Math & Puzzles: Progressions, Series & Word Problems

The series follows an arithmetic progression with a common difference of 0.5 (or 1/2). Starting from 0.5, the terms are 1.0, 1.5, and 2.0. Adding 0.5 to 2.0 yields 2.5. Option 90010 is correct because 2.5 fits the sequence before 3.0. Other options like 3 or 1/2 do not maintain the constant increment.

Working backwards: foxes = 1/3 deer, and wolves = foxes, so wolves:deer = 1:3. The other animals (elephants, tigers, rabbits) are extra information not needed for this ratio.

Let the numbers be $x$ and $y$. $(y-x):(y+x):xy = 1:7:24$. From $(y-x)/(y+x) = 1/7$, we get $7y-7x = y+x$, so $6y=8x$ or $y=4x/3$. Substituting into $(y+x)/xy = 7/24$ gives $(7x/3) / (4x^2/3) = 7/24$, which simplifies to $1/4x = 1/24$, so $x=6$. Then $y=8$. Product $xy = 48$.

Condition A says best and worst are the same age (twins). Condition B says the best singer's twin and the worst singer are opposite sex. If D and A are twins (son/daughter), and D is best, her twin A is male. If the worst is S (female), they are opposite sex. This fits the logic puzzles common in IQ tests.

Working backwards from the end (all have Rs 96), M had 216 after game 2. The game proceeds: K loses first, then L, then M, then N. Game 3 ends with everyone at 48, so game 2 ended with M=216, L=0, N=0, K=72.

Let Y's speed be $v$, so X's speed is $1.5v$. The relative speed is $0.5v$. To cover the 300m gap, it takes $300 / 0.5v = 600/v$ units of time. In that time, X travels distance = speed $ imes$ time = $1.5v imes (600/v) = 900$ meters.

Let $x$ be the number of years. Dhullar's salary: $500 + 25x$. Kalu's salary: $640 + 20x$. We need $500 + 25x > 640 + 20x$, which simplifies to $5x > 140$, or $x > 28$. Thus, in the 29th year, Dhullar's salary will cross Kalu's.

The series consists of 100 terms which can be grouped into 50 pairs: (1-6), (2-7), (3-8), and so on. Each pair simplifies to -5. Multiplying the value of one pair (-5) by the total number of pairs (50) results in -250. Other options are incorrect because they miscalculate the number of pairs or the value per pair.

The numbers form an arithmetic progression starting at 102 and ending at 396. Using $a=102$, $l=396$, and $d=6$, the number of terms $n$ is $\frac{396-102}{6} + 1 = 50$. The sum is $S_n = \frac{n}{2}(a + l) = \frac{50}{2}(102 + 396) = 25 \times 498 = 12450$. Other options result from calculation errors.

The series 1, 8, 15 is an arithmetic progression with common difference 7. Sum of n terms = n/2 × (2a + (n-1)d). For n=10: 10/2 × (2×1 + 9×7) = 5 × (2 + 63) = 5 × 65 = 325.

This is an arithmetic progression with first term $a = -12$ and common difference $d = 3$. Using the sum formula $S_n = \frac{n}{2}[2a + (n-1)d]$, we set $54 = \frac{n}{2}[-24 + (n-1)3]$. Simplifying gives $n^2 - 9n - 36 = 0$. Factoring $(n-12)(n+3) = 0$ yields $n=12$. Other options fail this quadratic equation.

Let the 5 numbers in AP be a-2d, a-d, a, a+d, a+2d. Their sum is 5a = 30, giving a = 6. The third term (middle term) is a = 6. Verifying with sum of squares: (a-2d)² + (a-d)² + a² + (a+d)² + (a+2d)² = 5a² + 10d² = 180 + 10d² = 220, so d² = 4. The third term remains 6 regardless of d.

If x, y, z are in GP, then y² = xz, so 2log y = log x + log z. Let a = log x, b = log y, c = log z. Then 2b = a + c. For the sequence 1/(1+a), 1/(1+b), 1/(1+c) to be in HP, their reciprocals (1+a), (1+b), (1+c) must be in AP. Since 2(1+b) = 2 + 2b = 2 + a + c = (1+a) + (1+c), the reciprocals are in AP, so the original sequence is in HP.

With only two terms given (1 and 4), we cannot uniquely determine the type of progression. They could be in AP (common difference 3), GP (common ratio 4), or HP (reciprocals 1 and 1/4 with common difference -3/4). At least three terms are required to definitively classify a progression.

For squares, perimeter P = 4s where s is the side length. If perimeters are in GP with common ratio r, then P₁, P₂, P₃... form a GP, meaning 4s₁, 4s₂, 4s₃... form a GP. Dividing each term by 4 gives s₁, s₂, s₃... which will also be in GP with the same common ratio r.

The pay scale notation '7000-500-12,500' indicates: starting salary 7000, annual increment 500, maximum ceiling 12,500. Number of years = (12500 - 7000) / 500 = 5500 / 500 = 11 years. After 11 increments, the salary reaches 7000 + (11 × 500) = 12,500.

The sequence follows the pattern 2n + 1. Starting with 3: 3×2+1=7, 7×2+1=15, 15×2+1=31. Therefore, 31×2+1=63, and 63×2+1=127, confirming the pattern.

Replacing symbols: 5 X (4 / 2) + 7 becomes 5 + (4 × 2) - 7. Following order of operations: 4 × 2 = 8, then 5 + 8 = 13, and 13 - 7 = 6. The key is maintaining correct precedence after symbol translation.

CAPITAL shifts forward: C→A (-2), A→Y (+2), P→N (-2), I→G (+2), T→R (-2), A→Y (+2), L→J (-2). The pattern alternates -2, +2. Applying this to FORGET: F→D (-2), O→M (-2), R→E (+2), G→E (-2), E→C (-2), T→R (-2), resulting in DMPECR.

CHEERS transforms to EHCSRE through a specific rearrangement: first half reversed (EH), second half reversed (SRE). Applying this to BASKET: first half (BAS) reverses to SAB, second half (KET) reverses to TEK, yielding SABTEK.