SSC CGL Tier 1 2023 Quantitative Aptitude Questions with Solutions

Explanation

Let the initial number of men be M. The total work to be completed is the product of the number of men and the time, which is 11M man-days.

The actual work done follows an arithmetic progression where the number of men on the first day is M, and 16 men leave at the end of each day for 15 days. The number of men working each day is M, M-16, M-32, … for n = 15 days. The total work done is the sum of this progression: Sₙ = ⁿ⁄₂ [2a + (n-1)d]

Substituting n = 15, a = M, and d = -16: Total Work = ¹⁵⁄₂ [2M + (15-1)(-16)] Total Work = ¹⁵⁄₂ [2M - 224] = 15(M - 112)

Equating the total work to the initial estimate: 11M = 15M - 1680 4M = 1680 M = 420

Explanation

From the given equation cos A + cos² A = 1, we can rearrange it to find cos A = 1 - cos² A. Using the trigonometric identity sin² A + cos² A = 1, we have 1 - cos² A = sin² A. Therefore, cos A = sin² A.

Squaring both sides of this relation gives cos² A = sin⁴ A, and cubing both sides gives cos³ A = sin⁶ A.

To find the value of the expression sin⁴ A + sin⁶ A, substitute the derived equivalents in terms of cosine: sin⁴ A + sin⁶ A = cos² A + cos³ A

Now, take the original equation cos A + cos² A = 1 and multiply the entire equation by cos A: cos A (cos A + cos² A) = cos A (1) cos² A + cos³ A = cos A

By comparing the two results, we see that sin⁴ A + sin⁶ A is equal to cos A.

Explanation

For a triangle with sides a, b, and c, where c is the longest side, the triangle is obtuse-angled if the sum of the squares of the two shorter sides is less than the square of the longest side, expressed as a² + b² < c².

In the set 15, 62, and 64, the sides satisfy the triangle inequality since 15 + 62 > 64. Testing the obtuse condition with a = 15, b = 62, and c = 64:

a² + b² = 15² + 62² = 225 + 3844 = 4069 c² = 64² = 4096

Comparing the values, 4069 < 4096, which confirms that a² + b² < c². This relationship between the squares of the sides satisfies the specific geometric requirement for an obtuse-angled triangle.

Explanation

The curved surface area (CSA) of a cone is given by the formula CSA = π r l, where r is the radius of the base and l is the slant height.

Given values: Radius r = 21 cm Slant height l = 60 cm π = ²²⁄₇

Substituting these values into the formula: CSA = ²²⁄₇ × 21 × 60

First, simplify the expression by dividing 21 by 7: CSA = 22 × 3 × 60

Next, multiply the numbers: 22 × 3 = 66 66 × 60 = 3960

The curved surface area of the cone is 3960 cm².

Explanation

The trigonometric value of csc 30° is the reciprocal of sin 30°. Given that sin 30° = ½, then csc 30° = 2. The standard value for cos 60° is ½. Substituting these specific values into the expression csc 30° - cos 60° gives:

2 - ½

To solve this, convert the integer into a fraction with a common denominator:

⁴⁄₂ - ½ = ³⁄₂

Explanation

The thief's lead when the policeman starts chasing is calculated by the distance covered in 6 minutes: 20 × ⁶⁄₆₀ = 2 km. After the policeman chases for 24 minutes (0.4 hours), the remaining gap is 400 m or 0.4 km. The reduction in the gap is 2 - 0.4 = 1.6 km.

The rate at which the gap is reduced is the relative speed between the policeman and the thief. Let the speed of the policeman be vₚ. Relative speed is vₚ - 20 km/h. Using the formula Distance = Speed × Time: 1.6 = (vₚ - 20) × 0.4 4 = vₚ - 20 vₚ = 24 km/h

To catch the thief completely, the policeman must close the initial gap of 2 km. The time required for this is: t = (Initial Gap)/(Relative Speed) = (2)/(24 - 20) = ²⁄₄ = 0.5 hours

The total distance from the spot of the crime where the catch occurs is the distance the policeman travels in 0.5 hours: D = 24 × 0.5 = 12 km

Explanation

The relationship between Marked Price (MP) and Cost Price (CP) is given by the formula (MP)/(CP) = (100 + Profit%)/(100 - Discount%).

Using the first set of conditions where the discount is 18% and the profit is 30%, the ratio is (MP)/(CP) = (100 + 30)/(100 - 18) = ¹³⁰⁄₈₂ = ⁶⁵⁄₄₁.

When the discount is changed to 12%, let the new profit be x%. Using the same ratio, we have ⁶⁵⁄₄₁ = (100 + x)/(100 - 12). Solving for x gives 100 + x = (65 × 88)/(41), which simplifies to 100 + x = ⁵⁷²⁰⁄₄₁. Dividing 5720 by 41 results in approximately 139.512. Therefore, x = 139.512 - 100 = 39.512. Rounding to two decimal places, the profit percentage is 39.51%.

Explanation

To determine the least value among the given expressions, we calculate the numerical result of each. For 12% of 625, the value is 0.12 × 625 = 75. For 15% of 555, the value is 0.15 × 555 = 83.25. For 10% of 720, the value is 0.10 × 720 = 72. For 9% of 845, the value is 0.09 × 845 = 76.05. Comparing these results, 72 is the smallest number.

Explanation

Let Ravi initial monthly income be I₁ = 100. Since he saves 20% of his income, his initial nominal savings are S₁ = 20.

Following the increase, his new monthly income becomes I₂ = 100 × (1 + 0.25) = 125. His nominal savings increased by 5%, so the new nominal savings are S₂ = 20 × (1 + 0.05) = 21.

The percentage of his present nominal income that he currently saves is calculated using the ratio of present savings to present income:

Percentage = S₂{I₂} × 100

Percentage = ²¹⁄₁₂₅ × 100

Percentage = 21 × ⅘ = ⁸⁴⁄₅ = 16.8%

Explanation

To find the total number of females below poverty line in States B and C, we calculate the number for each state individually using the population and the given proportions.

For State B, the total population is 6,000 and the percentage below poverty line is 13%. The total number of people below poverty line in State B is 6,000 × 0.13 = 780. The ratio of males to females below poverty line is 8 : 5. The number of females in this category is (5)/(8 + 5) × 780 = ⁵⁄₁₃ × 780 = 5 × 60 = 300.

For State C, the total population is 6,000 and the percentage below poverty line is 26%. The total number of people below poverty line in State C is 6,000 × 0.26 = 1,560. The ratio of males to females below poverty line is 9 : 4. The number of females in this category is (4)/(9 + 4) × 1,560 = ⁴⁄₁₃ × 1,560 = 4 × 120 = 480.

The total number of females below poverty line in both states is 300 + 480 = 780.

Explanation

The weighing machine shows 400 g for an actual weight of 350 g. The ratio of actual weight to shown weight is 350 / 400 = 7/8. When the machine shows a weight of 200 g for a packet, the actual quantity of almonds in that packet is: Actual Weight = 200 × ⅞ = 175 g

The cost price of almonds is ₹880 per kg, which translates to ₹0.88 per gram. The cost price (CP) for the actual quantity in one packet is: CP = 175 × 0.88 = 154

To obtain a profit of 25%, the selling price (SP) is calculated using the formula SP = CP × (1 + Profit% / 100): SP = 154 × (1 + 0.25) = 154 × 1.25 SP = 154 × ⁵⁄₄ = 192.50

Explanation

The relationship between the sum and difference of reciprocal terms is given by the algebraic identity (x + ¹⁄ₓ)² = (x - ¹⁄ₓ)² + 4. Based on the condition x - ¹⁄ₓ = 3, we substitute the value into the identity:

(x + ¹⁄ₓ)² = (3)² + 4 (x + ¹⁄ₓ)² = 9 + 4 = 13

Since the problem specifies x > 0, the value of x + ¹⁄ₓ must be the positive square root:

x + ¹⁄ₓ = √(13)

Explanation

For a sum of money to triple itself under simple interest, the interest earned must be exactly twice the principal amount. Let the principal be P. If the sum triples, the total amount becomes 3P, and the simple interest SI is 3P - P = 2P.

Using the simple interest formula SI = (P × R × T)/(100), where R is the rate of interest and T is the time period in years, we can substitute the known values: 2P = (P × R × 16)/(100)

By canceling P from both sides, the equation simplifies to: 2 = (16R)/(100) 200 = 16R R = ²⁰⁰⁄₁₆ R = 12.5

This can also be solved using the direct relationship R = (100(n - 1))/(T), where n is the number of times the principal multiplies. Substituting n = 3 and T = 16 gives R = (100(3 - 1))/(16) = ²⁰⁰⁄₁₆ = 12.5%.

Explanation

The ratio of boys to the total number of students (boys + girls) is given as 7:17. This implies that if there are 17 total parts, 7 parts represent the boys and the remaining 17 - 7 = 10 parts represent the girls. Therefore, the ratio of boys to girls is 7:10.

Given that the number of boys is 1099, we can set up the proportion where 7 units correspond to 1099. Dividing the number of boys by its corresponding ratio part gives the value of one unit:

1099 / 7 = 157

Since there are 10 units representing the girls, the total number of girls is:

157 × 10 = 1570

Explanation

Congruent triangles are identical in both shape and size, which implies that all their corresponding parts are equal. According to the property of CPCT (Corresponding Parts of Congruent Triangles), if △ ABC ≅ △ DEF, then the lengths of the corresponding sides must be equal, such that AB = DE, BC = EF, and AC = DF. The perimeter of any triangle is defined as the sum of the lengths of its three sides. For △ ABC, the perimeter is AB + BC + AC, and for △ DEF, the perimeter is DE + EF + DF. Since each individual side of △ ABC is equal to its corresponding side in △ DEF, the sum of these sides must also be equal. Consequently, AB + BC + AC = DE + EF + DF, establishing that the perimeters of congruent triangles are always equal.

Explanation

In a circular race, when two participants start from the same point and move in the same direction, they meet for the first time when the faster person gains a lead of exactly one full lap over the slower person. Let the speed of Shyam be v. Since Ram is thrice as fast, his speed is 3v. Their relative speed is 3v - v = 2v.

The time taken for the first meeting is given by the formula T = (L)/(Vᵣₑₗ₎, where L is the track length and Vᵣₑₗ is the relative speed. Substituting the given values:

8 = ¹⁴⁴⁰⁄₂ᵥ

Solving for v: 16v = 1440 v = 90 m/min

The length of the race is equal to the length of the track, which is 1440 m. The time taken by Shyam to finish the race is calculated by dividing the total distance by his speed:

Time = ¹⁴⁴⁰⁄₉₀ = 16 minutes

Explanation

Let the time taken by Pipe A to fill the cistern be x minutes. According to the problem, Pipe B takes 5 minutes more than Pipe A, so the time taken by Pipe B is x + 5 minutes.

The combined work rate of both pipes is the sum of their individual rates. Given that they fill the cistern together in 6 minutes, the equation is:

¹⁄ₓ + (1)/(x + 5) = ⅙

Combining the fractions on the left side:

((x + 5) + x)/(x(x + 5)) = ⅙ (2x + 5)/(x² + 5x) = ⅙

Cross-multiplying to solve for x:

6(2x + 5) = x² + 5x 12x + 30 = x² + 5x x² - 7x - 30 = 0

Factoring the quadratic equation:

x² - 10x + 3x - 30 = 0 x(x - 10) + 3(x - 10) = 0 (x - 10)(x + 3) = 0

The solutions are x = 10 or x = -3. Since time cannot be negative, we take x = 10. Thus, Pipe A takes 10 minutes and Pipe B takes 10 + 5 = 15 minutes to fill the cistern separately.

Explanation

The algebraic identity for the sum of cubes is a³ + b³ + c³ - 3abc = ½(a + b + c)[(a - b)² + (b - c)² + (c - a)²]. Substituting the given values into this equation gives 405 = ½(a + b + c)(54). Simplifying the expression on the right leads to 405 = 27(a + b + c). To find the value of the sum, we divide 405 by 27, resulting in a + b + c = 15.

Explanation

The surface area A of a sphere is calculated using the formula A = 4 π r², where r is the radius. Given the radius r = 35 cm and using π = ²²⁄₇, the calculation is as follows:

A = 4 × ²²⁄₇ × 35 × 35

By simplifying the expression, dividing 35 by 7 gives 5: A = 4 × 22 × 5 × 35 A = 88 × 175 A = 15400

The surface area of the spherical sculpture is 15400 cm².

Explanation

Pipe A fills 50% of the tank in 6 hours, so it fills the full tank in 12 hours. Pipe B fills the full tank in 18 hours. To find the total time taken when both pipes work together, use the formula for combined work: T = (xy)/(x+y).

Substituting the values, we get T = (12 × 18)/(12 + 18) = ²¹⁶⁄₃₀ = 7.2 hours.

To convert the time into minutes, multiply the result by 60: 7.2 × 60 = 432 minutes.

Explanation

A number is divisible by 9 if the sum of its digits is a multiple of 9. For the number 846523X7Y, the sum of the digits is 8 + 4 + 6 + 5 + 2 + 3 + X + 7 + Y = 35 + X + Y.

For 35 + X + Y to be divisible by 9, the possible values for X + Y are 1 (35 + 1 = 36) or 10 (35 + 10 = 45). Since X and Y are single digits and their difference is given as Y - X = 6, we test these cases:

If X + Y = 1 and Y - X = 6, adding the equations gives 2Y = 7, which does not yield an integer.

If X + Y = 10 and Y - X = 6, adding the equations gives 2Y = 16, which means Y = 8. Substituting Y = 8 into the equation X + Y = 10 gives X = 2. Both are valid single-digit integers.

Now, substitute X = 2 and Y = 8 into the expression √(2X + 4Y): √(2(2) + 4(8)) = √(4 + 32) = √(36) = 6

Explanation

In △ ABC, it is given that ∠B = 90°, which implies that the sum of the remaining two angles is A + C = 90°. From the given condition tan A = 1/√3, we find that A = 30°. Consequently, C = 90° - 30° = 60°. In a right triangle, the trigonometric ratios of complementary angles satisfy the identities sin A = cos C and cos A = sin C.

The expression sin A cos C + cos A sin C follows the sine addition formula sin(A + C). Substituting the values for A = 30° and C = 60°: sin 30° = 1/2 cos 60° = 1/2 cos 30° = √3/2 sin 60° = √3/2

Placing these values into the expression, we get: (1/2 × 1/2) + (√3/2 × √3/2) = 1/4 + 3/4 = 1

This result is consistent with the identity sin(A + C) = sin(90°) = 1.

Explanation

In a chase where both parties start and stop at the same time, the time of travel is constant. When time is constant, the distance covered is directly proportional to the speed, expressed as d ∝ v. The ratio of the speeds of the policeman (vₚ) and the thief (vₜ) is:

Ratio of speeds = 9.2 : 8 = 92 : 80 = 23 : 20

Since distance is proportional to speed, the ratio of the distances covered by the policeman and the thief is also 23 : 20. Let the distance covered by the policeman be 23x and the distance covered by the thief be 20x. The difference between these distances must equal the initial gap of 360 metres:

23x - 20x = 360 3x = 360 x = 120

The total distance run by the policeman to catch the thief is 23x:

Distance = 23 × 120 = 2760 m

Explanation

The net effective discount for two successive discounts of x% and y% is determined by the formula x + y - ˣʸ⁄₁₀₀. For the first scenario of 5% and 5%, the total discount is 5 + 5 - ²⁵⁄₁₀₀ = 9.75%. For the third scenario of 8% and 2%, the total discount is 8 + 2 - ¹⁶⁄₁₀₀ = 9.84%.

In the second scenario, a single discount of 10% is equivalent to 10 + 0 - (10 × 0)/(100) = 10%. Because the product term ˣʸ⁄₁₀₀ is subtracted from the sum in successive discounts, a single discount will always be greater than or equal to successive discounts that sum to the same value. Since 10% > 9.84% > 9.75%, the single discount of 10% yields the maximum savings on the amount of 7,500.

Explanation

To find the percentage of students who passed from college B for all the years put together, we first calculate the total number of students enrolled and the total number of students who passed in college B from 2010 to 2014.

Total enrolled in college B: 550 + 450 + 600 + 650 + 700 = 2950

Total passed in college B: 530 + 420 + 550 + 620 + 680 = 2800

The required percentage is calculated using the formula: Percentage = (Total Passed)/(Total Enrolled) × 100

Substituting the values: Percentage = ²⁸⁰⁰⁄₂₉₅₀ × 100 = ²⁸⁰⁄₂₉₅ × 100 = ⁵⁶⁄₅₉ × 100

Percentage = ⁵⁶⁰⁰⁄₅₉ ≈ 94.9152...%

Rounding to two decimal places, we get 94.92%.

Explanation

The algebraic identity for the square of a trinomial is (a + b + c)² = a² + b² + c² + 2(ab + bc + ca). Substituting the given values into this formula, we get 13² = (a² + b² + c²) + 2(54). This simplifies to 169 = a² + b² + c² + 108. By rearranging the equation to solve for the sum of squares, we have a² + b² + c² = 169 - 108 = 61.

Explanation

Given the equation (21cos A + 3sin A)/(3cos A + 4sin A) = 2, we cross-multiply to solve for the trigonometric terms:

21cos A + 3sin A = 2(3cos A + 4sin A)

Expanding the right side:

21cos A + 3sin A = 6cos A + 8sin A

Subtracting 6cos A and 3sin A from both sides to group like terms:

21cos A - 6cos A = 8sin A - 3sin A 15cos A = 5sin A

To find cot A, we use the identity cot A = (cos A)/(sin A). Dividing both sides by 15sin A:

(cos A)/(sin A) = ⁵⁄₁₅ cot A = ⅓

Explanation

The time taken by B to complete the work alone can be calculated using the combined work rate formula or the Least Common Multiple (LCM) method. To find the total work, we take the LCM of 120 and 72, which is 360 units.

Efficiency of A = Total Work / Time taken by A = 360 / 120 = 3 units per day. Combined efficiency of A and B = Total Work / Time taken by both = 360 / 72 = 5 units per day.

Efficiency of B = Combined efficiency - Efficiency of A = 5 - 3 = 2 units per day. Time taken by B working alone = Total Work / Efficiency of B = 360 / 2 = 180 days.

Alternatively, using the direct formula T_B = T_A × T_A₊_B{T_A - T_A₊_B}: T_B = (120 × 72)/(120 - 72) T_B = (120 × 72)/(48) T_B = 120 × 1.5 = 180 days.

Explanation

The rate of interest per annum is calculated using the formula for simple interest SI = (P × R × T)/(100), where P is the principal, R is the rate of interest, and T is the time period in years. Substituting the given values P = 21000, T = 3, and SI = 6400 into the formula, we get:

6400 = (21000 × R × 3)/(100)

Simplifying the expression by canceling the zeros:

6400 = 210 × R × 3 6400 = 630 × R

Solving for R:

R = ⁶⁴⁰⁰⁄₆₃₀ = ⁶⁴⁰⁄₆₃

Converting the improper fraction ⁶⁴⁰⁄₆₃ into a mixed fraction:

640 = (63 × 10) + 10 R = 10¹⁰⁄₆₃%

Explanation

Let the capacity of the cistern be the Least Common Multiple (LCM) of the individual filling and emptying times. Since 84 = 21 × 4 and 105 = 21 × 5, the LCM is 21 × 4 × 5 = 420 units.

The efficiency of one inlet pipe is 420 / 84 = 5 units per hour. The efficiency of one drain pipe is 420 / 105 = 4 units per hour. When 9 inlet pipes and 10 drain pipes are opened simultaneously, the net efficiency per hour is calculated as follows:

Net Efficiency = (9 × 5) - (10 × 4) = 45 - 40 = 5 units per hour

The time required to completely fill the cistern is the total capacity divided by the net efficiency:

Time = 420 / 5 = 84 hours

Explanation

The fundamental trigonometric identity states that sin²x + cos²x = 1 for any angle x. For an acute angle, cos x is positive and can be expressed as cos x = √(1 - sin²⁾ˣ.

Given that sin x = ³⁄₇, we substitute this value into the formula: cos x = √(1 - (³⁄₇)²⁾ cos x = √(1 - ⁹⁄₄₉) cos x = √((49 - 9)/(49)) cos x = √(⁴⁰⁄₄₉)

To simplify √(40), we factor it into its perfect square components: √(40) = √(4 × 10) = 2√(10). Taking the square root of the denominator, √(49) = 7. Therefore, the value is: cos x = 2√(10){7}

Explanation

The total weight of the class is the product of the average weight of the entire class and the total number of students, which is 60 × 46.5 = 2790 kg. The total weight contributed by the boys is the number of boys multiplied by their average weight, giving 35 × 42 = 1470 kg. By subtracting the total weight of the boys from the total weight of the class, the total weight of the girls is determined to be 2790 - 1470 = 1320 kg. Given that there are 25 girls in the class (60 - 35 = 25), their average weight is calculated by dividing their total weight by their count, resulting in ¹³²⁰⁄₂₅ = 52.8 kg.

Explanation

The area of a sector of a circle is calculated using the formula A = (θ)/(360°) × π r², where θ is the central angle and r is the radius. Substituting the given values A = 110 cm², θ = 56°, and π = ²²⁄₇ into the formula, we get:

110 = ⁵⁶⁄₃₆₀ × ²²⁄₇ × r²

Simplifying the fraction ⁵⁶⁄₇ results in 8, so the equation becomes:

110 = ⁸⁄₃₆₀ × 22 × r²

Further simplifying ⁸⁄₃₆₀ to ¹⁄₄₅ gives:

110 = ²²⁄₄₅ × r²

To find r², rearrange the terms:

r² = (110 × 45)/(22)

Since 110 divided by 22 is 5, we have:

r² = 5 × 45 = 225

Taking the square root of both sides, we find the radius:

r = √(225) = 15 cm

Explanation

If a value is x% less than another, the second value is (x)/(100 - x) × 100% more than the first. Here, x = 19, as Sam's marks are 19% less than Peter's. Substituting this value into the formula:

Percentage More = (19)/(100 - 19) × 100 Percentage More = ¹⁹⁄₈₁ × 100 Percentage More = ¹⁹⁰⁰⁄₈₁ = 23.4567...%

When rounding to two decimal places, 23.4567... becomes 23.46%. This percentage represents the increase required to get from Sam's marks (81% of Peter's) back to Peter's marks (100%).

Explanation

The percentage for Department 5 is calculated by subtracting the sum of percentages of Departments 1 through 4 from 100 percent. For the year 2020, the percentage share of Department 5 is 100 - (24 + 21 + 20 + 19) = 16%. Given the total number of employees is 1200, the absolute strength in 2020 is 0.16 × 1200 = 192.

For the year 2021, the percentage share of Department 5 is 100 - (21 + 22 + 18 + 19) = 20%. With the total number of employees increasing to 1600, the absolute strength in 2021 is 0.20 × 1600 = 320.

The variation in the strength of employees for Department 5 is the absolute difference between the two years: |320 - 192| = 128. Calculating the variations for the other departments yields lower values: Department 1 has a variation of 48 (|336 - 288|), Department 2 has 100 (|352 - 252|), Department 3 has 48 (|288 - 240|), and Department 4 has 76 (|304 - 228|). Thus, Department 5 exhibits the maximum variation.

Explanation

To find the value of the expression (x² + 7x + 1)/(x² + 11x + 1) given x + ¹⁄ₓ = 1, we simplify the fraction by dividing both the numerator and the denominator by x:

(x² + 7x + 1)/(x){(x² + 11x + 1)/(x)} = (x + 7 + 1)/(x){x + 11 + ¹⁄ₓ}

By rearranging the terms, the expression becomes:

((x + 1)/(x)) + 7{(x + ¹⁄ₓ) + 11}

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

(1 + 7)/(1 + 11) = ⁸⁄₁₂

Dividing both the numerator and the denominator by their greatest common divisor, 4, results in:

(8 ÷ 4)/(12 ÷ 4) = ⅔

Explanation

The value is calculated using the BODMAS rule, which dictates the order of operations: Brackets, Orders, Division and Multiplication, and Addition and Subtraction. First, simplify the expression within the first set of parentheses: 6 ÷ 2 - 3 + 7 - 2. Performing division first gives 3 - 3 + 7 - 2, which simplifies to 5. Next, simplify the expression within the square brackets [(3 - 2 ÷ 2) × 5 - 6]. Solving the inner parentheses first, 3 - 2 ÷ 2 becomes 3 - 1 = 2. Substituting this back into the square brackets results in 2 × 5 - 6, which equals 10 - 6 = 4. The original expression now reduces to 6 - 5 × 4. Following the order of operations, the multiplication is performed before the subtraction, resulting in 6 - 20 = -14.

Explanation

The length of a shadow is directly proportional to the height of the object when the angle of the sun is constant. Let H be the number of storeys and S be the length of the shadow. The ratio of height to shadow length remains constant:

(H₁)/(S₁) = (H₂)/(S₂)

Given H₁ = 7 storeys and S₁ = 28m, the ratio is ⁷⁄₂₈ = ¼. This indicates that every 4 meters of shadow corresponds to 1 storey of the building. For a shadow length S₂ = 48m, the number of storeys H₂ is determined by:

H₂ = (S₂)/(4) = ⁴⁸⁄₄ = 12

Explanation

The total work can be calculated based on the man-days required. Given that 6 men can complete the work in 8 days, the total work is 6 × 8 = 48 units. The same work is completed by 10 women in 16 days, meaning 10 × 16 = 160 woman-days is equal to 48 man-days. This establishes the efficiency ratio where 1 woman performs ⁴⁸⁄₁₆₀ = 0.3 units of work per day, assuming 1 man performs 1 unit per day.

In the first 2 days, 12 men were working, completing 12 × 2 = 24 units of work. The remaining work to be done is 48 - 24 = 24 units. After 2 days, 6 men left, leaving 12 - 6 = 6 men. They were joined by 5 women. The combined daily output of this new group is 6 × 1 + 5 × 0.3 = 6 + 1.5 = 7.5 units per day. The time taken to complete the remaining work is (24)/(7.5) = ²⁴⁰⁄₇₅ = ¹⁶⁄₅ days. To find the total time taken for the work to be completed, the initial 2 days are added to the time for the remaining work, resulting in 2 + ¹⁶⁄₅ = (10 + 16)/(5) = ²⁶⁄₅ days.

Explanation

The present worth (PW) is the principal amount that, when invested at a given simple interest rate (r) for a specific time (t), grows to the future amount (A). The relationship is defined by the formula A = PW + (PW × r × t)/(100), which can be rearranged to PW = (100 × A)/(100 + (r × t)).

Given the amount A = 1100, the rate r = 5%, and the time t = 2 years, the calculation is as follows:

PW = (100 × 1100)/(100 + (5 × 2))

PW = (110000)/(100 + 10)

PW = ¹¹⁰⁰⁰⁰⁄₁₁₀

PW = 1000

The principal value of ₹1,000 earns ₹100 in simple interest over 2 years at 5% per annum, resulting in the total due amount of ₹1,100.

Explanation

The expression can be simplified using standard trigonometric identities. The numerator, cos² 15° - sin² 15°, follows the double angle identity cos 2θ = cos² θ - sin² θ. By substituting θ = 15°, the numerator becomes cos(2 × 15°) = cos 30°. The value of cos 30° is √3{2}.

The denominator, cos² 145° + sin² 145°, follows the fundamental Pythagorean identity sin² θ + cos² θ = 1, which is valid for any angle θ. Therefore, the denominator equals 1. Combining these results, the value of the entire expression is {√3{2}}{1} = √3{2}.

Explanation

Since 45 is a multiple of 15, the remainder when the number is divided by 15 is equal to the remainder of 21 divided by 15. Calculating 21 modulo 15 gives 6.

Explanation

Dividing both the numerator and denominator of the expression by cos A gives (3 tan A + 2) / (3 tan A - 2). Substituting tan A = 3/8, the expression becomes (3(3/8) + 2) / (3(3/8) - 2) = (9/8 + 2) / (9/8 - 2) = (25/8) / (-7/8) = -25/7.

Explanation

To find the time taken to cross the standing train, the total distance (108 m + 148 m = 256 m) is divided by the speed of the Rajdhani Express. Converting the speed from 128 km/h to m/s gives 320/9 m/s, and dividing 256 m by 320/9 m/s results in exactly 7.2 seconds.

Explanation

Following the BODMAS rule, we first perform the division and multiplication from left to right: (5/6) × (4/5) ÷ (4/3) = (2/3) ÷ (4/3) = (2/3) × (3/4) = 1/2. Then, perform the addition and subtraction: 2/3 - 1/2 + 1/2 = 2/3. Therefore, the value of the expression is 2/3.

Explanation

The value of tan 4384 degrees is equal to tan(12 x 360 + 64) = tan 64 degrees. The value of cot 6814 degrees is equal to cot(18 x 360 + 334) = cot 334 degrees, which is equivalent to -cot 26 degrees or -tan 64 degrees. Adding these results together gives tan 64 - tan 64 = 0.

Explanation

Let the cost price be 100. The marked price is 140 (40 percent above cost price), and the selling price is 73.5 percent of 140, which equals 102.9. Since the selling price is 102.9 and the cost price is 100, the profit is 2.9, which is a 2.9 percent profit.

Explanation

Solving the equations x + 2y = 170 (from the sum of angles in a triangle) and 2x - 4y = 20 (given) results in x = 90 and y = 40. Since the three angles of the triangle are 90, 40, and 50 degrees, the presence of a 90-degree angle makes it a right-angled triangle.

Explanation

Efficiency of X is 6 units/day (30/5), Y is 2 units/day (30/15), and Z is 1 unit/day (30/30) based on a total work of 30 units. On day one X completes 6 units, and on day two X, Y, and Z together complete 9 units (6+2+1), resulting in 15 units of work every 2 days. To complete 30 units of work, it takes 2 cycles of 2 days, which equals 4 days.

Explanation

The difference between a 15% profit and a 10% profit is 5% of the cost price. Since this 5% difference is equal to Rs. 20, the total cost price (100%) is calculated as (20 / 5) * 100 = 400.