Gate 2000_
Duration: 3 min
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The video presents a solution to a Boolean algebra problem from the GATE-2000 exam. The problem asks for the values of four Boolean variables, x, y, z, and w, that satisfy a system of four simultaneous equations. The equations are: x + y + z = 1, xy = 0, xz + w = 1, and xy + z'w' = 0. The instructor explains that since this is a multiple-choice question, the most efficient method is to substitute the given options into the equations to verify which set of values satisfies all conditions. He systematically evaluates each option against the equations one by one.
Chapters
0:00 – 2:00 00:00-02:00
The instructor introduces the problem statement displayed on the screen, listing four Boolean equations involving variables x, y, z, and w. He identifies the four options: (A) 0 1 0 0, (B) 1 1 0 1, (C) 1 0 1 1, and (D) 1 0 0 0. He begins verification with the first equation, x + y + z = 1. Substituting values from each option, he finds that all four options satisfy this equation. For instance, option (A) gives 0 + 1 + 0 = 1, and option (B) gives 1 + 1 + 0 = 1. Since all options pass the first test, he proceeds to the second equation, xy = 0. He tests option (B) where x=1 and y=1, resulting in 1*1=1. This violates the equation xy=0. Thus, option (B) is eliminated. Options (A), (C), and (D) remain under consideration.
2:00 – 2:31 02:00-02:31
The instructor moves to the third equation, xz + w = 1, to narrow down the remaining options. He tests option (A) where x=0, z=0, and w=0. Substituting these values gives 0*0 + 0 = 0, which does not equal 1. Therefore, option (A) is eliminated. Next, he tests option (D) where x=1, z=0, and w=0. Substituting these values gives 1*0 + 0 = 0, which also fails. Thus, option (D) is eliminated. This leaves only option (C) as the potential correct answer. He verifies option (C) with x=1, z=1, and w=1. Substituting these gives 1*1 + 1 = 1, which satisfies the equation. Finally, he briefly checks the fourth equation, xy + z'w' = 0, for option (C). With x=1, y=0, z=1, and w=1, the expression becomes 0 + 0*0 = 0, which is correct. He concludes that option (C) is the correct solution and marks it on the screen.
The lecture effectively demonstrates a practical strategy for solving Boolean algebra problems in competitive exams. By utilizing the substitution method, the instructor efficiently eliminates incorrect options without needing to solve the system of equations algebraically from scratch. This step-by-step verification ensures accuracy and saves time during the exam.