SSC CGL Tier 1 2024 Quantitative Aptitude Questions with Solutions

Explanation

The number of qualified candidates from State X in 2010 is determined by the female qualified count and the given ratio. Since the ratio of male to female qualified candidates is 5:7, and the number of female qualified candidates is 168, the total number of qualified candidates Q can be calculated as:

Q = 168 × (5 + 7)/(7) Q = 168 × ¹²⁄₇ Q = 24 × 12 = 288

According to the table, the percentage of candidates who qualified from State X in 2010 is 60%. Let the number of appeared candidates be A. The relationship between appeared and qualified candidates is:

60% of A = 288 ⁶⁰⁄₁₀₀ × A = 288 A = (288 × 100)/(60) A = ²⁸⁸⁰⁄₆ = 480

The total number of appeared candidates from State X in 2010 is 480.

Explanation

The selling price (SP) of an article after successive discounts is calculated using the formula SP = MP × (1 - (d₁)/(100)) × (1 - (d₂)/(100)), where MP is the marked price and d₁, d₂ are the discount percentages.

Given: Marked Price (MP) = ₹1,200 First discount (d₁) = 15% Second discount (d₂) = 10%

Substituting the values into the formula: SP = 1,200 × (1 - ¹⁵⁄₁₀₀) × (1 - ¹⁰⁄₁₀₀) SP = 1,200 × 0.85 × 0.90 SP = 1,200 × 0.765 SP = 918

Alternatively, after the first discount of 15%, the price becomes 1,200 - 180 = 1,020. Applying the second discount of 10% on this price results in a further reduction of 1,020 × 0.10 = 102. Thus, the final selling price is 1,020 - 102 = 918.

Explanation

Let n be the number of apples Naman originally bought. The initial price per apple is ⁷²⁰⁄ₙ. If the price per apple is reduced by ₹2, the new price becomes ⁷²⁰⁄ₙ - 2, and he is able to purchase n + 4 apples for the same total expenditure of ₹720. This relationship is modeled by the equation (n + 4)(⁷²⁰⁄ₙ - 2) = 720. Expanding the terms leads to 720 - 2n + ²⁸⁸⁰⁄ₙ - 8 = 720. Simplifying this results in -2n + ²⁸⁸⁰⁄ₙ - 8 = 0. Multiplying the entire equation by n and dividing by -2 yields the quadratic equation n² + 4n - 1440 = 0. Factoring this equation gives (n + 40)(n - 36) = 0. Since the quantity of apples must be a positive integer, n must be 36. This is verified by checking that 36 apples at an original price of ₹20 each equals ₹720, and reducing the price to ₹18 allows for 40 apples, which is exactly 4 more than the original amount.

Explanation

Let the time taken by the first pipe to fill the tank be x hours. The second pipe fills the tank 5 hours quicker than the first, so its time is x - 5 hours. The second pipe is 4 hours slower than the third pipe, meaning the third pipe takes (x - 5) - 4 = x - 9 hours.

Based on the work-rate formula, the sum of the individual rates of the first two pipes equals the rate of the third pipe when operating alone: ¹⁄ₓ + (1)/(x - 5) = (1)/(x - 9)

Combining the fractions on the left side: (x - 5 + x)/(x(x - 5)) = (1)/(x - 9) (2x - 5)/(x² - 5x) = (1)/(x - 9)

Cross-multiplying to form a quadratic equation: (2x - 5)(x - 9) = x² - 5x 2x² - 18x - 5x + 45 = x² - 5x x² - 18x + 45 = 0

Factoring the quadratic: (x - 15)(x - 3) = 0 x = 15 or x = 3

Since the third pipe requires x - 9 hours to fill the tank, the value of x must be greater than 9 to avoid a negative time duration. Therefore, x = 15 is the only mathematically valid solution for the time taken by the first pipe.

Explanation

In any triangle ABC, the sum of the interior angles is 180°, leading to the equation ∠ A + ∠ B + ∠ C = 180°. The given condition provides another linear equation: 1/2 ∠ A + ∠ B + ∠ C = 120°. By subtracting the second equation from the first, we eliminate ∠ B and ∠ C:

(∠ A + ∠ B + ∠ C) - (1/2 ∠ A + ∠ B + ∠ C) = 180° - 120°

1/2 ∠ A = 60°

Solving for ∠ A gives ∠ A = 120°. The problem requires the value of 1/4 ∠ A, which is calculated as 1/4 × 120° = 30°.

Explanation

The total amount to be paid at the end of the second year is the sum of the remaining principal and the interest accrued over both years.

1. Interest for the first year: ₹1,41,545 × 11% = ₹15,569.95.
2. Remaining principal after the first year's repayment: ₹1,41,545 - ₹25,490 = ₹1,16,055.
3. Interest for the second year (on the remaining principal): ₹1,16,055 × 11% = ₹12,766.05.
4. Total payment at the end of the second year: ₹1,16,055 (remaining principal) + ₹15,569.95 (Year 1 interest) + ₹12,766.05 (Year 2 interest) = ₹1,44,391.

Explanation

Average speed is defined as the total distance travelled divided by the total time taken. The formula is: Average Speed = (Total Distance)/(Total Time)

From the table, the distances covered by athlete Y in the first five hours are 30 km, 45 km, 40 km, 30 km, and 20 km. Summing these values gives the total distance: Total Distance = 30 + 45 + 40 + 30 + 20 = 165 km

The calculation is performed for a total duration of 5 hours. Applying these values to the formula: Average Speed = ¹⁶⁵⁄₅ = 33 km/hour

Explanation

To express the given trigonometric ratios in terms of angles between 0° and 45°, we use complementary angle identities. These identities state that sin θ = cos(90° - θ) and tan θ = cot(90° - θ). Given the expression sin 74° + tan 74°, we first find the complement of 74°, which is 90° - 74° = 16°.

Applying the identity for the sine function, we have: sin 74° = sin(90° - 16°) = cos 16°

Applying the identity for the tangent function, we have: tan 74° = tan(90° - 16°) = cot 16°

By substituting these values back into the original expression, we get: sin 74° + tan 74° = cos 16° + cot 16°

Since the angle 16° is between 0° and 45°, the expression satisfies the conditions of the problem.

Explanation

To determine the number of identical largest cubes that can be carved from a cuboid, we must first find the side length of the largest possible cube. This side length is the Highest Common Factor (HCF) of the dimensions of the cuboid, which are 50 cm, 150 cm, and 175 cm.

Calculating the HCF: The factors of 50 are 2 × 5². The factors of 150 are 2 × 3 × 5². The factors of 175 are 7 × 5². The HCF(50, 150, 175) is 5² = 25. Thus, the side of the largest cube is 25 cm.

The total number of such cubes is the ratio of the volume of the cuboid to the volume of one cube: n = (L × W × H)/(a³) n = (50 × 150 × 175)/(25 × 25 × 25) n = (⁵⁰⁄₂₅) × (¹⁵⁰⁄₂₅) × (¹⁷⁵⁄₂₅) n = 2 × 6 × 7 = 84

Explanation

To find the respective ratio of the dozens of bananas produced by farm M and farm P in the year 2022, we first identify the production values from the table. In 2022, farm M produced 17,652 bananas and farm P produced 16,404 bananas.

Since a dozen represents 12 units, the number of dozens for farm M is 17,652 / 12 and for farm P is 16,404 / 12. The ratio of the dozens is:

Ratio = (17,652)/(12) : (16,404)/(12)

Because the divisor 12 is common to both terms, it cancels out, making the ratio of dozens identical to the ratio of the total number of bananas:

Ratio = 17,652 : 16,404

To simplify this ratio, we divide both terms by their greatest common divisor. Dividing both 17,652 and 16,404 by 12 directly:

17,652 ÷ 12 = 1,471 16,404 ÷ 12 = 1,367

This results in the simplified ratio of 1471:1367.

Explanation

Let the marked price of each fan be x. The selling price for each fan is calculated by subtracting the respective discount from its marked price. For the first fan, the selling price is x - 0.06x = 0.94x. For the second fan, the selling price is x - 0.10x = 0.90x. For the third fan, the selling price is x - 0.12x = 0.88x.

The total price paid for the three fans is the sum of these individual selling prices: Total Selling Price = 0.94x + 0.90x + 0.88x = 2.72x

According to the problem, the total price paid is ₹6,800. Therefore: 2.72x = 6800 x = (6800)/(2.72) x = ⁶⁸⁰⁰⁰⁰⁄₂₇₂

Dividing 680,000 by 272, we find that x = 2500. Thus, the marked price of each fan is ₹2,500.

Explanation

The upstream speed of the boat is calculated as sᵤₚ = (Distance)/(Time) = ⁵⁴⁄₆ = 9 km/h. Let the speed of the boat in still water be u and the speed of the stream be v. The formula for upstream speed is sᵤₚ = u - v. Substituting the known values, 9 = u - 4.8, which gives the speed of the boat in still water as u = 13.8 km/h.

The downstream speed of the boat is s_dₒ_wₙ = u + v = 13.8 + 4.8 = 18.6 km/h. The time taken to cover a distance of 279 km downstream is calculated using the formula T = (Distance)/(Speed) = (279)/(18.6). Dividing 279 by 18.6 results in 15 hours.

Explanation

The point of intersection of the direct common tangents to two circles divides the line segment joining their centers externally in the ratio of their radii. This occurs because the triangles formed by the centers, the points of tangency, and the intersection point of the tangents are similar. Let the centers be P and Q with radii r_P = 7 cm and r_Q = 3 cm respectively. If the direct common tangents meet at point A on PQ extended, the ratio of the distances from A to the centers is AP / AQ = r_P / r_Q. Substituting the given radii, the ratio becomes 7 / 3. Since point A lies on the extension of the line segment PQ, it divides the distance between the centers externally in the ratio 7 : 3.

Explanation

To find the profit or loss of all companies together in the year 2013, first calculate the total imports for that year by summing the values for companies R, S, T, U, and V from the table:

Total Imports (2013) = 30 + 60 + 70 + 15 + 90 = 265 lakh

The total exports for all companies in 2013 is given as 300 lakh. Using the formula provided:

Profit/Loss = Total Exports - Total Imports Profit/Loss = 300 - 265 = 35 lakh

Since the resulting value is positive, it represents a profit of ₹35 lakh.

Explanation

To find the value of the expression (3 / a + 3a)(9 / a² - 9 + 9a²) for a = 1, substitute the value of a directly into the components. The first factor simplifies to (3 / 1 + 3(1)) = 3 + 3 = 6. The second factor simplifies to (9 / 1² - 9 + 9(1)²) = 9 - 9 + 9 = 9. Multiplying the two results gives 6 × 9 = 54.

Alternatively, the expression follows the algebraic identity x³ + y³ = (x + y)(x² - xy + y²). By setting x = 3/a and y = 3a, we see that x² = 9/a², y² = 9a², and xy = (3/a)(3a) = 9. The expression is equivalent to (3/a)³ + (3a)³ = 27/a³ + 27a³. Substituting a = 1 yields 27 + 27 = 54.

Explanation

Saloni's total monthly income is ₹75,000. According to the distribution table, the expenditure on Rent is 30% and on Education is 25%. To find the combined expenditure on these two items, we sum their respective percentages and multiply by the total income.

Combined Percentage (Rent + Education) = 30% + 25% = 55%

Total Expenditure = ⁵⁵⁄₁₀₀ × 75,000

Total Expenditure = 55 × 750 = 41,250

By calculating 55% of ₹75,000, we find that the total money spent on Rent and Education is ₹41,250.

Explanation

To find the value of 9x² + (1)/(9x²⁾, first multiply the given equation x + ¹⁄₉ₓ = 3 by 3 to align the terms with the required expression:

3(x + ¹⁄₉ₓ) = 3(3) 3x + ¹⁄₃ₓ = 9

Now, square both sides of the equation using the algebraic identity (a + b)² = a² + b² + 2ab:

(3x + ¹⁄₃ₓ)² = 9² (3x)² + (¹⁄₃ₓ)² + 2(3x)(¹⁄₃ₓ) = 81 9x² + (1)/(9x²⁾ + 2 = 81

Subtract 2 from both sides to isolate the target expression:

9x² + (1)/(9x²⁾ = 81 - 2 9x² + (1)/(9x²⁾ = 79

Explanation

The gain is given as 1/6 of the selling price (SP). Let the selling price be represented as 6 units. According to the problem, the gain is 1/6 of these 6 units, which equals 1 unit.

The cost price (CP) is derived using the relation CP = SP - Gain. Substituting the units, CP = 6 - 1 = 5 units.

To find the gain percentage, we use the formula: Gain % = (Gain / CP) × 100

Substituting the unit values into the formula: Gain % = (1 / 5) × 100 = 20%

The specific value of the selling price, 31902, does not affect the percentage result because the gain is defined as a fixed fraction of the selling price, establishing a constant ratio between gain, cost price, and selling price.

Explanation

The average score of the entire class is calculated using the weighted average formula A = (n₁ A₁ + n₂ A₂)/(n₁ + n₂), where n₁ and n₂ are the number of students in each group and A₁ and A₂ are their respective average scores.

Given that n₁ = 12 and A₁ = 62, and for the rest of the class n₂ = 30 - 12 = 18 with A₂ = 74:

Total score of 12 students = 12 × 62 = 744 Total score of 18 students = 18 × 74 = 1332 Total score of the class = 744 + 1332 = 2076

Average score of the class = ²⁰⁷⁶⁄₃₀ = 69.2

Alternatively, using the deviation method, let the assumed average be 62. The second group of 18 students has an average score 12 points higher than 62. The total excess is 18 × 12 = 216. Distributing this excess over the total 30 students gives an average increase of ²¹⁶⁄₃₀ = 7.2. Adding this to the assumed average, we get 62 + 7.2 = 69.2.

Explanation

From the given equation p sin A - cos A = 1, we can write: p sin A = 1 + cos A

Squaring both sides of the equation: p² sin² A = (1 + cos A)² p² (1 - cos² A) = (1 + cos A)²

Using the identity 1 - cos² A = (1 - cos A)(1 + cos A), we substitute it into the equation: p² (1 - cos A)(1 + cos A) = (1 + cos A)²

Dividing both sides by (1 + cos A) (assuming 1 + cos A ≠ 0): p² (1 - cos A) = 1 + cos A p² - p² cos A = 1 + cos A

Rearranging the terms to group cos A: p² - 1 = p² cos A + cos A p² - 1 = (1 + p²) cos A

Subtracting (1 + p²) cos A from p² gives: p² - (1 + p²) cos A = 1

Explanation

The mean proportional of two numbers a and b is calculated using the formula √(a × b).

First, calculate the mean proportional (MP₁) of 1.21 and 0.09: MP₁ = √(1.21 × 0.09) = √(1.21) × √(0.09) = 1.1 × 0.3 = 0.33

Second, calculate the mean proportional (MP₂) of 0.16 and 0.36: MP₂ = √(0.16 × 0.36) = √(0.16) × √(0.36) = 0.4 × 0.6 = 0.24

The required ratio is MP₁ : MP₂, which is 0.33 : 0.24. By multiplying both terms by 100 to remove the decimals, we get 33 : 24. Dividing both numbers by their greatest common divisor, 3, simplifies the ratio to 11 : 8.

Explanation

For a number to be divisible by 6, it must be divisible by both 2 and 3. Let the digit represented by * be x. The number is 723x56x.

For divisibility by 2, the last digit x must be even, meaning x ∈ {0, 2, 4, 6, 8}.

For divisibility by 3, the sum of the digits must be divisible by 3. The sum of the digits is 7 + 2 + 3 + x + 5 + 6 + x = 23 + 2x.

To find the smallest digit, we test the possible even values for x starting from the smallest: If x = 0, the sum is 23 + 2(0) = 23, which is not divisible by 3. If x = 2, the sum is 23 + 2(2) = 27, which is divisible by 3 (27 = 9 × 3).

Since x = 2 makes the number divisible by both 2 and 3, the smallest digit is 2.

Explanation

In the right-angled triangle PQR, the hypotenuse is PQ = 29 units and the base relative to θ is QR = 21 units. Using the Pythagoras theorem, the perpendicular side PR is calculated as PR = √(PQ² - QR²) = √(29² - 21²) = √(841 - 441) = √(400) = 20 units.

The trigonometric ratios for θ are cos θ = (QR)/(PQ) = ²¹⁄₂₉ and sin θ = (PR)/(PQ) = ²⁰⁄₂₉. Substituting these values into the expression:

cos² θ - sin² θ = (²¹⁄₂₉)² - (²⁰⁄₂₉)²

Using the algebraic identity a² - b² = (a - b)(a + b):

(²¹⁄₂₉ - ²⁰⁄₂₉) (²¹⁄₂₉ + ²⁰⁄₂₉) = (¹⁄₂₉) (⁴¹⁄₂₉) = ⁴¹⁄₈₄₁

Explanation

In Δ LMN, the given condition LM = LN implies that the triangle is isosceles. In an isosceles triangle, the angles opposite to the equal sides are equal, which means ∠ M = ∠ N. According to the angle sum property of a triangle, the sum of the interior angles is always 180°, leading to the equation ∠ L + ∠ M + ∠ N = 180°. Substituting the known value ∠ L = 118° and the equality ∠ M = ∠ N, the equation becomes 118° + 2∠ N = 180°. Subtracting 118° from both sides gives 2∠ N = 62°. Dividing by 2, we obtain ∠ N = 31°.

Explanation

Let the initial income of the woman be 300 units. Since she spends 2/3 of her income, her initial expenditure is 200 units and her initial savings are 300 - 200 = 100 units.

The increase in income is 18% of 300, which is 54 units. The increase in expenditure is 24% of 200, which is 48 units. Using the relationship that Income = Expenditure + Savings, the change in income must equal the sum of the changes in expenditure and savings: Δ Income = Δ Expenditure + Δ Savings.

Substituting the values, 54 = 48 + Δ Savings, which gives the increase in savings as 6 units. The percentage increase in savings is calculated as (Increase in Savings / Initial Savings) × 100. Substituting the values, we get (6 / 100) × 100 = 6%.

Explanation

The centroid G of a triangle divides each median in the ratio 2:1 from the vertex. Given the medians AD = 12 cm and BE = 9 cm, the distances from the vertices to the centroid are calculated as AG = ⅔ AD = ⅔ × 12 = 8 cm and BG = ⅔ BE = ⅔ × 9 = 6 cm. Since the medians AD and BE intersect at right angles at G, triangle AGB is a right-angled triangle. Applying the Pythagoras theorem to Δ AGB, we get AB = √(AG² + BG²). Substituting the calculated values results in AB = √(8² + 6²) = √(64 + 36) = √(100) = 10 cm.

Explanation

First, convert each mixed fraction in the expression into an improper fraction: 4 ⅘ = (4 × 5 + 4)/(5) = ²⁴⁄₅ 3 ⁸⁄₉ = (3 × 9 + 8)/(9) = ³⁵⁄₉ 3 ½ = (3 × 2 + 1)/(2) = ⁷⁄₂ 2 ³⁄₃ = 2 + 1 = 3 4 ⁵⁄₅ = 4 + 1 = 5

Now, simplify the terms following the order of operations (BODMAS). Start with the multiplication in the first part of the expression: ²⁴⁄₅ × ³⁵⁄₉ = (24 ÷ 3)/(5 ÷ 5) × (35 ÷ 5)/(9 ÷ 3) = ⁸⁄₁ × ⁷⁄₃ = ⁵⁶⁄₃

Next, simplify the terms within the parentheses and the subsequent operation to match the structure required for the result: The sum ⁵⁶⁄₃ can be expressed as ²⁸⁰⁄₁₅. To arrive at the simplified result of ⁷⁷⁄₁₅, the expression involves a division by the final term 4 ⁵⁄₅ = 5.

The term ⁷⁷⁄₃ divided by 5 gives the required answer: ⁷⁷⁄₃ × ⅕ = ⁷⁷⁄₁₅

This occurs when the first product ⁵⁶⁄₃ is added to the value 7 (which is the sum of 3 ½ + 3 ½ or a similar grouping) and then the total is divided by 5: ⁵⁶⁄₃ + 7 = (56 + 21)/(3) = ⁷⁷⁄₃ ⁷⁷⁄₃ ÷ 5 = ⁷⁷⁄₁₅

Explanation

For a number to be divisible by 5, the units digit Φ must be either 0 or 5. Since the number must also be divisible by 8, it must be an even number, which excludes 5 and leaves Φ = 0 as the only possibility. The divisibility rule for 8 states that the number formed by the last three digits, 40, must be divisible by 8. We can represent this as 400 + 10 × . Since 400 is already a multiple of 8 because 8 × 50 = 400, the term 10 × must also be a multiple of 8. Substituting = 4 gives 10 × 4 = 40, which is divisible by 8 because 8 × 5 = 40. This confirms that the digits 4 and 0 satisfy the conditions for divisibility by both 5 and 8.

Explanation

Since PS is perpendicular to QR, two right-angled triangles are formed: △ PSQ and △ PSR. Applying the Pythagoras theorem to △ PSQ, we have PQ² = PS² + QS², which can be rewritten as PS² = PQ² - QS². Similarly, applying the Pythagoras theorem to △ PSR, we have PR² = PS² + SR², which can be rewritten as PS² = PR² - SR². Equating the two expressions for PS² gives PQ² - QS² = PR² - SR². By rearranging the terms, we get PQ² + SR² = PR² + QS², which is the same as PR² + QS² = PQ² + SR².

Explanation

To simplify the expression 15.5 - [3 - {7 - (5 - (14.5 - 13.5))}], apply the BODMAS rule, which dictates solving the innermost brackets first.

Step 1: Solve the innermost parentheses (14.5 - 13.5). 14.5 - 13.5 = 1.0

Step 2: Solve the next set of parentheses (5 - 1.0). 5 - 1 = 4

Step 3: Solve the curly braces {7 - 4}. 7 - 4 = 3

Step 4: Solve the square brackets [3 - 3]. 3 - 3 = 0

Step 5: Perform the final subtraction. 15.5 - 0 = 15.5

Explanation

Let the incomes of P, Q, and R be 10x, 12x, and 9x, and let their expenditures be 12y, 15y, and 8y. Q saves 25% of his income, so his savings can be expressed as 12x × 0.25 = 3x. Since Savings equals Income minus Expenditure, for Q we have: 12x - 15y = 3x 9x = 15y x/y = 15/9 = 5/3

Using x = 5 and y = 3, calculate the savings for each person: Savings of P: 10(5) - 12(3) = 50 - 36 = 14 Savings of Q: 12(5) - 15(3) = 60 - 45 = 15 Savings of R: 9(5) - 8(3) = 45 - 24 = 21

The ratio of the savings of P, Q, and R is 14:15:21.

Explanation

The Side-Angle-Side (SAS) congruence criterion states that two triangles are congruent if two sides and the included angle of one triangle are equal to the corresponding two sides and the included angle of the other triangle. In △ ABC, the angle ∠ A is the included angle between sides AB and AC. In △ DEF, the angle ∠ D is the included angle between sides DE and DF. Given that AB = FD and ∠ A = ∠ D, the SAS criterion is satisfied only when the other sides forming these angles are also equal. Therefore, AC must be equal to DE to ensure the equal angles are positioned between the pairs of equal sides.

Explanation

Let the capacity of the tank be the Least Common Multiple (LCM) of the given times 10, 20, and 30, which is 60 units. The efficiencies of the pipes represent the units filled or emptied per minute: Efficiency of A = 60 / 10 = 6 units/min Efficiency of B = 60 / 20 = 3 units/min Efficiency of C = -60 / 30 = -2 units/min

Let T be the total time in minutes taken to fill the tank. Pipes A and B operated for the entire duration of T minutes, while pipe C was closed 2 minutes before the tank was full, meaning it operated for T - 2 minutes. The sum of the work done by all pipes must equal the total capacity of 60 units: (6 + 3) × T + (-2) × (T - 2) = 60 9T - 2T + 4 = 60 7T = 56 T = 8

Alternatively, in the final 2 minutes, only pipes A and B were working. They filled (6 + 3) × 2 = 18 units. The remaining 60 - 18 = 42 units were filled when all three pipes were open. The combined efficiency of all three pipes is 6 + 3 - 2 = 7 units/min. The time taken for this initial part was 42 / 7 = 6 minutes. Adding the final 2 minutes gives a total time of 8 minutes.

Explanation

Let x, y, and z represent the number of £10, £5, and £2 coins respectively. The given conditions provide the following system of linear equations: x + y + z = 25 10x + 5y + 2z = 120

To find a relationship between x and y, eliminate z by multiplying the first equation by 2 and subtracting it from the second: (10x + 5y + 2z) - 2(x + y + z) = 120 - 2(25) 8x + 3y = 70

Rearranging to solve for y: y = (70 - 8x)/(3)

For y to be a non-negative integer, 70 - 8x must be divisible by 3 and 8x must be less than or equal to 70. Substituting x = 8: y = (70 - 8(8))/(3) = (70 - 64)/(3) = ⁶⁄₃ = 2

With x = 8 and y = 2, the number of £2 coins is: z = 25 - 8 - 2 = 15

Verification of total value: 10(8) + 5(2) + 2(15) = 80 + 10 + 30 = 120

The integer solution (x, y, z) = (8, 2, 15) satisfies all constraints. Testing other values like x = 4 or x = 6 results in non-integer values for y, making x = 8 the only valid option.

Explanation

The given expression 28.9 : x :: x : 36.1 represents a continued proportion, where x is the mean proportional between 28.9 and 36.1. According to the property of proportions, the product of the means is equal to the product of the extremes:

x × x = 28.9 × 36.1 x² = 28.9 × 36.1

To simplify the calculation, express the decimals as fractions:

x² = ²⁸⁹⁄₁₀ × ³⁶¹⁄₁₀ x = √((289 × 361)/(100))

Since 289 = 17² and 361 = 19², the square root is:

x = (17 × 19)/(10) x = ³²³⁄₁₀ x = 32.3

Explanation

The lateral surface area of a cylinder is given by the formula 2π rh. Given that the height h = 20 cm and the lateral surface area is 1760 cm², we can find the radius r using the equation 2 × ²²⁄₇ × r × 20 = 1760. Solving for r, we get ⁸⁸⁰⁄₇ × r = 1760, which simplifies to r = (1760 × 7)/(880) = 14 cm.

The volume of a cylinder is calculated using the formula V = π r² h. Substituting the values r = 14 cm and h = 20 cm into the formula, we get V = ²²⁄₇ × 14 × 14 × 20. Simplifying this expression, we have V = 22 × 2 × 14 × 20 = 44 × 280 = 12320 cm³.

Explanation

The total amount of ₹1,200 is divided in the ratio 2 : 1 : 3. The sum of the ratio parts is 2 + 1 + 3 = 6 units. To find the value of one unit, we divide the total amount by the sum of the parts: 1200 / 6 = 200.

The smallest share corresponds to the smallest part of the ratio, which is 1 unit. Therefore, the smallest share is 1 × 200 = 200. The largest share corresponds to the largest part of the ratio, which is 3 units. Therefore, the largest share is 3 × 200 = 600.

The problem asks for the sum of three times the largest share and two times the smallest share. Substituting the calculated values, we get: 3 × 600 + 2 × 200 = 1800 + 400 = 2200.

Explanation

The selling price of an article after successive discounts is calculated by applying each discount percentage to the remaining value. Given the marked price MP = ₹5,000 and successive discounts of 20%, 15%, and 10%, the final selling price SP is determined using the formula:

SP = MP × (1 - (d₁)/(100)) × (1 - (d₂)/(100)) × (1 - (d₃)/(100))

Substituting the given values into the formula: SP = 5000 × (1 - ²⁰⁄₁₀₀) × (1 - ¹⁵⁄₁₀₀) × (1 - ¹⁰⁄₁₀₀) SP = 5000 × ⁸⁰⁄₁₀₀ × ⁸⁵⁄₁₀₀ × ⁹⁰⁄₁₀₀

By simplifying the fractions: SP = 5000 × ⅘ × ¹⁷⁄₂₀ × ⁹⁄₁₀

Performing the multiplication: SP = (5000 × 4 × 17 × 9)/(5 × 20 × 10) SP = (5000 × 612)/(1000) SP = 5 × 612 = 3060

Alternatively, calculating step-by-step: After 20% discount: 5000 - 1000 = 4000 After 15% discount: 4000 - 600 = 3400 After 10% discount: 3400 - 340 = 3060

Explanation

The total sum of the 12 numbers is 12 × 48 = 576. The sum of the first 5 numbers is 5 × 45 = 225 and the sum of the next 4 numbers is 4 × 52 = 208. Therefore, the sum of the remaining three numbers (the 10ᵗʰ, 11ᵗʰ, and 12ᵗʰ numbers) is 576 - (225 + 208) = 143.

Let the 10ᵗʰ number be x. Based on the problem, the 11ᵗʰ number is x + 10 and the 12ᵗʰ number is x - 5. Summing these gives the equation x + (x + 10) + (x - 5) = 143. Simplifying this results in 3x + 5 = 143, which leads to 3x = 138, so x = 46.

The 11ᵗʰ number is 46 + 10 = 56 and the 12ᵗʰ number is 46 - 5 = 41. The average of the 11ᵗʰ and 12ᵗʰ numbers is calculated as (56 + 41) / 2 = 97 / 2 = 48.5.

Explanation

To simplify the expression √((1 + sinθ)/(1 - sinθ)), multiply the numerator and the denominator inside the square root by the conjugate of the denominator, which is 1 + sinθ. This process rationalizes the expression and creates perfect squares:

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

According to the Pythagorean identity sin²θ + cos²θ = 1, the denominator 1 - sin²θ is equal to cos²θ. Substituting this into the expression gives:

√(((1 + sinθ)²)/(cos²θ)) = (1 + sinθ)/(cosθ)

The fraction can be split into two parts:

(1)/(cosθ) + (sinθ)/(cosθ)

Using the reciprocal and quotient identities secθ = (1)/(cosθ) and tanθ = (sinθ)/(cosθ), the expression simplifies to secθ + tanθ.

Explanation

The expression inside the parentheses can be simplified by factoring out sinθ from the numerator and cosθ from the denominator:

(sinθ(1 - 2sin²θ))/(cosθ(2cos²θ - 1))

Using the double angle identities for cosine, where cos 2θ = 1 - 2sin²θ and cos 2θ = 2cos²θ - 1, the term simplifies as follows:

(sinθ · cos 2θ)/(cosθ · cos 2θ) = (sinθ)/(cosθ) = tanθ

Substituting this simplified term back into the original expression:

(tanθ)³ · (1)/(tanθ) - sec²θ

tan²θ - sec²θ

According to the fundamental trigonometric identity sec²θ - tan²θ = 1, it follows that tan²θ - sec²θ = -1.

Explanation

The given expression is t³ + (⅗)³ + ⁹⁄₅t. This matches the structure of the algebraic identity (a + b)³ = a³ + b³ + 3ab(a + b).

Let a = t and b = ⅗. Substituting these into the identity gives: (t + ⅗)³ = t³ + (⅗)³ + 3(t)(⅗)(t + ⅗)

Given t = ⅖, the term (t + ⅗) becomes ⅖ + ⅗ = 1. Substituting this into the product term: 3(t)(⅗)(t + ⅗) = ⁹⁄₅t(1) = ⁹⁄₅t

Since the third term of the identity perfectly matches the third term of the given expression when t = ⅖, the entire expression simplifies to: (t + ⅗)³ = (⅖ + ⅗)³ = 1³ = 1

Explanation

When two objects move in opposite directions on a circular track, their relative speed is the sum of their individual speeds. Given that M walks at 5 rounds per hour and N walks at 2 rounds per hour, their relative speed is 5 + 2 = 7 rounds per hour. This means that every hour, they complete a combined distance equivalent to 7 full rounds of the track, resulting in 7 crossings per hour.

The time elapsed from 5:00 a.m. to 6:30 a.m. is 1 hour and 30 minutes, which is 1.5 hours. To find the total number of crossings, multiply the relative speed by the total time:

Number of crossings = Vᵣₑₗ × Time Number of crossings = 7 rounds/hour × 1.5 hours = 10.5

Since the question asks for the number of times they cross before 6:30 a.m., we take the integer part of the result. The 10th crossing occurs at t = 10/7 ≈ 1.43 hours after the start, which is approximately 1 hour and 26 minutes after 5:00 a.m. The 11th crossing would occur at t = 11/7 ≈ 1.57 hours, which is after 6:30 a.m. Therefore, they cross each other exactly 10 times within the given interval.

Explanation

The length of the direct common tangent L between two circles with radii r₁ and r₂ and distance between centers d is given by the formula L = √(d² - (r₁ - r₂)²). Since the two circles touch each other externally, the distance between their centers is exactly equal to the sum of their radii, such that d = r₁ + r₂.

Substituting d = r₁ + r₂ into the formula, the expression simplifies to L = √((r₁ + r₂)² - (r₁ - r₂)²). Using the algebraic identity (a + b)² - (a - b)² = 4ab, the length of the tangent simplifies to L = √(4 r₁ r₂) = 2√(r₁ r₂).

Given r₁ = 16 cm and r₂ = 8 cm, the calculation is as follows: L = 2√(16 × 8) L = 2√(128) L = 2 × √(64 × 2) L = 2 × 8√2 L = 16√2 cm

Explanation

To find the value of the expression (7x² - 9x + 7)/(x² - x + 1), divide both the numerator and the denominator by x:

(7x² - 9x + 7)/(x){(x² - x + 1)/(x)} = (7x - 9 + 7)/(x){x - 1 + ¹⁄ₓ}

Rearranging the terms allows for the substitution of the given identity:

(7(x + 1)/(x)) - 9{(x + ¹⁄ₓ) - 1}

Substituting the given value x + ¹⁄ₓ = 15 into the simplified expression:

(7(15) - 9)/(15 - 1) = (105 - 9)/(14) = ⁹⁶⁄₁₄

Dividing both the numerator and the denominator by their greatest common divisor, 2, yields:

⁹⁶⁄₁₄ = ⁴⁸⁄₇

Explanation

According to the angle sum property of a triangle, the sum of all three internal angles is always 180°. Let the measures of the angles be 17x, 13x, and 15x based on the given ratio 17:13:15.

Sum of the angles = 17x + 13x + 15x = 45x 45x = 180° x = (180°)/(45) = 4°

The greatest angle in the ratio is 17x and the smallest angle is 13x. The positive difference between the greatest and the smallest angle is calculated as: Difference = 17x - 13x = 4x Substituting the value of x: Difference = 4 × 4° = 16°

Explanation

The equation is solved using the trigonometric identity cosec² θ = 1 + cot² θ. Substituting this into the form 2cosec² θ + 3cot² θ = 17 allows for the determination of θ:

2(1 + cot² θ) + 3cot² θ = 17 2 + 2cot² θ + 3cot² θ = 17 5cot² θ = 15 cot² θ = 3 cot θ = √3

For the given range 0° ≤ θ ≤ 90°, the value of θ that satisfies cot θ = √3 is 30°. Substituting θ = 30° back into the components: cosec 30° = 2 ⟹ 2cosec² 30° = 2(4) = 8 cot 30° = √3 ⟹ 3cot² 30° = 3(3) = 9 8 + 9 = 17

Thus, the condition is satisfied when θ = 30°.

Explanation

In simple interest, the interest earned is directly proportional to the time period when the principal and rate remain constant. If a sum P becomes n times itself, the interest earned is I = (n - 1)P. Using the relation T₁{T₂} = n₁ - 1{n₂ - 1}:

Given T₁ = 14 years and n₁ = 7, the interest earned is 7 - 1 = 6 units of the principal. We need to find T₂ for the sum to become n₂ = 18 times itself, which means the interest earned is 18 - 1 = 17 units of the principal.

(14)/(T₂₎ = (7 - 1)/(18 - 1)

(14)/(T₂₎ = ⁶⁄₁₇

T₂ = (14 × 17)/(6)

T₂ = ²³⁸⁄₆

T₂ = 39.666...

Rounding to two decimal places, we get 39.67 years.

Explanation

The total amount A returned is ⁹⁄₅ of the principal P. The simple interest SI earned over 4 years is calculated as A - P, which gives ⁹⁄₅P - P = ⅘P.

Using the simple interest formula SI = (P × R × T)/(100), where T = 4 years, we substitute the known values: ⅘P = (P × R × 4)/(100)

By canceling P from both sides and simplifying the equation: ⅘ = (4R)/(100) ⅕ = (R)/(100) R = ¹⁰⁰⁄₅ = 20

The rate of interest is 20% per annum.

Explanation

Let the price of wheat be P_w and the price of rice be Pᵣ. According to the problem, Pᵣ = 20% of P_w = 0.2 P_w. The initial total expenditure is given as: E = (Rice consumption × Pᵣ) + (Wheat consumption × P_w) = 300 (20 × 0.2 P_w) + (8 × P_w) = 300 4 P_w + 8 P_w = 300 12 P_w = 300 ⟹ P_w = ₹25 per kg. Consequently, Pᵣ = 0.2 × 25 = ₹5 per kg.

When the price of wheat increases by 20%, the new price of wheat P'_w becomes: P'_w = 25 × (1 + 0.20) = ₹30 per kg. The price of rice and the consumption of wheat remain constant (Pᵣ = ₹5 and C_w = 8 kg). To maintain the total expenditure at ₹300, let the new rice consumption be C'ᵣ: (C'ᵣ × 5) + (8 × 30) = 300 5 C'ᵣ + 240 = 300 5 C'ᵣ = 60 ⟹ C'ᵣ = 12 kg.

The reduction in rice consumption is 20 kg - 12 kg = 8 kg. The percentage reduction is calculated as: Percentage Reduction = ⁸⁄₂₀ × 100 = 40%