HackerRank Diagrammatic Reasoning Quiz-1

Circle - Word gets inverted with first alphabet last and last alphabet first
Square - fist and last letters get omitted
Triangle - letters at odd places omitted
Cross - All alphabets replaced by next alphabet
So answer is (b)

Solution:

Looking at the diagram in rows, the central circle equals half the sum of the numbers in the other circles to the left and right of the center.

Solution:
The sequence progresses circle, triangle, a diamond with alternate horizontal/vertical lines.

Solution:

Circle - Word gets inverted with first alphabet last and last alphabet first
Square - fist and last letters get omitted
Triangle - letters at odd places omitted
Cross - All alphabets replaced by next alphabet
So answer is (a)

Solution:

915 - 364 = 551
and 789 - 543 = 246
Similarly, 863 - 241 = 622.

Circle - Word gets inverted
MISSISSIPPI ⇒ IPPISSISSIM
Square - fist and last letters get omitted
IPPISSISSIM ⇒ PPISSISSI
Triangle - letters at odd places omitted
PPISSISSI ==> PSIS
Cross - All alphabets replaced by next alphabet
PSIS ⇒ QTJT
Alphabets remaining are Q, T and J. (3 alphabets)
So answer is (c)

Circle - Word gets inverted with first alphabet last and last alphabet first
Square - fist and last letters get omitted
Triangle - letters at odd places omitted
Cross - All alphabets replaced by next alphabet
So answer is (b)

Solution:

Set A: Each box contains one striped square, one shaded square, and two unshaded squares.

Set B: Each box contains two striped squares, one shaded square, and one unshaded square.

Figure: The figure contains two shaded squares and two striped squares.

Circle - Word gets inverted with first alphabet last and last alphabet first
Square - fist and last letters get omitted
Triangle - letters at odd places omitted
Cross - All alphabets replaced by next alphabet
So answer is (a)

Solution:
The given logic circuit is the combination of OR and NOT gates. So it should be the NOR gate. Besides, the NAND gate is the combination of AND and NOT gates.
The output of NOR gate will be O = \overline{X + Y} and output of the NAND gate will be O = \overline{XY} = \overline{X} + \overline{Y} (by de Morgan's law)

Solution:
Deriving the given circuit , we will get ,
(\overline{A}.\overline{B})+AB
= \overline{AB}+AB , which is a XNOR operation.

Solution:
Output of upper AND gate = \overline{A}B
Output of lower AND gate = A \overline{B}
Thus, output of OR gate = \overline{A}B + A \overline{B}
This is the Boolean expression for the XOR gate.

Solution:
We know that , for XOR gate operations ,
0 \oplus 0 = 0
0 \oplus 1 = 1
1 \oplus 0 = 1
1 \oplus 1 = 0
Therefore, from the circuit ,
10111100 will give the output Y = 1 .
The rest combinations will give the output Y = 0 .

Solution:

a- represent OR gate
b- represent the AND gate
c- represent the combination of OR and NOT gate, NOR gate.
d- represent, the combination of AND and NOT gate, NAND gate