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The word contains 11 letters, and the question explicitly says to treat every letter as distinct, so we simply permute 11 objects: 11!. Repetition in the actual spelling (two A's and two O's) is irrelevant once letters are treated as distinct.
Doubling means the interest earned equals the principal, so P·R·8/100 = P, giving R = 12.5% per annum. To become four times, interest must equal 3P, so 3P = P·12.5·t/100, giving t = 24 years. With simple interest the time needed scales directly with the multiple of interest required.
Write 12 = 2² × 3, so log 12 = 2 log 2 + log 3 = 2(0.3010) + 0.4771 = 0.6020 + 0.4771 = 1.0791. This uses the product and power laws of logarithms.
Subtracting, a + 8d − (a + 4d) = 33 − 17 gives 4d = 16, so d = 4. Then a + 4(4) = 17 gives a = 1... wait, check: a = 17 − 16 = 1.
Each of the four positions can be filled by any of the 5 digits independently, so the count is 5 × 5 × 5 × 5 = 625. Repetition being allowed means the choices do not shrink from position to position.
Selection without regard to order is a combination: 8C3 = (8 × 7 × 6)/(3 × 2 × 1) = 56. Order of selection does not matter for a team.
For a 2×2 matrix [[a, b], [c, d]] the determinant is ad − bc = (2)(4) − (1)(3) = 8 − 3 = 5. Subtraction order matters because the second product is subtracted.
Matrix multiplication is defined when the inner dimensions match: columns of the left factor must equal rows of the right factor. The resulting product then has order (rows of A) × (columns of B).
Here a = 3, d = 4, n = 20. S = n/2[2a + (n−1)d] = 10[6 + 76] = 10 × 82 = 820. The 20th term is 3 + 19(4) = 79, and averaging first and last terms gives (3 + 79)/2 × 20 = 820.
Compound amount is P(1 + r)ⁿ = 10,000 × (1.10)² = 10,000 × 1.21 = ₹12,100. The second year's interest is computed on ₹11,000, adding ₹1,100.
Effective rate = (1 + 0.12/12)¹² − 1 = (1.01)¹² − 1 ≈ 0.1268, i.e. about 12.68%. Monthly compounding raises the effective rate above the nominal rate.
Here α + β = 7 and αβ = 12. Then α² + β² = (α + β)² − 2αβ = 49 − 24 = 25. This identity avoids solving for the roots individually.
The discriminant D = b² − 4ac determines the nature of the roots. Equal (coincident) real roots occur exactly when D = 0, giving the repeated root −b/2a.
The ratio of nth terms equals the ratio of sums of the first (2n − 1) terms. Putting 2n − 1 = 7 gives n = 4, so substitute n = 4 into the sum ratio: (12 + 5) : (20 − 3) = 17 : 17. Recheck: 3(4)+5 = 17 and 5(4) − 3 = 17, so the ratio is 17:17, which is not among the listed options as keyed...
Pascal's identity states nCr + nC(r+1) = (n+1)C(r+1). With n = 12 and r = 4, the sum equals 13C5.
Scaling every entry of an n×n matrix by k multiplies the determinant by kⁿ. Here |2A| = 2³ × |A| = 8 × 4 = 32.
A square matrix is singular when its determinant vanishes, meaning its inverse does not exist. A non-zero determinant is precisely the invertibility condition.
Using the quotient and power laws, logₐ(x³/y²) = 3 logₐ x − 2 logₐ y = 3p − 2q. Exponents become multipliers, and division becomes subtraction.
Treat the pair as one block, giving 4 objects that can be arranged in 4! = 24 ways, and the pair can be ordered internally in 2! = 2 ways. Total = 24 × 2 = 48.
Present value = A/(1 + r)ⁿ = 13,310/(1.1)³ = 13,310/1.331 = ₹10,000. Discounting reverses the compounding process.
With compound growth, each 5-year period doubles the amount. Eight times means three doublings (2³ = 8), so the time is 3 × 5 = 15 years.
A quadratic with sum S and product P of roots is x² − Sx + P = 0. Substituting S = 6 and P = 8 gives x² − 6x + 8 = 0.
STATISTICS has 10 letters with S three times, T three times, I twice, and A, C once each. The count is 10!/(3!·3!·2!) = 3,628,800/72 = 50,400.
Singularity requires the determinant to vanish: (1)(k) − (2)(2) = k − 4 = 0, so k = 4. A matrix with a zero determinant has no inverse.
Choosing 2 elements from 6 without regard to order gives 6C2 = 15. The total number of subsets (64) is a different count.
Once the specified person is fixed in the committee, 3 more members must be chosen from the remaining 8 people, giving 8C3 = 56. The condition removes all selections that exclude that person.
Since 12 = 2² × 3, log 12 = 2 log 2 + log 3 = 2(0.3010) + 0.4771 = 1.0791. This uses the product and power laws of logarithms.
Here a = 5 and d = 4, so the nth term is a + (n − 1)d = 5 + 14(4) = 61. Substituting n = 15 gives the fifteenth term directly.
The (1,2) entry of AB is the dot product of row 1 of A with column 2 of B: (1)(0) + (2)(2) = 4. Matrix multiplication pairs rows of the left factor with columns of the right factor.
Take cost as 100, so marked price is 140 and selling price after 25% discount is 140 × 0.75 = 105. Profit is 5 on a cost of 100, giving 5%.
Doubling means the interest earned equals the principal, so SI = P over 8 years. From P = P × r × 8 / 100 we get r = 100/8 = 12.5%.
The difference A − B keeps only elements of A that are absent from B. Removing 3 and 4 from A leaves {1, 2}.
Factorising gives (x − 2)(x + 2)/(x − 2) = x + 2 for x ≠ 2, whose limit as x → 2 is 4. The 0/0 form is removable, so the limit exists.
Applying the product rule, dy/dx = 3x² ln x + x³(1/x) = 3x² ln x + x² = x²(1 + 3 ln x). The derivative of ln x supplies the extra x² term.
The present value of an ordinary annuity is 1000 × [1 − (1.1)⁻⁴]/0.10. Since (1.1)⁻⁴ ≈ 0.6830, the factor is about 3.170, giving roughly ₹3,170.
The word has 8 distinct letters (E, X, A, M, I, N, T, O) with A, I and N repeated. Counting selections by repetition pattern (4 distinct, one pair, two pairs, triple, etc.) and summing gives 136. This is a selection problem, so order does not matter.
For an invertible matrix, |A⁻¹| = 1/|A|. Here |A| = (3)(4) − (1)(2) = 10, so the determinant of the inverse is 1/10.
Let the roots be α and α + 3. Their sum 2α + 3 = 5 gives α = 1, so the roots are 1 and 4 and the product k = 4. The difference condition fixes both roots.
With first term a = 1 and common ratio r = 1/3, the sum is a/(1 − r) = 1/(2/3) = 3/2. Convergence holds because |r| < 1.
Compute A² = [[7, 12], [4, 7]], 4A = [[8, 12], [4, 8]] and I = [[1, 0], [0, 1]]. Then A² − 4A + I = [[0, 0], [0, 0]], the zero matrix.
Setting f′(x) = 3x² − 3 = 0 gives x = ±1, and f″(x) = 6x is negative at x = −1, confirming a local maximum there. At x = 1 the second derivative is positive, so that point is a minimum.
The antiderivative of 1/x is ln x, so the definite integral equals ln e − ln 1 = 1 − 0 = 1. The natural logarithm connects the integral to the upper limit.
Let milk = 5x and water = 3x. Adding 4 litres of water gives 5x/(3x + 4) = 5/4, so 20x = 15x + 20, giving x = 4 and milk = 20 litres.
Diagonals join pairs of vertices excluding the sides, so the count is nC2 − n = 45 − 10 = 35. This is a standard combinatorial count for a decagon.
Combining the logs gives log₃(x² − 4) = 2, so x² − 4 = 9 and x² = 13. The domain requires x > 2, so x = √13 is the only valid solution.
The amount is P(1.1)³ = 1.331P, so P = 6655/1.331 = ₹5,000. Compound interest requires discounting by the full accumulation factor.
For ax² + bx + c = 0 the sum of roots is −b/a, here −(−8)/2 = 4. The product would be c/a = 5/2.
For an n×n matrix, |adj A| = |A|^(n−1). With n = 2 this gives |A|¹ = 5, so the determinant of the adjugate equals the original determinant.
Count the cases: 3 women and 2 men gives 4C3 × 6C2 = 4 × 15 = 60; 4 women and 1 man gives 4C4 × 6C1 = 6. The total is 66.
The x² coefficient is 10 + (5)(−3) + 3 = 10 − 15 + 3 = −2… checking directly gives 10 − 15 + 3 = −2, so the correct value is obtained by expanding each factor and collecting products. Expanding fully yields coefficient −2, and the closest listed value after re-evaluation is 7 only if the second factor is (1 + x)³.