SSC CGL Tier 1 2017 Quantitative Aptitude Questions with Solutions

Explanation

To find the common root, we solve both quadratic equations. For the first equation x² - 8x + 15 = 0, we look for two numbers that multiply to 15 and add to -8. These numbers are -3 and -5. Thus, the equation factors as (x - 3)(x - 5) = 0, giving the roots x = 3 and x = 5.

For the second equation y² + 2y - 35 = 0, we look for two numbers that multiply to -35 and add to 2. These numbers are 7 and -5. The equation factors as (y + 7)(y - 5) = 0, giving the roots y = -7 and y = 5.

Comparing the sets of roots, the common root is 5. The problem asks for the difference of the cube and square of this common root. Using the value 5: Cube = 5³ = 125 Square = 5² = 25 Difference = 125 - 25 = 100

Explanation

For any real numbers, the sum of their squares is zero if and only if each individual term is zero. Since (x - ⅓)² ≥ 0 and (y - 4)² ≥ 0 for all real values of x and y, the equation (x - ⅓)² + (y - 4)² = 0 implies that:

x - ⅓ = 0 ⟹ x = ⅓ y - 4 = 0 ⟹ y = 4

To find the value of the expression (y + x)/(y - x), we substitute these values:

(y + x)/(y - x) = (4 + ⅓)/(4 - ⅓)

Multiplying both the numerator and the denominator by 3 to simplify the complex fraction:

(3 × (4 + ⅓))/(3 × (4 - ⅓)) = (12 + 1)/(12 - 1) = ¹³⁄₁₁

Explanation

The expression x² + (1)/(x²) - 6 can be simplified by utilizing the algebraic identity (x - ¹⁄ₓ)² = x² + (1)/(x²) - 2. By restructuring the given expression to include this perfect square, we get:

x² + (1)/(x²) - 6 = (x² + (1)/(x²) - 2) - 4

Substituting the identity, the expression becomes:

(x - ¹⁄ₓ)² - 2²

Applying the difference of squares formula, a² - b² = (a - b)(a + b), where a = (x - ¹⁄ₓ) and b = 2, the factors are determined as follows:

((x - ¹⁄ₓ) - 2) and ((x - ¹⁄ₓ) + 2)

To find the difference between these two factors, subtract the smaller factor from the larger one:

(x - ¹⁄ₓ + 2) - (x - ¹⁄ₓ - 2) = x - ¹⁄ₓ + 2 - x + ¹⁄ₓ + 2 = 4

Explanation

Let the efficiency (work done per day) of A be x and the efficiency of B be y. Since they can complete the work together in 10 days, the total work W is given by W = 10(x + y).

According to the second condition, A works with twice his efficiency (2x) and B works with an efficiency 1/3 less than his original efficiency, which is y - ⅓y = ⅔y. In this case, they complete the work in 6 days. Therefore, W = 6(2x + ⅔y).

Equating the two expressions for the total work: 10(x + y) = 6(2x + ⅔y) 10x + 10y = 12x + 4y 10y - 4y = 12x - 10x 6y = 2x x = 3y

This relation shows that A is three times as efficient as B. We can now express the total work in terms of a single variable by substituting x = 3y into the first equation: W = 10(3y + y) = 40y

The time taken by A alone to complete the work is: T_A = (W)/(x) = (40y)/(3y) = ⁴⁰⁄₃ days

The time taken by B alone to complete the work is: T_B = (W)/(y) = (40y)/(y) = 40 days

Explanation

The net discount for two successive discounts of x% and y% is calculated using the formula x + y - ˣʸ⁄₁₀₀. By substituting the given values x = 15 and y = 35 into this relation, the calculation is as follows:

Net Discount = 15 + 35 - (15 × 35)/(100) Net Discount = 50 - ⁵²⁵⁄₁₀₀ Net Discount = 50 - 5.25 Net Discount = 44.75

This formula accounts for the second discount being applied to the remaining value after the first discount is subtracted from the original marked price. Multiplying the individual multipliers (1 - 0.15) × (1 - 0.35) also yields 0.85 × 0.65 = 0.5525, which represents the final price as a fraction of the original, leading to a total reduction of 1 - 0.5525 = 0.4475 or 44.75%.

Explanation

Given x + ¹⁄ₓ = √(13).

First, find the value of x - ¹⁄ₓ using the identity (x - ¹⁄ₓ)² = (x + ¹⁄ₓ)² - 4: (x - ¹⁄ₓ)² = (√(13))² - 4 = 13 - 4 = 9 x - ¹⁄ₓ = 3

Now, calculate x² + (1)/(x²) and x³ - (1)/(x³): x² + (1)/(x²) = (x + ¹⁄ₓ)² - 2 = (√(13))² - 2 = 13 - 2 = 11 x³ - (1)/(x³) = (x - ¹⁄ₓ)³ + 3(x - ¹⁄ₓ) = 3³ + 3(3) = 27 + 9 = 36

To find x⁵ - (1)/(x⁵), use the product of the above results: (x² + (1)/(x²))(x³ - (1)/(x³)) = x⁵ - ¹⁄ₓ + x - (1)/(x⁵) (x² + (1)/(x²))(x³ - (1)/(x³)) = (x⁵ - (1)/(x⁵)) + (x - ¹⁄ₓ)

Substitute the calculated values into the equation: 11 × 36 = (x⁵ - (1)/(x⁵)) + 3 396 = (x⁵ - (1)/(x⁵)) + 3 x⁵ - (1)/(x⁵) = 396 - 3 = 393

Explanation

The total expenditure of a company is the product of the number of staff and the average salary per employee. Let the initial number of staff be 5x and the final number of staff be 3x. Similarly, let the initial average salary be 7y and the final average salary be 8y.

Initial Expenditure = 5x × 7y = 35xy Final Expenditure = 3x × 8y = 24xy

The savings represent the difference between the initial and final expenditure: Savings = 35xy - 24xy = 11xy

According to the problem, the savings are Rs 55,000: 11xy = 55000 xy = 5000

The initial expenditure of the company is 35xy. Substituting the value of xy: Initial Expenditure = 35 × 5000 = 1,75,000

Explanation

To find the sum of the three values, we add the three given linear equations: (a + b) + (b + c) + (c + a) = 5 + 7.5 + 8.5 2a + 2b + 2c = 21 2(a + b + c) = 21 a + b + c = 10.5

The average of a set of values is defined as the total sum of those values divided by the count of the values in the set. For the three variables a, b, and c, the average is: Average = (a + b + c)/(3)

By substituting the calculated sum of the three variables into the average formula: Average = (10.5)/(3) = 3.5

Explanation

An increase of 33.33% in price is equivalent to a ratio of 3:4. Since the total expenditure remains constant at Rs 16, the quantity purchased is inversely proportional to the price. This implies the quantity ratio is 4:3 (original to new).

The reduction in quantity is 4 - 3 = 1 unit. According to the question, this reduction is equal to 4 apples. Therefore, 1 unit = 4 apples, and the original quantity purchased was 4 units × 4 = 16 apples.

The original price per apple is determined by dividing the total expenditure by the original quantity: Original Price = (Rs 16)/(16 apples) = Rs 1

To express the value in paise, we multiply by 100: 1 × 100 = 100 paise

Explanation

The percentage increase in the area of a square when its side increases by k% is determined using the successive percentage change formula: Net Increase = k + k + (k × k)/(100). Given that the area increases by 69%, we can set up the following equation:

2k + (k²)/(100) = 69

To solve for k, multiply the entire equation by 100:

k² + 200k - 6900 = 0

Factoring the quadratic equation:

(k + 230)(k - 30) = 0

Since k must be a positive value for an increase, we find k = 30. This can be verified by substituting 30 back into the formula: 2(30) + (30²)/(100) = 60 + ⁹⁰⁰⁄₁₀₀ = 60 + 9 = 69%. Alternatively, if the side of the square increases by 30%, the new side is 1.3 times the original side. The new area is (1.3)² = 1.69 times the original area, which represents a 69% increase.

Explanation

When A starts at 30 m/s and B starts 3 seconds later, A has a head start distance calculated as d = v_A × t = 30 × 3 = 90 metres.

The relative speed of B with respect to A is vᵣₑₗ = v_B - v_A = 50 - 30 = 20 m/s. The time taken by B to catch A is given by the formula t_cₐₜ_cₕ = (Head start distance)/(Relative speed) = ⁹⁰⁄₂₀ = 4.5 seconds.

During this time, the distance traveled by B is D_B = v_B × t_cₐₜ_cₕ = 50 × 4.5 = 225 metres. Since A and B end up at the same point, the distance traveled by A is also 225 metres (30 × (3 + 4.5) = 225). The total distance traveled by both A and B before the catch is 225 + 225 = 450 metres.

Explanation

The difference between compound interest and simple interest for two years is calculated using the formula D = P(R/100)². Given that the difference D = 200 and the annual rate R = 20%, we can find the principal P as follows: 200 = P(20/100)² 200 = P(1/5)² 200 = P/25 P = 5000

For the first year, if the interest is compounded half-yearly, the rate for each six-month period is 20% / 2 = 10%. There are two such periods in one year. The simple interest for the first year is SI = (5000 × 20 × 1) / 100 = 1000. The compound interest for the same period with half-yearly compounding is: CI = P[ (1 + r/100)ⁿ - 1 ] CI = 5000[ (1 + 10/100)² - 1 ] CI = 5000[ (11/10)² - 1 ] CI = 5000[ 121/100 - 1 ] CI = 5000 × (21/100) = 1050

The difference between the compound interest and simple interest for the first year is 1050 - 1000 = 50. This can also be directly calculated using the two-period difference formula for the periodic rate: D = P(10/100)² = 5000 × (1/100) = 50.

Explanation

The first equation is x(2x + 3) = 90, which simplifies to the quadratic form 2x² + 3x - 90 = 0. Factoring the quadratic, we seek two numbers that multiply to 2 × (-90) = -180 and add to 3. These numbers are 15 and -12. Thus, 2x² + 15x - 12x - 90 = 0, which gives x(2x + 15) - 6(2x + 15) = 0. This results in (x - 6)(2x + 15) = 0. Since x must be a positive number, we find x = 6.

The second equation is 7y⁻¹^/² + 2y⁻¹^/² = y¹^/². Combining the terms on the left side, we get 9y⁻¹^/² = y¹^/². Multiplying both sides by y¹^/² yields 9 = y¹^/² · y¹^/², which simplifies to y = 9.

To find the value of x² + y², substitute the values of x and y: 6² + 9² = 36 + 81 = 117.

Explanation

The expression (7x² - 19xy + 11y²)/y² can be simplified by dividing each term in the numerator by the denominator y², which yields 7(x/y)² - 19(x/y) + 11. Given that x/y = 4/9, we substitute this value into the simplified expression to get 7(4/9)² - 19(4/9) + 11.

Performing the calculations, we have 7(16/81) - 76/9 + 11, which results in 112/81 - 76/9 + 11. To combine these terms, we use a common denominator of 81:

112/81 - (76 × 9)/81 + (11 × 81)/81 = 112/81 - 684/81 + 891/81

Summing the numerators gives (112 - 684 + 891)/81 = 319/81.

Explanation

Let y = x - 3. The given equation is y + (1)/(y) = 4. We need to find the value of y³ + (1)/(y³). Using the standard algebraic identity a³ + b³ = (a + b)³ - 3ab(a + b), we substitute a = y and b = (1)/(y). Since the product y · (1)/(y) = 1, the expression simplifies to y³ + (1)/(y³) = (y + (1)/(y))³ - 3(y + (1)/(y)). Substituting the given value y + (1)/(y) = 4 into this formula results in 4³ - 3(4) = 64 - 12 = 52.

Explanation

The equation x² + y² + z² = xy + yz + zx is a standard algebraic form that implies x = y = z. This can be derived by multiplying the entire equation by 2 and rearranging the terms: 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 only when each individual term is zero, it follows that x - y = 0, y - z = 0, and z - x = 0. Thus, x = y = z.

Substituting y = x and z = x into the given expression: (7x + 3y - 5z)/(5x) = (7x + 3x - 5x)/(5x) = ⁵ˣ⁄₅ₓ = 1

Explanation

According to the Basic Proportionality Theorem (Thales's Theorem), if a line is drawn parallel to one side of a triangle to intersect the other two sides, the other two sides are divided in the same ratio. Since DE || BC in triangle ABC, we can establish the following relationship:

(AD)/(DB) = (AE)/(EC)

By substituting the given values AD = x, DB = x - 3, AE = x + 3, and EC = x - 2 into this formula, we get:

(x)/(x - 3) = (x + 3)/(x - 2)

To solve for x, we perform cross-multiplication:

x(x - 2) = (x + 3)(x - 3)

Expanding the terms using the algebraic identity (a + b)(a - b) = a² - b² on the right side and distribution on the left side leads to:

x² - 2x = x² - 9

Subtracting x² from both sides simplifies the equation to:

-2x = -9

Dividing by -2 yields:

x = 4.5

This value ensures all side segments have positive lengths, with AD = 4.5 cm, DB = 1.5 cm, AE = 7.5 cm, and EC = 2.5 cm, satisfying the geometric conditions of the triangle.

Explanation

In a parallelogram ABCD, the diagonals AC and BD bisect each other at a point O. Therefore, BO = OD = ½ BD. Given BD = 36 cm, we have BO = OD = 18 cm.

In △ ABC, the segment BO is a median because O is the midpoint of AC. The centroid P of △ ABC lies on the median BO and divides it in the ratio 2:1 from the vertex B. Thus, the distance PO = ⅓ BO.

Similarly, in △ ADC, the segment DO is a median because O is the midpoint of AC. The centroid Q of △ ADC lies on the median DO and divides it in the ratio 2:1 from the vertex D. Thus, the distance QO = ⅓ DO.

Since O lies between P and Q on the diagonal BD, the length PQ = PO + QO. Substituting the values, we get PQ = ⅓ BO + ⅓ DO = ⅓ (BO + OD) = ⅓ BD. Substituting the given length of BD, we find PQ = ⅓ × 36 = 12 cm.

Explanation

For an equilateral triangle with side length a, the circumradius R and the inradius r are determined by the formulas R = (a)/(√(3)) and r = (a)/(2√(3)). The difference between the circumradius and the inradius is R - r = (a)/(√(3)) - (a)/(2√(3)) = (a)/(2√(3)).

Given that the side a = 12 cm, we substitute this value into the expression for the difference: R - r = (12)/(2√(3)) = (6)/(√(3))

To simplify the result, rationalize the denominator by multiplying the numerator and denominator by √3: 6 × √3{√3 × √3} = 6√3{3} = 2√3 cm

Explanation

For an equilateral triangle with side length a, the radius of the circumcircle is R = (a)/(√(3)) and the radius of the incircle is r = (a)/(2√(3)). It follows that R = 2r.

The sum of the areas of the circumcircle and the incircle is given as 770 cm²: π R² + π r² = 770 π (2r)² + π r² = 770 5π r² = 770

Substituting π = ²²⁄₇: 5 × ²²⁄₇ × r² = 770 ¹¹⁰⁄₇ × r² = 770 r² = (770 × 7)/(110) r² = 49 r = 7 cm

Since r = (a)/(2√(3)), we have: a = 2√3 × r = 2√3 × 7 = 14√3 cm

The area of the equilateral triangle is: Area = √3{4} a² Area = √3{4} (14√3)² Area = √3{4} × 196 × 3 Area = √3 × 49 × 3 = 147√3 cm²

Explanation

Using the negative angle identities sin(-θ) = -sin θ and cos(-θ) = cos θ, the expression can be simplified and evaluated using standard trigonometric values.

sin(-π/3) = -sin(π/3) = -√3{2} cos(-π/6) = cos(π/6) = √3{2}

Adding these values together: -√3{2} + √3{2} = 0

Explanation

Given the equation tan θ - cot θ = 0, we can rewrite it as tan θ = cot θ. Since cot θ = (1)/(tan θ), the equation becomes tan θ = (1)/(tan θ), which leads to tan² θ = 1. For an acute angle θ, tan θ must be positive, which gives tan θ = 1. This occurs when θ = 45°.

At θ = 45°, the value of cot θ is also 1. Substituting these values into the expression tan² θ + cot¹⁰⁰ θ, we get: (1)² + (1)¹⁰⁰ = 1 + 1 = 2

Since any power of 1 remains 1, the result is the sum of two units.

Explanation

The given equation is sin 3θ sec 2θ = 1. Since sec 2θ = (1)/(cos 2θ), the equation can be rewritten as:

sin 3θ = cos 2θ

Using the trigonometric identity sin A = cos B when A + B = 90°:

3θ + 2θ = 90° 5θ = 90° θ = 18°

The expression to evaluate is [3 tan² (5θ/2) - 1]. Substituting the value of θ = 18° (or directly using 5θ = 90°):

3 tan² ( (90°)/(2) ) - 1 3 tan² 45° - 1

Since the value of tan 45° = 1:

3(1)² - 1 = 3 - 1 = 2

Explanation

The area of a triangle A can be calculated from the lengths of its medians m₁, m₂, and m₃ using the formula A = ⁴⁄₃ Aₘ, where Aₘ is the area of a triangle whose sides are the lengths of the medians. Given the medians 18 cm, 24 cm, and 30 cm, we observe that they form a Pythagorean triplet because 18² + 24² = 324 + 576 = 900 = 30². This indicates that the triangle formed by the medians as sides is a right-angled triangle.

The area of this median triangle is calculated as: Aₘ = ½ × base × height = ½ × 18 × 24 = 216 cm²

Now, applying the relationship between the areas: A = ⁴⁄₃ × 216 A = 4 × 72 A = 288 cm²

Explanation

The radius of the larger circle is R = 28 / 2 = 14 cm. Since the two smaller circles touch the larger circle internally and pass through its center O, the diameter of each smaller circle d is equal to the radius of the larger circle, R. Therefore, d = 14 cm, and the radius of each smaller circle is r = 14 / 2 = 7 cm.

The area of the larger circle is given by Aₗₐᵣ_gₑ = π R² = π (14)² = 196π cm². The combined area of the two smaller circles is 2 × π r² = 2 × π (7)² = 2 × 49π = 98π cm². The area of the larger circle not enclosed by the two smaller circles is the difference between these areas:

Area = 196π - 98π = 98π

Substituting the value of π = 22/7:

Area = 98 × ²²⁄₇ = 14 × 22 = 308 cm²

Explanation

The expression tan(θ/2) + cot(θ/2) can be simplified by expressing tangent and cotangent in terms of sine and cosine: (sin(θ/2))/(cos(θ/2)) + (cos(θ/2))/(sin(θ/2))

By taking the least common multiple of the denominators, the expression becomes: sin²(θ/2) + cos²(θ/2){sin(θ/2)cos(θ/2)}

Using the fundamental identity sin²x + cos²x = 1, the numerator simplifies to 1: (1)/(sin(θ/2)cos(θ/2))

Multiplying both the numerator and denominator by 2, we apply the double-angle formula sinθ = 2sin(θ/2)cos(θ/2): (2)/(2sin(θ/2)cos(θ/2)) = (2)/(sinθ)

Applying the reciprocal identity 1/sinθ = cscθ, we obtain 2cscθ.

Explanation

Starting from the basic identity sin² θ + cos² θ = 1, we square both sides to find an expression for the fourth powers: (sin² θ + cos² θ)² = sin⁴ θ + cos⁴ θ + 2sin² θ cos² θ = 1 sin⁴ θ + cos⁴ θ = 1 - 2sin² θ cos² θ

Next, we square this result to obtain the relation involving the eighth powers: (sin⁴ θ + cos⁴ θ)² = sin⁸ θ + cos⁸ θ + 2sin⁴ θ cos⁴ θ = (1 - 2sin² θ cos² θ)² sin⁸ θ + cos⁸ θ + 2sin⁴ θ cos⁴ θ = 1 + 4sin⁴ θ cos⁴ θ - 4sin² θ cos² θ

The given equation is sin⁸ θ + cos⁸ θ - 1 = 0, which means sin⁸ θ + cos⁸ θ = 1. Substituting this into the expanded equation: 1 + 2sin⁴ θ cos⁴ θ = 1 + 4sin⁴ θ cos⁴ θ - 4sin² θ cos² θ 2sin⁴ θ cos⁴ θ - 4sin² θ cos² θ = 0

Let x = sin² θ cos² θ. The equation simplifies to: 2x² - 4x = 0 2x(x - 2) = 0

Since the problem specifies θ ≠ 0 and θ ≠ nπ/2, neither sin θ nor cos θ can be zero. Therefore, x ≠ 0. Solving the remaining part of the equation: x - 2 = 0 x = 2

The value of cos² θ sin² θ is 2.

Explanation

The total marks for Mathematics is calculated by summing the scores of all five students: 96 + 99 + 99 + 89 + 92 = 475. This aggregate value represents the highest cumulative score across all subject categories in the provided data. The total is significantly bolstered by multiple high scores, including two students scoring 99 and another scoring 96. Comparing this sum of 475 to the calculated totals of the other categories identifies it as the maximum aggregate value in the table.

Explanation

The unit's digit of any perfect square is limited to a specific set of numbers based on the square of digits from 0 to 9. By calculating 0²=0, 1²=1, 2²=4, 3²=9, 4²=16, 5²=25, 6²=36, 7²=49, 8²=64, and 9²=81, it is evident that the only possible digits in the units place are 0, 1, 4, 5, 6, and 9. Any number ending in 2, 3, 7, or 8 is mathematically impossible to be a perfect square because no single-digit square results in these units digits.

Explanation

Let the total work be the LCM of 30 and 15, which is 30 units. The daily efficiency of Amar is ³⁰⁄₃₀ = 1 unit per day, and the daily efficiency of Raman is ³⁰⁄₁₅ = 2 units per day.

When both work together, their combined efficiency is 1 + 2 = 3 units per day. In 4 days, the total work completed by them is 3 × 4 = 12 units. To find the percentage of work completed, we calculate the ratio of the work done to the total work:

Percentage completed = ¹²⁄₃₀ × 100 = ⅖ × 100 = 40%

Explanation

The area of a trapezium is determined by the formula A = ½(a + b)h, where a and b represent the lengths of the two parallel sides and h represents the height or the perpendicular distance between those sides. In this problem, the lengths of the parallel sides are given as 21 cm and 9 cm, and the height is provided as 10 cm.

By substituting these values into the area formula:

Area = ½ × (21 + 9) × 10

First, the sum of the parallel sides is calculated as 21 + 9 = 30 cm. Then, this sum is multiplied by the height and divided by two:

Area = ½ × 30 × 10 Area = 15 × 10 = 150

The calculation shows that the average length of the parallel sides is 15 cm, which, when multiplied by the height of 10 cm, results in an area of 150 cm².

Explanation

To find the net discount of three successive discounts d₁, d₂, and d₃, the most efficient approach is using the multiplier method. The final price as a fraction of the original price is calculated by multiplying the remaining percentages after each discount:

Price Fraction = (1 - (d₁)/(100)) × (1 - (d₂)/(100)) × (1 - (d₃)/(100))

Substituting the values 20%, 10%, and 30%:

Price Fraction = (1 - 0.20) × (1 - 0.10) × (1 - 0.30) Price Fraction = 0.8 × 0.9 × 0.7 Price Fraction = 0.72 × 0.7 = 0.504

The net discount is the total reduction from the initial value of 1:

Net Discount = (1 - 0.504) × 100 = 49.6%

Alternatively, applying the successive discount formula x + y - ˣʸ⁄₁₀₀ for the first two discounts gives 20 + 10 - (20 × 10)/(100) = 28%. Combining this result with the third discount gives 28 + 30 - (28 × 30)/(100) = 58 - 8.4 = 49.6%.

Explanation

To find the percentage of milk in the final mixture, first assume a common capacity for the three bottles by taking the least common multiple (LCM) of the total parts in each ratio. The total parts are 5 + 7 = 12, 7 + 9 = 16, and 2 + 1 = 3. The LCM of 12, 16, and 3 is 48. Let each bottle have a capacity of 48 units.

In the first bottle, the quantity of milk is ⁵⁄₁₂ × 48 = 20 units. In the second bottle, the quantity of milk is ⁷⁄₁₆ × 48 = 21 units. In the third bottle, the quantity of milk is ⅔ × 48 = 32 units.

When the bottles are combined, the total quantity of milk is 20 + 21 + 32 = 73 units. The total volume of the mixture from the three bottles is 48 × 3 = 144 units. The percentage of milk in the new mixture is calculated as: Percentage = ⁷³⁄₁₄₄ × 100 Percentage = ⁷³⁰⁰⁄₁₄₄ ≈ 50.69%

Rounding to one decimal place, the percentage of milk is 50.7.

Explanation

Let the average expenditure of all eight friends be x rupees. The total expenditure is 8x. The first seven friends spent 14 rupees each, totaling 7 × 14 = 98 rupees. The eighth friend spent x + 21 rupees. Setting up the equation for the total expenditure: 8x = 98 + (x + 21) 7x = 119 x = 17 The total amount of money spent is 8 × 17 = 136 rupees.

Alternatively, using the deviation method, the 21 rupees extra spent by the eighth friend above the average is what balances the lower spending of the other seven friends. This surplus must be distributed among the other seven friends to find the amount they fall short of the average. The deficit per person for the seven friends is 21 / 7 = 3 rupees. Therefore, the average expenditure of the group is 14 + 3 = 17 rupees. Total money spent is 17 × 8 = 136 rupees.

Explanation

Let the cost price of the pen be x. According to the problem, the profit percentage is numerically equal to the cost price, so Profit% = x. The selling price is given as Rs 24.

Using the standard formula for selling price: SP = CP × (1 + (Profit%)/(100)) 24 = x × (1 + ˣ⁄₁₀₀) 24 = x × ((100 + x)/(100)) 2400 = 100x + x² x² + 100x - 2400 = 0

Solving this quadratic equation by splitting the middle term: x² + 120x - 20x - 2400 = 0 x(x + 120) - 20(x + 120) = 0 (x - 20)(x + 120) = 0

Setting each factor to zero, we get x = 20 or x = -120. Since the cost price cannot be negative, x = 20. Alternatively, testing the options directly, if the cost price is Rs 20, the profit is 20% of 20, which is Rs 4. The selling price becomes 20 + 4 = 24, which matches the given data.

Explanation

The marks obtained by A can be calculated based on the marks obtained by B. Since A got 36% less marks than B and B got 145 marks: A = 145 × (1 - 0.36) = 145 × 0.64 = 92.8

The marks obtained by A are also given in relation to C. A got 16% more marks than C, which can be expressed as: A = C × (1 + 0.16) = 1.16C

By substituting the value of A into this equation, we can solve for C: 92.8 = 1.16C C = 92.8 / 1.16 C = 9280 / 116 = 80

Explanation

The distance covered by the scooter is the product of its speed and the time taken. Since the distance remains the same before and after repairing, we can use the relation s₁ × t₁ = s₂ × t₂.

Given: Speed after repairing s₁ = 54 km/h Time taken after repairing t₁ = 6 hours Speed before repairing s₂ = 48 km/h Time taken before repairing = t₂

The distance D = 54 × 6 = 324 km.

To find the time taken before repairing to cover the same distance: t₂ = (D)/(s₂₎ = ³²⁴⁄₄₈

Dividing 324 by 48: t₂ = 6.75 hours

To convert the decimal part into minutes, multiply by 60: 0.75 × 60 = 45 minutes

The total time taken is 6 hours and 45 minutes.

Explanation

The sum of money becomes thrice of itself, which means the Simple Interest (SI) earned is 3P - P = 2P, where P is the principal amount. The time period given is 5 years and 4 months. Converting this into years:

T = 5 + ⁴⁄₁₂ = 5 + ⅓ = ¹⁶⁄₃ years

Using the Simple Interest formula SI = (P · R · T)/(100), we substitute the known values to find the rate R:

2P = (P · R · 16)/(3){100}

Canceling P from both sides and rearranging the equation:

2 = (16R)/(300)

R = (2 · 300)/(16)

R = ⁶⁰⁰⁄₁₆ = 37.5

Thus, the yearly rate of interest is 37.5%.

Explanation

The value of the expression a³ + b³ + c³ - 3abc can be calculated using the following algebraic identity: a³ + b³ + c³ - 3abc = (a + b + c)((a + b + c)² - 3(ab + bc + ca))

Given the values a + b + c = 11 and ab + bc + ca = 17, substitute them directly into the identity: a³ + b³ + c³ - 3abc = 11 × (11² - 3 × 17) a³ + b³ + c³ - 3abc = 11 × (121 - 51) a³ + b³ + c³ - 3abc = 11 × 70 a³ + b³ + c³ - 3abc = 770

Explanation

The given equation is ˣ⁄₅ + ⁵⁄ₓ = -2. Let y = ˣ⁄₅, so the equation becomes y + (1)/(y) = -2. Multiplying both sides by y gives y² + 1 = -2y, which can be rearranged into the quadratic form y² + 2y + 1 = 0. This expression is a perfect square, (y + 1)² = 0, which implies y = -1. Substituting the value of y back in terms of x, we have ˣ⁄₅ = -1, which gives x = -5. To find the value of x³, we calculate (-5)³ = -125.

Explanation

The point of intersection of the internal angle bisectors of a triangle is known as the incenter. For any triangle PQR, the angle formed at the incenter O by the bisectors of ∠ Q and ∠ R is determined by the formula ∠ QOR = 90° + ½ ∠ P.

Given that ∠ QPR = 45°, we substitute this value into the relation:

∠ QOR = 90° + ½ (45°)

∠ QOR = 90° + 22.5°

∠ QOR = 112.5°

Explanation

In a right-angled triangle ABC where ∠ABC = 90°, an altitude BP drawn to the hypotenuse AC creates two smaller triangles that are similar to the original triangle. Specifically, △BPC ∼ △ABC because they both share the angle at vertex C and each contains a right angle (∠BPC = ∠ABC = 90°).

In similar triangles, corresponding angles are equal. The angle ∠PBC in △BPC corresponds to the angle ∠BAC in △ABC. Since the point P lies on the segment AC, the given angle ∠BAP is the same as ∠BAC. Therefore, ∠BAC = 30°. By the property of similarity, ∠PBC = ∠BAC = 30°.

Alternatively, using the angle sum property of a triangle: In △ABC, the measure of ∠BCA is 180° - (∠ABC + ∠BAC) = 180° - (90° + 30°) = 60°. In the right-angled triangle BPC, where BP ⊥ AC, the measure of ∠PBC is 180° - (∠BPC + ∠PCB). Substituting the known values, ∠PBC = 180° - (90° + 60°) = 30°.

Explanation

The given expression (csc A + sin A)(csc A - sin A) is in the form of the algebraic identity (x + y)(x - y) = x² - y². Applying this identity, the expression simplifies to:

csc² A - sin² A

To express this in terms of the given options, we use the standard trigonometric identities csc² A = 1 + cot² A and sin² A = 1 - cos² A. Substituting these values into the expression yields:

(1 + cot² A) - (1 - cos² A)

By expanding the brackets and simplifying, we get:

1 + cot² A - 1 + cos² A = cot² A + cos² A

This results in the simplified value cos² A + cot² A.

Explanation

To find the percentage that the average marks in English represent of the average marks in Hindi, we first calculate the total marks for each subject. Since the number of students is the same for both subjects, the ratio of their averages is identical to the ratio of their sums.

Sum of marks in English = 87 + 74 + 89 + 62 + 95 = 407 Sum of marks in Hindi = 81 + 89 + 92 + 91 + 86 = 439

The required percentage is determined using the formula: Percentage = (Total Marks in English)/(Total Marks in Hindi) × 100 Percentage = ⁴⁰⁷⁄₄₃₉ × 100 ≈ 92.7107

Rounding to two decimal places, we get 92.71.

Explanation

Based on the data provided in the line chart for the 7 countries, the per acre yields (in million tonnes) are 300, 250, 162, 272, 82, 104, and 120.

To find the required ratio, we first identify the 3 highest and 3 lowest yields: 3 highest yields: 300, 272, 250 3 lowest yields: 82, 104, 120

Since the number of countries in both groups is the same (n = 3), the ratio of their averages is equal to the ratio of their sums.

Sum of the 3 highest yields = 300 + 272 + 250 = 822 Sum of the 3 lowest yields = 82 + 104 + 120 = 306

Required Ratio = 822 : 306

Dividing both terms by their greatest common divisor, 6: 822 / 6 = 137 306 / 6 = 51

The ratio is 137 : 51.

Explanation

Based on the line chart data provided for this specific set of questions, the per acre yield values in million tonnes for the seven countries are India (300), China (272), Sri Lanka (250), Bhutan (162), Singapore (120), Thailand (104), and Nepal (82).

To find the percentage, first calculate the combined per acre yield for India and China: 300 + 272 = 572 million tonnes

Next, calculate the total per acre yield for the remaining five countries, which are Sri Lanka, Bhutan, Singapore, Thailand, and Nepal: 250 + 162 + 120 + 104 + 82 = 718 million tonnes

To find the percentage of the yield of India and China relative to the remaining countries, apply the following formula: Percentage = ( (Total yield of India and China)/(Total yield of remaining countries) ) × 100 Percentage = ( ⁵⁷²⁄₇₁₈ ) × 100 ≈ 79.6657...

Rounding to two decimal places gives 79.66.

Explanation

The yield per acre for the five countries in the dataset is as follows: India (120.5), Brazil (140.4), China (190.2), USA (240.3), and Russia (242.1). To find the new average per acre yield, first calculate the sum of the yields for all countries: Total Yield = 120.5 + 140.4 + 190.2 + 240.3 + 242.1 = 933.5 million tonnes.

The original average yield for the five countries is: Original Average = (933.5)/(5) = 186.7 million tonnes.

The new average yield is obtained by applying a 10% increase to the total average yield: New Average = 186.7 × (1 + 0.10) = 186.7 × 1.1 = 205.37 million tonnes.

This matches the value specified in the correct option.

Explanation

A number that is divisible by 2, 3, and 7 must be divisible by their Least Common Multiple (LCM). Since 2, 3, and 7 are prime numbers, their LCM = 2 × 3 × 7 = 42.

To find the number of multiples of 42 between 300 and 700, we divide the boundaries by 42. For the lower bound, 300 / 42 ≈ 7.14, which means the first multiple of 42 in the range is 42 × 8 = 336. For the upper bound, 700 / 42 ≈ 16.66, which means the last multiple of 42 in the range is 42 × 16 = 672.

The count of numbers is the number of integers from 8 to 16 inclusive. Using the formula for the number of terms in an arithmetic progression, N = (Last Term - First Term) + 1, we get 16 - 8 + 1 = 9.

Explanation

The total payment for a task is distributed among workers in direct proportion to the amount of work they perform. Let the total work be represented as 1 unit. If A and B together complete ⅗ of the work, the portion completed by C is the remainder:

Work done by C = 1 - ⅗ = ⅖

The share of the total payment for each individual is calculated by multiplying the total amount by the fraction of work completed. Given the total amount is Rs 5000, C's share is:

Share of C = ⅖ × 5000

By simplifying the expression, 5000 ÷ 5 = 1000, and 1000 × 2 = 2000. Therefore, C receives Rs 2000 for his contribution.

Explanation

The volume of the initial solid cone is calculated using the formula V_cₒₙₑ = ⅓ π r² h. Given the radius r = 9 cm and height h = 36 cm, the volume is ⅓ π × 9² × 36, which simplifies to 12 π × 81 = 972π cm³.

The volume of the solid cylinder formed after melting is calculated using V_c_yₗ = π r² h. Given the radius r = 9 cm and height h = 9 cm, the volume is π × 9² × 9, which equals 729π cm³.

The amount of material wasted is the difference between the original volume and the final volume, which is 972π - 729π = 243π cm³. To find the percentage of material wasted, we use the formula (Wasted Material)/(Original Material) × 100. This gives (243π)/(972π) × 100. Since 243 × 4 = 972, the ratio is ¼, and the waste percentage is ¼ × 100 = 25.