SSC CHSL Tier 1 2013 Quantitative Aptitude Questions with Solutions

Explanation

Let the second number be y.

From the given conditions: The first number is twice the second, which is 2y. The third number is one-third of the first, which is 2y/3.

The sum of these three numbers is 132: 2y + y + 2y/3 = 132

Multiplying the entire equation by 3 to clear the fraction: 6y + 3y + 2y = 396 11y = 396 y = 36

Explanation

Correct Answer: D

To find the least perfect square number divisible by 3, 4, 5, 6, and 8, first find the Least Common Multiple (LCM) of these numbers.

The prime factorization of the given numbers is: 3 = 3¹ 4 = 2² 5 = 5¹ 6 = 2 × 3 8 = 2³

The LCM is the product of the highest power of each prime factor involved: LCM(3, 4, 5, 6, 8) = 2³ × 3¹ × 5¹ = 120

In the prime factorization of 120, which is 2² × 2 × 3 × 5, the prime factors 2, 3, and 5 do not have pairs. To make 120 a perfect square, each prime factor must have an even exponent. Therefore, 120 must be multiplied by 2 × 3 × 5 = 30 to complete the pairs.

Required least perfect square = 120 × 30 = 3600

Explanation

Let the first part be x and the second part be 24 - x.

According to the given condition, 7 times the first part added to 5 times the second part is equal to 146:

7x + 5(24 - x) = 146

Expanding and simplifying the equation:

7x + 120 - 5x = 146

2x + 120 = 146

2x = 146 - 120

2x = 26

x = 13

Explanation

When a number leaves remainders r₁, r₂, r₃ when divided by d₁, d₂, d₃ respectively, and the difference between each divisor and its remainder is a constant k, the general form of the number is: N = LCM(d₁, d₂, d₃) × m - k, where m is a positive integer.

Here, the divisors are 5, 9, and 13, and the remainders are 2, 6, and 10. The common difference is: 5 - 2 = 3 9 - 6 = 3 13 - 10 = 3 Thus, k = 3.

Since 5, 9, and 13 are co-prime, their LCM is their product: LCM(5, 9, 13) = 5 × 9 × 13 = 585

For the least such number, setting m = 1: N = 585 - 3 = 582

Explanation

The area of the equilateral triangular base with side a = 12 cm is given by the formula: Base Area = (√3/4) × a² Base Area = (√3/4) × 12² = 36√3 cm²

The volume of a prism is the product of its base area and height (h): Volume = Base Area × h 1080 = 36√3 × h h = 1080 / (36√3) = 10√3 cm

The perimeter of the triangular base is: Perimeter = 3 × a = 3 × 12 = 36 cm

The lateral surface area (LSA) of the prism is: LSA = Perimeter × h = 36 × 10√3 = 360√3 cm²

The total surface area (TSA) of the prism is the sum of its lateral surface area and the areas of the two bases: TSA = LSA + 2 × Base Area TSA = 360√3 + 2 × (36√3) TSA = 360√3 + 72√3 = 432√3 cm²

Explanation

The initial ratio of lead to tin is 5:6. The total parts of the alloy equal 5 + 6 = 11 parts.

The total mass of the alloy is 12.1 kg, which means 11 parts correspond to 12.1 kg. The value of 1 part is 12.1 / 11 = 1.1 kg.

To make the ratio of lead to tin equal to 1:1, both components must have an equal number of parts. Since the amount of tin remains unchanged at 6 parts, the lead must also be increased to 6 parts.

This requires adding 1 part of lead (6 - 5 = 1 part). Since 1 part is equal to 1.1 kg, the amount of lead that needs to be added is 1.1 kg.

Explanation

Using the complementary angle identity, cos 67½° is equal to cos(90° - 22½°), which simplifies to sin 22½°.

For any acute angle θ, we can express sin θ in terms of tan θ using basic trigonometric identities: sin θ = tan θ / sec θ

Since sec θ = √(1 + tan² θ), this becomes: sin θ = tan θ / √(1 + tan² θ)

Substituting θ = 22½° and tan 22½° = x into this relation gives: sin 22½° = x / √(1 + x²)

This directly yields the value of cos 67½° as x / √(1 + x²).

Explanation

The sequence can be solved by analyzing the differences between consecutive terms. Let the terms be a₁, a₂, a₃, and a₄.

The first difference is: d₁ = a₂ - a₁ = 13 - 4 = 9

The second difference is: d₂ = a₃ - a₂ = 38 - 13 = 25

The sequence of differences (9, 25, ...) follows a recursive pattern where each successive difference is multiplied by 3 and then decreased by 2: dₙ = 3 × dₙ₋₁ - 2

Using this rule, the third difference is: d₃ = 3 × 25 - 2 = 75 - 2 = 73

Adding this difference to the third term yields the fourth term: a₄ = a₃ + d₃ = 38 + 73 = 111

Alternatively, the sequence follows the direct term-to-term rule: aₙ = 3 × aₙ₋₁ + cₙ where the added constant cₙ is a sequence starting at 1 and decreasing by 2 each step (1, -1, -3, ...).

For the second term: a₂ = 3 × 4 + 1 = 13

For the third term: a₃ = 3 × 13 - 1 = 38

For the fourth term: a₄ = 3 × 38 - 3 = 114 - 3 = 111

Explanation

Using the algebraic identity: (a² + b²)(x² + y²) - (ax + by)² = (ay - bx)²

From the given equations, we can express x² and y² as: x² = 1 - a² y² = 1 - b²

Adding these two equations gives: x² + y² = 2 - (a² + b²)

Substituting x² + y² = 2 - (a² + b²) and ax + by = 1 into the identity: (a² + b²)(2 - (a² + b²)) - 1² = (ay - bx)²

Let S = a² + b²: S(2 - S) - 1 = (ay - bx)² 2S - S² - 1 = (ay - bx)² -(S² - 2S + 1) = (ay - bx)² -(S - 1)² = (ay - bx)²

Since the square of any real number is non-negative, the left side -(S - 1)² is less than or equal to 0, while the right side (ay - bx)² is greater than or equal to 0.

For the equality to hold, both sides must be equal to 0: (S - 1)² = 0 S = 1

Thus, the numerical value of a² + b² is 1.

Explanation

To find the individual profit shares, first convert the initial investment ratio of A, B, and C into integer values. The LCM of the denominators 2, 3, and 5 is 30. Multiplying each term of the ratio by 30 gives: A's initial share = (7/2) × 30 = 105 B's initial share = (4/3) × 30 = 40 C's initial share = (6/5) × 30 = 36

Next, calculate the total equivalent monthly investment of each partner for the year: A invests 105 for the first 4 months. After 4 months, A increases the investment by 50%, making the new investment 105 + 52.5 = 157.5 for the remaining 8 months. A's total equivalent investment = (105 × 4) + (157.5 × 8) = 420 + 1260 = 1680 B's total equivalent investment = 40 × 12 = 480 C's total equivalent investment = 36 × 12 = 432

The ratio of their total investments, which is also the ratio of their profit shares, is: A : B : C = 1680 : 480 : 432

Dividing each term by 48 simplifies the ratio to: 35 : 10 : 9

The sum of the ratio parts is 35 + 10 + 9 = 54.

With a total profit of 43,200, B's share is: B's share = (10 / 54) × 43,200 = 10 × 800 = 8,000

Explanation

When two trains run in opposite directions, their relative speed is the sum of their individual speeds.

Relative speed = 65 km/h + 85 km/h = 150 km/h

Converting the relative speed into meters per second by multiplying with 5/18:

Relative speed = 150 × 5/18 = 125/3 m/s

When crossing each other completely, the total distance covered is the sum of the lengths of both trains. Since both trains have the same length L, the total distance is 2L.

Using the distance formula, Distance = Speed × Time:

2L = (125/3) × 6

2L = 125 × 2

L = 125 meters

Explanation

Let the cost price of the article be CP.

The initial transaction results in a loss of 12½%, which is equivalent to 12.5%. The subsequent transaction results in a profit of 6%.

The total percentage difference between a 12.5% loss and a 6% profit is: 12.5% + 6% = 18.5%

This difference of 18.5% in the cost price corresponds to the given price increase of ₹51.80.

18.5% of CP = 51.80 (18.5 / 100) × CP = 51.80 CP = (51.80 × 100) / 18.5 CP = 5180 / 18.5 CP = 280

Explanation

When the cost price of M articles is equal to the selling price of N articles, the gain percentage is calculated using the standard relation:

Gain% = ((M - N) / N) × 100%

Here, the cost price of 18 articles is equal to the selling price of 15 articles, meaning M = 18 and N = 15. Substituting these values into the relation:

Gain% = ((18 - 15) / 15) × 100% Gain% = (3 / 15) × 100% Gain% = 1/5 × 100% Gain% = 20%

Explanation

The single equivalent discount for two successive discounts of d₁% and d₂% is given by the formula d₁ + d₂ - (d₁ × d₂)/100.

Substituting the given values of 50% and 50% into the formula:

Equivalent Discount = 50 + 50 - (50 × 50)/100 Equivalent Discount = 100 - 2500/100 Equivalent Discount = 100 - 25 Equivalent Discount = 75%

Alternatively, if the initial price is assumed to be 100, the first discount of 50% reduces the price to 50. The second discount of 50% on this remaining price reduces it by 25, leading to a final price of 25. The total reduction from 100 to 25 represents an overall discount of 75%.

Explanation

Using the algebraic identity (x + y + z)² = x² + y² + z² + 2(xy + yz + zx), we substitute the given values: 15² = x² + y² + z² + 2(75) 225 = x² + y² + z² + 150 x² + y² + z² = 75

Since x² + y² + z² = 75 and xy + yz + zx = 75, we can equate them: x² + y² + z² = xy + yz + zx x² + y² + z² - xy - yz - zx = 0

Multiplying both sides by 2 gives: 2x² + 2y² + 2z² - 2xy - 2yz - 2zx = 0 (x - y)² + (y - z)² + (z - x)² = 0

Since the sum of squares of real numbers is zero, each term must be zero: x - y = 0, y - z = 0, and z - x = 0 This means x = y = z.

Given x + y + z = 15, we get: 3z = 15 z = 5 Thus, x = y = z = 5.

Substituting these values into the required expression: (x + 4y + z) / (3z) = (5 + 4(5) + 5) / (3 × 5) = 30 / 15 = 2.

Explanation

The system of equations for the intersecting lines at point (p, q) is: 3p - 5q = -8 3p + 5q = 32

Adding these two equations eliminates q: (3p - 5q) + (3p + 5q) = -8 + 32 6p = 24 p = 4

Substituting p = 4 back into the second equation: 3(4) + 5q = 32 12 + 5q = 32 5q = 20 q = 4

The value of p - q is therefore 4 - 4 = 0.

Explanation

The product xy is calculated using the difference of squares identity (a - b)(a + b) = a² - b²: xy = (√3 + √2)(√3 - √2) = (√3)² - (√2)² = 3 - 2 = 1

The sum of squares x² + y² is calculated using the algebraic identity (a + b)² + (a - b)² = 2(a² + b²): x² + y² = (√3 + √2)² + (√3 - √2)² = 2((√3)² + (√2)²) = 2(3 + 2) = 10

Substituting these values into the given expression: 8xy(x² + y²) = 8 × 1 × 10 = 80

Explanation

The relation between the number of persons (M), number of days (D), and working hours per day (H) for a constant amount of work is given by the formula:

M₁ × D₁ × H₁ = M₂ × D₂ × H₂

Substituting the given values:

50 × 12 × 6 = 60 × 8 × x

3600 = 480 × x

x = 3600 / 480

x = 15/2

x = 7 ½

Explanation

From the first given equation, x + 1/y = 1: x = 1 - 1/y = (y - 1)/y Taking the reciprocal of x gives: 1/x = y/(y - 1)

From the second given equation, y + 1/z = 1: 1/z = 1 - y Taking the reciprocal of 1/z gives: z = 1/(1 - y) = -1/(y - 1)

To find the value of z + 1/x, substitute these expressions: z + 1/x = -1/(y - 1) + y/(y - 1) z + 1/x = (y - 1)/(y - 1) z + 1/x = 1

Explanation

We use the algebraic identity: a³ - b³ = (a - b)³ + 3ab(a - b)

By substituting a = 4x and b = 5z, we get: 64x³ - 125z³ = (4x - 5z)³ + 3(4x)(5z)(4x - 5z) 64x³ - 125z³ = (4x - 5z)³ + 60xz(4x - 5z)

Substituting the given values, 4x - 5z = 16 and xz = 12, into the equation: 64x³ - 125z³ = 16³ + 60 × 12 × 16 64x³ - 125z³ = 4096 + 11520 64x³ - 125z³ = 15616

Explanation

The lengths of tangents drawn from an external point to a circle are equal. From vertex A, the tangents to the circle are AQ and AR, which means AQ = AR = 6 cm. Similarly, from vertex B, the tangents are BQ and BP, so BQ = BP. From vertex C, the tangents are CP and CR, so CP = CR.

The perimeter of triangle ABC is AB + BC + AC. Since the side BC is divided at point P, BC = BP + PC. The perimeter can therefore be written as AB + BP + PC + AC. Substituting BP with BQ and PC with CR, the perimeter becomes AB + BQ + CR + AC. Since AB + BQ = AQ and AC + CR = AR, this simplifies to AQ + AR.

Using the equality of the tangents from A, AQ = AR, which means the perimeter of triangle ABC is equal to 2 × AQ. Substituting the given value of AQ, the perimeter is 2 × 6 = 12 cm.

Explanation

The point of intersection of the internal angle bisectors of a triangle is the incenter. The angle formed at the incenter D by the bisectors of ∠B and ∠C is given by the standard geometric relation:

∠BDC = 90° + ½ ∠A

Substituting the given value of ∠A = 80°:

∠BDC = 90° + ½ × 80° ∠BDC = 90° + 40° ∠BDC = 130°

Explanation

By the exterior angle theorem, the exterior angle of a triangle is equal to the sum of its two opposite interior angles.

In triangle ABC with side BC produced to D: ∠ACD = ∠ABC + ∠BAC

Given that ∠ACD = 112° and ∠ABC = 3/4 ∠BAC, we can write: ∠BAC = 4/3 ∠ABC

Substituting this into the exterior angle equation: 112° = ∠ABC + 4/3 ∠ABC 112° = 7/3 ∠ABC

Solving for ∠ABC: ∠ABC = 112° × 3/7 ∠ABC = 16 × 3 ∠ABC = 48°

Explanation

Since XY is parallel to AC, triangle XBY is similar to triangle ABC. For similar triangles, the ratio of their areas is equal to the square of the ratio of their corresponding sides: Area(ΔXBY) / Area(ΔABC) = (BX / AB)²

Because the line XY divides the triangle ABC into two parts of equal area, the area of triangle XBY is exactly half of the area of triangle ABC: Area(ΔXBY) / Area(ΔABC) = ½

Equating the two expressions: (BX / AB)² = ½ BX / AB = 1 / √2

The side AB is the sum of AX and BX, which means AX = AB - BX. Dividing both sides by AB yields: AX / AB = 1 - BX / AB AX / AB = 1 - 1 / √2 AX / AB = (√2 - 1) / √2

To rationalize the denominator, multiply the numerator and the denominator by √2: AX / AB = (2 - √2) / 2

Explanation

The volume of a solid sphere is given by V = (4/3) × π × r³, where r is the radius of the sphere. Here, the diameter is 6 cm, so the radius r = 3 cm. V_sphere = (4/3) × π × 3³ = 36π cm³

The volume of a hollow cylinder is given by V = π × (R² - r_in²) × h, where R is the external radius, r_in is the internal radius, and h is the length. Here, the external diameter is 10 cm, which gives an external radius R = 5 cm, and the length h = 4 cm.

Since the sphere is melted to form the hollow cylindrical pipe, their volumes must be equal: 36π = π × (5² - r_in²) × 4 36 = 4 × (25 - r_in²) 9 = 25 - r_in² r_in² = 16 r_in = 4 cm

The width of the sheet of the pipe is the difference between the external radius and the internal radius: Width = R - r_in = 5 - 4 = 1 cm

Explanation

Let the common radius of the sphere and the cylinder be r, and the height of the cylinder be h.

The volume of a sphere is given by the formula: V_sphere = 4/3 × π × r³

The volume of a cylinder is given by the formula: V_cylinder = π × r² × h

Since the volumes of both solids are equal: 4/3 × π × r³ = π × r² × h

Dividing both sides by π × r²: 4/3 × r = h

To find the ratio of the radius to the height of the cylinder (r : h): r/h = 3/4

This gives the ratio r : h = 3 : 4.

Explanation

The perpendicular drawn from the center of a circle to a chord bisects the chord. This forms a right-angled triangle with the radius of the circle R, the perpendicular distance from the center d, and half of the chord length L/2. By the Pythagorean theorem:

R² = d² + (L/2)²

For the smaller chord of length L = 6 cm, the distance d = 4 cm. Substituting these values to find the radius R:

R² = 4² + (6/2)² R² = 16 + 9 R² = 25 R = 5 cm

For the second chord of length L = 8 cm, using the radius R = 5 cm, the distance d from the center is calculated as:

5² = d² + (8/2)² 25 = d² + 16 d² = 9 d = 3 cm

Explanation

Let the parallel sides of the isosceles trapezium be 4x and 7x. The sum of the parallel sides is 11x.

The height of the trapezium is 2/11 of the sum of its parallel sides: h = (2/11) × 11x = 2x

The area of the trapezium is 176 cm². Using the formula for the area of a trapezium: Area = ½ × (sum of parallel sides) × height 176 = ½ × 11x × 2x 176 = 11x² x² = 16 x = 4

Thus, the parallel sides are: 4 × 4 = 16 cm 7 × 4 = 28 cm

The height of the trapezium is: h = 2 × 4 = 8 cm

In an isosceles trapezium, the diagonal forms a right-angled triangle with the height and the projection of the diagonal on the longer base. The length of this base projection is: (16 + 28) / 2 = 22 cm

Using the Pythagorean theorem, the length of the diagonal (d) is: d² = 8² + 22² d² = 64 + 484 d² = 548 d = √548 = 2√137 cm

Explanation

The area of a triangle ABC given the lengths of its medians is related to the area of the triangle formed by taking the medians as its sides by the formula:

Area of ABC = 4/3 × Area of median triangle

Let the medians be m₁ = 18 cm, m₂ = 24 cm, and m₃ = 30 cm. Since 18² + 24² = 324 + 576 = 900 = 30², the lengths of these medians satisfy the Pythagorean theorem, meaning they form a right-angled triangle.

The area of this median triangle is:

Area of median triangle = ½ × base × height = ½ × 18 × 24 = 216 cm²

Applying the formula:

Area of ABC = 4/3 × 216 = 288 cm²

Explanation

The angles in the series 1°, 3°, 5°, ..., 89° form an arithmetic progression with the first term a = 1, last term l = 89, and common difference d = 2.

The number of terms n is calculated as: n = (l - a)/d + 1 = (89 - 1)/2 + 1 = 45

Using the complementary angle identity sin(90° - θ) = cos θ, we can write: sin²θ + sin²(90° - θ) = sin²θ + cos²θ = 1

Pairing the terms from the beginning and the end: (sin²1° + sin²89°) + (sin²3° + sin²87°) + ... + (sin²43° + sin²47°) + sin²45°

Since there are 45 terms, we obtain 22 such pairs, leaving the middle term sin²45° unpaired.

The sum is: 22 × 1 + sin²45° = 22 + (1/√2)² = 22½

Explanation

By the Arithmetic Mean-Geometric Mean (AM-GM) inequality, for any two positive real numbers a and b, the relation (a + b)/2 ≥ √(ab) always holds.

By setting a = cos²θ and b = sec²θ, which are positive real numbers for all values of θ where they are defined, we get:

(cos²θ + sec²θ)/2 ≥ √(cos²θ × sec²θ)

Since sec²θ = 1/cos²θ, their product is 1:

(cos²θ + sec²θ)/2 ≥ √1

cos²θ + sec²θ ≥ 2

Thus, the minimum value of cos²θ + sec²θ is 2. This minimum is achieved when cos²θ = sec²θ = 1, which occurs when θ is an integer multiple of π.

Explanation

Given the equation: cos θ + sec θ = 2

Since sec θ = 1/cos θ, the equation becomes: cos θ + 1/cos θ = 2

Multiplying by cos θ on both sides: cos² θ + 1 = 2 cos θ cos² θ - 2 cos θ + 1 = 0 (cos θ - 1)² = 0 cos θ = 1

Since cos θ = 1, we also have sec θ = 1.

The terms in the expression cos 10 θ + sec 110 represent cos¹⁰ θ + sec¹¹ θ (or cos¹⁰ θ + sec¹¹⁰ θ, due to typical printing layouts of exponents). Substituting cos θ = 1 and sec θ = 1: cos¹⁰ θ + sec¹¹ θ = 1¹⁰ + 1¹¹ = 1 + 1 = 2

Explanation

The volume V of a pyramid is given by the formula: V = ⅓ × Area of the base × Height

For the triangular base with sides a = 21 cm, b = 20 cm, and c = 13 cm, the semi-perimeter s is calculated as: s = (a + b + c) / 2 = (21 + 20 + 13) / 2 = 27 cm

Using Heron's formula, the area of the base A is: A = √(s(s - a)(s - b)(s - c)) A = √(27 × (27 - 21) × (27 - 20) × (27 - 13)) A = √(27 × 6 × 7 × 14) A = √(3³ × (2 × 3) × 7 × (2 × 7)) A = √(3⁴ × 2² × 7²) A = 3² × 2 × 7 = 126 cm²

With an altitude of 30 cm, the volume is: V = ⅓ × 126 × 30 V = 126 × 10 = 1260 cm³

Explanation

The measure of each exterior angle of a regular polygon with n sides is given by the formula 360°/n.

For a polygon with n sides, the exterior angle is 360°/n, and for a polygon with (n + 1) sides, the exterior angle is 360°/(n + 1).

According to the question, the difference between these two angles is 5°: 360/n - 360/(n + 1) = 5

Dividing both sides of the equation by 5: 72/n - 72/(n + 1) = 1

Taking the common denominator: 72 × ((n + 1 - n) / (n(n + 1))) = 1 72 / (n(n + 1)) = 1 n(n + 1) = 72

This can be rewritten as a quadratic equation: n² + n - 72 = 0 (n + 9)(n - 8) = 0

Since the number of sides n must be a positive integer, we discard the negative solution (n = -9). Thus, n = 8.

Explanation

The volume of a right prism is calculated using the formula:

Volume = Area of the base × Height of the prism

Since the base of the prism is a rhombus, its area is given by the formula:

Area of the base = ½ × d₁ × d₂

where d₁ and d₂ are the lengths of the diagonals of the rhombus.

Substituting the given lengths of the diagonals, d₁ = 14 cm and d₂ = 18 cm:

Area of the base = ½ × 14 × 18 = 126 cm²

With the height of the prism given as 20 cm, the volume of the prism is calculated as:

Volume = 126 × 20 = 2520 cm³

Explanation

Let the two numbers be 33a and 33b, where a and b are co-prime integers.

According to the question, the sum of the two numbers is 528: 33a + 33b = 528 33(a + b) = 528 a + b = 16

To maintain the HCF as 33, the integers a and b must be co-prime, meaning they share no common factor other than 1.

The pairs of co-prime integers (a, b) whose sum is 16 are: (1, 15) (3, 13) (5, 11) (7, 9)

This gives exactly 4 valid pairs of numbers.

Explanation

Let the rates of taps A, B, and C to fill the tank be a, b, and c respectively.

Based on the given rates: a + b = 1/6 b + c = 1/10

The combined rate of A and C together is based on a time of 7½ hours (which is 15/2 hours): a + c = 2/15

Adding these three equations together: 2(a + b + c) = 1/6 + 1/10 + 2/15

Finding a common denominator of 30: 2(a + b + c) = 5/30 + 3/30 + 4/30 2(a + b + c) = 12/30 = 2/5 a + b + c = 1/5

Subtracting the combined rate of A and B (a + b) from the total rate (a + b + c) gives the individual rate of C: c = (a + b + c) - (a + b) c = 1/5 - 1/6 c = 6/30 - 5/30 = 1/30

Thus, tap C alone can fill the tank in 30 hours.

Explanation

Given the trigonometric equation sec θ - cosec θ = 0, we can rewrite it as:

sec θ = cosec θ

Expressing these terms in terms of sine and cosine gives:

1/cos θ = 1/sin θ

By cross-multiplying, we obtain:

sin θ = cos θ

Dividing both sides by cos θ yields:

tan θ = 1

Since cot θ is the reciprocal of tan θ, we have:

cot θ = 1/tan θ = 1/1 = 1

Substituting these values into the expression tan θ + cot θ:

tan θ + cot θ = 1 + 1 = 2

Explanation

Let the height of the tree be h = 120 m. The two boats are on the same side of the tree, in line with its foot. The angles of depression to the boats are 45° and 60°, which are equal to their respective angles of elevation from the ground.

For the closer boat, the angle of elevation is 60°. Let its distance from the foot of the tree be x. Using the trigonometric tangent relation: tan 60° = 120 / x √3 = 120 / x x = 120 / √3 = 40√3 m

For the farther boat, the angle of elevation is 45°. Let its distance from the foot of the tree be y. Using the trigonometric tangent relation: tan 45° = 120 / y 1 = 120 / y y = 120 m

The distance between the two boats is the difference between their distances from the foot of the tree: y - x = 120 - 40√3 = 40 (3 - √3) m

Explanation

Let x = a + b - 2c, y = b + c - 2a, and z = c + a - 2b.

Adding these three terms: x + y + z = (a + b - 2c) + (b + c - 2a) + (c + a - 2b) = 0

According to the algebraic identity, if x + y + z = 0, then x³ + y³ + z³ = 3xyz.

Substituting the expressions for x, y, and z back into this identity yields: (a + b - 2c)³ + (b + c - 2a)³ + (c + a - 2b)³ = 3(a + b - 2c)(b + c - 2a)(c + a - 2b)

Explanation

Let the length and breadth of the rectangular field be 3x and 4x respectively.

A walks along the diagonal of the rectangle. By Pythagoras theorem, the length of the diagonal is √((3x)² + (4x)²) = √(9x² + 16x²) = 5x. Thus, the distance covered by A is 5x.

B walks along the sides of the rectangle to reach the diagonally opposite corner. Therefore, the distance covered by B is the sum of the length and the breadth, which is 3x + 4x = 7x.

Since both A and B start and finish at the same time, the time taken by both of them is equal. When time is constant, speed is directly proportional to the distance covered.

Ratio of speed of A to speed of B = Distance covered by A : Distance covered by B = 5x : 7x = 5 : 7.

Explanation

To find the increase in the number of candidates qualified under the Engineering discipline from 2009 to 2010, the calculation is as follows:

First, identify the percentage of candidates in the Engineering discipline, which is represented by the fifth column under the discipline percentage distribution (Arts, Science, Commerce, Agriculture, Engineering).

For the year 2009: The percentage of candidates qualified under Engineering is 10%. The total number of qualified candidates is 620. Number of Engineering candidates qualified = 10% of 620 = (10/100) × 620 = 62.

For the year 2010: The percentage of candidates qualified under Engineering is 12%. The total number of qualified candidates is 900. Number of Engineering candidates qualified = 12% of 900 = (12/100) × 900 = 108.

The increase in the number of qualified Engineering candidates from 2009 to 2010 is: 108 - 62 = 46.

Explanation

To determine the year with the lowest number of candidates qualified under the Agriculture discipline, we calculate the absolute number of candidates for the given years:

In 2008, the number of candidates is 7% of 500, which equals 35. In 2006, the number of candidates is 8% of 780, which equals 62.4. In 2009, the number of candidates is 10% of 620, which equals 62. In 2011, the number of candidates is 14% of 850, which equals 119.

The value of 35 in the year 2008 is the lowest among all the evaluated years.

Explanation

From the given data, the sale value of finished goods in 1999 is 600 Lakh Rs.

The amounts invested in purchasing raw materials during the years 1997, 1998, and 1999 are 375 Lakh Rs., 330 Lakh Rs., and 525 Lakh Rs., respectively.

The sum of the amount invested in raw materials during these three years is: 375 + 330 + 525 = 1230 Lakh Rs.

The required percentage is calculated as: (600 / 1230) × 100% = 48.78% ≈ 49%

Explanation

The difference between the value of sales of finished goods and the amount invested in raw materials (in ₹ lakhs) for each year is:

In 1995: 200 - 120 = 80 lakh In 1996: 300 - 225 = 75 lakh In 1997: 500 - 375 = 125 lakh In 1998: 400 - 330 = 70 lakh

The maximum difference among these years is 125 lakh, which occurs in 1997.

Explanation

Volume of a cylinder is given by V_cylinder = π r² h, where r is the radius and h is the length of the cylinder.

Volume of a sphere is given by V_sphere = (4/3) π R³, where R is the radius of the sphere.

Since the cylindrical rod is melted into 6 spherical balls of the same radius, we have R = r. Equating the volumes: V_cylinder = 6 × V_sphere π r² h = 6 × (4/3) π r³ h = 8 r

Substituting the given radius r = 50 cm = 0.5 m: h = 8 × 0.5 = 4 m

Explanation

The volume of a hemisphere with radius R is given by the formula (2/3)πR³. The volume of a right-circular cylinder with base radius R and height h is given by the formula πR²h.

Equating the two volumes as given in the problem: (2/3)πR³ = πR²h

Dividing both sides of the equation by πR²: (2/3)R = h

To find the ratio of the height to the base radius (h : R), we rearrange the equation: h/R = 2/3

This gives the ratio of the height and the radius of the base of the cylinder as 2 : 3.

Explanation

The volume of a pyramid is given by the formula V = ⅓ × Base Area × Height.

The base of the pyramid is a triangle with sides 5 cm, 12 cm, and 13 cm. Since 5² + 12² = 25 + 144 = 169 = 13², the base is a right-angled triangle with perpendicular sides of length 5 cm and 12 cm.

The area of this right-angled triangular base is calculated as ½ × 5 × 12 = 30 cm².

Substituting the volume of 330 cm³ and the base area of 30 cm² into the volume formula yields 330 = ⅓ × 30 × Height, which simplifies to 330 = 10 × Height. Solving for height gives 33 cm.

Explanation

The perimeter of a rhombus is 4a = 100 cm, which gives the length of each side as a = 25 cm.

The diagonals of a rhombus bisect each other at right angles. Given that one diagonal d₁ = 40 cm, its half-length is 20 cm. Using the Pythagorean theorem for the right-angled triangle formed by the halves of the diagonals and a side of the rhombus:

(d₂/2)² = a² - (d₁/2)² (d₂/2)² = 25² - 20² (d₂/2)² = 625 - 400 = 225 d₂/2 = √225 = 15 cm

This gives the length of the second diagonal d₂ = 30 cm.

The area of the rhombus is calculated using the formula: Area = ½ × d₁ × d₂ Area = ½ × 40 × 30 = 600 cm²

Explanation

The total area of the paper used to make the kite is the sum of the area of the square and the area of the equilateral triangle.

For the square portion with a diagonal of 32 cm: Area of the square = ½ × diagonal² = ½ × 32² = ½ × 1024 = 512 cm²

For the equilateral triangle portion with a side of 8 cm: Area of the equilateral triangle = (√3 / 4) × side² = (√3 / 4) × 8² = (√3 / 4) × 64 = 16√3 cm²

Using the value of √3 = 1.732: Area of the equilateral triangle = 16 × 1.732 = 27.712 cm²

Total area of paper used = Area of the square + Area of the equilateral triangle = 512 + 27.712 = 539.712 cm²