SSC CGL Tier 1 2024 Quantitative Aptitude Questions with Solutions

Explanation

To solve the expression 15÷ 3 of 2× 2+9÷ 18 of 2× 4-4÷ 8× 2, the BODMAS (Brackets, Of, Division, Multiplication, Addition, Subtraction) rule must be followed. This rule dictates that the "of" operation is performed before division and multiplication.

First, identify and solve the "of" components in the expression: 3 of 2 = 3 × 2 = 6 18 of 2 = 18 × 2 = 36

Substituting these values back into the expression gives: 15 ÷ 6 × 2 + 9 ÷ 36 × 4 - 4 ÷ 8 × 2

Next, perform the division operations from left to right: 15 ÷ 6 = ¹⁵⁄₆ = ⁵⁄₂ 9 ÷ 36 = ⁹⁄₃₆ = ¼ 4 ÷ 8 = ⁴⁄₈ = ½

The expression now becomes: ⁵⁄₂ × 2 + ¼ × 4 - ½ × 2

Now, perform the multiplication operations: ⁵⁄₂ × 2 = 5 ¼ × 4 = 1 ½ × 2 = 1

The simplified expression is: 5 + 1 - 1

Finally, performing the addition and subtraction results in: 5 + 1 - 1 = 5

Explanation

The equivalent discount for two successive discounts of x% and y% is given by the formula x + y - ˣʸ⁄₁₀₀. For successive discounts of 30% and 40%, the equivalent single discount is 30 + 40 - (30 × 40)/(100) = 70 - 12 = 58%.

The difference between the first discount of 65% and the equivalent second discount of 58% is 65% - 58% = 7%. Since the selling price is the marked price minus the discount, the difference in selling prices is equal to the difference in the discount amounts.

Given that this difference is ₹105, we can set up the equation 7% of Marked Price = 105. This can be written as ⁷⁄₁₀₀ × Marked Price = 105. Solving for the marked price, we get Marked Price = (105 × 100)/(7) = 15 × 100 = 1,500.

Explanation

The pole, the vertical wall, and the ground form a right-angled triangle where the pole serves as the hypotenuse. The length of the pole is L = 10 m, and the angle it makes with the horizontal ground is θ = 30°. The distance from the foot of the pole to the wall represents the side adjacent to the given angle.

Using the trigonometric ratio for cosine: cos(θ) = (Adjacent side)/(Hypotenuse)

Substituting the known values: cos(30°) = (Distance)/(10)

Since cos(30°) = √3{2}, the equation becomes: √3{2} = (Distance)/(10) Distance = 10 × √3{2} = 5√3 m

Explanation

The value of a² + (1)/(a²) can be determined using the algebraic identity (x + y)² = x² + y² + 2xy. By setting x = a and y = ¹⁄ₐ, the identity is expressed as:

(a + ¹⁄ₐ)² = a² + (¹⁄ₐ)² + 2(a)(¹⁄ₐ)

Simplifying the term 2(a)(¹⁄ₐ) gives 2, resulting in the formula:

(a + ¹⁄ₐ)² = a² + (1)/(a²) + 2

Rearranging this to solve for a² + (1)/(a²) gives the shortcut formula:

a² + (1)/(a²) = (a + ¹⁄ₐ)² - 2

Substituting the given value a + ¹⁄ₐ = 12 into the equation:

a² + (1)/(a²) = 12² - 2

Calculating the square of 12:

a² + (1)/(a²) = 144 - 2

Subtracting 2 from 144 results in:

a² + (1)/(a²) = 142

Explanation

The area of an equilateral triangle with side a is given by A_T = √3{4} a². Given the side a = 28 cm and √3 = 1.73, we calculate the total area of the triangle as: A_T = (1.73)/(4) × 28² = (1.73)/(4) × 784 = 1.73 × 196 = 339.08 cm²

The three sectors formed at the vertices of the equilateral triangle each have an angle of 60° and a radius r equal to half the side length (r = 14 cm). The combined area of these three sectors is: A_S = 3 × ⁶⁰⁄₃₆₀ × π r² = ½ × ²²⁄₇ × 14² = 11 × 2 × 14 = 308 cm²

The area of the triangle not included in the circles is the difference between the area of the triangle and the combined area of the three sectors: Area = A_T - A_S = 339.08 - 308 = 31.08 cm²

Explanation

Let S_P and S_Q represent the salaries of P and Q respectively. Since expenditure is the portion of salary not saved, P's expenditure is 100% - 64% = 36% of S_P, and Q's expenditure is 100% - 44% = 56% of S_Q. According to the problem, P's expenditure is 1.5 times Q's expenditure, which gives the following equation:

0.36 S_P = 1.5 × 0.56 S_Q

Simplifying the right side of the equation:

0.36 S_P = 0.84 S_Q

To find the ratio of P's salary to Q's salary, rearrange the equation to solve for S_P / S_Q:

(S_P)/(S_Q) = (0.84)/(0.36) = ⁸⁴⁄₃₆

Dividing both the numerator and the denominator by their greatest common divisor, 12:

(S_P)/(S_Q) = ⁷⁄₃

Explanation

To solve the expression 9÷ [1 + (4× (⅚ -⅓ +½))], follow the BODMAS rule, which mandates solving the innermost parentheses first.

First, calculate the value inside the smallest parentheses by finding a common denominator for the fractions: ⅚ - ⅓ + ½ = (5 - 2 + 3)/(6) = ⁶⁄₆ = 1

Next, substitute this value back into the expression and solve the multiplication within the brackets: 9 ÷ [1 + (4 × 1)] = 9 ÷ [1 + 4]

Now, simplify the terms within the square brackets: 1 + 4 = 5

Finally, perform the division operation: 9 ÷ 5 = ⁹⁄₅

Explanation

The given expression is 2(tan 54°)/(cot 36°) - (cot 41°)/(tan 49°).

Using the trigonometric identity for complementary angles, tan(90° - θ) = cot θ or cot(90° - θ) = tan θ.

For the first term, since 36° = 90° - 54°, we have cot 36° = cot(90° - 54°) = tan 54°. Substituting this into the first term: 2(tan 54°)/(tan 54°) = 2(1) = 2

For the second term, since 49° = 90° - 41°, we have tan 49° = tan(90° - 41°) = cot 41°. Substituting this into the second term: (cot 41°)/(cot 41°) = 1

Combining the results: 2 - 1 = 1

Explanation

A unique circle can be drawn through any three non-collinear points in a plane. Geometrically, if we have three non-collinear points A, B, and C, we can construct the perpendicular bisectors of the line segments AB and BC. Because the points are not on the same line, these two perpendicular bisectors are not parallel and will intersect at exactly one point, O. This point O is equidistant from A and B, and also equidistant from B and C, meaning OA = OB = OC. This intersection point serves as the unique center of the circle, and the constant distance to the points serves as the radius r. Mathematically, the general equation of a circle x² + y² + 2gx + 2fy + c = 0 contains three independent parameters (g, f, and c). Therefore, exactly three independent equations, provided by three non-collinear points, are required to solve for these constants and define the circle uniquely.

Explanation

The Selling Price (SP) is determined by applying the 40% discount to the Marked Price (MP) of ₹2,000. Using the formula SP = MP × (100 - Discount%)/(100): SP = 2000 × (100 - 40)/(100) = 2000 × 0.60 = 1200

The Cost Price (CP) is ₹1,700. Since the Selling Price (₹1,200) is less than the Cost Price, there is a loss. Loss = CP - SP = 1700 - 1200 = 500

The loss percentage is calculated on the Cost Price using the formula Loss Percentage = (Loss)/(CP) × 100: Loss Percentage = ⁵⁰⁰⁄₁₇₀₀ × 100 = ⁵⁰⁰⁄₁₇ ≈ 29.4117%

Rounding to two decimal places, the loss percentage is 29.41%.

Explanation

The sum of the squares of the first n natural numbers is given by the formula ⁿ⁽ⁿ⁺¹⁾⁽²ⁿ⁺¹⁾⁄₆. To calculate the average, we divide this sum by n, resulting in the formula:

Average = ⁽ⁿ⁺¹⁾⁽²ⁿ⁺¹⁾⁄₆

Substituting n = 48 into this formula:

Average = ((48 + 1)(2 × 48 + 1))/(6) Average = (49 × 97)/(6) Average = ⁴⁷⁵³⁄₆

Performing the division, we get 792.166…, which, when rounded to two decimal places, equals 792.17.

Explanation

The net rate of work done when both the inlet and outlet pipes are open simultaneously is the difference between their individual rates. The filling rate of the inlet pipe is ¹⁄₁₁ tank per hour, and the emptying rate of the outlet pipe is ¹⁄₁₅ tank per hour.

Using the combined work formula for an inlet and an outlet: Net Rate = ¹⁄₁₁ - ¹⁄₁₅

To solve this, find the common denominator, which is 11 × 15 = 165: Net Rate = (15 - 11)/(165) = ⁴⁄₁₆₅

The time taken to fill the empty tank is the reciprocal of the net rate: Time = ¹⁶⁵⁄₄

Converting this improper fraction into a mixed fraction: 165 ÷ 4 = 41 with a remainder of 1 Time = 41¼ hours

Explanation

The given equation a³ + (1)/(a³) = 2 for a > 0 follows the Arithmetic Mean-Geometric Mean (AM-GM) inequality, which states that for any positive real number x, the expression x + ¹⁄ₓ ≥ 2. The equality x + ¹⁄ₓ = 2 holds if and only if x = 1. By setting x = a³, the condition a³ + (1)/(a³) = 2 implies a³ = 1. Since a must be a positive real number, we find a = 1. Substituting this value into the expression to be evaluated, we get a + ¹⁄ₐ = 1 + ¹⁄₁ = 2. Alternatively, using the algebraic identity a³ + (1)/(a³) = (a + ¹⁄ₐ)³ - 3(a + ¹⁄ₐ), let k = a + ¹⁄ₐ. The equation becomes k³ - 3k = 2, or k³ - 3k - 2 = 0. Testing k = 2 as a root gives 2³ - 3(2) - 2 = 8 - 6 - 2 = 0, which satisfies the condition.

Explanation

From the given equation x² + 7x + 8 = 0, we can isolate the terms x² + 8 to find a substitution for the denominator of the required expression. Rearranging the terms gives: x² + 8 = -7x

The expression to be evaluated is: (4x)/(x²⁾⁻⁵ˣ⁺⁸ = (4x)/((x²⁾⁺⁸⁾⁻⁵ˣ

By substituting -7x for x² + 8 in the denominator, we get: (4x)/(-7x - 5x) = ⁴ˣ⁄₋₁₂ₓ

In the equation x² + 7x + 8 = 0, if x = 0, then 8 = 0, which is a contradiction. Therefore, x ≠ 0, and we can divide the numerator and the denominator by x: ⁴⁄₋₁₂ = -⅓

Explanation

The average production is calculated by dividing the total production of a car type by the number of years (4).

Total production of B-type cars = 680 + 750 + 620 + 540 = 2590 Average production of B-type cars = ²⁵⁹⁰⁄₄ = 647.5

Total production of A-type cars = 840 + 900 + 760 + 800 = 3300 Average production of A-type cars = ³³⁰⁰⁄₄ = 825

Total production of C-type cars = 890 + 960 + 1000 + 1200 = 4050 Average production of C-type cars = ⁴⁰⁵⁰⁄₄ = 1012.5

The combined average production of A-type and C-type cars is the average of their individual averages: Combined Average (A, C) = (825 + 1012.5)/(2) = (1837.5)/(2) = 918.75

The difference between the combined average of A and C and the average of B is: Difference = 918.75 - 647.5 = 271.25

Explanation

The formula for Simple Interest is SI = (P × R × T)/(100), where P is the principal sum, R is the rate of interest per annum, and T is the time in years.

According to the problem, the simple interest for 6 years is ⅗ of the principal sum. This can be expressed as SI = ⅗P. Substituting the given values into the formula, we get:

⅗P = (P × R × 6)/(100)

By cancelling P from both sides, the equation becomes:

⅗ = (6R)/(100)

Solving for R:

R = (3 × 100)/(5 × 6)

R = ³⁰⁰⁄₃₀

R = 10%

Alternatively, total interest as a percentage of the principal is ⅗ × 100 = 60%. Since this interest is accumulated over 6 years, the annual rate of interest is (60%)/(6) = 10%.

Explanation

In an isosceles triangle, two sides must be of equal length. Given that two sides are 6 cm and 12 cm, the third side must be either 6 cm or 12 cm. According to the Triangle Inequality Theorem, the sum of the lengths of any two sides of a triangle must be strictly greater than the length of the third side. If the third side were 6 cm, the sides would be 6 cm, 6 cm, and 12 cm. Checking the inequality, 6 + 6 > 12 results in 12 > 12, which is false, meaning these dimensions cannot form a triangle. If the third side is 12 cm, the sides are 6 cm, 12 cm, and 12 cm. Checking the inequalities, 6 + 12 > 12 and 12 + 12 > 6 are both true. Thus, the only mathematically possible length for the third side is 12 cm.

Explanation

Let x be the number of years ago when the ratio of their ages was 3:4. At that time, Nihira's age was 25 - x and Punith's age was 30 - x. Based on the given ratio, we can establish the following equation:

(25 - x)/(30 - x) = ¾

By cross-multiplying to solve for x:

4(25 - x) = 3(30 - x) 100 - 4x = 90 - 3x 100 - 90 = 4x - 3x x = 10

Ten years ago, Nihira was 15 years old and Punith was 20 years old. The ratio of their ages was 15:20, which simplifies to 3:4.

Explanation

The circumference of the base of the cylinder is given by the formula C = 2 π r. Given C = 154 cm and using π = 22/7, we can find the radius r as follows: 2 × ²²⁄₇ × r = 154 r = (154 × 7)/(44) = 3.5 × 7 = 24.5 cm

The height of the vessel is given as 49 mm. Converting this to centimeters to maintain unit consistency, we get h = 49 / 10 = 4.9 cm.

The volume V of a cylinder is calculated using the formula V = π r² h. Substituting the known values: V = ²²⁄₇ × (24.5)² × 4.9 V = 22 × 24.5 × 24.5 × 0.7 V = 15.4 × 600.25 = 9243.85 cm³

To convert the volume from cubic centimeters to litres, we use the relation 1 litre = 1000 cm³. Volume in litres = (9243.85)/(1000) = 9.24385 litres

Rounding to three places of decimals, the capacity of the vessel is 9.244 litres.

Explanation

The length of a direct common tangent (DCT) is given by the formula DCT = √(d²⁾⁻⁽ʳ^^¹ ⁻ ʳ^₂⁾^² and the length of a transverse common tangent (TCT) is given by the formula TCT = √(d²⁾⁻⁽ʳ^^¹ ⁺ ʳ^₂⁾^², where d is the distance between the centers of the circles and r₁, r₂ are their radii.

Given d = 36 cm, r₁ = 15 cm, and r₂ = 9 cm: DCT = √(36²⁾⁻⁽¹⁵⁻⁹⁾^^^² = √(36²⁾⁻⁶^^^² = √((36 - 6)(36 + 6)) = √(30 × 42) = √(1260) = 6√(35) cm. TCT = √(36²⁾⁻⁽¹⁵⁺⁹⁾^^^² = √(36²⁾⁻²⁴^^^² = √((36 - 24)(36 + 24)) = √(12 × 60) = √(720) = 12√5 cm.

To find the sum of these lengths: Sum = 6√(35) + 12√5 Sum = 6√(7 × 5) + 12√5 Sum = 6√5 × √7 + 6√5 × 2 Sum = 6√5 (√7 + 2) cm.

Explanation

For a system of linear equations a₁x + b₁y = c₁ and a₂x + b₂y = c₂ to have infinitely many solutions, the ratio of the coefficients must be equal: (a₁)/(a₂) = (b₁)/(b₂) = (c₁)/(c₂). In this system, the coefficients are a₁ = 3, b₁ = 1, c₁ = 3 and a₂ = a - b, b₂ = a + b, c₂ = 3a + b - 3.

Applying the condition for infinite solutions: (3)/(a - b) = (1)/(a + b) = (3)/(3a + b - 3)

Equating the first and second ratios: 3(a + b) = 1(a - b) 3a + 3b = a - b 2a = -4b a = -2b

Equating the second and third ratios: 1(3a + b - 3) = 3(a + b) 3a + b - 3 = 3a + 3b b - 3 = 3b -3 = 2b b = -³⁄₂

Substituting b = -³⁄₂ into the relation a = -2b: a = -2(-³⁄₂) = 3

Explanation

Before B starts moving, P has already been walking for 8 hours. The lead distance created by P is calculated as Distanceₗₑₐ_d = 8 km/h × 8 hours = 64 km. B must close this gap using the relative speed difference. Since both are moving in the same direction, the relative speed is 24 km/h - 8 km/h = 16 km/h. The time taken by B to catch P is Time = Distanceₗₑₐ_d{Relative Speed} = (64 km)/(16 km/h) = 4 hours. The distance from the starting point where B catches P is the total distance B travels during this time, which is 24 km/h × 4 hours = 96 km.

Explanation

The total profit percentage can be calculated using the weighted average of the profit percentages for each part of the transaction. This method is efficient as it remains independent of the total number of books.

First, identify the profit percentage for each segment based on the cost price (CP) of ₹200: Profit percentage on half the books sold at ₹300: p₁ = (300 - 200)/(200) × 100 = 50% Profit percentage on one-fifth of the books sold at ₹350: p₂ = (350 - 200)/(200) × 100 = 75% Profit percentage on the remaining books sold at CP: p₃ = 0%

Next, determine the weights (fractions) for these segments: Weight for the first segment: w₁ = ½ Weight for the second segment: w₂ = ⅕ Weight for the third segment: w₃ = 1 - (½ + ⅕) = 1 - ⁷⁄₁₀ = ³⁄₁₀

Applying the weighted average formula for total profit percentage: Profit % = (w₁ × p₁) + (w₂ × p₂) + (w₃ × p₃) Profit % = (½ × 50) + (⅕ × 75) + (³⁄₁₀ × 0) Profit % = 25 + 15 + 0 = 40%

Explanation

To find when X and Y meet for the first time on a circular track, we calculate their relative speed. First, convert the speeds from km/h to m/s by multiplying by ⁵⁄₁₈. The speed of X is 9 × ⁵⁄₁₈ = 2.5 m/s, and the speed of Y is 13.5 × ⁵⁄₁₈ = 3.75 m/s.

Since they are running in the same direction, their relative speed is the difference between their individual speeds. The relative speed vᵣₑₗ = 3.75 - 2.5 = 1.25 m/s. In a circular race, they meet for the first time when the faster runner gains exactly one full lap of distance over the slower runner.

Using the formula Time = (Distance)/(Relative Speed), where the distance is the length of the track L = 750 m: t = (750)/(1.25) t = ⁷⁵⁰⁰⁰⁄₁₂₅ t = 600 seconds

Explanation

The expression can be simplified by converting all trigonometric terms into their sine and cosine equivalents.

The first part, csc θ - sin θ , simplifies as: csc θ - sin θ = (1)/(sin θ) - sin θ = 1 - sin² θ{sin θ} = cos² θ{sin θ}

The second part, sec θ - cos θ , simplifies as: sec θ - cos θ = (1)/(cos θ) - cos θ = 1 - cos² θ{cos θ} = sin² θ{cos θ}

The third part, tan θ + cot θ , simplifies as: tan θ + cot θ = (sin θ)/(cos θ) + (cos θ)/(sin θ) = sin² θ + cos² θ{sin θ cos θ} = (1)/(sin θ cos θ)

Now, multiplying these three results together: ( cos² θ{sin θ} ) × ( sin² θ{cos θ} ) × ( (1)/(sin θ cos θ) )

Combining the terms into a single fraction: sin² θ cos² θ{sin² θ cos² θ} = 1

Explanation

By subtracting 5 from both sides of the equation m + 5 = sec² A csc² A (1 - cos² A)(1 - sin² A) + 5, we obtain the expression m = sec² A csc² A (1 - cos² A)(1 - sin² A). We apply the Pythagorean identities sin² A = 1 - cos² A and cos² A = 1 - sin² A to rewrite the terms in the parentheses. This yields m = sec² A csc² A sin² A cos² A. Given that the reciprocal trigonometric identities are sec² A = (1)/(cos² A) and csc² A = (1)/(sin² A), substituting these into the equation results in m = (1)/(cos² A) · (1)/(sin² A) · sin² A · cos² A. All terms in the numerator and denominator cancel out, leading to m = 1.

Explanation

In triangle ABC, since AC = BC, the angles opposite these sides are equal, so angle ABC = angle BAC = 70 degrees. The exterior angle ACD is the sum of the interior opposite angles, so angle ACD = 70 + 70 = 140 degrees. Therefore, angle ACD - angle ABC = 140 - 70 = 70 degrees.

Explanation

The fraction 41/80 is the largest because its value is 0.5125. By using 0.5 as a benchmark, it can be seen that 41/80 is the only fraction in the group that is greater than 1/2. For any fraction n/d, the value exceeds 0.5 if n > d/2. In this instance, the denominator is 80, and half of that is 40. Since the numerator 41 is greater than 40, the fraction 41/80 is greater than 0.5. The exact decimal calculation is:

41 ÷ 80 = 0.5125

All other fractions in the group have numerators that are less than half of their denominators, which results in values less than 0.5.

Explanation

The average speed of a journey where equal distances are covered at different speeds is calculated using the harmonic mean of those speeds. Since the bus covers three equal segments of distance (d/3 each) at speeds v₁, v₂, and v₃, the formula for average speed Vₐᵥ_g is:

Vₐᵥ_g = (3)/(1){v₁ + (1)/(v₂) + (1)/(v₃)}

Given the speeds v₁ = 40 km/h, v₂ = 50 km/h, and v₃ = 60 km/h, substitute these values into the formula:

Vₐᵥ_g = (3)/(1){40 + ¹⁄₅₀ + ¹⁄₆₀}

To simplify the denominator, find the least common multiple (LCM) of 40, 50, and 60, which is 600. Express the fractions with the common denominator:

¹⁄₄₀ + ¹⁄₅₀ + ¹⁄₆₀ = ¹⁵⁄₆₀₀ + ¹²⁄₆₀₀ + ¹⁰⁄₆₀₀ = ³⁷⁄₆₀₀

Now, substitute this back into the average speed equation:

Vₐᵥ_g = (3)/(37){600} = (3 × 600)/(37) = ¹⁸⁰⁰⁄₃₇

Performing the division:

1800 ÷ 37 ≈ 48.6486...

Rounding the result to two decimal places gives 48.65 km/h.

Explanation

Let the capacity of the cistern be the Least Common Multiple (LCM) of the individual filling times: LCM(40, 80, 120) = 240 units.

The efficiency (work per minute) of each pipe is: Efficiency of P = 240 / 40 = 6 units/min Efficiency of Q = 240 / 80 = 3 units/min Efficiency of R = 240 / 120 = 2 units/min

Pipe P remains open for the entire duration of 30 minutes (half an hour), while pipes Q and R are turned off after t minutes. The sum of the work done by all pipes must equal the total capacity: (Efficiency of P × 30) + (Efficiency of Q + Efficiency of R) × t = 240 (6 × 30) + (3 + 2) × t = 240 180 + 5t = 240 5t = 60 t = 12

The pipes Q and R should be turned off after 12 minutes.

Explanation

The algebraic identity relating the sum of three numbers, the sum of their squares, and the sum of their cubes minus thrice their product is a³ + b³ + c³ - 3abc = (a + b + c)(a² + b² + c² - (ab + bc + ca)).

To find the value of ab + bc + ca, use the identity (a + b + c)² = a² + b² + c² + 2(ab + bc + ca). Substituting the given values a + b + c = 18 and a² + b² + c² = 36:

18² = 36 + 2(ab + bc + ca) 324 = 36 + 2(ab + bc + ca) 288 = 2(ab + bc + ca) ab + bc + ca = 144

Now, substitute these values into the cubic identity:

a³ + b³ + c³ - 3abc = 18(36 - 144) a³ + b³ + c³ - 3abc = 18(-108) a³ + b³ + c³ - 3abc = -1944

Explanation

First, evaluate the first bracket: 25 - (48 ÷ 6) + (12 × 2) = 25 - 8 + 24 = 41. Next, simplify the second bracket: 5 + 3 of (25 - 20) = 5 + 3 of 5 = 5 + 15 = 20. Finally, perform the remaining operations: 41 + 78 - 20 = 119 - 20 = 99.

Explanation

For company S in 2015, exports are 51 and the ratio of exports to imports is 3:2, so imports = (51 * 2) / 3 = 34. For company U in 2015, exports are 70 and the ratio of exports to imports is 1:2, so imports = 70 * 2 = 140. The total imports for companies S and U together are 34 + 140 = 174 lakhs.

Explanation

Let the initial savings be 100 and expenditure be 400, resulting in an income of 500. A 20 percent increase in income (500 to 600) and a 10 percent increase in expenditure (400 to 440) leaves a new savings balance of 160. The increase from 100 to 160 is a 60 percent increase.

Explanation

The cost price is calculated as 144 / (1 - 0.28) = 144 / 0.72 = 200. Selling the article for 288 results in a gain of 88, and the gain percentage is (88 / 200) * 100 = 44%.

Explanation

The circumference of the new circle is the sum of the circumferences of the two given circles, so 2πR = 2π(12) + 2π(16), which simplifies to R = 12 + 16 = 28 cm. The diameter of the circle is 2R, which equals 2 × 28 = 56 cm.

Explanation

In a kite ABCD where AB = BC and AD = DC, the diagonal BD acts as an axis of symmetry, making triangles ABD and CBD congruent. This implies that angle ABD = angle CBD (68 = 3x + 2, so x = 22) and angle ADB = angle CDB (2y - 7 = 33, so y = 20). The value of 2x + 3y is 2(22) + 3(20) = 44 + 60 = 104.

Explanation

The given trigonometric equation is cot² θ - 2 cos² θ = 0. Since cot θ = (cos θ)/(sin θ), the equation can be rewritten as:

(cos² θ)/(sin² θ) - 2 cos² θ = 0

Factoring out cos² θ from both terms:

cos² θ ( (1)/(sin² θ) - 2 ) = 0

Given the condition 0° < θ < 90°, cos θ cannot be zero because cos θ = 0 only at θ = 90°. Therefore, the term inside the parentheses must equal zero:

(1)/(sin² θ) - 2 = 0 sin² θ = ½ sin θ = (1)/(√(2))

In the interval 0° < θ < 90°, the sine function takes the value (1)/(√(2)) when θ = 45°.

Explanation

For a system of two linear equations a₁x + b₁y = c₁ and a₂x + b₂y = c₂ to represent coincident lines, the ratio of their coefficients and constants must be equal: a₁{a₂} = b₁{b₂} = c₁{c₂}.

Substituting the values from the given equations K₁x + 3y = 8 and 4x + K₂y = 16, we get: K₁{4} = (3)/(K₂₎ = ⁸⁄₁₆

Simplifying the constant ratio: ⁸⁄₁₆ = ½

By equating the first ratio to the constant ratio: K₁{4} = ½ ⟹ K₁ = 2

By equating the second ratio to the constant ratio: (3)/(K₂₎ = ½ ⟹ K₂ = 6

Comparing the values of K₁ = 2 and K₂ = 6, it is evident that 6 = 3 × 2, which gives the relation K₂ = 3K₁.

Explanation

When two pipes work together to fill a tank, the total time taken T can be calculated using the formula T = (xy)/(x + y), where x and y are the individual times taken by each pipe. Substituting the given values x = 14 hours and y = 21 hours into the formula, we get T = (14 × 21)/(14 + 21). This results in T = ²⁹⁴⁄₃₅. Dividing both the numerator and the denominator by their greatest common divisor, 7, the fraction simplifies to ⁴²⁄₅. Converting this improper fraction into a mixed fraction, we get 8 ⅖ hours.

Alternatively, using the efficiency method, let the total capacity of the tank be the Least Common Multiple (LCM) of 14 and 21, which is 42 units. The efficiency of pipe X is ⁴²⁄₁₄ = 3 units per hour, and the efficiency of pipe Y is ⁴²⁄₂₁ = 2 units per hour. Their combined efficiency when both are opened together is 3 + 2 = 5 units per hour. The time taken to fill the tank is the total capacity divided by the combined efficiency, which is ⁴²⁄₅ = 8 ⅖ hours.

Explanation

According to the geometric property of circles, the angle subtended by a diameter at any point on the circumference is always a right angle, also known as the angle in a semicircle. Since PR is a diameter of the first circle and point Q is a point of intersection lying on its circumference, ∠ PQR must be 90°. Similarly, in the second circle where PS is the diameter, ∠ PQS is also 90°. Since the sum of ∠ PQR and ∠ PQS equals 180°, the points R, Q, and S are collinear, forming a straight line segment RS.

Explanation

To find the final payment, we calculate the interest and the remaining principal step by step:

1. The interest for the first year is 8% of the initial principal: 3,48,020 * 0.08 = ₹27,841.60.
2. At the end of the first year, Naman pays ₹15,090 towards the principal, leaving a new principal of 3,48,020 - 15,090 = ₹3,32,930.
3. The interest for the second year is 8% of the remaining principal: 3,32,930 * 0.08 = ₹26,634.40.
4. The total amount to be paid at the end of the second year is the remaining principal plus the interest from both years: 3,32,930 + 27,841.60 + 26,634.40 = ₹3,87,406.

Explanation

From the given equations, we isolate the trigonometric ratios: ˣ⁄ₚ = sec A cos B (y)/(q) = sec A sin B ᶻ⁄ᵣ = tan A

Squaring these ratios, we obtain: (x²)/(p²) = sec² A cos² B (y²)/(q²) = sec² A sin² B (z²)/(r²) = tan² A

Now, substitute these values into the required expression: (x²)/(p²) + (y²)/(q²) - (z²)/(r²) = sec² A cos² B + sec² A sin² B - tan² A

Factoring sec² A from the first two terms: sec² A (cos² B + sin² B) - tan² A

Using the fundamental trigonometric identity sin² θ + cos² θ = 1 , the expression becomes: sec² A (1) - tan² A = sec² A - tan² A

According to the identity 1 + tan² A = sec² A , it follows that sec² A - tan² A = 1 . Therefore, the value of the expression is 1.

Explanation

The result is obtained by applying the standard algebraic identity for the sum of the squares of a binomial sum and a binomial difference: (A + B)² + (A - B)² = 2(A² + B²) Substituting A = x and B = y into this identity gives: (x + y)² + (x - y)² = 2(x² + y²) This relationship is derived by expanding the two squares into x² + 2xy + y² and x² - 2xy + y² respectively. When these two expressions are added, the middle terms +2xy and -2xy are additive inverses and cancel out, leaving x² + y² + x² + y². Combining the like terms results in 2x² + 2y², which simplifies to the factored form 2(x² + y²).

Explanation

The initial distance between the policeman and the thief can be calculated using the concept of relative speed. Since both are running in the same direction, the relative speed is the difference between their individual speeds.

Relative Speed = Speed of Policeman - Speed of Thief = 12 km/h - 10 km/h = 2 km/h.

The time taken to catch the thief is given as 30 minutes. Converting this into hours: Time = 30 / 60 = 0.5 hours.

The distance between them is the product of their relative speed and the time taken to close the gap: Distance = Relative Speed × Time = 2 km/h × 0.5 hours = 1 km.

To convert this distance into metres: Distance in metres = 1 × 1000 = 1000 metres.

Explanation

The perimeter of a sector of a circle is the sum of the length of the circular arc and the two radii that bound the sector. The formula for the perimeter P is given by P = (θ)/(360°) × 2π r + 2r, where r is the radius and θ is the angle in degrees subtended at the centre.

Given the radius r = 14 cm, the central angle θ = 45°, and π = ²²⁄₇, we first calculate the length of the arc l: l = ⁴⁵⁄₃₆₀ × 2 × ²²⁄₇ × 14 l = ⅛ × 2 × 22 × 2 l = ⁸⁸⁄₈ = 11 cm

To find the total perimeter of the sector, we add the two radii to the arc length: P = l + 2r P = 11 + 2(14) P = 11 + 28 = 39 cm

Explanation

For any positive integer n, the algebraic identity xⁿ - yⁿ is always divisible by x - y. Since m is specified as an even number, we can let m = 2k, where k is a positive integer. Substituting this into the given expression, we have 8²ᵏ - 1. This can be rewritten in the form (8²)ᵏ - 1ᵏ, which simplifies to 64ᵏ - 1ᵏ. Applying the divisibility rule where x = 64 and y = 1, the expression 64ᵏ - 1 is necessarily divisible by 64 - 1 = 63 for all integer values of k. To verify this with the smallest even positive integer, setting m = 2 results in 8² - 1 = 64 - 1 = 63, which is exactly divisible by 63.

Explanation

A system of linear equations represented as a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0 has a unique solution if the lines intersect at exactly one point. This occurs when the ratio of the coefficients of x is not equal to the ratio of the coefficients of y, which is expressed by the condition (a₁)/(a₂) ≠ (b₁)/(b₂).

From the given equations, 4x + 6y - 7 = 0 and 6x + (k + 4)y - 21 = 0, the coefficients are a₁ = 4, b₁ = 6, a₂ = 6, and b₂ = k + 4. Substituting these into the unique solution condition gives:

⁴⁄₆ ≠ (6)/(k + 4)

Simplifying the fraction on the left to ⅔ and cross-multiplying the terms, we get:

2(k + 4) ≠ 6 × 3 2k + 8 ≠ 18 2k ≠ 18 - 8 2k ≠ 10 k ≠ 5

Explanation

In a Buy x Get y Free offer, the effective discount is calculated based on the total number of items handed over to the customer. Let the marked price of each sweater be M. In this specific scheme, the customer pays for 5 sweaters but receives a total of 7 sweaters.

Total Marked Price of 7 sweaters = 7M Total Discount given (value of 2 free sweaters) = 2M

The formula for the effective discount percentage is: Discount % = (Number of free items)/(Total number of items) × 100

Substituting the given values: Discount % = (2)/(5 + 2) × 100 = ²⁄₇ × 100

Using the fractional equivalent where ⅐ = 14 ²⁄₇%, we calculate: Discount % = 2 × 14 ²⁄₇% = 28 ⁴⁄₇%

Explanation

The average score of the whole class is determined using the weighted average formula A = (n₁ a₁ + n₂ a₂)/(n₁ + n₂). In this class of 30 students, the first group consists of n₁ = 10 students with an average score of a₁ = 90. The remaining students form the second group, where n₂ = 30 - 10 = 20, with an average score of a₂ = 75. To simplify the math, the number of students can be expressed as a ratio n₁:n₂ = 10:20, which reduces to 1:2. Using these ratios as weights, the calculation becomes ((1 × 90) + (2 × 75))/(1 + 2). This further simplifies to (90 + 150)/(3) = ²⁴⁰⁄₃ = 80.