CDS II 2018 Elementary Mathematics Question Paper with Answers & Explanations

Explanation

The LCM of 16, 36, 45, and 48 is calculated as follows: 16=2^4, 36=2^23^2, 45=3^25, 48=2^43. LCM = 2^4 3^2 5 = 16 9 5 = 720. The highest four-digit number is 9999. Dividing 9999 by 720 gives 13 with a remainder of 639. The highest four-digit number divisible by 720 is 720 13 = 9360.

Explanation

Given the equations x = yᵃ, y = zᵇ, and z = xᶜ, we can find the value of abc by using the method of successive substitution. Starting with the first equation, we have x = yᵃ. Substituting the value of y from the second equation, y = zᵇ, into this expression gives x = (zᵇ)ᵃ. According to the laws of exponents, (xᵐ)ⁿ = xᵐⁿ, so this simplifies to x = zᵃᵇ. Now, substituting the value of z from the third equation, z = xᶜ, into the equation we just derived, we get x = (xᶜ)ᵃᵇ. Applying the power rule again, we have x¹ = xᵃᵇᶜ. For this equality to hold for all positive values of x, the exponents must be equal. Therefore, 1 = abc. This proves that the value of abc is 1, which corresponds to option a.

Explanation

Let a = 2^(1/3), then x = 2 + a^2 + a. This implies x - 2 = a^2 + a. Cubing both sides: (x - 2)^3 = (a^2 + a)^3 = a^6 + 3a^5 + 3a^4 + a^3. Substituting a^3 = 2, a^4 = 2a, a^5 = 2a^2, and a^6 = 4, we get (x - 2)^3 = 4 + 3(2a^2) + 3(2a) + 2 = 6 + 6a^2 + 6a = 6(1 + a^2 + a). Since x = 2 + a^2 + a, then a^2 + a = x - 2, so (x - 2)^3 = 6(1 + x - 2) = 6(x - 1). Expanding this, x^3 - 6x^2 + 12x - 8 = 6x - 6, which simplifies to x^3 - 6x^2 + 6x = 2.

Explanation

A number is divisible by 33 if it is divisible by both 3 and 11. For the five digit number of the form XXYXX, let the digits be X and Y. The condition for divisibility by 11 is that the alternating sum of its digits must be divisible by 11. The alternating sum is X - X + Y - X + X = Y. Since Y is a single digit from 0 to 9, the only value that satisfies Y ≡ 0 (mod 11) is Y = 0.

Next, the condition for divisibility by 3 is that the sum of the digits must be divisible by 3. The sum of the digits is X + X + Y + X + X = 4X + Y. Substituting Y = 0 into this expression, we get 4X + 0 = 4X. For 4X to be divisible by 3, X must be a multiple of 3 because 4 and 3 are coprime. Since X is the leading digit of a five digit number, X cannot be 0. Thus, the possible values for X from the set {1, 2, ..., 9} are 3, 6, and 9. This gives us three possible numbers: 33033, 66066, and 99099. Therefore, there are 3 such numbers, which corresponds to option b.

Explanation

For XY235 to be divisible by 3, the sum of its digits (X + Y + 2 + 3 + 5 = X + Y + 10) must be divisible by 3. This means X + Y must be 2, 5, 8, etc. Given X + Y ≤ 5 and X is a leading digit (X ≥ 1), the possibilities are: if X + Y = 2, pairs (X, Y) are (1, 1) and (2, 0); if X + Y = 5, pairs (X, Y) are (1, 4), (2, 3), (3, 2), (4, 1), and (5, 0). Total possible pairs = 2 + 5 = 7.

Explanation

The inequality x^2 - 6x - 27 > 0 can be factored as (x - 9)(x + 3) > 0. The roots of the quadratic are x = 9 and x = -3. For the product to be positive, x must lie outside the interval [-3, 9]. Therefore, x > 9 or x < -3.

Explanation

Prime factorization of 38808: 38808 = 2^3 3^2 7^2 11^1. Total number of divisors = (3+1)(2+1)(2+1)(1+1) = 4 3 3 2 = 72. Exclusive of 1 and itself, the number of divisors is 72 - 2 = 70.

Explanation

The relationship between two polynomials P₁(x) and P₂(x), their Highest Common Factor (HCF), and Least Common Multiple (LCM) is given by the formula: P₁(x) × P₂(x) = HCF × LCM

Given: P₁(x) = x² - 4x - 21 HCF = x + 3 LCM = x³ - 9x² - x + 105

First, factorize P₁(x) to simplify the calculation: x² - 4x - 21 = (x - 7)(x + 3)

Now, substitute these into the formula to find P₂(x): P₂(x) = (HCF × LCM)/P₁(x) = ((x + 3)(x³ - 9x² - x + 105))/((x - 7)(x + 3)) P₂(x) = (x³ - 9x² - x + 105)/(x - 7)

Performing polynomial division of (x³ - 9x² - x + 105) by (x - 7): x²(x - 7) = x³ - 7x², leaving a remainder of -2x² - x + 105 -2x(x - 7) = -2x² + 14x, leaving a remainder of -15x + 105 -15(x - 7) = -15x + 105, leaving a remainder of 0

The quotient is x² - 2x - 15, which corresponds to option (c).

Explanation

Substitute the given values α + β = -q/p and αβ = r/p into the expressions provided in the options. For option (a), 1/(α + β) = 1/(-q/p) = -p/q. For option (b), 1/α + 1/β = (α + β)/(αβ) = (-q/p)/(r/p) = -q/r. For option (c), -1/(αβ) = -1/(r/p) = -p/r. For option (d), (αβ)/(α + β) = (r/p)/(-q/p) = -r/q.

Given the condition 1 < p < q < r, we evaluate the magnitude of these four negative values. We compare the positive ratios p/q, q/r, p/r, and r/q. Since p is the smallest and r is the largest of the three variables, the fraction p/r has the smallest numerator and the largest denominator, making it the smallest positive value among those less than 1. Specifically, because p < q, then p/r < q/r, and because q < r, then p/r < p/q. Additionally, r/q > 1, so it is the largest magnitude.

In the case of negative numbers, the value with the smallest absolute magnitude is the greatest. Since p/r is the smallest magnitude, -p/r is the greatest value. This corresponds to option (c).

Explanation

Let the efficiencies of A and B (work done per day) be a and b respectively. The total work W is given by the formula Work = Efficiency × Time.

From the first condition, working together they complete the job in 5 days: W = (a + b) × 5

From the second condition, where A is twice as efficient and B is one-third as efficient, they complete the job in 3 days: W = (2a + 1/3b) × 3

Equating the two expressions for W: 5(a + b) = 3(2a + 1/3b) 5a + 5b = 6a + b 4b = a

This relationship indicates that A is 4 times as efficient as B. To find the total work in terms of b, substitute a = 4b into the first equation: W = (4b + b) × 5 = 5b × 5 = 25b

The time taken by A alone to complete the job is: Time = W/a = 25b/4b = 25/4 = 6 1/4 days.

Thus, option (d) is the correct answer.

Explanation

Using the identity a^3 + b^3 = (a + b)(a^2 - ab + b^2), let a = x^2 and b = 1/x^2. Then x^6 + 1/x^6 = (x^2 + 1/x^2)((x^2)^2 - (x^2)(1/x^2) + (1/x^2)^2) = (x^2 + 1/x^2)(x^4 - 1 + 1/x^4). Comparing this with x^6 + 1/x^6 = k(x^2 + 1/x^2), we find k = x^4 - 1 + 1/x^4.

Explanation

Let the three consecutive natural numbers be n-1, n, and n+1. Then (n-1)^2 + n^2 + (n+1)^2 = 110, which simplifies to 3n^2 + 2 = 110, so 3n^2 = 108, n^2 = 36, and n = 6. The numbers are 5, 6, and 7. The sum of their cubes is 5^3 + 6^3 + 7^3 = 125 + 216 + 343 = 684.

Explanation

Let the integers be p = 16a and q = 16b, where a and b are coprime. p*q = 256ab = 7168, so ab = 28. Possible coprime pairs (a, b) are (1, 28) and (4, 7). If a=1 and b=28, p=16 (not > 60). If a=4 and b=7, p=64 and q=112 (both > 60). The sum p + q = 64 + 112 = 176.

Explanation

To find the value of log₁₀₀(72), we use the base change formula logᵦ a = (log꜀ a)/(log꜀ b). By choosing the new base c = 10, the expression becomes log₁₀₀(72) = (log₁₀ 72)/(log₁₀ 100). First, we calculate the denominator: log₁₀ 100 = log₁₀ 10² = 2 log₁₀ 10 = 2 × 1 = 2. Next, we evaluate the numerator log₁₀ 72. Since 72 can be factored into prime powers as 2³ × 3², we apply the product and power rules of logarithms: log₁₀ 72 = log₁₀ (2³ × 3²) = 3 log₁₀ 2 + 2 log₁₀ 3. Substituting the given values log₁₀ 2 = 0.3010 and log₁₀ 3 = 0.4771, we get log₁₀ 72 = 3(0.3010) + 2(0.4771) = 0.9030 + 0.9542 = 1.8572. Finally, we divide the result by the denominator: log₁₀₀(72) = 1.8572/2 = 0.9286. This matches option a.

Explanation

Let a^x = b^y = c^z = k. Then a = k^(1/x), b = k^(1/y), and c = k^(1/z). Given abc = 1, we have k^(1/x + 1/y + 1/z) = 1. Since k^0 = 1, it follows that 1/x + 1/y + 1/z = 0.

Explanation

For a quadratic equation ax² + bx + c = 0 with roots α and β, the sum of roots is α + β = -b/a and the product of roots is αβ = c/a. We need to find the value of the expression 1/(aα + b) + 1/(aβ + b).

Combining the fractions by taking a common denominator, we get: ((aβ + b) + (aα + b))/((aα + b)(aβ + b)) = (a(α + β) + 2b)/(a²αβ + ab(α + β) + b²)

Now, substitute the known values of the sum and product of the roots into the numerator and denominator: Numerator: a(-b/a) + 2b = -b + 2b = b Denominator: a²(c/a) + ab(-b/a) + b² = ac - b² + b² = ac

Therefore, the expression simplifies to b/ac, which corresponds to option (b).

Explanation

The three 3-digit numbers can be expressed in expanded form as follows: XYZ = 100X + 10Y + Z, YZX = 100Y + 10Z + X, and ZXY = 100Z + 10X + Y. To find the sum of these three numbers, we add the expressions together: Sum = (100X + 10Y + Z) + (100Y + 10Z + X) + (100Z + 10X + Y) Sum = (100X + 10X + X) + (100Y + 10Y + Y) + (100Z + 10Z + Z) Sum = 111X + 111Y + 111Z Sum = 111(X + Y + Z)

From this result, we can evaluate the statements. Statement 2 is correct because the sum is a multiple of 111, meaning it is perfectly divisible by 111. Statement 1 is incorrect because the sum is also a multiple of (X + Y + Z), meaning it is divisible by (X + Y + Z). Since only statement 2 is correct, the answer is option (b).

Explanation

The equation 1/m + 1/n - 1/mn = 2/5 can be rewritten as (m + n - 1) / mn = 2/5, which simplifies to 2mn - 5m - 5n = -5. Multiplying by 2 gives 4mn - 10m - 10n = -10. Adding 25 to both sides allows factoring: (2m - 5)(2n - 5) = 15. The factor pairs of 15 are (1, 15), (3, 5), (5, 3), and (15, 1). These lead to (m, n) pairs: (3, 10), (4, 5), (5, 4), and (10, 3). There are 4 such pairs of positive integers.

Explanation

To find the value of the expression aᵍ⁻ʳbʳ⁻ᵖcᵖ⁻ᵍ, we substitute the given values of a, b, and c into the expression.

Given: a = xyᵖ⁻¹ b = yzᵍ⁻¹ c = zxʳ⁻¹

The target expression is E = aᵍ⁻ʳbʳ⁻ᵖcᵖ⁻ᵍ. Substituting the values: E = (xyᵖ⁻¹)ᵍ⁻ʳ (yzᵍ⁻¹)ʳ⁻ᵖ (zxʳ⁻¹)ᵖ⁻ᵍ

Using the laws of indices (uv)ⁿ = uⁿvⁿ and (uᵐ)ⁿ = uᵐⁿ, we expand each term: E = xᵍ⁻ʳ (yᵖ⁻¹)ᵍ⁻ʳ · yʳ⁻ᵖ (zᵍ⁻¹)ʳ⁻ᵖ · zᵖ⁻ᵍ (xʳ⁻¹)ᵖ⁻ᵍ E = xᵍ⁻ʳ y⁽ᵖ⁻¹⁾⁽ᵍ⁻ʳ⁾ yʳ⁻ᵖ z⁽ᵍ⁻¹⁾⁽ʳ⁻ᵖ⁾ zᵖ⁻ᵍ x⁽ʳ⁻¹⁾⁽ᵖ⁻ᵍ⁾

Now, we group the powers of x, y, and z using the rule uᵐ · uⁿ = uᵐ⁺ⁿ: Exponent of x = (q-r) + (r-1)(p-q) = q - r + rp - rq - p + q = 2q - r - p + rp - rq Exponent of y = (p-1)(q-r) + (r-p) = pq - pr - q + r + r - p = pq - pr - q + 2r - p Exponent of z = (q-1)(r-p) + (p-q) = qr - qp - r + p + p - q = qr - qp - r + 2p - q

The simplified expression is x²ᵍ⁻ʳ⁻ᵖ⁺ʳᵖ⁻ʳᵍ yᵖᵍ⁻ᵖʳ⁻ᵍ⁺²ʳ⁻ᵖ zᵍʳ⁻ᵍᵖ⁻ʳ⁺²ᵖ⁻ᵍ.

Comparing this result with the options: Option a) abc = (xyᵖ⁻¹)(yzᵍ⁻¹)(zxʳ⁻¹) = xʳ yᵖ zᵍ Option b) xyz = x¹ y¹ z¹ Option c) 0

The derived expression does not match any of the options a, b, or c. Therefore, the correct option is d) None of the above.

Explanation

Let the number of sides be 5x and 4x. The interior angle of a regular polygon is ((n-2)180)/n. The difference is ((5x-2)180/5x) - ((4x-2)*180/4x) = 9. Simplifying: 180(1 - 2/5x) - 180(1 - 2/4x) = 9, which leads to 180(2/4x - 2/5x) = 9, so 20(1/2x - 1/2.5x) = 1, and 20(0.5/x - 0.4/x) = 1, so 2/x = 1, which means x = 2. The polygons have 10 and 8 sides (decagon and octagon), making statement 2 correct. Statement 3 is correct as the sum of exterior angles of any two polygons is 360 + 360 = 720 degrees. Statement 1 is incorrect.

Explanation

For the quadratic 2x^2 + 5x + 5, the minimum value occurs at x = -b/2a = -5/(2*2) = -5/4. Substituting this value back: 2(-5/4)^2 + 5(-5/4) + 5 = 2(25/16) - 25/4 + 5 = 25/8 - 50/8 + 40/8 = 15/8.

Explanation

The harmonic mean H of two positive numbers P and Q is given by the formula H = 2PQ/(P+Q). To find the value of the expression H/P + H/Q, we first factor out H from the sum to get H(1/P + 1/Q). By finding a common denominator for the fractions inside the bracket, the expression becomes H((Q+P)/PQ). Now, substitute the formula for the harmonic mean into this expression: 2PQ/(P+Q) × (P+Q)/PQ. Upon simplification, the term (P+Q) in the numerator cancels with (P+Q) in the denominator, and PQ in the numerator cancels with PQ in the denominator. This leaves only the constant 2. Thus, the value of H/P + H/Q is 2, which matches option (b).

Explanation

To find the sum of all possible products of a set of numbers taken two at a time, we use the algebraic identity: Σ₁ ≤ ᵢ < ⱼ ≤ ₙ aᵢ aⱼ = 1/2 [(Σᵢ₌₁ⁿ aᵢ)² - Σᵢ₌₁ⁿ aᵢ²]

The given set of numbers is {-1, 1, -2, 2, -3, 3, -4, 4}. First, calculate the sum of the numbers: Σ aᵢ = (-1 + 1) + (-2 + 2) + (-3 + 3) + (-4 + 4) = 0

Next, calculate the sum of the squares of these numbers: Σ aᵢ² = (-1)² + 1² + (-2)² + 2² + (-3)² + 3² + (-4)² + 4² Σ aᵢ² = 1 + 1 + 4 + 4 + 9 + 9 + 16 + 16 = 60

Now, substitute these values into the formula: Sum of products = 1/2 [0² - 60] = 1/2 [-60] = -30

Therefore, the sum of all possible products taken two at a time is -30, which corresponds to option b.

Explanation

By the Remainder Theorem, the remainder when P(x) is divided by (x - 2y) is P(2y). Substituting x = 2y into 3x^3 - 2x^2y - 13xy^2 + 10y^3: 3(2y)^3 - 2(2y)^2y - 13(2y)y^2 + 10y^3 = 24y^3 - 8y^3 - 26y^3 + 10y^3 = 0.

Explanation

Let X = a^2 - bc, Y = b^2 - ca, and Z = c^2 - ab. The expression is (YZ + XZ + XY) / XYZ = 1/X + 1/Y + 1/Z. Given ab + bc + ca = 0, testing with a=2, b=2, c=-1 (which satisfies the condition), we get X=6, Y=6, Z=-3. Then 1/6 + 1/6 - 1/3 = 0.

Explanation

To find the principal amount when the compound interest for a specific year is given, we use the formula for interest earned in the n-th year: Interest for n-th year = P(1 + R/100)ⁿ⁻¹ × R/100. For the second year (n=2), the formula becomes CI for 2nd year = P(1 + R/100) × R/100. Substituting the given values into the equation, we get 210 = P(1 + 5/100) × 5/100. This simplifies to 210 = P(1.05) × 0.05, which further simplifies to 210 = P × 0.0525. Solving for P, we calculate P = 210/0.0525 = 4000. Alternatively, using a shortcut, the interest for the second year is the interest on the principal plus the interest on the first year's interest. Interest for the 1st year is 5P/100 = P/20. The interest for the 2nd year is therefore P/20 + (P/20 × 5/100) = P/20 + P/400 = 21P/400. Setting 21P/400 = 210, we find P = 210 × 400/21 = 10 × 400 = 4000. Thus, option (c) is the correct value.

Explanation

To solve this problem, we use the principle of inclusion-exclusion from set theory. Let E represent the percentage of candidates who failed in English and H represent the percentage of candidates who failed in Hindi. The given data is n(E) = 50%, n(H) = 40%, and the intersection n(E ∩ H) = 15%, which represents those who failed in both.

The percentage of candidates who failed in at least one of the two subjects is given by the union formula n(E ∪ H) = n(E) + n(H) - n(E ∩ H). Substituting the given values into the formula, we get n(E ∪ H) = 50% + 40% - 15% = 75%.

Since 75% of the candidates failed in at least one subject (either English, Hindi, or both), the percentage of candidates who passed in both subjects is the complement of this union. Therefore, the percentage of candidates who passed in both English and Hindi is 100% - n(E ∪ H) = 100% - 75% = 25%. This corresponds to option (b).

Explanation

To calculate the speed of the train, we use the fundamental relation Speed = Total Distance/Time. When a train passes a platform, the total distance covered is the sum of the length of the train and the length of the platform.

Given: Length of train = 100 m Length of platform = 100 m Total distance = 100 m + 100 m = 200 m Time taken = 10 s

First, find the speed in meters per second: Speed = 200 m/10 s = 20 m/s

To convert the speed from m/s to kmph, multiply the value by 18/5: Speed in kmph = 20 × 18/5 Speed in kmph = 4 × 18 = 72 kmph

Thus, the speed of the train is 72 kmph, which corresponds to option d).

Explanation

To find the average speed for the entire journey, we use the standard formula for average speed: Average Speed = Total Distance / Total Time.

First, calculate the total distance covered by the cyclist: Total Distance = 20 km + 10 km + 30 km = 60 km.

Next, calculate the time taken for each segment of the journey using the formula Time = Distance / Speed: Time for the first part t₁ = 20/40 = 0.5 hours. Time for the second part t₂ = 10/10 = 1 hour. Time for the third part t₃ = 30/40 = 0.75 hours.

The total time taken for the entire journey is: Total Time = 0.5 + 1 + 0.75 = 2.25 hours.

Now, substitute these values into the average speed formula: Average Speed = 60/2.25 = 60/(9/4) = (60 × 4)/9 = 240/9 = 80/3 ≈ 26.67 kmph.

Thus, the average speed of the entire journey is 26.67 kmph, which corresponds to option b.

Explanation

The speed of a runner is proportional to the distance covered in a given time. From the first race, when A covers 1000 m, B covers 1000 - 150 = 850 m. Thus, the ratio of their speeds is Vₐ / Vᵦ = 1000 / 850 = 20 / 17. From the second race, when C covers 3000 m, D covers 3000 - 400 = 2600 m. Thus, the ratio of their speeds is V_C / V_D = 3000 / 2600 = 15 / 13.

Since the speed of B is equal to that of D (Vᵦ = V_D), we can find the ratio of the speeds of A and C as follows: Vₐ / V_C = (Vₐ / Vᵦ) / (V_C / V_D) = (20 / 17) / (15 / 13) = (20 / 17) × (13 / 15) = 52 / 51.

In a race of 6000 m, let D_C be the distance covered by C when A completes the race. Using the speed ratio: D_C = 6000 × (V_C / Vₐ) = 6000 × (51 / 52) = 1500 × (51 / 13) = 76500 / 13 ≈ 5884.62 m.

The distance by which A beats C is: 6000 - 5884.62 = 115.38 m.

Therefore, A beats C by 115.38 m, which matches option c.

Explanation

Let the present age of Sonu be x years. There are 5 members in the family: father, mother, Sonu, Savita, and Sonia. The current sum of their ages is 96 years. When Sonu was born x years ago, the other 4 members of the family were each x years younger, and Sonu's age was 0.

The formula for the sum of ages x years ago is: Sum then = (Present Sum) - (Number of members × x) 66 = 96 - (5 × x) 5x = 96 - 66 5x = 30 x = 6

Sonu's present age is 6 years. According to the given condition, the father's present age is 6 times that of Sonu's present age: Father's present age = 6 × 6 = 36 years.

To find the father's age 12 years hence, we add 12 to his current age: Father's age after 12 years = 36 + 12 = 48 years.

Thus, option d is correct.

Explanation

Let the time taken by B to complete the work be T days. Then A takes T/3 days. Given T - T/3 = 10, we have 2T/3 = 10, so T = 15.

Explanation

Total children = 85. Boys (B) + Girls (G) = 85. G = 0.7B, so 1.7B = 85, B = 50 and G = 35. Boys playing badminton = 0.6 50 = 30. Boys playing only badminton = 0.5 50 = 25. Thus, boys playing both = 30 - 25 = 5. Total children playing both = 12, so girls playing both = 12 - 5 = 7. Children playing only table tennis = 0.4 * 85 = 34. Boys playing only table tennis = Total Boys - Boys playing badminton = 50 - 30 = 20. Girls playing only table tennis = 34 - 20 = 14. Girls playing only badminton = Total Girls - (Girls playing both + Girls playing only table tennis) = 35 - (7 + 14) = 14.

Explanation

Let X and Y be the prices before tax. X + Y = 130. Total amount = X + Y + 0.09Y = 136.75. Substituting X + Y = 130: 130 + 0.09Y = 136.75, so 0.09Y = 6.75, and Y = 6.75 / 0.09 = 75.

Explanation

Let total property be P. Wife gets P/2, Ravi gets P/4, Raj gets P/4. Ravi dies: Widow gets P/8, Raj gets P/8. Raj's total = P/4 + P/8 = 3P/8. Raj dies: Widow gets 3P/16, Mother gets 3P/16. Mother's total = P/2 + 3P/16 = 11P/16. 11P/16 = 88,000, so P = (88,000 16) / 11 = 8,000 16 = 128,000.

Explanation

Let cost of one lemon bottle = L. Orange bottle = 2L. Total cost = 4L + 2L = 6L. Three people (X, Y, Z) share equally, and Z pays ₹ 50 for his share, so each share = ₹ 50. Total = 3 × 50 = ₹ 150 ⇒ 6L = 150 ⇒ L = 25. Cost of one orange bottle = 2L = ₹ 50.

Explanation

Let the ages of the mother and daughter ten years ago be 3x and x respectively. Their present ages can therefore be represented as 3x + 10 and x + 10. In another ten years, their ages will be (3x + 10) + 10 = 3x + 20 and (x + 10) + 10 = x + 20. According to the given condition, the ratio of these future ages is 13:7.

We can set up the equation as (3x + 20)/(x + 20) = 13/7. Cross-multiplying the terms, we get 7(3x + 20) = 13(x + 20). Expanding the brackets gives 21x + 140 = 13x + 260. Rearranging the terms to solve for x, we have 21x - 13x = 260 - 140, which simplifies to 8x = 120, resulting in x = 15.

Substituting the value of x back into the expressions for present ages, the mother's present age is 3(15) + 10 = 45 + 10 = 55 years, and the daughter's present age is 15 + 10 = 25 years. This corresponds to option b.

Explanation

Total boys = 60. Boys playing chess (C) = 45, Boys playing carrom (K) = 30. Since every boy plays at least one: n(C ∪ K) = n(C) + n(K) - n(C ∩ K). 60 = 45 + 30 - n(C ∩ K), so n(C ∩ K) = 15. Boys playing carrom only = n(K) - n(C ∩ K) = 30 - 15 = 15.

Explanation

Let the equal amount borrowed be P. Total Interest = (P 5 4 / 100) + (P 4 4 / 100) = 405. (20P + 16P) / 100 = 405, so 36P = 40500, P = 1125. The option refers to the principal P.

Explanation

This problem can be solved using the Chain Rule formula for work and time, which states that the total man-hours required to complete a task remains constant regardless of the distribution of labor and time. The formula is given by M₁ · D₁ · H₁ = M₂ · D₂ · H₂, where M represents the number of men, D represents the number of days, and H represents the number of hours worked per day.

Substituting the given values from the first scenario: M₁ = 12 men, D₁ = 10 days, and H₁ = 8 hours per day. For the second scenario: M₂ = 8 men, D₂ = 8 days, and H₂ is the unknown number of hours per day.

The calculation is as follows: 12 × 10 × 8 = 8 × 8 × H₂ 960 = 64 × H₂ H₂ = 960/64 H₂ = 15

Therefore, the 8 men must work 15 hours per day to complete the wall in 8 days, which corresponds to option (c).

Explanation

When a vendor sells a mixture of milk and water at the cost price of pure milk, the profit is derived solely from the added water. The shortcut formula for this scenario is: Profit % = (Quantity of water)/(Quantity of milk) × 100

Given that the profit is 12.5%, the quantity of milk is 28 litres, and letting the quantity of water be Vᵥᵥ, we substitute these values into the formula: 12.5 = (Vᵥᵥ/28) × 100

Rearranging to solve for Vᵥᵥ: Vᵥᵥ = (12.5 × 28)/100 Vᵥᵥ = 0.125 × 28

Recognizing that 0.125 is equivalent to the fraction 1/8: Vᵥᵥ = 1/8 × 28 Vᵥᵥ = 28/8 = 3.5 litres

Thus, the amount of water added is 3.5 litres, which corresponds to option (c).

Explanation

A correct clock's hands meet every 65 5/11 (720/11) minutes. This clock meets every 72 minutes, so it is slow. Loss per 72 minutes = 72 - 720/11 = 72/11 minutes. Loss in 24 hours = (72/11) * (1440 / 72) = 1440/11 = 130 10/11 minutes. Note: The answer key states A (121 9/11), but the calculation for T=72 yields 130 10/11.

Explanation

To solve this problem, we use the concept of relative speed. The thief travels at vₜ = 40 kmph and the owner at vₒ = 60 kmph. The thief has a head start of 0.5 hours.

The initial distance gap created by the thief is: Distance gap = vₜ × time = 40 kmph × 0.5 h = 20 km.

The relative speed of the owner with respect to the thief is: Relative speed = vₒ - vₜ = 60 kmph - 40 kmph = 20 kmph.

The time taken by the owner to bridge this gap and catch the thief is: t꜀ₐₜ꜀ₕ = Distance gap/Relative speed = 20 km/20 kmph = 1 hour.

The total distance from the house where the meeting occurs is: Distance = vₒ × t꜀ₐₜ꜀ₕ = 60 kmph × 1 h = 60 km.

The total time elapsed from the moment of the theft is: Total time = Head start + t꜀ₐₜ꜀ₕ = 0.5 h + 1 h = 1.5 hours.

Thus, the owner meets the thief 60 km from the house and 1.5 hours after the theft, which corresponds to option (b).

Explanation

To find the time taken by Y alone to complete the job, we use the standard work-rate formula for two people working together. Let Tₓᵧ be the time taken by X and Y together and Tₓ be the time taken by X alone. The time taken by Y alone, Tᵧ, can be calculated using the formula:

Tᵧ = (Tₓᵧ × Tₓ)/(Tₓ - Tₓᵧ)

Substituting the given values, where Tₓᵧ = 6 days and Tₓ = 12 days:

Tᵧ = (6 × 12)/(12 - 6) Tᵧ = 72/6 Tᵧ = 12 days

Alternatively, using the efficiency method, if we assume the total work is the LCM of 6 and 12, which is 12 units, the combined efficiency of X and Y is 12 / 6 = 2 units per day. The efficiency of X alone is 12 / 12 = 1 unit per day. Therefore, the efficiency of Y is 2 - 1 = 1 unit per day. The time taken by Y alone is 12 / 1 = 12 days. This matches option b.

Explanation

Using the formula (M1 D1) / W1 = (M2 D2) / W2: (12 16) / 10 = (8 D2) / 20. 192 / 10 = 8D2 / 20, so 19.2 = 0.4D2, and D2 = 19.2 / 0.4 = 48.

Explanation

(0.2)^25 = (2 10^-1)^25 = 2^25 10^-25. Log10(2^25) = 25 0.30103 = 7.52575. Thus, 2^25 = 10^7.52575 = 10^0.52575 10^7. The value is 10^0.52575 10^7 10^-25 = 10^0.52575 * 10^-18. Since 10^0.52575 is between 1 and 10, the number of zeros after the decimal point is 18 - 1 = 17.

Explanation

Using the property of Componendo and Dividendo, if (a+b)/(a-b) = x/y, then a/b = (x+y)/(x-y). Given the ratio of the sum and difference of the ages of the father F and the son S is 11:3, we have (F+S)/(F-S) = 11/3. Applying the formula, the ratio of their present ages is F/S = (11+3)/(11-3) = 14/8 = 7/4. Since the ratio is 7:4 and not 8:5, statement 1 is incorrect.

For statement 2, let the present ages be F = 7x and S = 4x. The son attains twice his present age when he reaches 2 × 4x = 8x. This occurs after a time interval of 8x - 4x = 4x years. At that time, the father's age will be 7x + 4x = 11x. The ratio of their ages at that time will be 11x/8x = 11/8. Thus, statement 2 is correct. Therefore, only option b is the right choice.

Explanation

The inequalities are y ≥ 5 - x and y ≥ x - 3. For y to be negative, we would need x > 5 and x < 3 simultaneously, which is impossible. Thus, y must be positive. For any positive y, there are values of x in both the first and second quadrants that satisfy these conditions (e.g., (4,1) in first, (-1,10) in second).

Explanation

The equation x² - y² = 0 factors as (x - y)(x + y) = 0, which represents two straight lines, x - y = 0 and x + y = 0, passing through the origin. The second equation (x - a)² + y² = 1 represents a circle with center (a, 0) and radius r = 1. For the system to have a single positive solution, the circle must be tangent to the line x - y = 0 in the first quadrant.

The condition for tangency is that the perpendicular distance from the center of the circle (a, 0) to the line x - y = 0 must be equal to the radius of the circle. Using the distance formula d = |Ax₀ + By₀ + C|/√(A² + B²), we substitute the coefficients of the line A = 1, B = -1, C = 0 and the center (a, 0):

1 = |1(a) - 1(0) + 0|/√(1² + (-1)²)

1 = |a|/√2

|a| = √2

This gives a = ±√2. Since the problem specifies a positive solution, the center of the circle must be on the positive x-axis, so we take a = √2. This corresponds to option (a).

Explanation

For a cubic polynomial f(x) = ax³ + bx² + cx + d with zeros α, β, and γ, we use the relations between the coefficients and the roots (Vieta's formulas). The sum of the roots is α + β + γ = -b/a and the sum of the products of the roots taken two at a time is αβ + βγ + γα = c/a.

To find the sum of the squares of the roots, we use the algebraic identity: (α + β + γ)² = α² + β² + γ² + 2(αβ + βγ + γα)

Rearranging the formula to isolate the sum of squares, we get: α² + β² + γ² = (α + β + γ)² - 2(αβ + βγ + γα)

Substituting the values from Vieta's formulas: α² + β² + γ² = (-b/a)² - 2(c/a) α² + β² + γ² = b²/a² - 2c/a

To combine these terms over a common denominator a²: α² + β² + γ² = (b² - 2ac) / a²

This matches option d, which is the correct result.

Explanation

Pass at least one = 100 - 2 = 98%. Using the principle of inclusion-exclusion: P(B ∪ P ∪ C) = P(B) + P(P) + P(C) - [P(B ∩ P) + P(P ∩ C) + P(B ∩ C)] + P(B ∩ P ∩ C). 98 = 55 + 62 + 60 - [25 + 30 + 28] + x, so 98 = 177 - 83 + x, 98 = 94 + x, x = 4%.

Explanation

Using the percentages from Question 24: Bio only = 55 - (25 + 28) + 4 = 6%. Phys only = 62 - (25 + 30) + 4 = 11%. Chem only = 60 - (30 + 28) + 4 = 6%. Total exactly one = 6 + 11 + 6 = 23%.

Explanation

Given Biology (B)=55%, Physics (P)=62%, Chemistry (C)=60%. Intersections: P and B=25%, P and C=30%, B and C=28%. Failed all=2%, so Union (P or B or C)=98%. Using the formula Union = P + B + C - (P∩B + P∩C + B∩C) + (P∩B∩C), we get 98 = 62 + 55 + 60 - (25 + 30 + 28) + (P∩B∩C), which simplifies to 98 = 94 + (P∩B∩C), so P∩B∩C = 4%. Students passed in at least two subjects = (P∩B + P∩C + B∩C) - 2(P∩B∩C) = (25 + 30 + 28) - 2(4) = 83 - 8 = 75%. For 360 students, 75% of 360 = 270.

Explanation

To solve this problem, we use the principle of inclusion-exclusion for three sets. Let B, P, and C represent the sets of students who passed in Biology, Physics, and Chemistry respectively. The percentage of students who passed in at least one subject is n(B ∪ P ∪ C) = 100% - 2% = 98%.

The formula for the union of three sets is: n(B ∪ P ∪ C) = n(B) + n(P) + n(C) - [n(B ∩ P) + n(P ∩ C) + n(C ∩ B)] + n(B ∩ P ∩ C)

Substituting the given values: 98 = 55 + 62 + 60 - [25 + 30 + 28] + n(B ∩ P ∩ C) 98 = 177 - 83 + n(B ∩ P ∩ C) 98 = 94 + n(B ∩ P ∩ C) n(B ∩ P ∩ C) = 4%

Now, we identify the values required for the ratio: Number of students who passed in both Physics and Chemistry: n(P ∩ C) = 30% Number of students who passed in both Biology and Physics but not Chemistry: n(P ∩ B) - n(B ∩ P ∩ C) = 25% - 4% = 21%

The required ratio is 30 : 21. Dividing both sides by 3, we get 10 : 7, which corresponds to option b.

Explanation

Hotel ratings are a classic example of an ordinal scale of measurement. In statistics, an ordinal scale is used when data can be categorized and placed in a meaningful order or rank, but the exact difference between the ranks is not quantifiable or equal. For instance, a 5 star hotel is considered better than a 4 star hotel, and a 4 star hotel is better than a 3 star hotel. While the order of quality is clearly defined, the qualitative difference in luxury or service between a 1 star and 2 star hotel is not necessarily the same as the difference between a 4 star and 5 star hotel. Since the data reflects a relative ranking rather than a precise numerical interval or a ratio with a true zero point, it is classified as an ordinal scale. This distinguishes it from nominal scales, which only label data, and interval or ratio scales, which require precise mathematical distances between values.

Explanation

The fastest way to solve this problem is by using the Rule of Alligation, which is a shortcut for finding the ratio in which two components of different values are mixed to produce a mixture of a specific mean value. The ratio of the number of students in the two sections is inversely proportional to the difference between their respective averages and the combined average.

Using the alligation formula: n₇ / n₈ = (Avg₈ - Avg꜀ₒₘ₆ᵢₙₑ₀) / (Avg꜀ₒₘ₆ᵢₙₑ₀ - Avg₇)

Given values: Average of Section A (Avg₇) = 65 Average of Section B (Avg₈) = 70 Combined Average (Avg꜀ₒₘ₆ᵢₙₑ₀) = 67

Substituting the values into the formula: n₇ / n₈ = (70 - 67) / (67 - 65) n₇ / n₈ = 3 / 2

The calculation shows that the ratio of the number of students in section A to section B is 3:2, which matches option a).

Explanation

For a dataset with n = 19 observations, the median is the value at the (n+1)/2 = (19+1)/2 = 10th position. Given that the median is 30, the 10th observation in the sorted list is 30. This indicates that there are 9 observations less than or equal to 30 and 9 observations greater than or equal to 30. When two new observations, 8 and 32, are added, the total number of observations becomes n = 21. The new median is the value at the (21+1)/2 = 11th position. Since the value 8 is less than 30, it is placed in the lower half of the data, and since 32 is greater than 30, it is placed in the upper half. Because one value was added to both sides of the original median, the original 10th observation shifts to the 11th position in the new sorted list. Thus, the 11th observation is still 30, making the new median 30. This corresponds to option b).

Explanation

A frequency polygon is constructed by plotting the midpoints of class intervals against their corresponding frequencies and joining these points with straight line segments. When the number of observations increases and the number of classes is increased, the width of each class interval typically decreases. As the class width becomes smaller, the distance between the plotted midpoints reduces, and the density of the data points on the graph increases. This reduction in interval size minimizes the sharp fluctuations and steep angles between the connected line segments. Mathematically, as the number of classes tends to infinity and the class width tends to zero, the frequency polygon converges toward the continuous probability density function of the underlying distribution. Consequently, the resulting shape transitions from a jagged appearance to a more continuous and smooth curve, providing a better approximation of the actual population distribution.

Explanation

The pooled mean x̄ of two sets of observations is calculated using the weighted average formula: x̄ = (n₁x̄₁ + n₂x̄₂)/(n₁ + n₂). By the fundamental property of weighted averages, the resulting mean must always lie between the minimum and maximum of the individual means being averaged, provided the weights are positive.

Given that x̄₂ > x̄₁, the combined mean x̄ must satisfy x̄₁ < x̄ < x̄₂. We can verify this by rewriting the formula as x̄ = x̄₁ + n₂(x̄₂ - x̄₁)/(n₁ + n₂). Since n₁, n₂ are positive and x̄₂ > x̄₁, the term n₂(x̄₂ - x̄₁)/(n₁ + n₂) is strictly positive, which proves x̄ > x̄₁. Similarly, rewriting it as x̄ = x̄₂ - n₁(x̄₂ - x̄₁)/(n₁ + n₂) shows that x̄ < x̄₂.

The additional information that n₂ < n₁ implies that x̄ will be numerically closer to x̄₁ than to x̄₂, but the inequality x̄₁ < x̄ < x̄₂ remains true regardless of the relative sizes of n₁ and n₂. Therefore, option a is correct.

Explanation

Statement I is true because the median is determined by the middle position of the data. In a grouped frequency distribution, the median is calculated using the formula Median = L + ((n/2 - cf)/f) × h, where L is the lower limit of the median class, n is the total frequency, cf is the cumulative frequency of the preceding class, f is the frequency of the median class, and h is the class width. Since this calculation only relies on the properties of the median class and the cumulative frequency, the exact boundaries of the first or last intervals do not matter.

Statement II is true because the median is defined as a positional average, meaning it is the value that divides a distribution into two equal halves based on its position. This inherent property allows it to ignore the magnitude of extreme values and the specifics of open end intervals, provided the median class itself is closed. Therefore, the fact that the median is a positional average is the reason why it can be computed for open end distributions, making Statement II the correct explanation of Statement I.

Explanation

The given expression (2 tan θ)/(1 - tan² θ) is the standard trigonometric identity for tan(2θ). To find its value, we first need to determine tan θ using the given value of cos θ = 1/√5 for 0 < θ < π/2.

Since tan² θ + 1 = sec² θ, we have sec θ = 1/cos θ = √5. Substituting this into the identity gives tan² θ = sec² θ - 1 = (√5)² - 1 = 5 - 1 = 4. Because θ is in the first quadrant, tan θ must be positive, so tan θ = √4 = 2.

Now, substitute tan θ = 2 into the double angle formula: tan(2θ) = 2(2)/(1 - (2)²) = 4/(1 - 4) = 4/-3 = -4/3.

Thus, the value of the expression is -4/3, which corresponds to option (b).

Explanation

The cosine function cos x is a strictly decreasing function in the interval (0, 90°). This means that as the value of the angle x increases from 0° to 90°, the value of cos x continuously decreases from 1 toward 0. Mathematically, for any two angles α and β in the first quadrant, if α < β, then cos α > cos β. Conversely, if cos α < cos β, then it must be that α > β. Given the condition cos θ < cos ϕ where both θ and ϕ are between 0° and 90°, the angle θ must be greater than the angle ϕ to result in a smaller cosine value. For example, if cos θ = 0.5 and cos ϕ = 0.866, then θ = 60° and ϕ = 30°, which clearly shows θ > ϕ. Therefore, option (b) is the correct relationship.

Explanation

Let the center of the hemispherical dome be O and the top of the flag be F. The height of the top of the flag from the ground is OF = r + h. Let P be the first point of observation and Q be the second point.

From point P, the angle of elevation is 30°, so tan 30° = (r+h)/OP, which gives OP = (r+h)√3. From point Q, the angle of elevation is 45°, so tan 45° = (r+h)/OQ, which gives OQ = r+h. The distance moved is d = OP - OQ = (r+h)√3 - (r+h) = (r+h)(√3-1).

At point Q, the flag is just visible, meaning the line of sight QF is tangent to the dome at some point S. In the right triangle ΔOSQ, where OS is the radius r perpendicular to the tangent QF, we have sin(∠OQS) = OS/OQ. Since the elevation ∠OQS = 45°, we get sin 45° = r/OQ, which implies OQ = r√2.

Equating the two expressions for OQ, we have r + h = r√2, which leads to h = r(√2-1). Substituting r+h = r√2 into the equation for d: d = r√2(√3-1) r = d/(√2(√3-1))

Now, substitute r back into the expression for h: h = (d(√2-1))/(√2(√3-1)) Rationalizing the term 1/(√3-1) by multiplying the numerator and denominator by √3+1: h = (d(√2-1)(√3+1))/(√2(3-1)) = ((√3+1)(√2-1))/(2√2)d

This matches the expression in option (d).

Explanation

Given the trigonometric values for standard angles, we can directly determine the values of A and B. From the equation cos B = √3/2, we know that B = 30° because B is an acute angle. Substituting this value into the first equation, sin (A + B) = √3/2, we get sin (A + 30°) = √3/2. Since sin 60° = √3/2 and A is an acute angle, we have A + 30° = 60°, which implies A = 30°. Now, we substitute these values into the required expression tan (2A - B) to calculate the final result. This yields tan (2 × 30° - 30°) = tan (60° - 30°) = tan 30°. The value of tan 30° is 1/√3, which matches option (c).

Explanation

For statement 1, we simplify the expression cos θ / (1 - sin θ) + cos θ / (1 + sin θ) = 4 by taking cos θ as a common factor: cos θ ( 1 / (1 - sin θ) + 1 / (1 + sin θ) ) = 4. Combining the fractions using a common denominator gives cos θ ( (1 + sin θ + 1 - sin θ) / ((1 - sin θ)(1 + sin θ)) ) = 4, which simplifies to cos θ ( 2 / (1 - sin² θ) ) = 4. Using the trigonometric identity 1 - sin² θ = cos² θ, we get 2 cos θ / cos² θ = 4, which further simplifies to 2 / cos θ = 4. This implies cos θ = 1/2. Since 0 < θ < 90°, the value of θ is 60°. Thus, statement 1 is correct.

For statement 2, we test if θ = 60° satisfies the equation 3 tan θ + cot θ = 5 cos θ. Substituting θ = 60° into the left hand side (LHS), we get 3 tan 60° + cot 60° = 3√3 + 1/√3 = (9 + 1)/√3 = 10/√3 ≈ 5.77. The right hand side (RHS) is 5 cos 60° = 5 × 1/2 = 2.5. Since 5.77 ≠ 2.5, statement 2 is incorrect. Therefore, only statement 1 is correct, making option a the correct choice.

Explanation

For statement 1, we simplify the expression cos²θ = 1 - (p² + q²)/2pq to obtain cos²θ = (2pq - (p² + q²))/2pq = -(p² + q² - 2pq)/2pq = -(p-q)²/2pq. For any real angle θ, the value of cos²θ must satisfy 0 ≤ cos²θ ≤ 1. For the expression -(p-q)²/2pq to be non negative, either (p-q)² = 0 or 2pq must be negative. If p=q, then cos²θ = 0, which is possible. If 2pq < 0, we can let q = -k where k > 0 and p > 0. Then cos²θ = -(p+k)²/-2pk = (p+k)²/2pk. By the AM-GM inequality, (p+k)² ≥ 4pk, which implies (p+k)²/2pk ≥ 4pk/2pk = 2. This contradicts the condition that cos²θ ≤ 1. Thus, the equation is possible only when p=q.

For statement 2, we simplify tan²θ = 4pq/(p+q)² - 1 to get tan²θ = (4pq - (p+q)²)/(p+q)² = (4pq - (p² + 2pq + q²))/(p+q)² = -(p² - 2pq + q²)/(p+q)² = -(p-q)²/(p+q)². Since tan²θ ≥ 0 for all θ where it is defined and (p+q)² > 0 for non zero real numbers p, q where p ≠ -q, the RHS must be non negative. However, since (p-q)² ≥ 0, the term -(p-q)² is always non positive. The only way the equation can hold is if -(p-q)² = 0, which implies p=q. In this case, tan²θ = 0, which is possible. Thus, statement 2 is also possible only when p=q.

Both statements 1 and 2 are correct.

Explanation

For statement 1, let x = cos θ. The expression is x + 1/x. According to the Arithmetic Mean-Geometric Mean (AM-GM) inequality, for any positive real number x, x + 1/x ≥ 2. Similarly, for any negative real number x, x + 1/x ≤ -2. Since cos θ ranges from -1 to 1, the value of cos θ + sec θ will always be in the range (-∞, -2] ∪ [2, ∞). Because 1.5 lies between -2 and 2, the expression can never equal 1.5. Thus, statement 1 is correct.

For statement 2, we use the trigonometric identity sec² θ + csc² θ = 1/cos² θ + 1/sin² θ. Combining the fractions, we get (sin² θ + cos² θ)/(sin² θ cos² θ) = 1/(sin² θ cos² θ). Using the double angle formula sin 2θ = 2 sin θ cos θ, we can rewrite this as 1/((sin 2θ)/2)² = 4/sin² 2θ. Since the maximum value of sin² 2θ is 1, the minimum value of 4/sin² 2θ is 4/1 = 4. Therefore, the value can never be less than 4, making statement 2 correct. Since both statements are correct, option (c) is the right choice.

Explanation

Given sin^2x + sin x = 1, which implies sin x = 1 - sin^2x = cos^2x. The expression cos^12x + 3cos^10x + 3cos^8x + cos^6x is a binomial expansion of (cos^4x + cos^2x)^3. Substituting cos^2x = sin x, the expression becomes ( (sin x)^2 + sin x )^3. Since sin^2x + sin x = 1, the value is (1)^3 = 1.

Explanation

Let 3sin theta + 5cos theta = X = 4 and 3cos theta - 5sin theta = Y. Square both: X^2 = 9sin^2 theta + 25cos^2 theta + 30sin theta cos theta and Y^2 = 9cos^2 theta + 25sin^2 theta - 30sin theta cos theta. Adding them: X^2 + Y^2 = 9(sin^2 theta + cos^2 theta) + 25(sin^2 theta + cos^2 theta) = 9 + 25 = 34. Since X = 4, X^2 = 16. Thus, 16 + Y^2 = 34, which gives Y^2 = 18.

Explanation

Given cot theta (1 + sin theta) = 4m and cot theta (1 - sin theta) = 4n. This gives cot theta + cos theta = 4m and cot theta - cos theta = 4n. Adding gives 2 cot theta = 4(m + n) => cot theta = 2(m + n). Subtracting gives 2 cos theta = 4(m - n) => cos theta = 2(m - n). Multiplying the original equations: cot^2 theta (1 - sin^2 theta) = 16mn => cot^2 theta cos^2 theta = 16mn. Substituting the values: [2(m + n)]^2 [2(m - n)]^2 = 16mn => 16(m + n)^2 (m - n)^2 = 16mn => (m^2 - n^2)^2 = mn.

Explanation

In a right triangle with base u² - v² and hypotenuse u² + v², the height h is found using the Pythagorean theorem: h² = (u² + v²)² - (u² - v²)². Using the identity a² - b² = (a-b)(a+b), we get h² = (2v²)(2u²) = 4u²v², which simplifies to h = 2uv. This is a standard Pythagorean triple.

The area of the triangle is given by the formula Area = 1/2 × base × height. Substituting the values, we have 2016 = 1/2 × (u² - v²) × 2uv = uv(u² - v²) = uv(u-v)(u+v). Factoring 2016 as 32 × 63 or 32 × 9 × 7, we look for integers u and v that satisfy this equation. By trial and error or prime factorization, let u = 9 and v = 7. Checking the product: 9 × 7 × (9-7) × (9+7) = 63 × 2 × 16 = 2016. This satisfies the condition.

The perimeter P is the sum of all sides: P = (u² - v²) + 2uv + (u² + v²) = 2u² + 2uv = 2u(u + v). Substituting u = 9 and v = 7, we get P = 2 × 9 × (9 + 7) = 18 × 16 = 288 units. Thus, option (b) is correct.

Explanation

For an equilateral triangle with side length l, the radius r of the inscribed circle (inradius) is given by the formula r = l/(2√3). The diameter of this circle is d = 2r = l/√3. When a square is inscribed in a circle, the diameter of the circle is equal to the diagonal of the square. Let the side of the square be s. The diagonal of the square is s√2. Therefore, s√2 = l/√3. Squaring both sides gives 2s² = l²/3. The area of the square is s², so we solve for s² to get s² = l²/(3 × 2) = l²/6. This corresponds to option (d).

Explanation

Walls area per room = 2(L + B) H = 2(6 + 4) 2.5 = 50 sq m. Total for 5 rooms = 5 50 = 250 sq m. Area of two square windows = 2 (2.5 2.5) = 2 6.25 = 12.5 sq m. Net area to paint = 250 - 12.5 = 237.5 sq m. Cans required = 237.5 / 20 = 11.875. Since paint comes in 1 litre cans, 12 cans are required.

Explanation

According to Varignon's Theorem, the quadrilateral formed by joining the mid-points of the sides of any quadrilateral is a parallelogram, and its area is exactly half the area of the original quadrilateral. In this case, since T is a parallelogram, the resulting figure S is also a parallelogram with Area of S = 1/2 Area of T. This establishes that the ratio of the area of T to that of S is 2:1, making statement 1 correct. To evaluate statement 2, we apply the Mid-point Theorem. In ΔABC, the line segment joining the mid-points of sides AB and BC is parallel to the diagonal AC and has a length of 1/2 AC. Similarly, the opposite side of parallelogram S also has a length of 1/2 AC. The other two sides of S are each half the length of the other diagonal BD of parallelogram T. Consequently, the perimeter of S is 1/2 AC + 1/2 BD + 1/2 AC + 1/2 BD = AC + BD. This proves that the perimeter of S is equal to the sum of the diagonals of T, not half of the sum. Therefore, statement 2 is incorrect, and only option a is correct.

Explanation

To find the area of a triangle when the lengths of all three sides are known, we use Heron's formula. Given the sides a = 5 cm, b = 6 cm, and c = 7 cm, we first calculate the semi-perimeter s using the formula s = (a + b + c) / 2. Substituting the values, we get s = (5 + 6 + 7) / 2 = 18 / 2 = 9 cm.

Now, we apply Heron's formula for the area: Area = √s(s - a)(s - b)(s - c). Substituting the values into the formula gives Area = √9(9 - 5)(9 - 6)(9 - 7) = √9 × 4 × 3 × 2. Simplifying the expression, we get Area = √216. Since 216 = 36 × 6, the area is 6√6 cm².

Using the approximate value of √6 ≈ 2.449, the area becomes 6 × 2.449 = 14.694 cm². Rounding this to one decimal place, we obtain 14.7 cm², which matches option b.

Explanation

Area of plot = 144pi = pir^2 => r = 12 m. Path width = 5 m, so total radius R = 12 + 5 = 17 m. Total area = pi R^2 = pi * 17^2 = 289pi sq m.

Explanation

The lateral surface area of a cone is calculated using the formula LSA = πrl, where r is the radius of the base and l is the slant height. Given the lateral surface area is 462cm² and the slant height l = 35cm, we substitute these values into the formula as follows:

462 = 22/7 × r × 35

Dividing 35 by 7 gives 5, so the equation becomes:

462 = 22 × r × 5 462 = 110r

To find the radius r, we divide 462 by 110:

r = 462/110 = 4.2cm

Thus, the radius of the base of the cone is 4.2cm, which corresponds to option (c).

Explanation

When a semi-circular plate of radius R is rolled to form a cone, the radius of the plate becomes the slant height l of the resulting cone, so l = R. The arc length of the semi-circular plate becomes the circumference of the base of the cone. Since the arc length of a semi-circle is πR and the circumference of the cone base is 2πr, where r is the base radius, we have the relation 2πr = πR, which simplifies to r = R/2.

The angle α between the generator (slant height) and the axis of the cone is the semi-vertical angle. Using the trigonometric relation for the right triangle formed by the radius, height, and slant height of the cone, we use the formula sin(α) = r/l. Substituting the derived values, we get sin(α) = (R/2) / R = 1/2. Since sin(30°) = 1/2, the angle α is 30°, which corresponds to option (c).

Explanation

The lateral surface area (LSA) of a right cylinder is given by the formula 2πrh and the total surface area (TSA) is given by 2πrh + 2πr², where r is the radius and h is the height. According to the given condition, the lateral surface area is half of the total surface area, which can be expressed as 2πrh = 1/2(2πrh + 2πr²). Simplifying this equation, we get 2πrh = πrh + πr². Subtracting πrh from both sides yields πrh = πr². Dividing both sides by πr (since the radius cannot be zero), we find that h = r. Given that the height h = π cm, it follows that the radius r = π cm. This corresponds to option (b).

Explanation

To find the least number of cubes, the side of the cube must be the Greatest Common Divisor (GCD) of 20, 15, and 10. GCD(20, 15, 10) = 5 cm. Total cubes = (Volume of block) / (Volume of cube) = (20 15 10) / (5 5 5) = 3000 / 125 = 24.

Explanation

Let the side of the cube be denoted by a. The formula for the length of the diagonal d of a cube in terms of its side is given by d = a√3. According to the question, the length of the diagonal is l, so we have the relation l = a√3. From this, we can express the side of the cube as a = l/√3.

The total surface area (TSA) of a cube is the sum of the areas of its six equal square faces, which is given by the formula TSA = 6a². Substituting the value of a derived from the diagonal length, we get TSA = 6(l/√3)². Simplifying the expression, we have TSA = 6 × l²/3 = 2l². Thus, the total surface area of the cube is 2l², which corresponds to option d.

Explanation

Let the equal perimeter of the three figures be P. For the equilateral triangle, the side a = P/3, so the area T = (√3/4) a² = (√3/4) (P/3)² = (√3 P²)/36 ≈ (1.732 P²)/36 ≈ 0.0481 P². For the square, the side s = P/4, so the area S = s² = (P/4)² = P²/16 = 0.0625 P². For the circle, the radius r = P/2π, so the area C = π r² = π (P/2π)² = P²/4π ≈ P²/12.566 ≈ 0.0796 P². Comparing the coefficients, we find 0.0481 < 0.0625 < 0.0796, which implies T < S < C. This result is consistent with the isoperimetric theorem, which states that for a given perimeter, the area increases as the number of sides of a regular polygon increases, with the circle representing the limit of a polygon with infinite sides and thus providing the maximum area. Therefore, option (a) is correct.

Explanation

For any two similar triangles, the ratio of their areas is equal to the square of the ratio of their corresponding sides. Let the ratio of their corresponding sides be s₁/s₂ and the areas be A₁ and A₂. The applicable formula is A₁/A₂ = (s₁/s₂)² .

Given A₁ = 7 - 4√3 and A₂ = 7 + 4√3, we have (s₁/s₂)² = (7 - 4√3)/(7 + 4√3). To find the ratio of sides, we take the square root of both sides: s₁/s₂ = √((7 - 4√3)/(7 + 4√3)). To simplify this expression, we rationalize the denominator by multiplying both the numerator and the denominator inside the square root by the conjugate of the denominator, which is 7 - 4√3.

This gives s₁/s₂ = √(((7 - 4√3)(7 - 4√3))/((7 + 4√3)(7 - 4√3))) = √((7 - 4√3)²/(7² - (4√3)²)) = √((7 - 4√3)²/(49 - 48)) = √((7 - 4√3)²). Thus, the ratio of the corresponding sides is 7 - 4√3, which matches option (a).

Explanation

Let the radius of the circle be r and the length of the chord be L = √3r. The angle subtended by a chord at the center of the circle is given by the formula θ = 2 sin⁻¹(L/2r). Substituting the given values, we get θ = 2 sin⁻¹(√3r/2r) = 2 sin⁻¹(√3/2). Since sin⁻¹(√3/2) = 60°, the central angle θ = 2 × 60° = 120°.

The angle subtended by the chord at any point on the major arc, denoted as α, is half the angle subtended at the center. Therefore, α = θ/2 = 120°/2 = 60°.

The angle subtended by the chord at any point on the minor arc, denoted as β, and the angle subtended at the major arc are opposite angles of a cyclic quadrilateral. The sum of opposite angles in a cyclic quadrilateral is 180°. Thus, β = 180° - α = 180° - 60° = 120°.

Given that β = kα, we have 120° = k × 60°, which simplifies to k = 120/60 = 2. Thus, the value of k is 2, which corresponds to option b.

Explanation

The point where the external bisectors of two angles of a triangle intersect is known as the excenter of the triangle. For any triangle ABC, the angle formed by the intersection of the external bisectors of ∠ABC and ∠ACB at point D is given by the standard geometric formula ∠BDC = 90° - 1/2∠BAC. This formula is derived from the fact that the sum of angles in a triangle is 180° and the external angle is the sum of the two opposite interior angles.

Given that ∠BAC = 50°, we substitute this value into the formula: ∠BDC = 90° - 1/2(50°) ∠BDC = 90° - 25° ∠BDC = 65°

Thus, the angle ∠BDC is 65°, which corresponds to option (b). This shortcut is the most efficient way to solve such problems in competitive exams, bypassing the need to calculate individual interior and exterior angles of the triangle.

Explanation

The volume of a cone is given by the formula V = 1/3 πr²h. When comparing the volumes of two cones, we can establish the ratio as V₁/V₂ = (1/3 πr₁²h₁)/(1/3 πr₂²h₂) = (r₁/r₂)² × h₁/h₂. According to the problem, the ratio of the radii of their bases is r₁/r₂ = 3/1 and the ratio of their heights is h₁/h₂ = 1/3. Substituting these specific values into the ratio formula, we get V₁/V₂ = (3/1)² × 1/3 = 9 × 1/3 = 3/1. Consequently, the ratio of their volumes is 3:1, which corresponds to option (c).

Explanation

Since lines AB and CD intersect at point O, the angles ∠AOC and ∠AOD form a linear pair because they lie on the straight line CD. By the linear pair axiom, the sum of these adjacent angles is 180°. Therefore, we have the relation ∠AOC + ∠AOD = 180°. Given the condition that ∠AOC = 5 ∠AOD, we substitute this into the equation to get 5 ∠AOD + ∠AOD = 180°, which simplifies to 6 ∠AOD = 180°. Solving for ∠AOD, we find ∠AOD = 180°/6 = 30°. Consequently, ∠AOC = 5 × 30° = 150°. Because vertically opposite angles are equal, ∠BOC = ∠AOD = 30° and ∠BOD = ∠AOC = 150°. Thus, the four angles at the point of intersection O are 30°, 150°, 30°, and 150°, making option (b) the correct choice.

Explanation

Let the two fixed points be A(-a, 0) and B(a, 0) on the coordinate plane, so that the origin is the midpoint of AB. Let the moving point P have coordinates (x, y). According to the given condition, the sum of the squares of the distances from P to A and B is a constant k. Using the distance formula, we have PA² = (x + a)² + y² and PB² = (x - a)² + y².

The condition PA² + PB² = k can be written as: (x + a)² + y² + (x - a)² + y² = k

Expanding the squares: (x² + 2ax + a² + y²) + (x² - 2ax + a² + y²) = k

Combining like terms: 2x² + 2y² + 2a² = k 2(x² + y²) = k - 2a² x² + y² = (k - 2a²)/2

This equation is of the form x² + y² = r², where r² = (k - 2a²)/2. This is the standard equation of a circle centered at the origin (the midpoint of AB). Therefore, the locus of point P is a circle, which corresponds to option (b).

Explanation

In a right triangle with hypotenuse AC, AC^2 = AB^2 + BC^2. For any positive a, b, c where c^2 = a^2 + b^2, c is the largest side (c > a and c > b). Thus, c^3 = c * c^2 = c(a^2 + b^2) = ca^2 + cb^2. Since c > a, ca^2 > a^3, and since c > b, cb^2 > b^3. Therefore, c^3 > a^3 + b^3, so AC^3 > AB^3 + BC^3.

Explanation

The area of the region bounded externally by a square and internally by an inscribed circle is found by subtracting the area of the circle from the area of the square. Given the side of the square is s = 2a, the area of the square is calculated as s² = (2a)² = 4a². Since the circle touches all four sides of the square, the diameter of the circle is equal to the side of the square, meaning d = 2a. Therefore, the radius of the circle is r = d/2 = a. The area of the circle is πr² = πa². The required area is the difference between these two values: Area = 4a² - πa² = (4 - π)a². This matches option a.