Number Systems and Series

The Binary System uses only two digits: 0 and 1. This is the fundamental number system used in digital computing because electronic circuits can easily represent two states (on/off). Barter is an economic exchange system, not a number system.

The series J, _, M, A, _, J represents the first letter of months: January, February, March, April, May, June. The missing letters are F (February) and M (May), so F, M is the correct pair.

This is identical to question 86819. The series J, _, M, A, _, J represents months: January, February, March, April, May, June. The missing elements are F (February) and M (May), so F, M is correct.

This is the triangular number sequence. The pattern shows the differences: 3-1=2, 6-3=3, 10-6=4, 15-10=5. Each difference increases by 1, so the next difference is 6, making the next number 15+6=21. Alternatively, these are 1, 1+2, 1+2+3, 1+2+3+4, 1+2+3+4+5.

The decimal number system (base 10) is what we use in everyday life, with digits 0-9. Octal is base 8, hexadecimal is base 16, and binary is base 2 - all used in computing but not in daily calculations.

The sequence represents the first letters of number words: One, Two, Three, Four, Five, Six, Seven, Eight. The next letter in this pattern is 'e' for Eight, completing the sequence of number word initials.

The series represents the first letters of the days of the week: Monday, Tuesday, Wednesday, Thursday. The next two days are Friday and Saturday, represented by the letters F and S.

Kamla's pattern: differences are 2, 3, 4, 5... so day n = 1 + 2+3+...+(n-1) = 1 + (n-1)n/2. Bimla's pattern: differences are 3, 4, 5, 6... so day n = 1 + 3+4+...+(n+1) = 1 + (n+2)(n-1). They fast together when 1 + (n-1)n/2 = 1 + (n+2)(n-1). This gives (n-1)n/2 = (n+2)(n-1), so n/2 = n+2, which gives n = -4, impossible for positive n. They never fast on the same day.

If the number is x, then 100000 + x is the number with 1 in front, and 10x + 1 is the number with 1 behind. The equation is 3(100000 + x) = 10x + 1. Solving gives 7x = 299999, so x = 42857.

The years 2000, 2012, and 2008 are leap years (divisible by 4). The year 2010 is a non-leap year, making it the odd one out.

Counting frequency in: 2 3 4 2 1 3 4 2 8 7 5 2 1 3 4 6 2 3 4 1 9 5 3. 2 appears 5 times, 3 appears 5 times, 4 appears 4 times, 1 appears 3 times, 5 appears 2 times. Therefore, 2 and 3 have the maximum frequency (5 each), making Option D correct.

Let n be total steps, v be escalator speed (steps/sec), w be walking speed (steps/sec). When walking 26 steps: time = 30 sec, escalator moves 30v steps, so n = 26 + 30v. When walking 34 steps: time = 18 sec, escalator moves 18v steps, so n = 34 + 18v. Solving: 26 + 30v = 34 + 18v → 12v = 8 → v = 2/3 steps/sec. Then n = 26 + 30(2/3) = 26 + 20 = 46 steps. Answer D is correct.

Let N = total visible steps and e = escalator speed (steps/sec). Going up: N = 50(1+e). Going down: N = 125(5-e). Equating: 50(1+e) = 125(5-e). 50+50e = 625-25e. 75e = 575, so e = 23/3. Then N = 50(1+23/3) = 50(26/3) = 1300/3... wait, let me verify. Actually: time up = N/(1+e) = 50, time down = N/(5-e) = 125. From these: N = 50(1+e) and N = 125(5-e). Setting equal: 50+50e = 625-25e, so 75e = 575, e = 23/3. This gives N = 50(26/3) = 1300/3, which isn't 100. Let me reconsider: The steps counted are the steps the person takes. Total visible steps when escalator is still is constant. Let that be S. When going up at 1 step/sec for 50 seconds, escalator has moved up e*50 steps. So person covered 50 steps and escalator contributed e*50 steps, total S = 50 + 50e. When going down at 5 steps/sec for 25 seconds (125 steps at 5/sec = 25 sec), escalator moves down e*25 steps. Person covers 125 steps down but escalator takes away e*25, so total S = 125 - 25e. Equating: 50 + 50e = 125 - 25e, so 75e = 75, e = 1. Then S = 50 + 50 = 100 steps.

This is the 'look-and-say' sequence: 1 is read as 'one 1' -> 11, 11 is 'two 1s' -> 21, 21 is 'one 2, one 1' -> 1211, etc. Continuing: 13112221 is 'one 1, one 3, two 1s, three 2s, one 1' -> 1113213211. The sequence describes the digits in the previous term.

Let the 5-digit number be N. If we put 1 at the beginning, we get 100000 + N. If we put 1 at the end, we get 10N + 1. According to the problem: (10N + 1) = 3 × (100000 + N). Solving: 10N + 1 = 300000 + 3N → 7N = 299999 → N = 42857. Verification: 142857 ÷ 428571 = 1/3. The other options don't satisfy this mathematical relationship.