GATE CE 2026 Set 1 — Question 9

MCQ+2 / -0.67MediumDeduction & InductionAnalytical AptitudeGeneral Aptitude

General Aptitude → Analytical Aptitude → Deduction & Induction

Last updated 5 September 2026

Question

Consider a linear arrangement of seven bulbs, each of which can be in the ON or OFF states. The initial configuration of the bulbs is shown in the figure. In every Step, the states of the bulbs are changed based on the following rules: Any OFF bulb with exactly one ON neighbor at the end of the previous Step is turned ON. Any ON bulb with both neighbors ON at the end of the previous Step is turned OFF. The state of any bulb not meeting the conditions above is left unchanged. The state of bulbs at the end of Step 1 and Step 2 are also shown in the figure. The number of bulbs which are ON at the end of Step 8 is______

Correct answer (B) 4 Solution Let 1 represent ON and 0 represent OFF. Initial: 0 0 0 1 0 0 0 Step 1: 0 0 1 1 1 0 0 (Bulbs 3 and 5 turn ON as they had exactly one ON neighbor, bulb 4 stays ON as it didn't have both neighbors ON) Step 2: 0 1 1 0 1 1 0 (Bulbs 2 and 6 turn ON; bulb 4 turns OFF as both its neighbors were ON) Step 3: 1 1 1 0 1 1 1 (Bulbs 1 and 7 turn ON; bulbs 2, 3, 5, 6 stay ON as they don't have both neighbors ON) Step 4: 1 0 1 0 1 0 1 (Bulbs 2 and 6 turn OFF as both neighbors were ON; bulbs 1, 3, 5, 7 stay ON) Step 5: 1 0 1 0 1 0 1 (Bulb 2 has two ON neighbors, so stays OFF. Bulb 4 has two ON neighbors, so stays OFF. Bulb 6 has two ON neighbors, so stays OFF. End bulbs 1 and 7 stay ON. Bulbs 3 and 5 stay ON as they don't have both neighbors ON). Since Step 5 is identical to Step 4, the system has reached a steady state. Thus, at Step 8, the configuration remains 1 0 1 0 1 0 1. The number of ON bulbs is 4. Turn this into a strength. Explore AI-powered practice and doubt support with Success Tracker.

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Correct answer (B) 4 Solution Let 1 represent ON and 0 represent OFF. Initial: 0 0 0 1 0 0 0 Step 1: 0 0 1 1 1 0 0 (Bulbs 3 and 5 turn ON as they had exactly one ON neighbor, bulb 4 stays ON as it didn't have both neighbors ON) Step 2: 0 1 1 0 1 1 0 (Bulbs 2 and 6 turn ON; bulb 4 turns OFF as both its neighbors were ON) Step 3: 1 1 1 0 1 1 1 (Bulbs 1 and 7 turn ON; bulbs 2, 3, 5, 6 stay ON as they don't have both neighbors ON) Step 4: 1 0 1 0 1 0 1 (Bulbs 2 and 6 turn OFF as both neighbors were ON; bulbs 1, 3, 5, 7 stay ON) Step 5: 1 0 1 0 1 0 1 (Bulb 2 has two ON neighbors, so stays OFF. Bulb 4 has two ON neighbors, so stays OFF. Bulb 6 has two ON neighbors, so stays OFF. End bulbs 1 and 7 stay ON. Bulbs 3 and 5 stay ON as they don't have both neighbors ON). Since Step 5 is identical to Step 4, the system has reached a steady state. Thus, at Step 8, the configuration remains 1 0 1 0 1 0 1. The number of ON bulbs is 4.

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