NDA Mathematics · 12 questions · 2017 to 2026
NDA Binary Numbers previous year questions with solutions
Binary Numbers is the smallest topic in the Mathematics paper, about half a mark in a hundred. The level has not moved since 2017; the only new turn is an LCM question in 2026. Every question UPSC has set on it since 2017, solved. Grouped by the method each needs, newest first; open a question for its answer and worked solution.
4.1 Binary notation and conversion (3)4.2 Binary addition and subtraction (4)4.3 Binary multiplication, division, HCF and LCM (3)4.4 Algebraic identities in binary (2)
4.1 Binary notation and conversion
1. NDA 2023 (I)
What is the binary number equivalent to decimal number 1011?
Answer and solution
No correct option. This question is defective as printed in UPSC's paper.
Method. decimal to binary by powers of 2
1011 = 512 + 256 + 128 + 64 + 32 + 16 + 2 + 1 (= 1024 − 13).
So 1011 = (1111110011)2, a ten-digit binary number.
The options are worth: (a) 10112 = 11, (b) 1110112 = 59, (c) 111110012 = 249, (d) 1111100112 = 499.
None equals 1011.
Decimal 1011 is (1111110011)2, which matches no option; option (d) (111110011)2 = 499 is that string with one 1 missing, so if (d) was meant to print 1111110011 it would be the answer.
2. NDA 2019 (II)
A binary number is represented by (cdccddcccddd)2, where c > d. What is its decimal equivalent?
Answer and solution
Answer: (d) 2872
Method. binary digits with c > d force c = 1, d = 0; then place values
Binary digits are 0 or 1, and c > d, so c = 1 and d = 0.
(cdccddcccddd)2 = (101100111000)2.
Place values of the 1s (12 digits, leftmost 211): 211 + 29 + 28 + 25 + 24 + 23.
= 2048 + 512 + 256 + 32 + 16 + 8 = 2872.
3. NDA 2018 (I)
The binary number expression of the decimal number 31 is
Answer and solution
Answer: (d) 11111
Method. decimal to binary by powers of 2
31 = 16 + 8 + 4 + 2 + 1 = 24 + 23 + 22 + 21 + 20.
So 31 = (11111)2.
Shortcut. 31 = 25 − 1, and 2n − 1 is n ones in binary: 11111.
4.2 Binary addition and subtraction
4. NDA 2025 (II)
What is the sum of the binary numbers (101101101)2 and (100011)2?
Answer and solution
Answer: (a) (110010000)2
Method. binary addition (or convert, add, convert back)
(101101101)2 = 256 + 64 + 32 + 8 + 4 + 1 = 365; (100011)2 = 32 + 2 + 1 = 35.
365 + 35 = 400.
400 = 256 + 128 + 16 = (110010000)2.
5. NDA 2020 (I)
The number (1101101 + 1011011)2 can be written in decimal system as
Answer and solution
Answer: (c) (200)10
Method. convert each binary number to decimal and add
(1101101)2 = 64 + 32 + 8 + 4 + 1 = 109.
(1011011)2 = 64 + 16 + 8 + 2 + 1 = 91.
Sum = 109 + 91 = 200, i.e. (200)10.
6. NDA 2018 (II)
The sum of the binary numbers (11011)2, (10110110)2 and (10011x0y)2 is the binary number (101101101)2. What are the values of x and y?
Answer and solution
Answer: (b) x = 1, y = 0
Method. binary addition: find the missing addend
(11011)2 = 27; (10110110)2 = 128 + 32 + 16 + 4 + 2 = 182; (101101101)2 = 256 + 64 + 32 + 8 + 4 + 1 = 365.
So (10011x0y)2 = 365 − 27 − 182 = 156.
156 = 128 + 16 + 8 + 4 = (10011100)2.
Matching 1 0 0 1 1 x 0 y with 1 0 0 1 1 1 0 0: x = 1, y = 0.
7. NDA 2017 (I)
In the binary equation (1p101)2 + (10q1)2 = (100r00)2 where p, q and r are binary digits, what are the possible values of p, q and r respectively?
Answer and solution
Answer: (a) 0, 1, 0
Method. binary column addition with unknown digits
(10q1)2 has 4 digits and (1p101)2 has 5, so the sum (100r00)2 = 32 + 4r is at most 31 + 15 = 46; hence r can be 0 or 1.
In decimal: (1p101)2 = 16 + 8p + 5 = 21 + 8p, (10q1)2 = 8 + 2q + 1 = 9 + 2q, (100r00)2 = 32 + 4r.
So 21 + 8p + 9 + 2q = 32 + 4r, i.e. 8p + 2q = 2 + 4r.
p = 1 gives 8 + 2q = 2 + 4r, impossible for digits (left ≥ 8, right ≤ 6). So p = 0 and 2q = 2 + 4r, giving q = 1, r = 0.
Check: (10101)2 + (1011)2 = 21 + 11 = 32 = (100000)2. (holds)
p, q, r = 0, 1, 0.
Shortcut. Test the options: only (a) gives 10101 + 1011 = 100000.
4.3 Binary multiplication, division, HCF and LCM
8. NDA 2026 (II)
What is the LCM of (11110)2, (10100)2 and (11011)2?
Answer and solution
Answer: (b) (1000011100)2
Method. LCM by prime factorisation after converting to decimal
(11110)2 = 30 = 2 × 3 × 5; (10100)2 = 20 = 22 × 5; (11011)2 = 27 = 33.
LCM = 22 × 33 × 5 = 540.
540 = 512 + 16 + 8 + 4 = (1000011100)2.
(The other options are 524, 572 and 541.)
9. NDA 2022 (I)
What is (1110011)2 ÷ (10111)2 equal to?
Answer and solution
Answer: (a) (101)2
Method. binary division via decimal conversion
(1110011)2 = 64 + 32 + 16 + 2 + 1 = 115.
(10111)2 = 16 + 4 + 2 + 1 = 23.
115 ÷ 23 = 5 exactly.
5 = (101)2.
10. NDA 2017 (II)
The remainder and the quotient of the binary division (101110)2 ÷ (110)2 are respectively
Answer and solution
Answer: (b) (100)2 and (111)2
Method. binary division via decimal conversion
(101110)2 = 32 + 8 + 4 + 2 = 46; (110)2 = 6.
46 = 6 × 7 + 4.
Quotient 7 = (111)2; remainder 4 = (100)2.
Remainder and quotient, in that order: (100)2 and (111)2.
4.4 Algebraic identities in binary
11. NDA 2025 (I)
If x = (1111)2, y = (1001)2 and z = (110)2, then what is x3 - y3 - z3 - 3xyz equal to?
Answer and solution
Answer: (d) (0)2
Method. if a + b + c = 0 then a3 + b3 + c3 = 3abc, with a = x, b = −y, c = −z
x = (1111)2 = 15, y = (1001)2 = 9, z = (110)2 = 6.
x − y − z = 15 − 9 − 6 = 0.
Take a = x, b = −y, c = −z: a + b + c = 0, so a3 + b3 + c3 = 3abc, i.e. x3 − y3 − z3 = 3x(−y)(−z) = 3xyz.
Hence x3 − y3 − z3 − 3xyz = 0 = (0)2.
Check: 3375 − 729 − 216 − 2430 = 0. (holds)
12. NDA 2022 (I)
If x3 + y3 = (100010111)2 and x + y = (11111)2, then what is (x − y)2 + xy equal to?
Answer and solution
No correct option. This question is defective as printed in UPSC's paper.
Method. x3 + y3 = (x + y)(x2 − xy + y2), and (x − y)2 + xy = x2 − xy + y2
(100010111)2 = 256 + 16 + 4 + 2 + 1 = 279; (11111)2 = 31.
(x − y)2 + xy = x2 − 2xy + y2 + xy = x2 − xy + y2.
x3 + y3 = (x + y)(x2 − xy + y2), so x2 − xy + y2 = 279 ÷ 31 = 9.
9 = (1001)2.
Defective as printed: x + y = 31 and x3 + y3 = 279 give x2 − xy + y2 = 9 and xy = 952/3, and the discriminant 312 − 4 · 952/3 = −925/3 < 0, so no real x, y satisfy the data. Taken formally, the identity x3 + y3 = (x + y)(x2 − xy + y2) gives (x − y)2 + xy = 9 = (1001)2, option (b).