OR gates
(b) Universal gates
(c) NAND gates
(d) Exclusive-NOR gates
(e) AND gates

S15. Ans.(b)
Sol. Universal gates are the gates that can be used to implement any type of Boolean logic or circuit. There are two universal gates, namely, NAND gate and NOR gate, which can be used to implement any of the basic logic gates.