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Exercise 1.1 · Q1

Q.Write the following numbers in decimal notation: (1010101100110)2(1010101100110)_2, (101011000110)2(101011000110)_2, (101111100110)2(101111100110)_2, (1000000000110)2(1000000000110)_2.

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Each binary number is converted to decimal by summing di×2id_i \times 2^i over every bit that equals 1: the four results are 5478, 2758, 3046 and 4102.

Binary-to-decimal conversion: for a binary number (dndn−1…d1d0)2(d_n d_{n-1}\dots d_1 d_0)_2, the decimal value is

N10=∑i=0ndi×2iN_{10} = \sum_{i=0}^{n} d_i \times 2^i

where each di∈{0,1}d_i \in \{0,1\} is the digit at position ii, counted from the right starting at i=0i=0.

  1. (1010101100110)2(1010101100110)_2 — locate the 1-bits (positions counted from the right, starting at 0): positions 12,10,8,6,5,2,112,10,8,6,5,2,1. 212+210+28+26+25+22+21=4096+1024+256+64+32+4+2=54782^{12}+2^{10}+2^{8}+2^{6}+2^{5}+2^{2}+2^{1} = 4096+1024+256+64+32+4+2 = 5478.
  2. (101011000110)2(101011000110)_2 — 1-bits at positions 11,9,7,6,2,111,9,7,6,2,1. 211+29+27+26+22+21=2048+512+128+64+4+2=27582^{11}+2^{9}+2^{7}+2^{6}+2^{2}+2^{1} = 2048+512+128+64+4+2 = 2758.
  3. (101111100110)2(101111100110)_2 — 1-bits at positions 11,9,8,7,6,5,2,111,9,8,7,6,5,2,1. 211+29+28+27+26+25+22+21=2048+512+256+128+64+32+4+2=30462^{11}+2^{9}+2^{8}+2^{7}+2^{6}+2^{5}+2^{2}+2^{1} = 2048+512+256+128+64+32+4+2 = 3046.
  4. (1000000000110)2(1000000000110)_2 — 1-bits at positions 12,2,112,2,1. 212+22+21=4096+4+2=41022^{12}+2^{2}+2^{1} = 4096+4+2 = 4102.
  5. Self-check by the doubling method (start at 0; for each digit left-to-right compute v→2v+div \to 2v+d_i) on the first number: 0→1→2→5→10→21→42→85→171→342→684→1369→2739→54780\to1\to2\to5\to10\to21\to42\to85\to171\to342\to684\to1369\to2739\to5478 — matches step 1. ✓
✓Final answer

(1010101100110)2=5478, (101011000110)2=2758, (101111100110)2=3046, (1000000000110)2=4102(1010101100110)_2=5478,\ (101011000110)_2=2758,\ (101111100110)_2=3046,\ (1000000000110)_2=4102

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