SSC CGL Tier 1 2021 Quantitative Aptitude Questions with Solutions

Explanation

Since ∠ APB = 122° is obtuse, the point P must lie on the minor arc AB of the circle. Let Q be any point on the major arc AB. In the cyclic quadrilateral APBQ, the sum of opposite angles is 180°. Therefore, ∠ AQB = 180° - ∠ APB = 180° - 122° = 58°.

The angle subtended by a chord at the center is twice the angle subtended by it at any point on the major arc. Thus, the central angle ∠ AOB = 2 × ∠ AQB = 2 × 58° = 116°. In △ OAB, the sides OA and OB are radii of the circle, making it an isosceles triangle where ∠ OAB = ∠ OBA.

The sum of angles in △ OAB is 180°. Substituting the values, we get ∠ OAB + ∠ OBA + ∠ AOB = 180°, which simplifies to 2 ∠ OAB + 116° = 180°. Solving for the angle, 2 ∠ OAB = 64°, which gives ∠ OAB = 32°.

Explanation

In △ PQR, given that EF ∥ QR, △ PEF and △ PQR are similar triangles by the AA similarity criterion because ∠ P is common and ∠ PEF = ∠ PQR as corresponding angles. For similar triangles, the ratio of their areas is equal to the square of the ratio of their corresponding sides.

The length of side PQ is the sum of PE and EQ, so PQ = PE + EQ = 2 + 3 = 5 cm. The ratio of corresponding sides between the larger triangle PQR and the smaller triangle PEF is (PQ)/(PE) = ⁵⁄₂.

The ratio of the areas is calculated as:

(area(△ PQR))/(area(△ PEF)) = ((PQ)/(PE))² = (⁵⁄₂)² = ²⁵⁄₄

This results in the ratio 25:4.

Explanation

Let n be the initial number of students. Each of the 4 new students weighs 60 kg, which is 4.5 kg more than the initial average weight of 55.5 kg (60 - 55.5 = 4.5). Consequently, the total surplus weight brought into the class by these new students is 4 × 4.5 = 18 kg.

This total surplus weight is distributed across all the students now present in the class (n + 4) to result in an average weight increase of 360 g, or 0.36 kg. Using the deviation method, the total surplus must equal the total increase across all individuals: Total Surplus = (Total number of students) × Increase in average 18 = (n + 4) × 0.36

Rearranging the equation to solve for n: n + 4 = (18)/(0.36) n + 4 = ¹⁸⁰⁰⁄₃₆ n + 4 = 50 n = 46

The initial number of students in the class was 46.

Explanation

The sides of the triangular plot are 12 m, 35 m, and 37 m. Since 12² + 35² = 144 + 1225 = 1369 = 37², the triangle is right-angled. The area of this triangle is given by ½ × 12 × 35 = 210 m².

Let the sides of the rectangular field be 7x and 3x. The area of the rectangle is 7x × 3x = 21x². Given that the area of the rectangular field is equal to the area of the triangular plot, we have 21x² = 210, which simplifies to x² = 10, so x = √(10).

The perimeter of the rectangular field is 2(7x + 3x) = 2(10x) = 20x. Substituting the value of x, the perimeter is 20√(10) m.

Explanation

The line joining the centers of two intersecting circles is the perpendicular bisector of their common chord. Let C₁ and C₂ be the centers of the circles with radii r₁ = 15 cm and r₂ = 37 cm. Let the common chord AB have a length of 24 cm, and let M be the intersection point of the chord and the line connecting the centers.

Since C₁ C₂ bisects AB perpendicularly, AM = ½ AB = ²⁴⁄₂ = 12 cm. In the right-angled triangle AMC₁, using the Pythagorean theorem, the distance from the first center to the chord is C₁ M = √(r₁² - AM²) = √(15² - 12²) = √(225 - 144) = √(81) = 9 cm.

Similarly, in the right-angled triangle AMC₂, the distance from the second center to the chord is C₂ M = √(r₂² - AM²) = √(37² - 12²) = √(1369 - 144) = √(1225) = 35 cm. The distance between the centers is the sum of these segments: C₁ C₂ = C₁ M + C₂ M = 9 + 35 = 44 cm.

Explanation

Using the algebraic identity x³ - y³ = (x - y)((x - y)² + 3xy), we substitute the given values 316 = 4(4² + 3xy). Simplifying this gives 79 = 16 + 3xy, which leads to 3xy = 63 or xy = 21.

We now have a system of equations: x - y = 4 and xy = 21. Seeking two numbers whose difference is 4 and product is 21, we find x = 7 and y = 3. We verify that y > 0 and x³ - y³ = 7³ - 3³ = 343 - 27 = 316, which matches the problem statement.

The value of x⁴ - y⁴ is calculated as 7⁴ - 3⁴. Computing the powers, we get 2401 - 81 = 2320.

Explanation

A number that is exactly divisible by 7, 11, and 13 must be divisible by their least common multiple. Since 7, 11, and 13 are prime numbers, their LCM is 7 × 11 × 13 = 1001. Any six-digit number formed by repeating a three-digit number abc in the form abcabc is always divisible by 1001, because abcabc = abc × 1000 + abc = abc × (1000 + 1) = abc × 1001.

Given the six-digit number is 823p2q, it must follow the abcabc pattern to be divisible by 1001. By comparing 823p2q with the pattern abcabc, we identify a = 8, b = 2, and c = 3. This means the number is 823823. Comparing the digits, we find p = 8 and q = 3. The required value is p - q = 8 - 3 = 5.

Explanation

The percentage of sale relative to production is determined using the formula P = (Sale / Production) × 100. For the year 2015, the calculation is (1000 / 1250) × 100 = 80%. For the year 2019, the calculation is (1390 / 1600) × 100 = 86.875%. Both of these percentages satisfy the condition of being 80% or more but less than 90%.

Explanation

To evaluate the given expression, we simplify the numerator and denominator separately using trigonometric identities and standard values. The numerator is tan 13° tan 36° tan 45° tan 54° tan 77°. By applying the identity tan θ · tan(90° - θ) = 1, we can group complementary angles. Specifically, tan 13° tan 77° = tan 13° cot 13° = 1 and tan 36° tan 54° = tan 36° cot 36° = 1. Since tan 45° = 1, the numerator simplifies to 1 · 1 · 1 = 1.

For the denominator, we substitute the standard values sec 60° = 2, sin 60° = √3{2}, and cos 60° = ½ into the expression 2 sec² 60° (sin² 60° - 3 cos 60° + 2). This yields 2(2)² [ (√3{2})² - 3(½) + 2 ], which simplifies to 8 (¾ - ³⁄₂ + 2). Combining the terms inside the parentheses gives 8 ((3 - 6 + 8)/(4)) = 8 (⁵⁄₄) = 10. Dividing the simplified numerator by the simplified denominator results in ⅒.

Explanation

The table provides the percentage of marks obtained by each student. To find the average marks of students B, C, D, and F in Math, first calculate the average percentage for these four students.

Average percentage = (92 + 80 + 74 + 84)/(4) = ³³⁰⁄₄ = 82.5%

The maximum marks for Math is 150. Therefore, the average marks are calculated by applying the average percentage to the total marks:

Average Marks = 82.5% of 150 = (82.5)/(100) × 150 = 0.825 × 150 = 123.75

Alternatively, the individual marks for B, C, D, and F are 138 (92% of 150), 120 (80% of 150), 111 (74% of 150), and 126 (84% of 150) respectively. The sum of these marks is 138 + 120 + 111 + 126 = 495. Dividing the total marks by the number of students gives the average as ⁴⁹⁵⁄₄ = 123.75.

Explanation

The total percentage difference between an 8% loss and a 4% profit is 12%, as these represent deviations on opposite sides of the cost price. This 12% of the cost price (CP) is equal to the ₹300 increase in the selling price.

By setting the CP as 100%, the initial selling price at an 8% loss is 92%, and the second selling price at a 4% profit is 104%. The difference between these two states is 104% - 92% = 12%.

Using the formula for the cost price: CP = (Difference in Price)/(Total Percentage Change) × 100 CP = ³⁰⁰⁄₁₂ × 100 CP = 25 × 100 CP = 2500

This direct calculation shows that the 12% shift caused by the ₹300 increase identifies the base cost price.

Explanation

Let the two numbers be 2x and 3x. Based on the first condition where 5 is subtracted from the first number and 6 is added to the second number to yield a ratio of 5:12, we set up the equation:

(2x - 5)/(3x + 6) = ⁵⁄₁₂

Cross-multiplying gives: 12(2x - 5) = 5(3x + 6) 24x - 60 = 15x + 30 9x = 90 x = 10

The original numbers are 2(10) = 20 and 3(10) = 30. If 8 is added to each of these numbers, the first number becomes 20 + 8 = 28 and the second number becomes 30 + 8 = 38. The resulting ratio is 28:38. Dividing both terms by their greatest common divisor, 2, the simplified ratio is 14:19.

Explanation

The upstream speed of the boat is calculated by dividing the distance by the time taken, giving 30 km / 3 hours = 10 km/h. Similarly, the downstream speed is 30 km / 1 hour = 30 km/h.

The speed of the boat in still water is the average of the downstream and upstream speeds, which is (30 + 10)/(2) = 20 km/h. To find the time required to cover 60 km in still water, divide the distance by this speed: 60 km / 20 km/h = 3 hours.

Explanation

To simplify the expression 8÷ 4of 2 - 15÷ 2of 5 - 6÷ 5× (-7 + 5)of 2, follow the BODMAS rule (Brackets, Of, Division, Multiplication, Addition, Subtraction).

First, simplify the expression inside the bracket: (-7 + 5) = -2

The expression now becomes: 8 ÷ 4 of 2 - 15 ÷ 2 of 5 - 6 ÷ 5 × (-2) of 2

Next, evaluate the "of" operations, which have higher priority than division and multiplication: 4 of 2 = 4 × 2 = 8 2 of 5 = 2 × 5 = 10 (-2) of 2 = (-2) × 2 = -4

Substitute these values back into the expression: 8 ÷ 8 - 15 ÷ 10 - 6 ÷ 5 × (-4)

Now, perform the division operations: 8 ÷ 8 = 1 15 ÷ 10 = 1.5 6 ÷ 5 = 1.2

The expression is now: 1 - 1.5 - 1.2 × (-4)

Next, perform the multiplication: 1.2 × (-4) = -4.8

Substitute this result: 1 - 1.5 - (-4.8) 1 - 1.5 + 4.8

Finally, solve the addition and subtraction from left to right: -0.5 + 4.8 = 4.3

In fractional form, 4.3 is equal to ⁴³⁄₁₀.

Explanation

To find the districts where the number of female teachers exceeds the number of male teachers by more than 500, we calculate the difference Female - Male for each district and compare it to 500.

In the East district, the number of female teachers is 2375 and the number of male teachers is 1650. The difference is 2375 - 1650 = 725. Since 725 > 500, the East district satisfies the condition.

In the North district, the number of female teachers is 2651 and the number of male teachers is 1075. The difference is 2651 - 1075 = 1576. Since 1576 > 500, the North district also satisfies the condition.

Explanation

To find the required percentage, first calculate the actual marks obtained by students F and A in the specified subjects based on the percentage of maximum marks provided in the table.

Marks obtained by student F: English: 72% of 50 = 36 Science: 65% of 80 = 52 Hindi: 60% of 75 = 45 Total marks for student F = 36 + 52 + 45 = 133

Marks obtained by student A: English: 70% of 50 = 35 Mathematics: 90% of 150 = 135 Science: 65% of 80 = 52 Hindi: 64% of 75 = 48 Total marks for student A = 35 + 135 + 52 + 48 = 270

The required percentage is calculated as follows: Percentage = (Total marks of F)/(Total marks of A) × 100 Percentage = ¹³³⁄₂₇₀ × 100 = ¹³³⁰⁄₂₇ ≈ 49.259%

Rounding to one decimal place, we get 49.3%.

Explanation

The interest rate is 10% per annum, which is 1/10 in fractional form. For a loan repaid in equal annual installments, the present value P of the loan is calculated using the formula P = (I)/((1 + r)) + (I)/((1 + r)²), where I is the installment and r is the rate.

Using the ratio method, the relationship between principal and installment for two years at 10% is: Year 1: Principal 10 → Installment 11 Year 2: Principal 10² → Installment 11² (i.e., 100 → 121)

Since the installments are equal, we multiply the first year ratio by 11 to make both installments 121 units. Year 1: Principal 110 → Installment 121 Year 2: Principal 100 → Installment 121

Total Principal = 110 + 100 = 210 units Total Amount Paid = 2 × 121 = 242 units Total Interest = 242 - 210 = 32 units

Given each installment is ₹6534: 121 units = 6534 1 unit = ⁶⁵³⁴⁄₁₂₁ = 54

Total Interest = 32 × 54 = 1728

Explanation

The expansion of a binomial expression (ax + b)ⁿ is given by the formula ₖ₌₀ⁿ n{k} (ax)ⁿ⁻ᵏ (b)ᵏ. To find the coefficient of x in the expansion of (3x - 4)³, we look for the term where the power of x is 1.

Using the algebraic identity (A - B)³ = A³ - 3A²B + 3AB² - B³, we can substitute A = 3x and B = 4. The term that contains x to the first power is 3AB².

Substituting the values into this specific term: Term = 3 × (3x) × (4)² Term = 3 × 3x × 16 Term = 9 × 16x Term = 144x

By comparing this result to the general form, the numerical factor multiplying x is 144.

Explanation

In an equilateral △ ABC, the side length s = 3√7 cm and each interior angle, including ∠ B, is 60°. Since P divides the side BC in the ratio BP : PC = 1 : 2, the length of segment BP is ¹⁄₁₊₂ × BC = ⅓ × 3√7 = √7 cm.

To find the length of AP, we apply the Law of Cosines in △ ABP:

AP² = AB² + BP² - 2(AB)(BP)cos(∠ B)

Substituting the values AB = 3√7, BP = √7, and cos(60°) = ½:

AP² = (3√7)² + (√7)² - 2(3√7)(√7)(½) AP² = (9 × 7) + 7 - (3 × 7) AP² = 63 + 7 - 21 AP² = 70 - 21 = 49

Taking the square root of both sides, we find AP = √(49) = 7 cm.

Explanation

The expression is sec⁴ θ (1 - sin⁴ θ) - 2 tan² θ. Using the algebraic identity a² - b² = (a - b)(a + b), we rewrite the term (1 - sin⁴ θ) as (1 - sin² θ)(1 + sin² θ). Substituting the trigonometric identity 1 - sin² θ = cos² θ, the expression becomes sec⁴ θ · cos² θ (1 + sin² θ) - 2 tan² θ.

Since sec⁴ θ · cos² θ = (1)/(cos⁴ θ) · cos² θ = sec² θ, the expression simplifies to sec² θ (1 + sin² θ) - 2 tan² θ. Expanding this gives sec² θ + sec² θ sin² θ - 2 tan² θ.

Recognizing that sec² θ sin² θ = (1)/(cos² θ) · sin² θ = tan² θ, we can rewrite the expression as sec² θ + tan² θ - 2 tan² θ. This further simplifies to sec² θ - tan² θ. Applying the fundamental identity sec² θ - tan² θ = 1 yields the final result of 1.

Explanation

To find the value of the given expression, we use the standard algebraic identity x³ + y³ + z³ - 3xyz = (x + y + z)((x + y + z)² - 3(xy + yz + zx)).

Substituting the given values x + y + z = 3, xy + yz + zx = -12, and xyz = -16 into the identity: x³ + y³ + z³ - 3(-16) = 3 × (3² - 3(-12)) x³ + y³ + z³ + 48 = 3 × (9 + 36) x³ + y³ + z³ + 48 = 3 × 45 x³ + y³ + z³ + 48 = 135 x³ + y³ + z³ = 135 - 48 = 87

The problem asks for the value of √(x³ + y³ + z³ + 13). Substituting the calculated value of x³ + y³ + z³: √(87 + 13) = √(100) = 10

Explanation

The work rate of A is determined by the total hours taken to complete one unit of work, which is 2 × 10 = 20 hours. The work rate of B is calculated from the given information that 5 units of work are completed in 15 × 10 = 150 hours; thus, B takes ¹⁵⁰⁄₅ = 30 hours to complete one unit of work.

When working together, their combined hourly rate is the sum of their individual rates: Rateₜₒₜₐₗ = ¹⁄₂₀ + ¹⁄₃₀ = (3 + 2)/(60) = ⁵⁄₆₀ = ¹⁄₁₂ work per hour.

The total time required for them to complete the work together is the reciprocal of the combined rate, which is 12 hours. Since the problem defines a standard work day as 10 hours, the additional time required beyond one full day is 12 - 10 = 2 hours.

Explanation

The ratio of the number of girls to the number of boys in the class is 60:40, which simplifies to 3:2. To find the average marks of the whole class, we can use the weighted average formula based on these ratios:

Average = (Ratio_gᵢᵣₗₛ × Average_gᵢᵣₗₛ) + (Ratio_bₒ_yₛ × Average_bₒ_yₛ){Ratio_gᵢᵣₗₛ + Ratio_bₒ_yₛ}

Substituting the given average marks into the formula:

Average = ((3 × 70) + (2 × 63))/(3 + 2)

Average = (210 + 126)/(5)

Average = ³³⁶⁄₅ = 67.2

This approach effectively combines the individual group averages according to their respective strengths in the total population of 90 students.

Explanation

In a cyclic quadrilateral, angles subtended by the same arc at the circumference are equal. Therefore, ∠ BAC = ∠ BDC because both angles are subtended by the arc BC.

To find ∠ BDC, consider △ CED. The angle ∠ BEC is the exterior angle to △ CED at vertex E. According to the exterior angle theorem, the measure of an exterior angle of a triangle is equal to the sum of the measures of its two opposite interior angles. In this triangle, we have:

∠ BEC = ∠ ECD + ∠ EDC

By substituting the provided values into this equation:

138° = 35° + ∠ EDC

Subtracting 35° from both sides gives the measure of ∠ EDC:

∠ EDC = 138° - 35° = 103°

Since the points B, E, and D are collinear on the diagonal, ∠ BDC is the same as ∠ EDC. Thus, ∠ BDC = 103°. Since we established that ∠ BAC = ∠ BDC due to the properties of a circle, the measure of ∠ BAC is 103°.

Explanation

In the circle with centre O, AD is the diameter, so O lies on AD. Since OA and OC are radii, △ OAC is an isosceles triangle where OA = OC. The angle ∠ AOC is supplementary to ∠ DOC, hence ∠ AOC = 180° - ∠ DOC = 180° - 60° = 120°.

The base angles of isosceles △ OAC are ∠ OAC = ∠ OCA = (180° - 120°) / 2 = 30°. In △ OBA, point B lies on AC, and we are given ∠ OBA = 60°. The sum of angles in △ OBA is 180°, so ∠ AOB = 180° - (30° + 60°) = 90°. Using the trigonometric ratio in right-angled △ OBA: AB = (OB)/(cos 60°) = (7)/(1/2) = 14 cm OA = OB tan 60° = 7√3 cm

Since OA is the radius r, the length of chord AC in △ OAC can be calculated using the Sine Rule: (AC)/(sin ∠ AOC) = (OA)/(sin ∠ OCA) ⟹ (AC)/(sin 120°) = 7√3{sin 30°} AC = 7√3 × (√3/2){1/2} = 21 cm

Since point B is on the chord AC, the length BC is the difference between the total chord length and the segment AB: BC = AC - AB = 21 - 14 = 7 cm

Explanation

Let the two numbers be A = 5x and B = 8x, where x is a common multiplier. The sum of the two numbers is A + B = 5x + 8x = 13x. According to the problem, adding 5 to each number changes the ratio to 2:3, which is represented by the equation:

(5x + 5)/(8x + 5) = ⅔

By cross-multiplying the terms, we get:

3(5x + 5) = 2(8x + 5) 15x + 15 = 16x + 10

Rearranging the terms to solve for x:

16x - 15x = 15 - 10 x = 5

The sum of the numbers is calculated by substituting the value of x back into the sum expression:

Sum = 13x = 13 × 5 = 65

Explanation

Let the initial income of Radha be 100. Since she saves 25% of her income, her initial savings are 25 and her initial expenditure is 100 - 25 = 75.

The new income after a 29% increase is 100 × 1.29 = 129. The new expenditure after a 20% increase is 75 × 1.20 = 90.

New savings are calculated as New Income - New Expenditure: New Savings = 129 - 90 = 39

The increase in savings is 39 - 25 = 14. The percentage increase in savings is calculated using the formula: Percentage Increase = (Increase)/(Original Value) × 100 ¹⁴⁄₂₅ × 100 = 14 × 4 = 56%

Explanation

Total interest for 3 years is ₹5832, so the interest for 1 year is 5832 / 3 = 1944. Using the concept of deviation, if the entire sum of ₹25600 were invested at 7% per annum, the annual interest would be 25600 × 7 / 100 = 1792. The difference between the actual annual interest and this calculated interest is 1944 - 1792 = 152. This surplus occurs because a portion of the capital was invested at 9% instead of 7%, which is an extra 9% - 7% = 2%. Therefore, 2% of the amount invested at 9% equals ₹152. Letting the amount invested at 9% be x, we have 0.02x = 152. Solving for x gives x = 152 / 0.02 = 7600. Alternatively, using the rule of alligation, the average rate of interest is (1944 / 25600) × 100 = 243/32%. The ratio of the parts is (9 - 243/32) : (243/32 - 7), which simplifies to 45 : 19. The amount at 9% is (19 / 64) × 25600 = 19 × 400 = 7600.

Explanation

The values provided in the table represent the percentage of marks obtained by each student in the respective subjects. For Subject B, the maximum marks are 80. To find the average marks for Manju, Rekha, and Abhi, we first determine the average of their percentage scores in Subject B.

Average percentage of Manju, Rekha, and Abhi in Subject B = (85 + 75 + 65)/(3) = ²²⁵⁄₃ = 75%

Once the average percentage is known, the average marks can be calculated by applying this percentage to the total marks for the subject: Average marks = 75% of 80 = ⁷⁵⁄₁₀₀ × 80 = ¾ × 80 = 60

Alternatively, calculating individual marks for each student: Manju's marks = 85% of 80 = 68 Rekha's marks = 75% of 80 = 60 Abhi's marks = 65% of 80 = 52

Sum of marks = 68 + 60 + 52 = 180 Average marks = ¹⁸⁰⁄₃ = 60

Explanation

The relationship between income, expenditure, and savings can be solved using the rule of alligation, where the percentage increase in total income is the weighted average of the percentage increases in expenditure and savings.

Given: Increase in Expenditure (E%) = 20% Increase in Savings (S%) = 40% Overall increase in Income (I%) = 28%

The ratio of Expenditure (E) to Savings (S) is determined by: E : S = (S% - I%) : (I% - E%) E : S = (40 - 28) : (28 - 20) = 12 : 8 = 3 : 2

Total income consists of the sum of expenditure and savings parts, which is 3 + 2 = 5 units. Radha saves X% of her income, meaning X is the percentage of savings relative to the total income: X = (Savings)/(Income) × 100 X = ⅖ × 100 = 40

Explanation

Using the algebraic identity for the cube of a binomial, (a - b)³ = a³ - 3a²b + 3ab² - b³, expand the expression on the right-hand side (2 - 3x)³ by setting a = 2 and b = 3x:

(2 - 3x)³ = (2)³ - 3(2)²(3x) + 3(2)(3x)² - (3x)³ (2 - 3x)³ = 8 - 36x + 54x² - 27x³

The given equation is 8 + 2px² - 36x - 27x³ = (2 - 3x)³. By comparing the coefficients of the x² term on both sides of the equation:

2p = 54 p = 54 / 2 p = 27

Explanation

To find the total time, we first determine the capacity of the tank by taking the Least Common Multiple (LCM) of the given times (12, 16, and 24), which is 48 units.

The individual efficiencies of the pipes are calculated as: Efficiency of A = 48 / 12 = 4 units/hour Efficiency of B = 48 / 16 = 3 units/hour Efficiency of C = 48 / 24 = -2 units/hour (negative as it is an emptying pipe)

For the first 4 hours, all three pipes are open. Their combined efficiency is 4 + 3 - 2 = 5 units/hour. The amount of the tank filled in 4 hours is 5 × 4 = 20 units.

The remaining capacity to be filled is 48 - 20 = 28 units. Since pipe B is closed, only pipes A and C continue to work. Their combined efficiency is 4 - 2 = 2 units/hour. The time taken to fill the remaining 28 units is 28 / 2 = 14 hours.

The total time required to fill the empty tank completely is the sum of the initial duration and the remaining duration: 4 + 14 = 18 hours.

Explanation

The fourth proportional to three numbers a, b, and c is defined as the value x such that a : b = c : x, which yields the formula x = (b × c)/(a). For the numbers 12, 18, and 6, the fourth proportional is: x = (18 × 6)/(12) = ¹⁰⁸⁄₁₂ = 9.

The third proportional to two numbers a and b is the value y such that a : b = b : y, which yields the formula y = (b²)/(a). For the numbers 4 and k, the third proportional is: y = (k²)/(4).

Setting the fourth proportional equal to the third proportional as specified in the problem: 9 = (k²)/(4) k² = 9 × 4 = 36 k = √(36) = 6.

Explanation

To determine which city had the maximum number of trees planted in 2016 and 2019 combined, calculate the sum for each city using the values provided in the table.

For Chandigarh, the total is 1800 + 2440 = 4240. For Ahmedabad, the total is 2500 + 1950 = 4450. For Pune, the total is 1800 + 1900 = 3700. For Kolkata, the total is 2000 + 1600 = 3600.

Comparing these aggregate values, Ahmedabad recorded the highest number of trees planted across the two specific years.

Explanation

The relationship between the Marked Price (MP) and the Cost Price (CP) when a discount and a profit percentage are involved is given by the formula: (MP)/(CP) = (100 + Profit%)/(100 - Discount%)

Given that the shopkeeper makes a profit of 20% after allowing a discount of 10%, we substitute these values into the formula: (MP)/(CP) = (100 + 20)/(100 - 10) = ¹²⁰⁄₉₀ = ⁴⁄₃

When the shopkeeper does not give any discount, the Selling Price (SP) is equal to the Marked Price (MP). The profit percentage in this scenario is calculated by comparing the Marked Price directly to the Cost Price: Profit% = (MP - CP)/(CP) × 100 Profit% = ((MP)/(CP) - 1) × 100

Substituting the ratio (MP)/(CP) = ⁴⁄₃ into the equation: Profit% = (⁴⁄₃ - 1) × 100 Profit% = ⅓ × 100 = 33.333...

Corrected to two decimal places, the profit percentage is 33.33%.

Explanation

Since the number 593ab is divisible by 3, 7, and 11, it must be divisible by their least common multiple. Because 3, 7, and 11 are prime numbers, their LCM = 3 × 7 × 11 = 231.

To find the specific values of a and b, we consider the maximum possible value for the 5-digit number, which is 59399, and divide it by 231. Using long division, 59399 ÷ 231 gives a quotient of 257 and a remainder of 32. To find the exact multiple of 231 that fits the form 593ab, we subtract the remainder from the maximum value: 59399 - 32 = 59367

Comparing 59367 with 593ab, we identify a = 6 and b = 7.

Now, we substitute these values into the required expression (a² - b² + ab): 6² - 7² + (6 × 7) = 36 - 49 + 42 = -13 + 42 = 29

Explanation

In a triangle, the point of intersection of the internal angle bisectors is the incenter. The angle formed at the incenter by two vertices is related to the third interior angle of the triangle by the formula ∠ BOC = 90° + (∠ A)/(2).

Substituting the given value ∠ A = 50° into the formula: ∠ BOC = 90° + (50°)/(2) ∠ BOC = 90° + 25° ∠ BOC = 115°

This formula is derived from the angle sum property of triangles. In △ ABC, ∠ A + ∠ B + ∠ C = 180°, so ∠ B + ∠ C = 180° - ∠ A. In △ BOC, the angles at the base are half of the original angles, thus ∠ BOC = 180° - (∠ B + ∠ C)/(2). Replacing the sum of B and C results in 180° - (180° - ∠ A)/(2), which simplifies to 90° + (∠ A)/(2).

Explanation

The total increase in marks is the sum of the additional marks obtained in Mathematics, Computer Science, and History: Total Increase = 27 + 10 + 18 = 55

The increase in the average score is the difference between the new average and the initial average: Increase in Average = 72.6 - 67.6 = 5.0

The number of papers n can be calculated using the formula: n = (Total Increase in Marks)/(Increase in Average Score)

Substituting the values: n = ⁵⁵⁄₅ = 11

Explanation

To find the value of x³ + (125)/(8x³⁾, first rearrange the given equation 2x² - 7x + 5 = 0 by dividing every term by 2x. This results in x - ⁷⁄₂ + ⁵⁄₂ₓ = 0, which can be written as x + ⁵⁄₂ₓ = ⁷⁄₂.

The target expression x³ + (125)/(8x³⁾ is in the form of a³ + b³, where a = x and b = ⁵⁄₂ₓ. Using the algebraic identity a³ + b³ = (a + b)³ - 3ab(a + b), substitute the values of a and b: x³ + (⁵⁄₂ₓ)³ = (x + ⁵⁄₂ₓ)³ - 3(x)(⁵⁄₂ₓ)(x + ⁵⁄₂ₓ)

Substitute x + ⁵⁄₂ₓ = ⁷⁄₂ into the identity: x³ + (125)/(8x³⁾ = (⁷⁄₂)³ - 3(⁵⁄₂)(⁷⁄₂) x³ + (125)/(8x³⁾ = ³⁴³⁄₈ - ¹⁰⁵⁄₄ x³ + (125)/(8x³⁾ = ³⁴³⁄₈ - ²¹⁰⁄₈ = ¹³³⁄₈

Converting the improper fraction ¹³³⁄₈ into a mixed fraction, we get 16 ⅝ since 133 = (16 × 8) + 5. This value corresponds to the provided answer.

Explanation

The expression is solved using the VBODMAS rule, which dictates the order of operations as Brackets, Of, Division, Multiplication, Addition, and Subtraction.

First, solve the term in the brackets: (9 - 13) = -4. Next, evaluate the of operator, which takes precedence over division: 21 of 4 = 21 × 4 = 84. Substitute these values back into the expression: 7 × 4 ÷ 84 - 5 ÷ 4 × (-4) + 2 - 2 ÷ 8.

Now, perform division and multiplication from left to right. The term 7 × 4 ÷ 84 is calculated as 7 × (4 ÷ 84) = 7 × ¹⁄₂₁ = ⅓. The term 5 ÷ 4 × (-4) is calculated as (⁵⁄₄) × (-4) = -5. The term 2 ÷ 8 is ¼.

The expression now simplifies to: ⅓ - (-5) + 2 - ¼. Combining the constants and the fractions: ⅓ + 5 + 2 - ¼ = 7 + ⅓ - ¼. To subtract the fractions, find a common denominator of 12: 7 + ⁴⁄₁₂ - ³⁄₁₂ = 7 + ¹⁄₁₂. The final result is the mixed fraction 7 ¹⁄₁₂, which is represented by the integers in option a.

Explanation

According to the Internal Angle Bisector Theorem, the bisector of an angle of a triangle divides the opposite side into segments that are proportional to the adjacent sides of the triangle. Therefore, in triangle ABC, since AD is the bisector of ∠ A, we have the relation:

(BD)/(DC) = (AB)/(AC)

Substituting the given values AB = 5 cm and AC = 7.5 cm:

(BD)/(DC) = (5)/(7.5) = ⅔

Since BC = BD + DC = 10 cm, we can express BD as:

BD = (2)/(2 + 3) × BC = ⅖ × 10 = 4 cm

Let M be the mid-point of BC. The distance of the mid-point from vertex B is:

BM = (BC)/(2) = ¹⁰⁄₂ = 5 cm

The distance between the point D and the mid-point M is the absolute difference between their distances from vertex B:

DM = |BM - BD| = |5 - 4| = 1 cm

Explanation

Let the total number of bananas be 100 and the Cost Price (CP) of each banana be 1. Thus, Total CP = 100.

One-fifth of the total bananas are rotten and discarded: Rotten bananas = 1/5 × 100 = 20 Bananas remaining with the merchant = 100 - 20 = 80

The merchant sells two-fifths of the bananas currently with him at a 15% profit: Quantity sold at 15% profit = 2/5 × 80 = 32 Selling Price (SP₁) = 32 × 1.15 = 36.8

The rest of the bananas are sold at a 10% profit: Remaining quantity = 80 - 32 = 48 Selling Price (SP₂) = 48 × 1.10 = 52.8

Total Selling Price (SP) = 36.8 + 52.8 = 89.6

The overall loss is calculated as: Loss = Total CP - Total SP = 100 - 89.6 = 10.4 Loss Percentage = (Loss / Total CP) × 100 = (10.4 / 100) × 100 = 10.4%

Explanation

A quadrilateral whose vertices lie on a circle is called a cyclic quadrilateral. A fundamental property of a cyclic quadrilateral is that the sum of its opposite angles is 180°.

For the given quadrilateral ABCD, the pairs of opposite angles are (∠ A, ∠ C) and (∠ B, ∠ D). Therefore, ∠ A + ∠ C = 180° and ∠ B + ∠ D = 180°.

Given that ∠ A = 3∠ C, substituting this into the first equation gives 3∠ C + ∠ C = 180°, which simplifies to 4∠ C = 180°. Thus, ∠ C = 45°.

Similarly, it is given that ∠ D = 2∠ B, or ∠ B = ½∠ D. Substituting this into the second equation gives ½∠ D + ∠ D = 180°, which simplifies to ³⁄₂∠ D = 180°. Solving for ∠ D yields ∠ D = (180° × 2)/(3) = 120°.

The difference between the measures of ∠ D and ∠ C is calculated as 120° - 45° = 75°.

Explanation

The total simple interest earned over 9 years is the sum of the interest accrued in each specific time interval. The cumulative interest percentage is calculated as follows:

For the first 2 years at 8% p.a.: 2 × 8 = 16% For the next 4 years at 10% p.a.: 4 × 10 = 40% For the remaining period of 3 years (9 - 2 - 4 = 3) at 12% p.a.: 3 × 12 = 36%

Summing these percentages gives the total interest as a percentage of the principal P: Total interest percentage = 16% + 40% + 36% = 92%

Given that the total simple interest is 18358.60, we set up the equation: 92% of P = 18358.60 ⁹²⁄₁₀₀ × P = 18358.60 P = (18358.60 × 100)/(92) P = ¹⁸³⁵⁸⁶⁰⁄₉₂

Dividing 1835860 by 92 yields 19955.

Explanation

The side of the square shaped field s is calculated by taking the square root of its area. Given the area is 1764 m², we have s = √(1764) = 42 m.

The breadth b of the rectangular park is one-sixth of the side of the square field, so b = ⁴²⁄₆ = 7 m. The length l of the rectangular park is four times its breadth, which gives l = 4 × 7 = 28 m.

The area of the rectangular park is the product of its length and breadth: Area = l × b = 28 × 7 = 196 m²

The total cost of levelling the park is calculated by multiplying the area of the park by the rate of levelling per square meter: Total Cost = 196 × 30 = 5880

Explanation

The property of tangents to a circle states that tangents drawn from an external point to a circle are equal in length. For the quadrilateral ABCD circumscribing a circle, the vertices act as external points.

At vertex A, the tangents are AP and AS. Given AS = 6 cm, then AP = 6 cm. At vertex C, the tangents are CQ and CR. Given CR = 5 cm, then CQ = 5 cm. The side BC is the sum of the segments BQ and CQ, so BC = BQ + CQ. Substituting the known values, 12 = BQ + 5, which means BQ = 7 cm. At vertex B, the tangents are BP and BQ. Since BQ = 7 cm, then BP = 7 cm. The length of side AB is the sum of segments AP and BP. AB = AP + BP = 6 + 7 = 13 cm.

Explanation

To find the minimum value of the expression 25 cos² x + 9 sec² x, we apply the Arithmetic Mean-Geometric Mean (AM-GM) Inequality. The inequality states that for any non-negative real numbers A and B, the relationship (A+B)/(2) ≥ √(AB) holds, which simplifies to A + B ≥ 2√(AB).

Let A = 25 cos² x and B = 9 sec² x. Since cos² x and sec² x are always non-negative for real x (where defined), we substitute these into the formula:

25 cos² x + 9 sec² x ≥ 2 √((25 cos² x) · (9 sec² x))

Since sec x = (1)/(cos x), the product cos² x · sec² x equals 1. The expression becomes:

25 cos² x + 9 sec² x ≥ 2 √(25 · 9 · 1)

25 cos² x + 9 sec² x ≥ 2 √(225)

25 cos² x + 9 sec² x ≥ 2 · 15

25 cos² x + 9 sec² x ≥ 30

Equality occurs when A = B, which means 25 cos² x = 9 sec² x. This results in cos⁴ x = ⁹⁄₂₅, or cos² x = ⅗. Since this value lies within the valid range for cos² x (between 0 and 1), the minimum value is achievable and is equal to 30.

Explanation

The relationship between (x + ¹⁄ₓ) and (x - ¹⁄ₓ) is given by the algebraic identity (x - ¹⁄ₓ)² = (x + ¹⁄ₓ)² - 4. By substituting the given value:

(x - ¹⁄ₓ)² = (¹⁷⁄₄)² - 4 (x - ¹⁄ₓ)² = ²⁸⁹⁄₁₆ - 4 (x - ¹⁄ₓ)² = (289 - 64)/(16) (x - ¹⁄ₓ)² = ²²⁵⁄₁₆

Taking the square root on both sides gives x - ¹⁄ₓ = ± ¹⁵⁄₄. Since the condition x > 1 is provided, it implies x is greater than its reciprocal ¹⁄ₓ, making the difference x - ¹⁄ₓ positive. Alternatively, by observation, x + ¹⁄ₓ = ¹⁷⁄₄ can be written as 4 + ¼. Given x > 1, we can identify x = 4. Substituting this into the required expression, we get 4 - ¼ = ¹⁵⁄₄.

Explanation

Let u be the speed of the boat in still water and v be the speed of the river. The time T taken for a roundtrip journey between two points at distance D is the sum of the time taken downstream and upstream: T = (D)/(u+v) + (D)/(u-v) = (D(u-v) + D(u+v))/(u² - v²) = (2uD)/(u² - v²)

When the speed of the boat in still water is doubled, the new speed becomes 2u. The new time taken T' for the same roundtrip is: T' = (D)/(2u+v) + (D)/(2u-v) = (D(2u-v) + D(2u+v))/((2u)² - v²) = (4uD)/(4u² - v²)

Given that the new time relates to the original time by the factor ⁸⁰⁄₁₆₁, we equate the ratios: (T')/(T) = (4uD / (4u² - v²)) / (2uD / (u² - v²)) = ⁸⁰⁄₁₆₁ 2(u² - v²) / (4u² - v²) = ⁸⁰⁄₁₆₁

Cross-multiplying to solve for the ratio u:v: 161 × 2(u² - v²) = 80(4u² - v²) 322u² - 322v² = 320u² - 80v² 322u² - 320u² = 322v² - 80v² 2u² = 242v² u² = 121v²

Taking the square root on both sides: u/v = √(121) = ¹¹⁄₁

The ratio of the speed in still water to the speed of the river is 11:1.

Explanation

From the given equation ¹⁄ₓ - (1)/(y) = ¹¹⁄₂₄, we have (y - x)/(xy) = ¹¹⁄₂₄. Given that x - y = 11, it follows that y - x = -11. Substituting this into the fraction yields (-11)/(xy) = ¹¹⁄₂₄, which simplifies to xy = -24.

The expression to be evaluated is x³ - y³ + x² y². We can expand the first part using the algebraic identity x³ - y³ = (x - y)((x - y)² + 3xy). Substituting the values of x - y and xy gives: x³ - y³ = 11 × (11² + 3(-24)) = 11 × (121 - 72) = 11 × 49 = 539

Next, we find the value of the second part of the expression: x² y² = (xy)² = (-24)² = 576

Combining these results, the final value is x³ - y³ + x² y² = 539 + 576 = 1115.