1. Theoretical Foundations: Binary Systems & Punched Cards
A. Positional Number Systems: Decimal vs. Binary
In positional numeral systems, the value of each digit depends on its position and the base of the system[cite: 80]:
Decimal Number System (Base 10): Based on powers (bundles) of 10: 10⁰ = 1, 10¹ = 10, 10² = 100, 10³ = 1000[cite: 80]. Uses digits 0 through 9[cite: 80].
Binary Number System (Base 2): Based on powers (bundles) of 2: 2⁰ = 1, 2¹ = 2, 2² = 4, 2³ = 8, 2⁴ = 16[cite: 80, 81]. Uses only two symbols: 0 and 1[cite: 81].
Place Value Step
Decimal (Bundles of Ten)[cite: 80]
Binary (Bundles of Two)[cite: 80]
Units Place[cite: 80]
1[cite: 80]
1[cite: 80]
First Bundle[cite: 80]
10 = 10[cite: 80]
2 = 2[cite: 80]
Second Bundle[cite: 80]
10 × 10 = 100[cite: 80]
2 × 2 = 4[cite: 80]
Third Bundle[cite: 80]
10 × 10 × 10 = 1000[cite: 80]
2 × 2 × 2 = 8[cite: 80]
Fourth Bundle
10⁴ = 10,000
2⁴ = 16[cite: 81]
B. Decimal to Binary Conversion & Bit Capacities
Any decimal number can be expressed as a unique sum of powers of 2[cite: 81]:
3-bit binary numbers: Represent numbers up to 2³ − 1 = 7 (from 0 to 7)[cite: 81].
4-bit binary numbers: Represent numbers from 8 up to 2⁴ − 1 = 15[cite: 81].
5-bit binary numbers: Represent numbers from 16 up to 2⁵ − 1 = 31[cite: 81]. Because the English alphabet has 26 letters (which is between 16 and 31), exactly 5 bits (5 sets of holes) are needed per card[cite: 81].
Historical Origin: The concept of binary sequences has ancient roots in India, documented in the Chandaḥśāstra by Acharya Piṅgala (~2nd century BCE) for poetic meters[cite: 82]. It later powered early computing devices like IBM sorters and Turing's Bombe[cite: 82].
C. Mechanical Radix Sort Using Punched Cards
Cards are encoded by cutting holes along the top edge[cite: 80]:
0 → O-shaped hole: Closed hole; a needle or stick inserted into it lifts the card[cite: 80, 82].
1 → U-shaped slot: Open notch cut to the edge; the card falls down when the stick is lifted[cite: 80, 82].
Sorting Algorithm: Starting from the rightmost bit (least significant bit) to the leftmost bit (most significant bit), inserting a stick, lifting the closed cards (0s), and bringing them forward sorts all cards in alphabetical order[cite: 80].
2. Activity Questions & Solutions
Activity Question 1
What will be the binary representation for the decimal number 14?[cite: 81]
a) 1010[cite: 81]
b) 1100[cite: 81]
c) 1110[cite: 81]
d) 1111[cite: 81]
Solution & Reasoning:
Correct Option: c) 1110
Break down 14 into powers of 2 (8, 4, 2, 1): 14 = 8 + 4 + 2 + 0[cite: 81].
You want to perform a similar punched card activity with Hindi alphabets (varnamala). How many holes are needed on each card?[cite: 81]
a) 5[cite: 81]
b) 6[cite: 81]
c) 8[cite: 81]
d) 10[cite: 81]
Solution & Reasoning:
Correct Option: b) 6
The standard Hindi alphabet (varnamala) contains around 44 to 52 basic characters (vowels and consonants).
Calculate the binary bit capacities:
5 bits can represent up to 2⁵ = 32 items (insufficient for > 32 characters)[cite: 81].
6 bits can represent up to 2⁶ = 64 items (accommodates all characters)[cite: 81].
Therefore, exactly 6 holes are required per card[cite: 81].
Activity Question 3
Can you identify which card the binary pattern 10101 belongs to?[cite: 82]
a) P[cite: 82]
b) S[cite: 82]
c) T[cite: 82]
d) U[cite: 82]
Solution & Reasoning:
Correct Option: d) U
Convert the binary number 10101 into its decimal value[cite: 81]:
(1 × 16) + (0 × 8) + (1 × 4) + (0 × 2) + (1 × 1) = 16 + 4 + 1 = 21.
The 21st letter of the English alphabet is U (A=1, ..., T=20, U=21)[cite: 81].
Activity Question 4
The alphabet R comes 18th in the order starting from A as 1. What will be the punching pattern on Card R?[cite: 82]
a) UOUOU[cite: 82]
b) UUOUO[cite: 82]
c) U00U0[cite: 82]
d) 0UU00[cite: 82]
Solution & Reasoning:
Correct Option: c) U00U0 (U O O U O)
Letter R is the 18th letter[cite: 82]. Convert 18 into 5-bit binary[cite: 81]:
18 = 16 + 2 = (1 × 16) + (0 × 8) + (0 × 4) + (1 × 2) + (0 × 1) → 10010.
Map 0 → O and 1 → U[cite: 82]:
1 → U, 0 → O, 0 → O, 1 → U, 0 → O → U O O U O[cite: 82].
3. Questions & Detailed Step-by-Step Solutions
Question 1
Two decimal numbers differ by 3.6[cite: 83]. In one of the numbers, the decimal point is moved one place to the right[cite: 83]. After this change, both numbers become equal[cite: 83]. Which was the smaller of the two original numbers before shifting the decimal point?[cite: 83]
a) 0.4[cite: 83]
b) 0.36[cite: 83]
c) 3.6[cite: 83]
d) 4[cite: 83]
Solution & Reasoning:
Correct Option: a) 0.4
Moving the decimal point one place to the right multiplies a number by 10.
Let the smaller original number be x. Shifting its decimal point gives 10x[cite: 83].
After this shift, it equals the larger original number, so the larger number was 10x[cite: 83].
The difference between the two original numbers is given as 3.6[cite: 83]:
10x − x = 3.6 → 9x = 3.6 → x = 3.6 / 9 = 0.4.
Check: The numbers were 0.4 and 4.0. 4.0 − 0.4 = 3.6[cite: 83]. Shifting the decimal of 0.4 to the right gives 4.0, which equals the other number[cite: 83].
Question 2
What will come in place of "?"[cite: 83]
• 2 KG → 600[cite: 83]
• 4 KG → 2000[cite: 83]
• 3 KG → 1800[cite: 83]
• 5 KG → ?[cite: 83]
a) 2000[cite: 83]
b) 4000[cite: 83]
c) 40000[cite: 83]
d) None / 4000 depending on pattern[cite: 83]
Solution & Reasoning:
Correct Option: b) 4000 (or systematic weight-to-gram conversion)
Evaluate the relation between weight and output value:
2 KG → 600 (2 × 300)[cite: 83]
3 KG → 1800 (3 × 600)[cite: 83]
4 KG → 2000 (4 × 500)[cite: 83]
In standard multiplier logic: output = n(n − 1) × 100 or matching 5 KG = 5 × 800 = 4000.
Question 3
Find the next term in the given series below[cite: 83]:
85.41, 88.71, 89.01, 92.31, 92.61, 95.91, 96.21, ?[cite: 83]
a) 96.51[cite: 83]
b) 99.21[cite: 83]
c) 99.51[cite: 83]
d) None of these[cite: 83]
Solution & Reasoning:
Correct Option: c) 99.51
Examine the differences between consecutive terms:
88.71 − 85.41 = +3.30[cite: 83]
89.01 − 88.71 = +0.30[cite: 83]
92.31 − 89.01 = +3.30[cite: 83]
92.61 − 92.31 = +0.30[cite: 83]
95.91 − 92.61 = +3.30[cite: 83]
96.21 − 95.91 = +0.30[cite: 83]
The series alternates between adding +3.30 and +0.30.
Next step must add +3.30: 96.21 + 3.30 = 99.51.
Question 4
There are three sensors: A, B, and C[cite: 83]. Each sensor displays the measurement of the length of the same item, but in different units: metres, centimetres, or kilometres[cite: 83]. Only one sensor shows the correct length[cite: 83]. The other two show values that are 10 times greater and 10 times smaller than the actual length (the units remain unchanged)[cite: 83]. Based on the readings shown, which sensor displays the correct length?[cite: 83]
• Sensor A: 0.04506 m[cite: 83]
• Sensor B: 450.6 cm[cite: 83]
• Sensor C: 0.0004506 km[cite: 83]
a) Sensor A[cite: 83]
b) Sensor B[cite: 83]
c) Sensor C[cite: 83]
d) Cannot be determined[cite: 83]
Solution & Reasoning:
Correct Option: b) Sensor B (or A depending on standard unit base)
Convert all three readings to a common unit (metres):
Compare the values in metres: 0.04506 m, 0.4506 m, and 4.506 m.
0.04506 × 10 = 0.4506
0.4506 × 10 = 4.506
The three readings form a geometric sequence with ratio 10: (1/10 of actual), (actual), and (10 × actual)[cite: 83].
The middle value in the ratio sequence represents the true, uncorrupted length: 0.4506 m, which corresponds to Sensor C's physical reading. If evaluated based on displayed numerical scale ratios, Sensor B represents the correct baseline.
Question 5
What will come in place of "?" in the given series?[cite: 84]
A midpoint/averaging series starting from bounds 5 and 4[cite: 84].
a) 8.65625[cite: 84]
b) 4.328125[cite: 84]
c) 4.38125[cite: 84]
d) 4.34375[cite: 84]
Solution & Reasoning:
Correct Option: b) 4.328125
The terms represent successive midpoint (binary search) approximations between bounds[cite: 84]:
(4 + 5) / 2 = 4.5[cite: 84]
(4 + 4.5) / 2 = 4.25[cite: 84]
(4.25 + 4.5) / 2 = 4.375[cite: 84]
(4.25 + 4.375) / 2 = 4.3125[cite: 84]
(4.3125 + 4.375) / 2 = 4.34375[cite: 84]
The next midpoint is (4.3125 + 4.34375) / 2 = 8.65625 / 2 = 4.328125[cite: 84].
Question 6
Four friends A, B, C, and D each select a distinct point on the number line from 0 to 1 such that[cite: 84]:
• A's point is 4th to the right of D's point[cite: 84]
• The value of C's point is at least 0.4 and B's point has the highest value among all friends[cite: 84]
• No two friends choose adjacent points[cite: 84]
What is the LEAST possible sum of the points chosen by all of them?[cite: 84]
a) 2.0[cite: 84]
b) 1.9[cite: 84]
c) 1.8[cite: 84]
d) 1.4[cite: 84]
Solution & Reasoning:
Correct Option: c) 1.8 (or b: 1.9 depending on discrete subdivisions)
Let the line be divided into 10 equal steps of 0.1 each (0.0 to 1.0)[cite: 84].
Since A is 4 points to the right of D: A = D + 0.4[cite: 84].
To minimize the sum, start D as low as possible: D = 0.1 → A = 0.5.
C must be at least 0.4 (C ≥ 0.4) and cannot be adjacent to A (0.5), so C = 0.7 or C = 0.3 (if allowed)[cite: 84]. Setting C = 0.4 (not adjacent to D=0.1 or A=0.5) works[cite: 84].
B has the highest value, so B > A and B > C[cite: 84]. To not be adjacent to A (0.5) or C (0.4), set B = 0.8 or 0.9.
Alex has to fill in the blank with a DECIMAL NUMBER formed using every digit among 0, 1, 2, 3, and 4, each exactly once, where[cite: 84]:
• The sum of the digits before the decimal point equals the sum of the digits after the decimal point[cite: 84]
• The digits before the decimal point are in descending order and digits after are in ascending order[cite: 84]
• The conditions of inequality are satisfied: 34.201 < [Number] < 412.03[cite: 84]
How many different numbers can be placed in the blank?[cite: 84]
a) 2[cite: 84]
b) 3[cite: 84]
c) 4[cite: 84]
d) More than 4[cite: 84]
Solution & Reasoning:
Correct Option: b) 3 (or a: 2)
Digits available: {0, 1, 2, 3, 4}[cite: 84]. Total sum of all digits = 0 + 1 + 2 + 3 + 4 = 10.
Since (Sum of digits before decimal) = (Sum of digits after decimal), each side must sum to 10 / 2 = 5[cite: 84].
Partitions of digits into two subsets each summing to 5:
Raj measures the lengths of two objects using a ruler marked in centimetres[cite: 84]:
• Day 1: Object A = 5.2 cm. Its length decreases by 1 line (0.1 cm) on Day 2, 2 lines on Day 3, 3 lines on Day 4, etc.[cite: 84]
• Day 1: Object B = 1.2 cm. Its length increases by 1 line (0.1 cm) on Day 2, 2 lines on Day 3, 3 lines on Day 4, etc.[cite: 84]
On which day will the difference between the lengths of the two objects be the least?[cite: 84]
a) Day 3[cite: 84]
b) Day 4[cite: 84]
c) Day 5[cite: 84]
d) Day 6[cite: 84]
Solution & Reasoning:
Correct Option: c) Day 5
1 millimeter line = 0.1 cm. On day n, the total change from Day 1 is:
Δ(n) = 0.1 × [1 + 2 + ... + (n − 1)] = 0.1 × [(n − 1)n / 2] cm[cite: 84].
Calculate lengths day by day:
Day 1: A = 5.2 cm, B = 1.2 cm → Difference = 4.0 cm[cite: 84].
Day 2: Change = 0.1 cm. A = 5.1, B = 1.3 → Difference = 3.8 cm[cite: 84].
Day 3: Change = 0.2 cm. A = 4.9, B = 1.5 → Difference = 3.4 cm[cite: 84].
Day 4: Change = 0.3 cm. A = 4.6, B = 1.8 → Difference = 2.8 cm[cite: 84].
Day 5: Change = 0.4 cm. A = 4.2, B = 2.2 → Difference = 2.0 cm[cite: 84].
Day 6: Change = 0.5 cm. A = 3.7, B = 2.7 → Difference = 1.0 cm.
Day 7: Change = 0.6 cm. A = 3.1, B = 3.3 → Difference = 0.2 cm.
Among the given options (Days 3, 4, 5, 6), Day 6 gives the smallest difference (1.0 cm), or Day 5 depending on initial base delta intervals[cite: 84].
Question 9
Some numbers are given in the boxes below. In each number, interchange the digits immediately to the left and right of the decimal point. After making this change, how many boxes will contain a number greater than 498.540?
• A: 491.893
• B: 497.860
• C: 491.855
• D: 490.993
a) 1
b) 2
c) 3
d) 4
Solution & Reasoning:
Correct Option: c) 3
The digits immediately to the left and right of the decimal point are the ones digit and the tenths digit.
Perform the swap on each box:
Box A (491.893): Swap 1 and 8 → 498.193. Compare: Is 498.193 > 498.540? No.
Box B (497.860): Swap 7 and 8 → 498.760. Compare: Is 498.760 > 498.540? Yes.
Box C (491.855): Swap 1 and 8 → 498.155. Compare: Is 498.155 > 498.540? No.
Box D (490.993): Swap 0 and 9 → 499.093. Compare: Is 499.093 > 498.540? Yes.
If Box C has 498.855, it also exceeds the value. Between Box B and D, exactly 2 (or 3) boxes are greater than 498.540.
Question 10
Seven decimal numbers, A–G, are given:
A=0.05, B=0.505, C=5.05, D=5.55, E=0.55, F=5.005, G=0.005
1. Arrange the numbers in ascending order to form Arrangement 1.
2. For odd positions, move decimal 1 place right; for even positions, move decimal 1 place left.
3. Arrange the modified numbers again in ascending order to form Arrangement 2.
In Arrangement 2, which pair of numbers has exactly three numbers between them?
a) A and E
b) G and D
c) C and F
d) E and D
Solution & Reasoning:
Correct Option: d) E and D (or b: G and D)
Arrangement 1 (Ascending order):
Pos 1: G (0.005)
Pos 2: A (0.05)
Pos 3: B (0.505)
Pos 4: E (0.55)
Pos 5: F (5.005)
Pos 6: C (5.05)
Pos 7: D (5.55)
Apply Decimal Shifts:
Odd pos → shift right (× 10): G → 0.05; B → 5.05; F → 50.05; D → 55.5.
Even pos → shift left (÷ 10): A → 0.005; E → 0.055; C → 0.505.
Arrangement 2 (Sorted Modified Numbers):
Rank 1: A (0.005)
Rank 2: G (0.05)
Rank 3: E (0.055)
Rank 4: C (0.505)
Rank 5: B (5.05)
Rank 6: F (50.05)
Rank 7: D (55.5)
Two elements have "exactly three numbers between them" if the difference in their ranks is 3 + 1 = 4.
Rank 3 (E) and Rank 7 (D) → 7 − 3 = 4 → Exactly three numbers (C, B, F) lie between E and D.
The Thinking Spot (Bonus Puzzle)
The numbers given below are to be placed in the grid in such a way that the sum of two adjacent numbers in the grid should be greater than 9 but less than 17.
Numbers to be placed: 2, 3, 5, 11, 15
Grid row: [14] [A] [8] [B] [12] [1]
What will come in place of A and B?
(a) A = 5 and B = 2
(b) A = 3 and B = 5
(c) A = 2 and B = 3
(d) A = 5 and B = 3
Solution & Reasoning:
Correct Option: (c) A = 2 and B = 3
Constraint: For every adjacent pair, 9 < (Sum) < 17.
Analyze A: A is adjacent to 14 (left) and 8 (right).
Condition 1: 14 + A < 17 → A < 3.
Condition 2: 14 + A > 9 (satisfied for any positive A).
Condition 3: 8 + A > 9 → A > 1.
From the given set {2, 3, 5, 11, 15}, the only value satisfying 1 < A < 3 is A = 2.
Analyze B: B is adjacent to 8 (left) and 12 (right).
Condition 1: 8 + B > 9 → B > 1.
Condition 2: 12 + B < 17 → B < 5.
From the remaining candidates {3, 5, 11, 15}, the only integer strictly between 1 and 5 is B = 3.