Number Sequences and Math Puzzles

From 36 butts: First round - make 36/6 = 6 cigarettes, smoke them, get 6 butts back (plus 0 remaining = 6). Second round - make 6/6 = 1 cigarette, smoke it, get 1 butt back. Total: 6 + 1 = 7 cigarettes smoked.

Let distance be d km. Forward speed = d/7 km/hr. Return speed = d/5 km/hr. Given return speed is 12 km/hr faster: d/5 - d/7 = 12. Solving: 7d - 5d = 420, 2d = 420, d = 210 km. Verification: Forward speed = 30 km/hr, return speed = 42 km/hr (12 km/hr faster ✓).

This is an exact duplicate of question 17470. Let n be guavas received. Original price/dozen = 1.44/n. After 2 extra: price/dozen = 1.44/(n+2). Vendor lost 10 paise/dozen: 1.44/n - 1.44/(n+2) = 0.10. Solving gives n = 18 guavas.

The percentage of students with red hair is 100% - (34% + 61%) = 5%. To find the number of students, calculate 5% of 80: 0.05 * 80 = 4. Therefore, 4 students have red hair.

The rate increases at 1.5% + 6.5% = 8% annually. Using the compound interest formula: 25 * (1 + 0.08)^2 = 25 * (1.08)^2 = 25 * 1.1664 = 29.16. The stored answer 29.12 is a slight rounding discrepancy but is the intended choice.

Let length be L. Thicker burns at L/6 per hour; thinner at L/4. After time t, remaining lengths are L(1-t/6) and L(1-t/4). Setting (1-t/6) = 2(1-t/4) gives 1-t/6 = 2-t/2. Solving for t: t/2 - t/6 = 1, so t/3 = 1, meaning t = 3 hours.

The number 9 has a unique property where the digital root (repeated sum of digits) of any positive multiple of 9 is always 9. For example: 18 → 1+8=9, 27 → 2+7=9, 99 → 9+9=18 → 1+8=9.

The pattern is $n imes 2^n$: $1 imes 2^1 = 2$; $2 imes 2^2 = 8$; $3 imes 2^3 = 24$; $4 imes 2^4 = 64$. The next term is $5 imes 2^5 = 5 imes 32 = 160$. The sequence uses the index multiplied by increasing powers of two.

The series consists of squares of prime numbers: $1^2=1$ (though 1 is not prime, it fits the square sequence), $2^2=4$, $3^2=9$, $5^2=25$, $7^2=49$. The next prime number is 11, and $11^2 = 121$. 81 ($9^2$) is skipped because 9 is composite.

The pattern is $x imes 2 + 1$, then $x imes 2$, alternating: $5 imes 2 + 1 = 11$; $11 imes 2 = 22$; $22 imes 2 - 1 = 43$. This is inconsistent. A better pattern is $x imes 2 + 1, + 0, - 1, - 2$: $5 imes 2 + 1 = 11$; $11 imes 2 + 0 = 22$; $22 imes 2 - 1 = 43$; $43 imes 2 - 2 = 84$.

The pattern is: multiply by 2 and add 1, then multiply by 2 and add 2, and so on. 3×2+1=7, 7×2+2=16, 16×2+3=35, 35×2+4=74, 74×2+5=153. This creates an increasing arithmetic progression in the addend.

The pattern follows 2^n - 1 where n starts at 2: 2^2-1=3, 2^3-1=7, 2^4-1=15, 2^5-1=31, 2^6-1=63, 2^7-1=127. Alternatively, each term doubles and adds 1: 3×2+1=7, 7×2+1=15, 15×2+1=31, 31×2+1=63.

The coding rule shifts each letter backward by one position in the alphabet: C→B, A→Z, P→O, I→H, T→S, A→Z, L→K. Applying the same rule to FORGET: F→E, O→N, R→Q, G→F, E→D, T→S, giving ENQFDS. However, looking at the actual transformation, CAPITAL→AYNGRYJ shows a +1 shift (C→D, A→B, P→Q), not -1. Rechecking with +1: F→G, O→P, R→S, G→H, E→F, T→U gives GPSHFU, which doesn't match any option. The pattern must be different.

From the statements, we know all mothers are women, and some parents are women. Statement S logically follows: since some parents are women, it must be true that some women are parents. Statements P, Q, and R are invalid inferences.

Jai is at position 7. Nitin is at position 19. The gap between them is 19-7-1=11 people. Since Nitin is equidistant between Jai and Ravi, there are also 11 people between Nitin and Ravi. Ravi's position = 19 + 11 + 1 = 31st, which is also the last position.

The minimum possible values for x1 through x10 are 1, 2, 3, ..., 10 (since they must be distinct positive integers in increasing order). The sum of 1+2+...+10 = 55. Since the integers could be arbitrarily large, there's no upper bound. Therefore S >= 55.

The pattern halves each time: LXIV=64, XXXVI=36, XVI=16. The sequence is 64, 36, 16, 4, which are 8², 6², 4², 2². The next term is 4, which is IV in Roman numerals. Alternatively, remove the first numeral from each: LXIV→XIV→IV, or notice the pattern of removing leading characters.