SSC CGL Tier 1 2021 Quantitative Aptitude Questions with Solutions

Explanation

In similar triangles △ ABC ∼ △ RPQ, the ratio of the lengths of corresponding sides is equal to the square root of the ratio of their areas. According to the similarity correspondence:

1. A ↔ R
2. B ↔ P
3. C ↔ Q

Thus, the side AC corresponds to the side RQ. From the given information, RQ = QR = 6 cm. The ratio of the areas is given as (ar(△ ABC))/(ar(△ PQR)) = ¹⁶⁄₂₅. The ratio of the corresponding sides is (AC)/(RQ) = √(¹⁶⁄₂₅) = ⅘. Substituting the known value of RQ: (AC)/(6) = ⅘ ⟹ AC = (6 × 4)/(5) = ²⁴⁄₅ = 4.8 cm.

Explanation

The sum of the three-digit numbers is (100a + 10b + c) + (100c + 10a + b) + (100b + 10c + a), which simplifies to 111(a + b + c). Since 111 = 3 x 37, the sum is always divisible by 3, 37, and (a + b + c), making 31 the only number in the options that does not divide the sum.

Explanation

The first offer provides a flat 25 percent discount, whereas the second offer provides a successive discount equivalent to 24.32 percent (calculated as 14 + 12 - (14 * 12 / 100)). The difference between the two offers is 0.68 percent of 10000, which makes the first offer better by 68 rupees.

Explanation

Solving the equation tan θ + 3/tan θ - 2√3 = 0 leads to the quadratic equation tan² θ - 2√3tan θ + 3 = 0, which simplifies to (tan θ - √3)² = 0. This gives tan θ = √3, hence θ = 60° for the range 0° < θ < 90°. Substituting θ = 60° into the expression cosec² θ + cos² θ gives (2/√3)² + (1/2)² = 4/3 + 1/4 = 19/12.

Explanation

The problem asks for the measure of angle AFB in a cyclic quadrilateral ABCD where sides AB and DC produced meet at E, and sides AD and BC produced meet at F. Given that in triangle ADE, angle ADE = 80 degrees and angle AED = 50 degrees, we first find the third angle, angle EAD = 180 - (80 + 50) = 50 degrees. In the cyclic quadrilateral ABCD, these angles correspond to the interior angles at vertices D and A, so angle D = 80 degrees and angle A = 50 degrees. The opposite interior angles are therefore angle B = 180 - 80 = 100 degrees and angle C = 180 - 50 = 130 degrees. Angle AFB is formed by the intersection of lines AD and BC; using triangle ABF, where angle FAB = 50 degrees and the exterior angle at B (angle ABF) is 80 degrees (equal to the interior opposite angle D), we find angle AFB = 180 - (50 + 100) = 30 degrees (or using triangle FDC, 180 - (100 + 50) = 30 degrees).

Explanation

In a triangle right-angled at B, the angles A and C are complementary, meaning A + C = 90°. From the properties of complementary angles, we have sin C = sin(90° - A) = cos A and cos C = cos(90° - A) = sin A. Substituting these relations into the given expression (sin A(cos C + cos A))/(cos C(sin C - sin A)), we get:

(sin A(sin A + cos A))/(sin A(cos A - sin A))

The sin A terms in the numerator and denominator cancel out, leaving:

(sin A + cos A)/(cos A - sin A)

To express this in terms of cot A, we divide both the numerator and the denominator by sin A:

(sin A)/(sin A) + (cos A)/(sin A){(cos A)/(sin A) - (sin A)/(sin A)} = (1 + cot A)/(cot A - 1)

Given that cot A = 2, we substitute this value into the simplified expression:

(1 + 2)/(2 - 1) = ³⁄₁ = 3

Explanation

The area of an equilateral triangle is given by the formula (√3/4) × side². Setting (√3/4) × side² = 9√3 leads to side² = 36, which means the side length is 6 cm.

Explanation

The measure of the angle between the altitude AD and the circumdiameter AP in △ ABC is given by the formula ∠ DAP = |∠ B - ∠ C|. Substituting the given values ∠ B = 68° and ∠ C = 52°, we get ∠ DAP = |68° - 52°| = 16°.

Alternatively, this can be derived using geometric properties. In △ ABD, since AD ⊥ BC, ∠ BAD = 90° - ∠ B = 90° - 68° = 22°. Because AP is the diameter of the circumcircle, the angle subtended at the circumference ∠ ABP = 90°. In △ ABP, ∠ BAP = 90° - ∠ BPA. The angles ∠ BPA and ∠ BCA are subtended by the same arc AB, so ∠ BPA = ∠ C = 52°. Thus, ∠ BAP = 90° - 52° = 38°. Finally, ∠ DAP = ∠ BAP - ∠ BAD = 38° - 22° = 16°.

Explanation

The efficiency of each individual can be expressed relative to A. Let the efficiency of A be 1 unit per day. Since B is 25% more efficient than A, the efficiency of B is 1.25 units per day. C is 50% less efficient than B, making the efficiency of C equal to 1.25 × 0.5 = 0.625 units per day.

The combined efficiency of A, B, and C working together is: 1 + 1.25 + 0.625 = 2.875 units per day

Using the relation Total Work = Efficiency × Time, and taking the time for A to be 11 ½ days (or 11.5 days), the total work is: Work = 1 × 11.5 = 11.5 units

The time taken by A, B, and C working together to complete this work is: Time = (Total Work)/(Combined Efficiency) = (11.5)/(2.875)

Using the fractional equivalents 11.5 = ²³⁄₂ and 2.875 = ²³⁄₈: Time = ²³⁄₂ ÷ ²³⁄₈ = ²³⁄₂ × ⁸⁄₂₃ = 4 days

Explanation

The sum of all 23 numbers is 23 * 51 = 1173. The 12th number is included in both the first 12 and the last 12, so it can be calculated as (12 * 49 + 12 * 54) - 1173 = 1236 - 1173 = 63. Removing the 12th number leaves a sum of 1173 - 63 = 1110, and the new average for the remaining 22 numbers is 1110 / 22 = 50.4545..., which rounds to 50.45.

Explanation

The overall fail percentage can be determined by calculating the weighted average of the failure rates for boys and girls. Given that girls constitute 35% of the student population, boys constitute the remaining 65%. Since 80% of the boys passed, the failure rate for boys is 100% - 80% = 20%. Similarly, as 90% of the girls passed, their failure rate is 100% - 90% = 10%.

The percentage of students who failed is calculated as: Total Fail Percentage = (Percentage of Boys × Boy Fail Rate) + (Percentage of Girls × Girl Fail Rate) Total Fail Percentage = (0.65 × 20) + (0.35 × 10) Total Fail Percentage = 13 + 3.5 = 16.5

Alternatively, using the total number of students: Number of Girls = 0.35 × 1400 = 490 Number of Boys = 1400 - 490 = 910 Failed Girls = 10% of 490 = 49 Failed Boys = 20% of 910 = 182 Total Failed Students = 49 + 182 = 231 Percentage of Failed Students = ²³¹⁄₁₄₀₀ × 100 = 16.5

Explanation

By squaring both sides of the equation x + 1/x = 7, we get (x + 1/x)² = 7², which simplifies to x² + 2 + 1/x² = 49. Subtracting 2 from both sides gives x² + 1/x² = 47.

Explanation

Let the length of side AB be x. Then AC = 2x + 4 and BC = 3x - 4. Since the perimeter is 60 cm, x + (2x + 4) + (3x - 4) = 60, which simplifies to 6x = 60, so x = 10. The side lengths are 10 cm, 24 cm, and 26 cm; these form a right-angled triangle (10² + 24² = 26²), so the area is ½ * 10 * 24 = 120 square cm.

Explanation

Given 2x + 3y + 1 = 0, we can use the algebraic identity that if a + b + c = 0, then a³ + b³ + c³ = 3abc. Setting a = 2x, b = 3y, and c = 1, we get (2x)³ + (3y)³ + (1)³ = 3(2x)(3y)(1), which simplifies to 8x³ + 27y³ + 1 = 18xy. Rearranging this gives 8x³ + 27y³ - 18xy = -1, and adding 8 to both sides results in 8x³ + 27y³ - 18xy + 8 = 7.

Explanation

The relationship between the amount A, principal P, rate R, and time n for compound interest is given by the formula A = P(1 + (R)/(100))ⁿ. Substituting the given values, we have 11495 = 9500(1 + (R)/(100))². Dividing the amount by the principal gives (1 + (R)/(100))² = ¹¹⁴⁹⁵⁄₉₅₀₀ = ¹²¹⁄₁₀₀. Taking the square root of both sides, 1 + (R)/(100) = ¹¹⁄₁₀ = 1.1. This yields (R)/(100) = 0.1, hence the original rate R is 10 percent per annum.

The problem requires the simple interest on the same sum P = 9500 for the same time T = 2 years at double the original rate. The new rate R' is 2 × 10 = 20 percent per annum. Simple interest is calculated using the formula SI = (P × R' × T)/(100). Substituting the values, we get SI = (9500 × 20 × 2)/(100) = 95 × 40 = 3800.

Explanation

Let the number of coins Chetan has be C = 100. Gaurav has 40% more coins than Chetan, which means Gaurav has G = 100 × (1 + 0.40) = 140 coins. Avinash has 20% less coins than Gaurav, so Avinash has A = 140 × (1 - 0.20) = 140 × 0.8 = 112 coins.

To determine the percentage by which Chetan's coins are less than Avinash's, calculate the difference relative to Avinash's total coins: Difference = A - C = 112 - 100 = 12 Percentage less = (A - C)/(A) × 100 = ¹²⁄₁₁₂ × 100

Simplifying the fraction: ¹²⁄₁₁₂ × 100 = ³⁄₂₈ × 100 = ³⁰⁰⁄₂₈ = ⁷⁵⁄₇ Performing the division: 75 ÷ 7 ≈ 10.714

Correct to one decimal place, the percentage is 10.7%.

Explanation

To find the value of x² + (4)/(x²⁾ from the given equation x - ²⁄ₓ = 15, we apply the algebraic identity (a - b)² = a² + b² - 2ab.

Squaring both sides of the equation x - ²⁄ₓ = 15: (x - ²⁄ₓ)² = 15²

Expanding the left side where a = x and b = ²⁄ₓ: x² + (²⁄ₓ)² - 2 · x · ²⁄ₓ = 225

Simplifying the terms: x² + (4)/(x²⁾ - 4 = 225

By shifting -4 to the right side of the equation: x² + (4)/(x²⁾ = 225 + 4 x² + (4)/(x²⁾ = 229

Explanation

To simplify the expression, we find the square root of the given value 6724. We can determine this by identifying the range: since 80² = 6400 and 90² = 8100, the square root of 6724 must lie between 80 and 90.

Because the number 6724 ends in the digit 4, its square root must end in either 2 or 8, as 2² = 4 and 8² = 64. We test 82: 82² = (80 + 2)² 82² = 80² + 2(80)(2) + 2² 82² = 6400 + 320 + 4 = 6724

Thus, the square root of 6724 is 82.

Explanation

To find the required ratio, calculate the total number of boys in the year 2014 and the total number of girls in the year 2020 across all three streams.

The total number of boys in 2014 is the sum of boys in Arts, Science, and Commerce for that year: Total Boys₂₀₁₄ = 42 + 42 + 32 = 116

The total number of girls in 2020 is the sum of girls in Arts, Science, and Commerce for that year: Total Girls₂₀₂₀ = 43 + 30 + 41 = 114

The ratio of the total number of boys in 2014 to the total number of girls in 2020 is: Ratio = 116 : 114

Dividing both terms by their greatest common divisor, which is 2, we get: Ratio = ¹¹⁶⁄₂ : ¹¹⁴⁄₂ = 58 : 57

Explanation

The given equation sin² θ = 2sin θ - 1 can be rewritten as sin² θ - 2sin θ + 1 = 0, which is the perfect square (sin θ - 1)² = 0. Solving this gives sin θ = 1, which means θ = 90 degrees. Substituting this value into the expression 1 + cos θ gives 1 + cos 90 = 1 + 0 = 1.

Explanation

The SSS (Side-Side-Side) similarity theorem states that two triangles are similar if their corresponding sides are proportional. For two triangles to be similar, the order of vertices in the similarity statement must reflect the precise correspondence between proportional sides. Given the proportionality (AB)/(DF) = (BC)/(DE) = (AC)/(EF), we can identify the corresponding vertices by observing which letters are common to the ratios.

In the pair (AB)/(DF) = (BC)/(DE), the vertex B is common to the numerators and D is common to the denominators, which means B ↔ D. In the pair (BC)/(DE) = (AC)/(EF), the vertex C is common to the numerators and E is common to the denominators, meaning C ↔ E. In the pair (AB)/(DF) = (AC)/(EF), the vertex A is common to the numerators and F is common to the denominators, meaning A ↔ F. By arranging these corresponding vertices in the same sequence, we get the similarity △ BCA ∼ △ DEF. This notation correctly aligns the ratios as (BC)/(DE) = (CA)/(EF) = (BA)/(DF), which matches the given information.

Explanation

In the right-angled triangle OAP, where OA is the radius and AP is the tangent, the angle at A is 90°. Given OP = 2 OA, let the radius OA be r. Then OP = 2r. In △ OAP, cos(∠ AOP) = (OA)/(OP) = (r)/(2r) = ½, which implies ∠ AOP = 60°.

Let M be the point where the line OP intersects the chord AB. Since the line joining the center to the external point P bisects the chord AB perpendicularly, M is the midpoint of AB, so AM = 5 cm. In the right-angled triangle OAM, sin(∠ AOM) = (AM)/(OA). Since ∠ AOM = ∠ AOP = 60°, we have sin(60°) = (5)/(r), which gives (√3)/(2) = (5)/(r), so r = (10)/(√3) cm.

In △ OAP, using the tangent ratio, tan(∠ AOP) = (AP)/(OA). Substituting the known values, tan(60°) = (AP)/(r), which means √3 = (AP)/((10)/(√3)). Solving for AP, we get AP = ((10)/(√3)) × √3 = 10 cm.

Explanation

Let the initial monthly salaries of Anil and Kumud be 19x and 17x respectively. Following the hikes of ₹2000 for Anil and ₹1000 for Kumud, their new salaries are 19x + 2000 and 17x + 1000. Given the new ratio is 8 : 7, the equation is:

(19x + 2000)/(17x + 1000) = ⁸⁄₇

Cross-multiplying the terms: 7(19x + 2000) = 8(17x + 1000) 133x + 14000 = 136x + 8000 136x - 133x = 14000 - 8000 3x = 6000 x = 2000

The present salary of Kumud is represented by 17x. Substituting the value of x: 17 × 2000 = 34000

Explanation

The single discount offered by Trader A is 25%. For Trader B, the effective discount of two successive discounts of x% and y% is given by the formula x + y - ˣʸ⁄₁₀₀.

Substituting the values for Trader B: Effective Discount = 20 + 5 - (20 × 5)/(100) = 25 - ¹⁰⁰⁄₁₀₀ = 25 - 1 = 24%

The difference in the discount percentages offered by Trader A and Trader B is 25% - 24% = 1%.

Given that the discount provided by Trader A is ₹320 more than that of Trader B, this 1% of the marked price is equal to ₹320.

Let the marked price be M. 1% of M = 320 ¹⁄₁₀₀ × M = 320 M = 320 × 100 = 32000

The marked price of the item is ₹32,000.

Explanation

Let the work rate of one man be m and one woman be w. The given relation 3w = 2m implies that the work rate of one woman is w = ⅔m. The total work to be completed is defined by 18 men working for 14 days, which is 18m × 14 = 252m man-days.

In the first stage, a group of 5 men and 6 women works for 4 days. Expressing the women's capacity in terms of men, the group's efficiency is 5m + 6(⅔m) = 5m + 4m = 9m. The work completed during these first 4 days is 9m × 4 = 36m.

The remaining work is 252m - 36m = 216m. At this point, 3 more men join the group, resulting in a new group of 8 men and 6 women. The efficiency of this new group is 8m + 6(⅔m) = 8m + 4m = 12m. The time taken to finish the remaining work is 216m / 12m = 18 days.

The total time taken to complete the entire project is the sum of the initial 4 days and the additional 18 days, totaling 22 days.

Explanation

The average of the first n positive odd numbers is given by the formula A = n. This is because the sum of the first n odd numbers is n², and dividing by n yields n. For the first ten positive odd numbers (n = 10), the average is 10.

The average of the first n positive even numbers is given by the formula A = n + 1. This is derived from the sum of the first n even numbers, which is n(n + 1), and dividing by n results in n + 1. For the first fifteen positive even numbers (n = 15), the average is 15 + 1 = 16.

The product of the average of the first ten positive odd numbers and the average of the first fifteen positive even numbers is 10 × 16 = 160.

Explanation

Let the Cost Price (CP) of the article be 100. A loss of 3.3% (noting that 33% in the question is a clerical error for 3.3%) results in a Selling Price (SP) of: SP = 100 - 3.3 = 96.7

The problem states that this selling price is four-fifth of the Marked Price (MP): SP = ⅘ MP 96.7 = 0.8 × MP MP = (96.7)/(0.8) = 120.875

If the shopkeeper sells the article at the marked price (MP), the new selling price is 120.875. The profit is calculated as: Profit = MP - CP = 120.875 - 100 = 20.875

Profit percentage is: Profit % = (20.875)/(100) × 100 = 20.875%

Rounding to the nearest integer, the profit percentage is 21%.

Explanation

The angle subtended by an arc at the center of a circle is twice the angle subtended by it at any point on the remaining part of the circumference. Given that ∠ BOC = 84°, the angle ∠ BAC subtended by the arc BC at point A on the circle is calculated as ∠ BAC = ½ × ∠ BOC = (84°)/(2) = 42°.

Since the line segment AC is produced to point D, the points A, C, and D are collinear, making ∠ BAD = ∠ BAC = 42°. In the triangle ABD, the sum of the interior angles must equal 180°. The angles within this triangle are ∠ BAD, ∠ ADB, and ∠ ABD. We are given that ∠ BDC (which is the same as ∠ ADB) is 40°. Using the angle sum property of a triangle, we have ∠ ABD = 180° - (∠ BAD + ∠ ADB). Substituting the known values, ∠ ABD = 180° - (42° + 40°) = 180° - 82° = 98°.

Explanation

The man's speed is 3km/h, which can be converted to meters per minute as 3 × ¹⁰⁰⁰⁄₆₀ = 50m/min. The distance he covers while crossing the field diagonally in 5 minutes represents the length of the diagonal D of the square field. Therefore, D = speed × time = 50 × 5 = 250m.

The area A of a square field can be calculated using the length of its diagonal D with the formula A = D²{2}. Substituting the value of the diagonal into the formula yields A = 250²{2} = ⁶²⁵⁰⁰⁄₂ = 31250m².

Explanation

The given equation y = 2x + 1 can be rearranged as 2x - y + 1 = 0. We apply the algebraic identity which states that if a + b + c = 0, then a³ + b³ + c³ = 3abc. By substituting a = 2x, b = -y, and c = 1, the condition a + b + c = 2x - y + 1 = 0 is satisfied.

Substituting these values into the identity yields (2x)³ + (-y)³ + (1)³ = 3(2x)(-y)(1). Simplifying this expression results in 8x³ - y³ + 1 = -6xy. To find the required value, we rearrange the equation by moving the term involving xy to the left side and the constant term to the right side, which gives 8x³ - y³ + 6xy = -1.

Explanation

Let the speeds of the two cars be x km/h and y km/h, with x > y. When moving in the same direction, the relative speed is the difference between their speeds. Given they meet in 4.5 hours over a distance of 45 km: x - y = (45)/(4.5) = 10

When moving towards each other, the relative speed is the sum of their speeds. The time taken is 27 minutes, which is ²⁷⁄₆₀ = ⁹⁄₂₀ hours. Using the distance formula: x + y = (45)/(9){20} = (45 × 20)/(9) = 100

To find the speed of the faster car (x), add the two equations: (x - y) + (x + y) = 10 + 100 2x = 110 x = 55

The speed of the faster car is 55 km/h.

Explanation

In a right-angled triangle PQR with the right angle at Q, where points M and N lie on sides PQ and QR respectively, the relationship between the lengths of the segments is given by the theorem: PN² + MR² = PR² + MN²

This follows from applying the Pythagorean theorem to the four right-angled triangles formed with vertex Q. In △ PQN, PN² = PQ² + QN². In △ MQR, MR² = MQ² + QR². Adding these gives: PN² + MR² = (PQ² + QR²) + (MQ² + QN²)

Since PQ² + QR² = PR² (from △ PQR) and MQ² + QN² = MN² (from △ MQN), we substitute these values to get: PN² + MR² = PR² + MN²

Substituting the given values into the formula: 9² + 7² = PR² + 3² 81 + 49 = PR² + 9 130 = PR² + 9 PR² = 121 PR = √(121) = 11 cm.

Explanation

The divisibility rule for 11 states that the difference between the sum of digits at odd positions and the sum of digits at even positions must be a multiple of 11, including zero. For the number 4a067b, we identify the digits at different positions. Sum of digits at odd positions (from the right) is S₁ = b + 6 + a = a + b + 6. Sum of digits at even positions is S₂ = 7 + 0 + 4 = 11.

The condition for divisibility is S₁ - S₂ = 11k, where k is an integer. Substituting the sums, we get (a + b + 6) - 11 = 11k, which simplifies to (a + b) - 5 = 11k. Since a and b are single digits from 0 to 9, the range for the sum (a + b) is 0 ≤ a + b ≤ 18.

Setting k = 0 gives a + b - 5 = 0, resulting in a + b = 5. Setting k = 1 gives a + b - 5 = 11, resulting in a + b = 16. Setting k = 2 gives a + b - 5 = 22, resulting in a + b = 27, which is outside the possible range for the sum of two digits. Setting k = -1 gives a + b - 5 = -11, resulting in a + b = -6, which is also impossible.

The possible values for (a + b) are 5 and 16. The sum of all possible values is 5 + 16 = 21.

Explanation

To find the value of the expression, substitute the standard values of the trigonometric ratios: sin 60° = √3{2}, cos 30° = √3{2}, sin 45° = (1)/(√(2)), and sin 90° = 1.

The expression is given by: sin² 60° + cos² 30° - sin² 45° - 3sin² 90°

Substituting the values: = (√3{2})² + (√3{2})² - ((1)/(√(2)))² - 3(1)² = ¾ + ¾ - ½ - 3 = ⁶⁄₄ - ½ - 3 = ³⁄₂ - ½ - 3 = (3 - 1)/(2) - 3 = ²⁄₂ - 3 = 1 - 3 = -2

Explanation

The given equation is x⁴ - 62x² + 1 = 0. Dividing the entire equation by x², we obtain x² - 62 + (1)/(x²⁾ = 0, which simplifies to x² + (1)/(x²⁾ = 62.

To find the value of x + ¹⁄ₓ, we use the algebraic identity (x + ¹⁄ₓ)² = x² + (1)/(x²⁾ + 2. Substituting the known value of x² + (1)/(x²⁾ into this identity gives (x + ¹⁄ₓ)² = 62 + 2 = 64. Since the problem states x > 0, we take the positive square root to find x + ¹⁄ₓ = √(64) = 8.

The expression x³ + x⁻³ is equivalent to x³ + (1)/(x³⁾. Using the cubic identity x³ + (1)/(x³⁾ = (x + ¹⁄ₓ)³ - 3(x + ¹⁄ₓ), we substitute the value 8 for (x + ¹⁄ₓ). This results in 8³ - 3(8), which simplifies to 512 - 24. The final calculation yields 488.

Explanation

Given the equation x⁸ - 34x⁴ + 1 = 0, we divide the entire expression by x⁴ (since x > 0): x⁴ - 34 + (1)/(x⁴) = 0 x⁴ + (1)/(x⁴) = 34

To find the value of x² + (1)/(x²), we use the algebraic identity a² + b² = (a + b)² - 2ab: (x² + (1)/(x²))² - 2 = 34 (x² + (1)/(x²))² = 36 x² + (1)/(x²) = 6

Next, we find x - ¹⁄ₓ using the identity a² + b² = (a - b)² + 2ab: (x - ¹⁄ₓ)² + 2 = 6 (x - ¹⁄ₓ)² = 4 x - ¹⁄ₓ = 2

The value of x³ - x⁻³ (which is x³ - (1)/(x³)) is found using the cubic identity a³ - b³ = (a - b)³ + 3ab(a - b): x³ - (1)/(x³) = (x - ¹⁄ₓ)³ + 3(x)(¹⁄ₓ)(x - ¹⁄ₓ) Substituting x - ¹⁄ₓ = 2: x³ - (1)/(x³) = 2³ + 3(2) x³ - (1)/(x³) = 8 + 6 = 14

Explanation

Apply the Componendo and Dividendo rule to the equation (sinθ + cosθ)/(sinθ - cosθ) = ⁵⁄₁ to find the ratio of sinθ to cosθ: ((sinθ + cosθ) + (sinθ - cosθ))/((sinθ + cosθ) - (sinθ - cosθ)) = (5 + 1)/(5 - 1) (2sinθ)/(2cosθ) = ⁶⁄₄ ⟹ tanθ = ³⁄₂ Squaring both sides results in tan²θ = ⁹⁄₄.

To evaluate the target expression, divide both the numerator and the denominator by cos²θ: 4{sin²θ{cos²θ} + (3)/(cos²⁾^θ}{2cos²θ{cos²θ} + (2)/(cos²⁾^θ} = 4tan²θ + 3sec²θ{2 + 2sec²θ}

Using the trigonometric identity sec²θ = 1 + tan²θ, the expression simplifies to: 4tan²θ + 3(1 + tan²θ){2 + 2(1 + tan²θ)} = 7tan²θ + 3{2tan²θ + 4}

Substituting tan²θ = ⁹⁄₄ into the simplified formula: (7(9)/(4)) + 3{2(⁹⁄₄) + 4} = (63 + 12)/(4){(18 + 16)/(4)} = ⁷⁵⁄₃₄

Explanation

In a circle, angles subtended by the same arc at the circumference are equal. Since both ∠ BAC and ∠ BDC subtend the same arc BC, we have ∠ BDC = ∠ BAC = 22°. In △ OCD, sides OC and OD are both radii of the circle, which makes △ OCD an isosceles triangle with OC = OD. In an isosceles triangle, the angles opposite the equal sides are also equal, so ∠ OCD = ∠ ODC = 22°. The sum of the internal angles of any triangle is 180°. Therefore, in △ OCD, the measure of the central angle is calculated as ∠ COD = 180° - (∠ ODC + ∠ OCD) = 180° - (22° + 22°) = 136°.

Explanation

The binomial expansion of (a + b)ⁿ is given by the general term Tᵣ₊₁ = n{r} aⁿ⁻ʳ bʳ. In the expression (5 - x²{3})³, we have a = 5, b = -x²{3}, and n = 3. To find the coefficient of x², we identify the term where the exponent of x is 2. The general term is 3{r} (5)³⁻ʳ (-x²{3})ʳ, which simplifies to 3{r} (5)³⁻ʳ (-⅓)ʳ x²ʳ.

To obtain x², we set the exponent 2r = 2, which gives r = 1. Substituting r = 1 into the general term:

T₂ = 3{1} (5)³⁻¹ (-x²{3})¹ T₂ = 3 · 5² · (-x²{3}) T₂ = 3 · 25 · (-⅓) x² T₂ = -25x²

The coefficient of x² is -25.

Explanation

The relationship between the marked price (MP), cost price (CP), gain percentage, and discount percentage is given by the formula (MP)/(CP) = (100 + Gain%)/(100 - Discount%). Substituting the given values where CP = ₹920, MP = ₹1472, Gain% = 20, and Discount% = x:

¹⁴⁷²⁄₉₂₀ = (100 + 20)/(100 - x)

Simplifying the fraction ¹⁴⁷²⁄₉₂₀ by dividing both the numerator and denominator by 184 results in ⁸⁄₅. The equation becomes:

⁸⁄₅ = (120)/(100 - x)

Cross-multiplying leads to 8(100 - x) = 120 × 5, which simplifies to 800 - 8x = 600. Rearranging the terms gives 8x = 200, and solving for x yields x = 25.

Explanation

Simplifying the numerator by finding a common denominator for the terms inside the parentheses gives ((sinθ cosθ + 1)/cosθ)² + ((cosθ sinθ + 1)/sinθ)², which factors into (1 + sinθ cosθ)² * (1/cos²θ + 1/sin²θ). Further simplification leads to (1 + sinθ cosθ)² * (1 / (sin²θ cos²θ)), which is equivalent to (secθ cscθ + 1)², making the ratio of the numerator to the denominator equal to 1.

Explanation

In triangle ABD, which is right-angled at B, we apply the Pythagorean theorem: AB² + BD² = AD². Substituting the given values BD = 5 cm and AD = 13 cm, we get AB² + 5² = 13². This results in AB² = 169 - 25 = 144, so AB = 12 cm.

In the larger right-angled triangle ABC, we apply the Pythagorean theorem again: AB² + BC² = AC². Substituting AB = 12 cm and AC = 37 cm, we have 12² + BC² = 37². This gives 144 + BC² = 1369, which simplifies to BC² = 1225. Taking the square root, we find BC = 35 cm.

Since point D lies on the segment BC, the total length BC is the sum of BD and DC. Therefore, BC = BD + DC. Substituting the known values 35 = 5 + DC, we calculate DC = 30 cm.

Explanation

The total sum corresponds to the sum of the ratio units for A, B, C, and D, which is 4 + 12 + 30 + 45 = 91 units. The difference between the shares of A and D is represented by the difference in their ratio units, calculated as 45 - 4 = 41 units. According to the problem, this difference is ₹5,535. Therefore, the value of one unit is ⁵⁵³⁵⁄₄₁ = 135. To find the total sum, we multiply the value of a single unit by the total number of units: 91 × 135 = 12285.

Explanation

Let the speed of the river flow be vkm / h. The speed of the motor boat downstream is (20 + v)km / h and its speed upstream is (20 - v)km / h. The total time for the round trip is the sum of the time taken for the downstream and upstream journeys.

Using the formula Time = Distance / Speed, the total time T is given by: 150 / (20 + v) + 150 / (20 - v) = 16

Factoring out the distance and finding a common denominator: 150 [ (20 - v + 20 + v) / (20² - v²) ] = 16 150 [ 40 / (400 - v²) ] = 16 6000 / (400 - v²) = 16

Simplifying the equation: 6000 / 16 = 400 - v² 375 = 400 - v² v² = 400 - 375 v² = 25 v = 5

The speed of the flow of the river is 5km / h.

Explanation

For a number to be divisible by 72, it must be divisible by both 8 and 9, as these are co-prime factors of 72. According to the divisibility rule for 8, the last three digits of the number 7698x138y, which are 38y, must be divisible by 8. Dividing 38y by 8, we find that 384 is the only number in the 380s range divisible by 8 (since 384 = 8 × 48), which gives y = 4.

According to the divisibility rule for 9, the sum of the digits of the number must be a multiple of 9. The sum of the digits is 7 + 6 + 9 + 8 + x + 1 + 3 + 8 + y, which simplifies to 42 + x + y. Substituting the value of y = 4, the sum becomes 42 + x + 4 = 46 + x. For 46 + x to be divisible by 9, the nearest multiple of 9 greater than 46 is 54. Therefore, 46 + x = 54, which yields x = 8.

Now, substituting the values of x and y into the required expression: √(4x + y) = √(4(8) + 4) √(32 + 4) = √(36) = 6

Explanation

Let the efficiency of pipe C be 12 units/hour. Since C is 12 times as fast as B and 3 times as fast as A, the efficiency of B is 1 unit/hour and the efficiency of A is 4 units/hour. Their combined efficiency is 17 units/hour, so the total capacity of the tank is 17 units/hour multiplied by 4 hours, which equals 68 units. Therefore, pipe A alone will take 68 units divided by 4 units/hour, which is 17 hours.

Explanation

In right-angled triangle ABC, where BD is the altitude to the hypotenuse AC, the property of similarity between △ ADB and △ ABC yields the geometric relation AB² = AD × AC.

Since D lies on the segment AC, the length of AD is found by subtracting CD from AC: AD = AC - CD = 9 - 3 = 6 cm

Substituting the known values into the leg rule formula: AB² = 6 × 9 AB² = 54 AB = √(54) AB = √(9 × 6) AB = 3√6 cm

Explanation

Let the loss incurred in the first transaction be L. The problem states that selling the article for 490 results in a gain equal to four times this loss, which is 4L. Since the Cost Price (CP) is the same in both scenarios, the relationship is expressed as: CP = Selling Price₁ + Loss = Selling Price₂ - Gain 455 + L = 490 - 4L Rearranging the terms gives: 5L = 490 - 455 = 35 L = 7 Calculating the Cost Price: CP = 455 + 7 = 462 To find the selling price required to gain 25%, we multiply the Cost Price by 125%: Required Selling Price = 462 × 1.25 = 462 × ⁵⁄₄ Required Selling Price = ²³¹⁰⁄₄ = 577.50

Explanation

To find the average of numbers from 1 to 50 that are multiples of 2 or 5, we apply the Principle of Inclusion-Exclusion to determine the total sum and the total count of such numbers.

For multiples of 2 (2, 4, …, 50): The number of terms n₂ = ⁵⁰⁄₂ = 25. The sum S₂ = (n₂)/(2)(first + last) = ²⁵⁄₂(2 + 50) = 25 × 26 = 650.

For multiples of 5 (5, 10, …, 50): The number of terms n₅ = ⁵⁰⁄₅ = 10. The sum S₅ = ¹⁰⁄₂(5 + 50) = 5 × 55 = 275.

For multiples of both 2 and 5 (multiples of 10: 10, 20, …, 50): The number of terms n₁₀ = ⁵⁰⁄₁₀ = 5. The sum S₁₀ = ⁵⁄₂(10 + 50) = 5 × 30 = 150.

The total count of unique numbers is n = n₂ + n₅ - n₁₀ = 25 + 10 - 5 = 30. The total sum of unique numbers is S = S₂ + S₅ - S₁₀ = 650 + 275 - 150 = 775.

The average is calculated as: Average = (S)/(n) = ⁷⁷⁵⁄₃₀ = ¹⁵⁵⁄₆ = 25.833…

Correct to one decimal place, the average is 25.8.

Explanation

The expression is simplified using the BODMAS rule, which establishes the precedence of operations as Brackets, Of, Division, Multiplication, Addition, and Subtraction.

Begin with the first parenthetical expression: ¾ - ¼ ÷ ¼ of ⅖. According to the order of operations, the of operation is performed before division. Thus, ¼ of ⅖ = (1 × 2)/(4 × 5) = ⅒. The expression then reduces to ¾ - ¼ ÷ ⅒. Next, perform the division: ¼ ÷ ⅒ = ¼ × ¹⁰⁄₁ = ¹⁰⁄₄ = ⁵⁄₂. Finally, the subtraction yields ¾ - ⁵⁄₂ = (3 - 10)/(4) = -⁷⁄₄.

Next, evaluate the second parenthetical expression: ¾ ÷ ⅔ of ⅗. Again, the of operation precedes division: ⅔ of ⅗ = (2 × 3)/(3 × 5) = ⅖. The division calculation follows: ¾ ÷ ⅖ = ¾ × ⁵⁄₂ = ¹⁵⁄₈.

The final step is to divide the result of the first bracket by the result of the second: -⁷⁄₄ ÷ ¹⁵⁄₈ = -⁷⁄₄ × ⁸⁄₁₅. Multiplying these fractions gives -⁵⁶⁄₆₀. Reducing the fraction to its lowest terms by dividing both the numerator and the denominator by 4 results in -¹⁴⁄₁₅, which is mathematically equivalent to ¹⁴⁄₋₁₅.