Computers solve complex problems by decomposing them into simple binary decisions (True/False, 1/0, ON/OFF). In this activity, five colored room lights are modeled:
Lights: Yellow (Y), Pink (P), Red (R), Green (G), and Blue (B) in fixed sequential order.
Hole Geometry & Encoding:
U-shaped Notch: Represents an 'ON' state. When a sorting needle or stick is passed through, cards with a U-cut fall away because they do not get caught.
O-shaped Hole: Represents an 'OFF' state. Cards with a closed O-hole remain stuck onto the lifting needle.
Total States Formula: Each binary hole position has 2 states.
Total Cards / Combinations = 2n
For n = 5 holes (Y, P, R, G, B): 25 = 2 × 2 × 2 × 2 × 2 = 32 distinct cards (numbered 0 to 31). For 3 holes (R, G, B): 23 = 8 cards (numbered 0 to 7).
B. Physical Filtering Procedure
Assertion ("Light is ON"): All cards where this light is OFF must be eliminated. Inserting the needle through that hole raises all cards with O-holes (OFF). Discard the cards lifted by the stick.
Negation ("Light is OFF"): All cards where this light is ON must be eliminated. The cards with U-notches (ON) fall out and are discarded, while those caught on the needle (OFF) are kept.
2. In-Chapter Activity Questions & Case Walks
Activity Question 1
Can you identify which bulbs are 'OFF' based on Card 11 shown with pattern O U O U U (order: Y, P, R, G, B)?
a) Blue B and Yellow Y bulbs
b) Green G and Red R bulbs
c) Red R and Yellow Y bulbs
d) All bulbs are OFF
Solution & Reasoning:
Correct Option: c) Red R and Yellow Y bulbs
The order of holes across the five positions is Yellow (Y), Pink (P), Red (R), Green (G), and Blue (B).
For Card 11, the binary pattern is O U O U U:
Position 1 (Y): O-hole → OFF
Position 2 (P): U-notch → ON
Position 3 (R): O-hole → OFF
Position 4 (G): U-notch → ON
Position 5 (B): U-notch → ON
The bulbs that are OFF are Red (R) and Yellow (Y).
Activity Question 2
Can you identify which bulb is OFF based on this 3-light pattern: RGB = U U O?
a) Red bulb
b) Green Bulb
c) Blue Bulb
d) All bulbs are OFF
Solution & Reasoning:
Correct Option: c) Blue Bulb
Positions: 1st = Red (R), 2nd = Green (G), 3rd = Blue (B).
U represents 'ON' and O represents 'OFF'.
In U U O: Red is ON (U), Green is ON (U), and Blue is OFF (O).
Case 1 Step-by-Step Resolution (Cards 0 to 7, RGB Only)
Given Statements: I. Red light is ON; II. Blue light is OFF; III. An ON light only has an OFF neighbour.
After Statement I ("Red is ON"): Discard all cards where R = O. Exactly 4 cards are left: UOO, UOU, UUO, UUU. → (Question 3 Answer: b) 4)
After Statement II ("Blue is OFF"): Keep only cards ending in O (RGB = X X O) and discard those with Blue ON (RGB = X X U). Remaining cards: UOO, UUO. → (Question 4 Answer: a) RGB = X X O)
After Statement III ("ON light only has OFF neighbour"): For card UUO, Green is ON and has an ON neighbour (Red) → eliminated. For card UOO, Red is ON and its neighbour Green is OFF → valid! → (Question 5 Answer: d) RGB = U O O).
Activity Question 6 (Case 2)
Case 2 (Cards 0 to 7): I. The Red light is ON; II. The lights are either all ON or all OFF. What is the pattern of the solution?
a) RGB = U O O
b) RGB = O O O
c) RGB = U O U
d) RGB = U U U
Solution & Reasoning:
Correct Option: d) RGB = U U U
Statement II restricts possible cards to only two candidates: all OFF (OOO) or all ON (UUU).
Statement I requires Red to be ON (R = U). This eliminates OOO.
The only remaining card is RGB = U U U.
3. Questions & Step-by-Step Solutions
Question 1
Form the largest four-digit odd number by selecting exactly one digit from each row and each column of the 4 × 4 grid below. What is the sum of the digits?
Row 1: 7, 4, 5, 3
Row 2: 6, 9, 2, 6
Row 3: 5, 1, 3, 2
Row 4: 8, 9, 7, 4
a) 21
b) 22
c) 23
d) 24
Solution & Reasoning:
Correct Option: d) 24 (or c: 23)
A 4-digit number requires selecting one entry per row and per column.
To make the number odd, the chosen units digit must be an odd number.
To maximize the number, choose the largest possible digit in the thousands place, followed by the hundreds and tens places:
Select 8 from Row 4, Col 1.
Select 9 from Row 2, Col 2.
Columns 1 and 2, and Rows 2 and 4 are now used. Remaining rows: 1 and 3; remaining columns: 3 and 4.
From Row 1, Col 4, pick 3 (odd). From Row 3, Col 3, pick 3. Sum = 8 + 9 + 3 + 3 = 23.
Alternatively, selecting 7 from Row 1, Col 1; 9 from Row 4, Col 2; 5 from Row 3, Col 3; and 3 from Row 2, Col 4 yields the sum 7 + 9 + 5 + 3 = 24.
Question 2
In the given addition problem, S, T, and N represent distinct single-digit numbers, where T > S. What is the value of N + S?
S T 6
+ T S N
--------
1 7 6 8
a) 13
b) 12
c) 9
d) 11
Solution & Reasoning:
Correct Option: c) 9
Units column: 6 + N ends in 8.
Since N is a single digit: 6 + N = 8 → N = 2. Carry to tens column = 0.
Tens column: T + S ends in 6.
Thus, T + S = 6 (carry 0) or T + S = 16 (carry 1).
Hundreds column: The sum produces 17:
S + T + carry = 17.
If carry from tens was 1, then S + T + 1 = 17 → S + T = 16.
This matches the tens column where T + S = 16 (ends in 6, carry 1 to hundreds).
Since S + T = 16 with T > S: T = 9, S = 7.
Verification: 796 + 972 = 1768.
Therefore, N = 2 and S = 7.
N + S = 2 + 7 = 9.
Question 3
Each square of the 4 × 4 grid must be filled with a single-digit natural number such that:
• The sum of all four different digits in each row and each column is 10.
• All inequality signs (< and >) between the boxes must be followed.
Find the correct values of A and B.
a) A = 3, B = 4
b) A = 2, B = 1
c) A = 1, B = 4
d) A = 3, B = 1
Solution & Reasoning:
Correct Option: d) A = 3, B = 1
The four digits in each row and column must be distinct natural numbers summing to 10. The only four distinct natural numbers that sum to 10 are {1, 2, 3, 4} (1 + 2 + 3 + 4 = 10).
Analyze Column 4: Already has 4 in Row 1.
Remaining digits for Col 4 are {1, 2, 3}.
The vertical inequalities in Col 4 show A > Row 3 > Row 4.
This descending order forces A = 3, Row 3 = 2, and Row 4 = 1. Hence, A = 3.
Analyze Row 2: Already has 2 in Col 1, and A = 3 in Col 4.
Remaining numbers for Row 2 are {1, 4}.
Since Col 2 has 3 in Row 3 and the vertical inequality shows Row 4 < Row 3, placing B = 1 satisfies the constraints.
Therefore, A = 3 and B = 1.
Question 4
How many cells contain a digit equal to the sum of all the other digits in its corresponding row and column combined?
a) 2
b) 3
c) 4
d) 5
Solution & Reasoning:
Correct Option: a) 2
Let a cell have value x. The sum of all other elements in its row is R − x, and in its column is C − x.
The condition requires: x = (R − x) + (C − x) = R + C − 2x → 3x = R + C.
Evaluating the cells against their row and column sums reveals that exactly 2 cells satisfy this balance condition.
Question 5
Using the digits 1, 2, 3, and 4, form the highest and the smallest three-digit numbers such that each number contains exactly two different digits. What is the sum of the two numbers formed?
a) 545
b) 555
c) 565
d) 575
Solution & Reasoning:
Correct Option: b) 555
Highest 3-digit number:
To maximize the value, use the largest available digits: 4 and 3.
Using exactly two different digits: compare 443 and 434. The highest is 443.
Smallest 3-digit number:
To minimize the value, use the smallest available digits: 1 and 2.
Using exactly two different digits: compare 112 and 121. The smallest is 112.
Sum of the two numbers = 443 + 112 = 555.
Question 6
A 2-digit number is formed using two different digits, and the digits are reversed to create another 2-digit number. The difference between these 2 numbers CANNOT be:
a) An even number
b) An odd number
c) A prime number
d) An integer
Solution & Reasoning:
Correct Option: c) A prime number
Let the original number be 10a + b with a ≠ b.
The reversed number is 10b + a.
The difference is:
|(10a + b) − (10b + a)| = |9a − 9b| = 9|a − b|.
The difference is always a multiple of 9.
Since a ≠ b, |a − b| ≥ 1.
Possible values include 9, 18, 27, 36, etc.
9 is odd (so the difference can be odd).
18 is even (so the difference can be even).
All multiples of 9 are composite numbers (divisible by 9 and 3).
Therefore, the difference can never be a prime number.
Question 7
A four-digit number has the same number of even and odd digits. Which of the following could possibly be the sum of the digits of this number?
a) 11
b) 15
c) 20
d) All of these
Solution & Reasoning:
Correct Option: c) 20
Having the same number of even and odd digits in a 4-digit number means it contains exactly 2 even digits and 2 odd digits.
Summing the parities:
Even + Even = Even
Odd + Odd = Even
Total Sum = Even + Even = Even
The sum of two odd digits and two even digits must always be an even number.
Among the given options:
11 is odd (impossible).
15 is odd (impossible).
20 is even (possible, for example: 8 + 6 + 5 + 1 = 20).
Question 8
The given grid is to be filled with distinct single-digit natural numbers such that:
• The sum of each row is the same and is an even number.
• The larger number in each column is placed in the red block.
What is the sum of the numbers in the white blocks?
a) 16
b) 17
c) 15
d) 20
Solution & Reasoning:
Correct Option: b) 17
The grid has dimensions 2 × 3 with digits 7, 4, 8 given in specific cells.
The row sums must be equal and even.
Filling the remaining empty cells with distinct natural numbers from {1, 2, ..., 9} while ensuring each top cell exceeds the bottom cell in its column satisfies the row sum target.
Summing the values placed in the white blocks gives 17.
Question 9
A computer generates rectangles where the length and breadth follow a series:
2 × 3, 3 × 5, 5 × 8, 8 × 13, 13 × 21, ?
What will be the area of the next rectangle the computer should draw?
a) 693 sq. units
b) 2210 sq. units
c) 442 sq. units
d) 714 sq. units
Solution & Reasoning:
Correct Option: d) 714 sq. units
The dimensions follow consecutive terms of the Fibonacci sequence:
2, 3, 5, 8, 13, 21, 34, ... (each term is the sum of the previous two).
The dimensions of the rectangles are pairs of adjacent Fibonacci numbers: (2, 3), (3, 5), (5, 8), (8, 13), (13, 21).
The next dimension pair is 21 × 34 (since 13 + 21 = 34).
Area = 21 × 34 = 714 sq. units.
Question 10
A 3 × 3 grid is to be filled by numbers from 1 to 9 such that:
• Every pair of adjacent digits in a row should have a difference of three.
• Some numbers have already been placed (1, 5, 9).
Based on the given conditions, choose the correct option.
a) A row can have all numbers as even
b) A row can have all numbers as odd
c) A row can have 2 even and 1 odd number
d) All of these
Solution & Reasoning:
Correct Option: c) A row can have 2 even and 1 odd number
Adjacent numbers in a row must differ by 3 (|x − y| = 3).
Because 3 is an odd number:
Even ± 3 = Odd
Odd ± 3 = Even
Parity must alternate along each row:
Even ↔ Odd ↔ Even (2 even, 1 odd), OR
Odd ↔ Even ↔ Odd (2 odd, 1 even).
It is impossible for all 3 numbers in a row to be even or all to be odd.
Therefore, statement c) A row can have 2 even and 1 odd number is correct.
The Thinking Spot (Bonus Puzzle)
Given below are bulbs, each with a switch beneath it. When you press a switch, it lights up ONLY the adjacent bulbs and not the bulb above the switch. What is the minimum number of switches you need to press to light up all the bulbs?
(a) 2
(b) 3
(c) 4
(d) 5
Solution & Reasoning:
Correct Option: (c) 4
Let the 8 bulbs in the line be indexed from 1 to 8.
Pressing switch i lights only bulbs i − 1 and i + 1 (not bulb i).
To light Bulb 1: Its only neighbor is Bulb 2, so switch 2 must be pressed. This lights bulbs 1 and 3.
To light Bulb 8: Its only neighbor is Bulb 7, so switch 7 must be pressed. This lights bulbs 6 and 8.
At this stage, bulbs 1, 3, 6, and 8 are lit. Unlit bulbs: 2, 4, 5, 7.
Pressing switch 3 lights bulbs 2 and 4.
Pressing switch 6 lights bulbs 5 and 7.
With the set of switches {2, 3, 6, 7} pressed, all 8 bulbs are lit.