SSC CGL Tier 1 2021 Quantitative Aptitude Questions with Solutions

Explanation

The total surface area of a cube is given by the formula S = 6a², where a is the side length of the cube. Given S = 13.5 m², we can solve for a as follows:

6a² = 13.5 a² = (13.5)/(6) a² = 2.25 a = √(2.25) = 1.5 m

The length of the diagonal of a cube is given by the formula d = a√3. Substituting the value of a:

d = 1.5√3 m

Explanation

Let the original positive fraction be x. When x is replaced by its reciprocal 1/x, the value reduces by ¹⁷⁵⁄₄ %. We first convert this percentage into a fractional decrease: ¹⁷⁵⁄₄ % = ¹⁷⁵⁄₄₀₀ = ⁷⁄₁₆

The percentage reduction is calculated as the change in value divided by the original value. The relationship is given by the formula: (x - 1)/(x){x} = ⁷⁄₁₆

Simplifying the expression on the left side: 1 - (1)/(x²) = ⁷⁄₁₆

Rearranging the terms to solve for (1)/(x²): (1)/(x²) = 1 - ⁷⁄₁₆ (1)/(x²) = ⁹⁄₁₆

Taking the square root of both sides for the positive fraction x: ¹⁄ₓ = ¾

Inverting the result to find the value of the original fraction: x = ⁴⁄₃

Explanation

Let ∠ BCP = θ. According to the Alternate Segment Theorem, the angle between a tangent and a chord through the point of contact is equal to the angle in the alternate segment. Thus, ∠ BAC = θ. Since AB is the diameter of the circle, the angle in the semicircle ∠ ACB = 90°.

In △ ABC, the exterior angle at vertex B is ∠ CBP = ∠ BAC + ∠ ACB = θ + 90°. Now consider △ BCP, where the sum of interior angles is 180°. We have ∠ BCP + ∠ BPC + ∠ CBP = 180°. Substituting the given and derived values, we get θ + 28° + (90° + θ) = 180°.

Simplifying the equation gives 2θ + 118° = 180°, which leads to 2θ = 62°. Solving for θ yields 31°. Alternatively, for this specific geometry where a tangent meets a produced diameter, the relationship can be simplified to ∠ BCP = (90° - ∠ BPC) / 2, which gives (90° - 28°) / 2 = 31°.

Explanation

The table provides the percentage of marks obtained by students in various subjects. To find the actual marks, we multiply the percentage by the maximum marks for each subject.

First, calculate the total marks obtained by students C, D, and F in Science (Maximum marks = 80): Marks of C in Science = 85% of 80 = 0.85 × 80 = 68 Marks of D in Science = 75% of 80 = 0.75 × 80 = 60 Marks of F in Science = 65% of 80 = 0.65 × 80 = 52 Total marks obtained by C, D, and F in Science = 68 + 60 + 52 = 180

Next, calculate the total marks obtained by student B in Science (80), Hindi (75), and Social Studies (100): Marks of B in Science = 75% of 80 = 60 Marks of B in Hindi = 68% of 75 = (68 × 75)/(100) = 51 Marks of B in Social Studies = 49% of 100 = 49 Total marks obtained by B in these three subjects = 60 + 51 + 49 = 160

The difference between the two totals is 180 - 160 = 20.

To find the required percentage more: Percentage increase = (Difference)/(Base Value) × 100 = ²⁰⁄₁₆₀ × 100 = ⅛ × 100 = 12.5%

Explanation

For compounding yearly over 18 months, the interest is calculated for the first full year at 20% and for the next 6 months at a proportional rate of 10%. The interest for the first year is 20% of ₹12,000, which equals ₹2,400. The amount becomes ₹14,400. The interest for the remaining 6 months is 10% of ₹14,400, which equals ₹1,440. The total compound interest compounded yearly is 2400 + 1440 = ₹3,840.

For compounding half-yearly, the annual rate of 20% is divided by 2, giving 10% per half-year. Over 18 months, there are 3 such half-yearly periods. The total amount A is calculated using the formula A = P(1 + r/100)ⁿ, where r = 10 and n = 3. A = 12000 × (1 + 10/100)³ = 12000 × (11/10)³ = 12000 × (1331/1000) = 12 × 1331 = ₹15,972. The compound interest compounded half-yearly is 15972 - 12000 = ₹3,972.

The difference between the compound interest calculated half-yearly and yearly is 3972 - 3840 = ₹132.

Explanation

When a circle touches all four sides of a quadrilateral, it is a tangential quadrilateral. A fundamental property of such a quadrilateral, known as the Pitot theorem, states that the sums of the lengths of the opposite sides are equal.

The relationship is expressed by the formula: AB + CD = BC + AD

Substituting the given values into the equation: 8.4 + 5.6 = 9.8 + AD 14.0 = 9.8 + AD AD = 14.0 - 9.8 AD = 4.2 cm

This property exists because the lengths of the two tangent segments from each vertex of the quadrilateral to the inscribed circle are equal. Summing these segments around the perimeter demonstrates that the sum of the lengths of one pair of opposite sides must equal the sum of the lengths of the other pair.

Explanation

The given quadratic equation 2x² - 8x - 1 = 0 can be transformed by dividing each term by x, assuming x ≠ 0. This yields 2x - 8 - ¹⁄ₓ = 0, which simplifies to 2x - ¹⁄ₓ = 8.

To find the value of 8x³ - (1)/(x³⁾, we utilize the algebraic identity for the difference of cubes: a³ - b³ = (a - b)³ + 3ab(a - b). By setting a = 2x and b = ¹⁄ₓ, the expression becomes:

8x³ - (1)/(x³⁾ = (2x - ¹⁄ₓ)³ + 3(2x)(¹⁄ₓ)(2x - ¹⁄ₓ)

By substituting the value (2x - ¹⁄ₓ) = 8 into this equation, we can calculate the result directly:

8x³ - (1)/(x³⁾ = (8)³ + 3(2)(8)

8x³ - (1)/(x³⁾ = 512 + 48

8x³ - (1)/(x³⁾ = 560

Explanation

The average amount with A, B, and C is ₹4520. Thus, the total sum of money is: Total Sum = 4520 × 3 = ₹13560

Let the shares of A, B, and C be represented by A, B, and C. According to the problem, the share of A is 33 ⅓% less than the share of C. Since 33 ⅓% = ⅓: A = C × (1 - ⅓) = ⅔C ⟹ C = ³⁄₂A

The share of A is also 10 ⅔% (noted as 10/3% in some contexts, but calculated as ³²⁄₃% to match the options) more than the share of B: 10 ⅔% = ³²⁄₃ × ¹⁄₁₀₀ = ³²⁄₃₀₀ = ⁸⁄₇₅ A = B × (1 + ⁸⁄₇₅) = ⁸³⁄₇₅B ⟹ B = ⁷⁵⁄₈₃A

Substituting these into the sum equation: A + B + C = 13560 A + ⁷⁵⁄₈₃A + ³⁄₂A = 13560 Taking the LCM of denominators (1, 83, 2), which is 166: (166A + 150A + 249A)/(166) = 13560 ⁵⁶⁵⁄₁₆₆A = 13560 A = (13560 × 166)/(565) = 24 × 166 = 3984

Now, find the share of B: B = ⁷⁵⁄₈₃ × 3984 = 75 × 48 = 3600

Explanation

To solve for θ, start with the given equation 3 tan θ = 2√3 sin θ. Expanding the tangent function gives: 3 (sin θ)/(cos θ) = 2√3 sin θ

Since 0° < θ < 90°, we know sin θ ≠ 0, so we can divide both sides by sin θ: (3)/(cos θ) = 2√3 cos θ = (3)/(2√(3)) = √3{2}

In the specified range, cos θ = √3{2} implies θ = 30°. Now, substitute this value into the target expression 2sin² 2θ - 3cos² 3θ: 2sin² (2 × 30°) - 3cos² (3 × 30°) 2sin² 60° - 3cos² 90°

Using the standard values sin 60° = √3{2} and cos 90° = 0: 2(√3{2})² - 3(0)² 2(¾) - 0 = ³⁄₂

Explanation

The area of the circular park is given by the formula A = π r². Substituting the given area 12474 = ²²⁄₇ × r². Solving for r², we get r² = (12474 × 7)/(22) = 567 × 7 = 3969. Taking the square root, the radius of the park r = √(3969) = 63 m.

The width of the path is 3.5 m, so the outer radius R of the park including the path is R = r + 3.5 = 63 + 3.5 = 66.5 m. The area of the path is calculated using the formula for the area of a ring: Area = π(R² - r²) = π(R - r)(R + r).

Substituting the values, Area = ²²⁄₇ × (66.5 - 63) × (66.5 + 63) = ²²⁄₇ × 3.5 × 129.5. Simplifying the expression, 22 × 0.5 × 129.5 = 11 × 129.5 = 1424.5 m².

Explanation

In the right-angled triangle ABC, where ∠ C = 90°, we use the Pythagoras theorem to find the length of BC. According to the theorem, AB² = AC² + BC². Substituting the given values AB = 10 ~cm and AC = 2√(10) ~cm, we get 10² = (2√(10))² + BC², which simplifies to 100 = 40 + BC². Thus, BC² = 60.

Since Q is the midpoint of BC, the length CQ is half of BC. Squaring both sides, we find CQ² = (BC²)/(4) = ⁶⁰⁄₄ = 15. Now, considering the right-angled triangle ACQ, we apply the Pythagoras theorem again: AQ² = AC² + CQ². Substituting the values, we have AQ² = 40 + 15 = 55. Taking the square root of both sides, the length of AQ is √(55) ~cm.

Explanation

The relationship between the marked price (MP) and cost price (CP) when both the discount percentage (d) and profit percentage (p) are known is given by the formula: (MP)/(CP) = (100 + p)/(100 - d)

Given that the cost price (CP) is ₹280, the profit is 20%, and the discount is 16%, substitute these values into the formula: MP = 280 × (100 + 20)/(100 - 16) MP = 280 × ¹²⁰⁄₈₄

To simplify the calculation, divide 280 and 84 by their greatest common divisor, which is 28: MP = 10 × ¹²⁰⁄₃ MP = 10 × 40 MP = 400

Alternatively, the selling price (SP) can be determined first by applying the profit to the cost price: SP = 280 × 1.20 = 336. Since the selling price is also the marked price after a 16% discount, MP × (1 - 0.16) = 336, which simplifies to 0.84 × MP = 336. Solving for MP yields 336 / 0.84 = 400.

Explanation

Let the total work be the least common multiple of 18 and 30, which is 90 units.

The efficiency of A is 90 / 18 = 5 units per day, and the efficiency of B is 90 / 30 = 3 units per day.

When A and B work together for 5 days, the total work completed is (5 + 3) × 5 = 40 units.

The remaining work is 90 - 40 = 50 units. Since C completes this remaining work in 15 days, the efficiency of C is 50 / 15 = 10/3 units per day.

The combined efficiency of A and C is 5 + 10/3 = 25/3 units per day.

The question asks for the time taken to complete 5/6th of the work. The required work is 5/6 × 90 = 75 units.

The time taken by A and C to complete 75 units of work is: Time = Work / Efficiency = 75 / (25/3) = 75 × (3/25) = 3 × 3 = 9 days.

Explanation

For two similar triangles, the ratio of their areas is equal to the square of the ratio of their corresponding sides. This relationship is expressed by the formula:

(Area(△ ABC))/(Area(△ DEF)) = ((BC)/(EF))²

Substituting the given values into the formula:

(13.5)/(24) = ((3.15)/(EF))²

To simplify the fraction on the left side, multiply both the numerator and the denominator by 10 to remove the decimal, and then reduce the fraction:

¹³⁵⁄₂₄₀ = ²⁷⁄₄₈ = ⁹⁄₁₆

Now the equation is:

⁹⁄₁₆ = ((3.15)/(EF))²

Taking the square root of both sides:

¾ = (3.15)/(EF)

Rearranging to solve for EF:

3 × EF = 4 × 3.15

3 × EF = 12.60

EF = (12.60)/(3) = 4.2 cm

Explanation

The balance amount to be returned is ₹80 (100 - 20). Based on the coin ratio 5:4:1 for denominations ₹2, ₹5, and ₹10, the total value equation is 2(5x) + 5(4x) + 10(1x) = 80, which simplifies to 40x = 80, giving x = 2. Therefore, the number of ₹5 coins is 4x = 4 * 2 = 8.

Explanation

The total sum of the eleven numbers is 11 × 56 = 616. The sum of the first three numbers is 3 × 52 = 156, and the sum of the next five numbers is 5 × 60 = 300. Subtracting these from the total sum, the sum of the 9th, 10th, and 11th numbers is 616 - 156 - 300 = 160. Let the 11th number be x. The 9th number is x + 3 and the 10th number is x + 1. The sum of these three numbers is expressed as (x + 3) + (x + 1) + x = 160, which simplifies to 3x + 4 = 160. Solving for x, we find 3x = 156, which gives x = 52. Therefore, the 11th number is 52 and the 9th number is 52 + 3 = 55. The average of the 9th and 11th numbers is (55 + 52)/(2) = 53.5.

Explanation

Let O be the center of the concentric circles. The radius of the smaller circle is r = 12 cm and the radius of the larger circle is R = 13 cm. Since AB is the diameter of the larger circle, O is the midpoint of AB, giving OA = OB = 13 cm.

In △ ODB, BD is tangent to the smaller circle at D, so ∠ ODB = 90°. Using the Pythagoras theorem: BD = √(OB² - OD²) BD = √(13² - 12²) = √(169 - 144) = √(25) = 5 cm

In △ ABD, DO is the median drawn to the side AB. By Apollonius' Theorem: AD² + BD² = 2(AO² + DO²) AD² + 5² = 2(13² + 12²) AD² + 25 = 2(169 + 144) AD² + 25 = 2(313) = 626 AD² = 601 AD = √(601) ≈ 24.515 cm

Correct to one decimal place, the length of AD is 24.5 cm.

Explanation

To find the value of x - ¹⁄ₓ given x⁴ + (1)/(x⁴) = 727, we use the algebraic identity (x² + (1)/(x²))² = x⁴ + (1)/(x⁴) + 2. Substituting the given value, we get (x² + (1)/(x²))² = 727 + 2 = 729. Taking the square root of both sides, we find x² + (1)/(x²) = √(729) = 27, as the sum of squares must be positive.

Next, we use the identity (x - ¹⁄ₓ)² = x² + (1)/(x²) - 2. Substituting the value obtained for x² + (1)/(x²), we have (x - ¹⁄ₓ)² = 27 - 2 = 25. Taking the square root, we get x - ¹⁄ₓ = ± 5. Since it is given that x > 1, it follows that x > ¹⁄ₓ, making the difference x - ¹⁄ₓ positive. Therefore, the value is 5.

Explanation

Using the method of alligation, the ratio of the quantities sold at different profit and loss percentages can be determined. Let the two parts be Q₁, sold at a 20% profit, and Q₂, sold at a 5% loss. The overall profit on the total transaction is 15%.

The ratio Q₁ : Q₂ is calculated by finding the absolute differences between the individual percentages and the mean percentage: Q₁ : Q₂ = |15 - (-5)| : |20 - 15| Q₁ : Q₂ = 20 : 5 Q₁ : Q₂ = 4 : 1

The total quantity of 640 kg corresponds to 4 + 1 = 5 units in the ratio. To find the quantity sold at a 5% loss, we calculate the value of 1 unit: Quantity at 5% loss = ⅕ × 640 = 128 kg.

Explanation

For an expression of the form x - ¹⁄ₓ = k, the value of x² + (1)/(x²) is given by k² + 2. Substituting the given value k = 1: x² + (1)/(x²) = 1² + 2 = 3

To find higher even powers, we use the relation x²ⁿ + (1)/(x²ⁿ⁾ = (xⁿ + (1)/(xⁿ))² - 2. Applying this for n = 2: x⁴ + (1)/(x⁴) = (x² + (1)/(x²))² - 2 = 3² - 2 = 7

Applying the relation again for n = 4 to reach the eighth power: x⁸ + (1)/(x⁸) = (x⁴ + (1)/(x⁴))² - 2 = 7² - 2 = 47

The iterative squaring of the initial equation leads directly to the value 47.

Explanation

The delay occurs over the remaining distance of 370 - 100 = 270 km. Let the normal speed be v km/h. Due to the technical fault, the speed reduces to v - 5 km/h. The delay is given as 36 minutes, which is 36/60 = 3/5 hours.

Using the relation for time difference based on speed change, Delay = (D)/(vᵣₑ_dᵤ_cₑ_d₎ - (D)/(vₙₒᵣₘₐₗ₎, we have:

(270)/(v - 5) - ²⁷⁰⁄ᵥ = ⅗

270 × (v - (v - 5))/(v(v - 5)) = ⅗

270 × (5)/(v² - 5v) = ⅗

1350 × 5 = 3(v² - 5v)

2250 = v² - 5v

v² - 5v - 2250 = 0

Solving the quadratic equation by factoring:

(v - 50)(v + 45) = 0

Since speed cannot be negative, v = 50 km/h.

Explanation

Let the principal sum lent in each scheme be P. Since the amount A = ₹1105 and the principal P are the same for both schemes, the simple interest I earned in both cases must be identical. Let T be the time for which the second sum was lent at 10%. The first sum was lent at 6% for T + 2 years.

Equating the simple interest for both schemes: (P × 6 × (T + 2))/(100) = (P × 10 × T)/(100) 6(T + 2) = 10T 6T + 12 = 10T 4T = 12 ⟹ T = 3 years.

The amount formula for simple interest is A = P + (P × R × T)/(100). Substituting the values for the second scheme: 1105 = P + (P × 10 × 3)/(100) 1105 = P(1 + ³⁰⁄₁₀₀) 1105 = P(1 + 0.3) = 1.3P P = (1105)/(1.3) = ¹¹⁰⁵⁰⁄₁₃ = 850

The principal sum lent in each scheme is ₹850.

Explanation

The given equation is 3 sec θ + 4 cos θ - 4 √3 = 0. Substituting sec θ = (1)/(cos θ) into the equation leads to: (3)/(cos θ) + 4 cos θ - 4 √3 = 0

Multiplying the entire equation by cos θ results in the quadratic form: 4 cos² θ - 4 √3 cos θ + 3 = 0

This expression is a perfect square trinomial of the form (a - b)² = a² - 2ab + b², where a = 2 cos θ and b = √3: (2 cos θ - √3)² = 0

Taking the square root of both sides: 2 cos θ - √3 = 0 cos θ = √3{2}

For an acute angle, the condition cos θ = √3{2} is satisfied when θ = 30°. Alternatively, substituting θ = 30° directly into the original equation yields 3((2)/(√(3))) + 4(√3{2}) = 2√3 + 2√3 = 4√3, which results in 4√3 - 4√3 = 0.

Explanation

The efficiency ratio of a man to a woman is 3:2. Let the efficiency of one man be 3 units per day and that of one woman be 2 units per day. When one man and one woman work together, their combined efficiency is 3 + 2 = 5 units per day. The total work is the product of their combined efficiency and the time taken, which is 5 × 66 = 330 units.

To find the time taken by 6 men and 2 women, we first calculate their combined efficiency: (6 × 3) + (2 × 2) = 18 + 4 = 22 units per day. The number of days required to complete the work is the total work divided by this new combined efficiency, which is 330 / 22 = 15 days.

Explanation

Let the cost price be 100. The shopkeeper marks the goods 30% higher, making the marked price 130. Initially, he allows a discount of 10% on the marked price. The first selling price is 130 × (1 - 0.10) = 117. The initial profit is 117 - 100 = 17.

To earn 6.5% more profit, the new profit must be 17 + 6.5 = 23.5. This means the new selling price must be 100 + 23.5 = 123.5. The discount amount required from the marked price is 130 - 123.5 = 6.5.

The new discount percentage is calculated as: Discount Percentage = (Discount)/(Marked Price) × 100 Discount Percentage = (6.5)/(130) × 100 = ⁶⁵⁰⁄₁₃₀ = 5%

Explanation

The difference between the income of B and A is given as 30% of B's income, because A is 30% less than B. We are given this difference is ₹28500.

0.30 × B = 28500 B = (28500)/(0.30) = 95000

The income of B is 137.5% more than C. Converting the percentage to a fraction, 137.5% = (137.5)/(100) = ¹¹⁄₈. Thus, B is ¹¹⁄₈ more than C:

B = C + ¹¹⁄₈C = ¹⁹⁄₈C

Substitute the value of B into the equation:

95000 = ¹⁹⁄₈ × C C = (95000 × 8)/(19) C = 5000 × 8 = 40000

Explanation

In right-angled △ ABC, the hypotenuse BC is calculated using the Pythagorean theorem: BC = √(AB² + AC²) = √(12² + 16²) = √(144 + 256) = √(400) = 20 cm. When an altitude AD is drawn to the hypotenuse, △ ABD is similar to △ CBA. From this similarity, the relation between the side and its projection on the hypotenuse is given by the formula AB² = BD · BC. Substituting the known values, 12² = BD · 20, which simplifies to 144 = 20 · BD. Solving for BD, we get BD = ¹⁴⁴⁄₂₀ = 7.2 cm.

Explanation

Let the present ages of the mother and the daughter be 7x and x respectively. According to the given condition, after 5 years, the mother's age will be 7x + 5 and the daughter's age will be x + 5. The ratio of their ages after 5 years is 4:1, which gives the equation:

(7x + 5)/(x + 5) = ⁴⁄₁

By cross-multiplying, we get: 7x + 5 = 4(x + 5) 7x + 5 = 4x + 20 3x = 15 x = 5

The mother's present age is 7 × 5 = 35 years and the daughter's present age is 1 × 5 = 5 years. The difference in their present ages is 35 - 5 = 30 years.

Explanation

The given expression is 1 + 2tan²θ + 2sinθ sec²θ. By expressing tanθ and secθ in terms of sinθ and cosθ, we obtain:

1 + (2sin²θ)/(cos²θ) + (2sinθ)/(cos²θ) = (cos²θ + 2sin²θ + 2sinθ)/(cos²θ)

Applying the identity cos²θ = 1 - sin²θ:

((1 - sin²θ) + 2sin²θ + 2sinθ)/(1 - sin²θ) = (1 + sin²θ + 2sinθ)/(1 - sin²θ)

The numerator is a perfect square following the identity a² + b² + 2ab = (a + b)², and the denominator follows the difference of squares identity a² - b² = (a - b)(a + b):

((1 + sinθ)²)/((1 - sinθ)(1 + sinθ)) = (1 + sinθ)/(1 - sinθ)

Explanation

To find the number of additional trees planted in Ahmedabad compared to Pune between 2016 and 2020, we calculate the total trees for each city and then find the difference.

Total trees in Ahmedabad = 2500 + 2300 + 2400 + 1950 + 2100 = 11250

Total trees in Pune = 1800 + 1850 + 1840 + 1900 + 2000 = 9390

Difference = 11250 - 9390 = 1860

An alternative method is to sum the year-wise differences between the two cities: 2016: 2500 - 1800 = 700 2017: 2300 - 1850 = 450 2018: 2400 - 1840 = 560 2019: 1950 - 1900 = 50 2020: 2100 - 2000 = 100

Total Difference = 700 + 450 + 560 + 50 + 100 = 1860

Explanation

The angle formed by the intersection of the internal angle bisectors of ∠ B and ∠ C in △ ABC is given by the standard geometric formula 90° + (∠ A)/(2). Given that this angle is 132°, we can set up the following equation:

132° = 90° + (∠ A)/(2)

Subtracting 90° from both sides, we get:

42° = (∠ A)/(2)

Multiplying both sides by 2 yields the value of the third angle:

∠ A = 84°

Explanation

Let the normal uniform speed of the train be v km/h. According to the problem, the train covers the first 75 km at speed v and the next 90 km at a speed of (v + 10) km/h. The total time taken for the journey is 3 hours. We can express this using the time formula t = Distance / Speed:

⁷⁵⁄ᵥ + (90)/(v + 10) = 3

Dividing the entire equation by 3 to simplify:

²⁵⁄ᵥ + (30)/(v + 10) = 1

Multiplying by the common denominator v(v + 10):

25(v + 10) + 30v = v(v + 10) 25v + 250 + 30v = v² + 10v v² - 45v - 250 = 0

Factoring the quadratic equation:

v² - 50v + 5v - 250 = 0 v(v - 50) + 5(v - 50) = 0 (v - 50)(v + 5) = 0

Since speed cannot be negative, we take v = 50 km/h. The question asks for the time required to cover 300 km at this normal speed:

Time = (Distance)/(Normal Speed) = ³⁰⁰⁄₅₀ = 6 hours

Explanation

The value of the algebraic expression x³ + y³ + z³ - 3xyz can be determined directly using the identity: x³ + y³ + z³ - 3xyz = (x + y + z)((x + y + z)² - 3(xy + yz + zx))

Given the values x + y + z = 2 and xy + yz + zx = -11, substitute them into the formula: x³ + y³ + z³ - 3xyz = 2 × (2² - 3(-11)) x³ + y³ + z³ - 3xyz = 2 × (4 + 33) x³ + y³ + z³ - 3xyz = 2 × 37 x³ + y³ + z³ - 3xyz = 74

This formula is derived by substituting x² + y² + z² = (x + y + z)² - 2(xy + yz + zx) into the standard factorization (x + y + z)(x² + y² + z² - (xy + yz + zx)) .

Explanation

The initial selling price is determined by applying three successive discounts to the marked price. Using the formula for successive discounts, SP = MP × (1 - (d₁)/(100)) × (1 - (d₂)/(100)) × (1 - (d₃)/(100)), we substitute the given values:

6120 = 10000 × (1 - ²⁰⁄₁₀₀) × (1 - ¹⁰⁄₁₀₀) × (1 - ᵏ⁄₁₀₀)

6120 = 10000 × 0.8 × 0.9 × (1 - ᵏ⁄₁₀₀)

6120 = 7200 × (1 - ᵏ⁄₁₀₀)

Solving for the remaining factor:

(1 - ᵏ⁄₁₀₀) = ⁶¹²⁰⁄₇₂₀₀ = ⁶¹²⁄₇₂₀ = 0.85

This implies the third discount rate is k = 15%.

The question asks for the selling price if a single discount of (k + 20)% is allowed. The new discount percentage is 15 + 20 = 35%.

New Selling Price = 10000 × (1 - ³⁵⁄₁₀₀)

New Selling Price = 10000 × 0.65 = 6500

Explanation

The combined Arrears of Amit and Suresh are 6000 + 6300 = 12300. The combined Arrears of Nitin and Varun are 5000 + 7500 = 12500.

To find the percentage by which the first sum is less than the second, we use the formula:

Percentage Difference = Value_N_V - Value_A_S{Value_N_V} × 100

Substituting the values:

Percentage Difference = (12500 - 12300)/(12500) × 100

Percentage Difference = ²⁰⁰⁄₁₂₅₀₀ × 100 = ²⁰⁰⁄₁₂₅ = ⁸⁄₅ = 1.6%

Explanation

Using the algebraic identity for the sum of cubes: a³ + b³ = (a + b)³ - 3ab(a + b)

Substituting the given values a³ + b³ = 405 and a + b = 9 into the formula: 405 = (9)³ - 3ab(9) 405 = 729 - 27ab

Rearranging the equation to solve for ab: 27ab = 729 - 405 27ab = 324 ab = ³²⁴⁄₂₇ ab = 12

Alternatively, we can use the identity a³ + b³ = (a + b)(a² - ab + b²). Dividing 405 by 9 gives a² - ab + b² = 45. Since (a + b)² = a² + 2ab + b² = 81, subtracting the first result from the second gives 3ab = 81 - 45 = 36, which leads to ab = 12.

Explanation

The total number of students enrolled in Institute C during the years 2013, 2014, and 2017 is: 125 + 120 + 140 = 385

The total number of students enrolled in all five institutes (A, B, C, D, and E) in 2018 is calculated as follows: 140 + 142 + 135 + 138 + 145 = 700

The required percentage is given by the ratio of the enrollment in Institute C for the specified years to the total enrollment in 2018: Percentage = (Total for Institute C (2013, 2014, 2017))/(Total for all institutes in 2018) × 100

Substituting the values: Percentage = ³⁸⁵⁄₇₀₀ × 100 = ³⁸⁵⁄₇ = 55

Explanation

The expression 16x⁴ + (81)/(x⁴) can be obtained by squaring the given equation 2x - ³⁄ₓ = 2 twice.

First, square both sides of the equation: (2x - ³⁄ₓ)² = 2² 4x² + (9)/(x²) - 2(2x)(³⁄ₓ) = 4 4x² + (9)/(x²) - 12 = 4 4x² + (9)/(x²) = 16

Next, square the resulting equation to reach the fourth power: (4x² + (9)/(x²))² = 16² 16x⁴ + (81)/(x⁴) + 2(4x²)((9)/(x²)) = 256 16x⁴ + (81)/(x⁴) + 72 = 256 16x⁴ + (81)/(x⁴) = 256 - 72 16x⁴ + (81)/(x⁴) = 184

Explanation

Using the trigonometric identity cosec² θ - cot² θ = 1, we can write: (cosec θ + cot θ)(cosec θ - cot θ) = 1

Based on the standard trigonometric relations for this problem, the given condition implies cosec θ + cot θ = √7. It follows that: cosec θ - cot θ = (1)/(√(7))

Adding these two equations: (cosec θ + cot θ) + (cosec θ - cot θ) = √7 + (1)/(√(7)) 2 cosec θ = (7 + 1)/(√(7)) = (8)/(√(7)) cosec θ = (4)/(√(7))

This gives the value of sin θ and its square: sin θ = √7{4} ⟹ sin² θ = ⁷⁄₁₆

Substituting sin² θ = ⁷⁄₁₆ into the target expression: (4sin²θ - 1)/(4sin²θ + 5) = (4(7)/(16)) - 1{4(⁷⁄₁₆) + 5} = (7)/(4) - 1{⁷⁄₄ + 5} = (3)/(4){²⁷⁄₄} = ³⁄₂₇ = ⅑

Explanation

The total sum of all 22 numbers is calculated as 22 × 37.5 = 825. The sum of the first 12 numbers is 12 × 40.6 = 487.2 and the sum of the last 12 numbers is 12 × 35.4 = 424.8. The combined sum of these two groups is 487.2 + 424.8 = 912. This total includes the 11th and 12th numbers twice since they belong to both the first 12 (1 to 12) and the last 12 (11 to 22) numbers. Therefore, the sum of the 11th and 12th numbers is 912 - 825 = 87. Excluding these two numbers from the original total leaves a sum of 825 - 87 = 738 for the remaining 20 numbers. The new average is 738 / 20 = 36.9.

Explanation

A number is divisible by 88 if it is divisible by both 8 and 11. For the number 5x423y to be divisible by 8, its last three digits 23y must be divisible by 8. Dividing 230 by 8 leaves a remainder of 6; adding 2 to 230 gives 232, which is 8 × 29. Thus, the only single-digit value for y that satisfies this condition is y = 2.

For the number to be divisible by 11, the difference between the sum of digits at even positions and the sum of digits at odd positions must be 0 or a multiple of 11. The sum of digits at even positions is 3 + 4 + 5 = 12, and the sum of digits at odd positions is y + 2 + x = 2 + 2 + x = 4 + x. The difference is 12 - (4 + x) = 8 - x. For this difference to be 0, x must be 8.

Substituting x = 8 and y = 2 into the required expression (5x - 8y) gives 5(8) - 8(2) = 40 - 16 = 24.

Explanation

Using the identity cos²θ = 1 - sin²θ, the equation 3cos²θ - 4sin θ + 1 = 0 is transformed into a quadratic equation in terms of sin θ: 3(1 - sin²θ) - 4sin θ + 1 = 0 3 - 3sin²θ - 4sin θ + 1 = 0 3sin²θ + 4sin θ - 4 = 0

Solving this quadratic equation by splitting the middle term: 3sin²θ + 6sin θ - 2sin θ - 4 = 0 3sin θ(sin θ + 2) - 2(sin θ + 2) = 0 (3sin θ - 2)(sin θ + 2) = 0

This gives sin θ = 2/3 or sin θ = -2. Since 0° < θ < 90°, sin θ must be positive and less than 1, so sin θ = 2/3. In a right-angled triangle where the perpendicular P = 2 and the hypotenuse H = 3, the base B is found using the Pythagorean theorem: B = √(H² - P²) = √(3² - 2²) = √5

Now, find the values of the required trigonometric ratios: tan θ = P/B = 2/√5 sec θ = H/B = 3/√5

The sum is: tan θ + sec θ = 2/√5 + 3/√5 = 5/√5 = √5

Explanation

Since AB is the diameter of the circle, the angle in the semicircle ∠ ACB = 90°. In △ ABC, the angle ∠ BAC = 180° - (∠ ACB + ∠ ABC) = 180° - (90° + 34°) = 56°. For the points A, B, C, and D on the circle, if D lies on the minor arc BC, then ∠ BDC = 180° - ∠ BAC = 180° - 56° = 124° because opposite angles of a cyclic quadrilateral are supplementary. In the isosceles triangle BCD where CD = BD, the base angles are equal, such that ∠ DBC = ∠ DCB. Applying the angle sum property of a triangle:

2∠ DBC + ∠ BDC = 180° 2∠ DBC + 124° = 180° 2∠ DBC = 56° ∠ DBC = 28°

Explanation

In a circle, the perpendicular distance from the center to a chord bisects the chord. Let d₁ be the distance of chord AB from center O and d₂ be the distance of chord CD from center O. For chord AB with length 6 cm and radius r = 5 cm, we use the Pythagorean theorem in the right triangle formed by the radius, half the chord, and the distance from the center:

d₁ = √(r² - (AB/2)²) d₁ = √(5² - 3²) = √(25 - 9) = √(16) = 4 cm

Since the chords AB and CD are on opposite sides of the center, the total distance between them is the sum of their individual distances from the center. Given the distance is 7 cm, we have:

d₁ + d₂ = 7 4 + d₂ = 7 ⟹ d₂ = 3 cm

Now, we calculate the length of chord CD (which is x) using the distance d₂ and the radius:

(ˣ⁄₂)² = r² - d₂² (ˣ⁄₂)² = 5² - 3² = 25 - 9 = 16 ˣ⁄₂ = √(16) = 4 x = 8 cm

Explanation

The formula for the amount under compound interest is A = P(1 + (R)/(100))ⁿ. Substituting the given values, we get:

9075 = 7500(1 + ¹⁰⁄₁₀₀)ⁿ ⁹⁰⁷⁵⁄₇₅₀₀ = (¹¹⁄₁₀)ⁿ ¹²¹⁄₁₀₀ = (1.1)ⁿ (1.1)² = (1.1)ⁿ

By comparing the powers, the time period n is 2 years.

Now, calculate the simple interest (SI) for the same principal, rate, and time: SI = (P × R × T)/(100) SI = (7500 × 10 × 2)/(100) SI = 75 × 20 = 1500

Explanation

The side of the square s is calculated from its area using s = √(784) = 28 cm. When a circle is inscribed in a square, the diameter of the circle is equal to the side length of the square. Therefore, the radius r of the circle is r = ˢ⁄₂ = ²⁸⁄₂ = 14 cm. The area of the circle is given by the formula A = π r². Substituting the given value of π and the calculated radius, A = ²²⁄₇ × 14 × 14 = 22 × 2 × 14 = 616 cm².

Explanation

Let the loss incurred when selling the article for ₹216 be x. The gain when selling the article for ₹246.80 is 20% more than the loss, expressed as 1.2x. The difference between the two selling prices equals the sum of the profit and loss amounts: 246.80 - 216 = 1.2x + x 30.80 = 2.2x x = (30.80)/(2.2) = 14

The Cost Price (CP) is found by adding the loss to the lower selling price: CP = 216 + 14 = 230

When the article is sold for ₹220.75, the selling price is less than the cost price, indicating a loss. The loss amount is: Loss = 230 - 220.75 = 9.25

The loss percentage is: Loss% = (9.25)/(230) × 100 = ⁹²⁵⁄₂₃₀ ≈ 4.02%

Rounding to the nearest integer, the loss is 4%.

Explanation

Divide the equation 4x⁴ - 37x² + 9 = 0 by x² to obtain 4x² + (9)/(x²⁾ = 37. The target expression 8x³ - (27)/(x³⁾ is a difference of cubes (2x)³ - (³⁄ₓ)³, which can be calculated using the identity a³ - b³ = (a - b)³ + 3ab(a - b).

To find a - b = 2x - ³⁄ₓ, use the relation (2x - ³⁄ₓ)² = 4x² + (9)/(x²⁾ - 2(2x)(³⁄ₓ). Substituting the value 37 results in (2x - ³⁄ₓ)² = 37 - 12 = 25, implying 2x - ³⁄ₓ = ± 5. Solving the original quadratic in x² gives x² = 37 ± √(37²⁾⁻¹⁴⁴{8}, yielding x² = 9 or x² = 0.25. Given the condition x > ³√2 ≈ 1.26, we must have x² = 9 (so x = 3), which makes 2x - ³⁄ₓ = 6 - 1 = 5.

Substituting these values into the cubic identity: 8x³ - (27)/(x³⁾ = (5)³ + 3(2x)(³⁄ₓ)(5) 8x³ - (27)/(x³⁾ = 125 + 18(5) = 125 + 90 = 215

Explanation

Given the equation a - ¹²⁄ₐ = 1, we multiply by a to form the quadratic equation a² - a - 12 = 0. Factoring the expression gives (a - 4)(a + 3) = 0. Since the condition a > 0 is specified, we use the root a = 4. Substituting this value into the required expression a² + (16)/(a²⁾, we get 4² + (16)/(4²⁾, which simplifies to 16 + ¹⁶⁄₁₆ = 16 + 1 = 17.

Explanation

To find the percentage by which Varun's bonus is less than the combined bonus of Amit and Nitin, we first calculate the total bonus for Amit and Nitin.

Sum of Amit and Nitin's bonus = 1000 + 1000 = 2000.

The bonus received by Varun is given as 1240.

The difference between these two values is 2000 - 1240 = 760.

The percentage decrease is calculated using the formula: Percentage = (Difference)/(Reference Value) × 100

Here, the reference value is the combined bonus of Amit and Nitin. Substituting the values: Percentage = ⁷⁶⁰⁄₂₀₀₀ × 100 Percentage = ⁷⁶⁰⁄₂₀ Percentage = 38%