SSC CGL Tier 1 2011 Quantitative Aptitude Questions with Solutions

Explanation

The efficiency ratio of X and Y is 3:1. Because time taken is inversely proportional to efficiency, the ratio of time taken by X and Y is 1:3. Let X take k days and Y take 3k days. The difference in their time is given as 3k - k = 40 days, so 2k = 40, which means k = 20. Consequently, X takes 20 days and Y takes 60 days to complete the work alone. Using the formula for combined work T = Tₓ T_y{Tₓ + T_y}, the time taken to complete the work together is (20 × 60)/(20 + 60) = ¹²⁰⁰⁄₈₀ = 15 days.

Explanation

The area of the square is 81 cm², so the side s of the square is √(81) = 9 cm. The total length of the copper wire is equal to the perimeter of this square, which is 4 × s = 4 × 9 = 36 cm.

When the same wire is bent into a semicircle of radius r, the total length of the wire corresponds to the perimeter of the semicircle. This perimeter includes the curved arc length π r and the straight diameter 2r. Thus, the equation for the length of the wire is r(π + 2) = 36. Using the approximation π = 22/7, the relation becomes r(22/7 + 2) = 36. This simplifies to r(36/7) = 36. Solving for r, we find r = (36 × 7) / 36 = 7 cm.

Explanation

The volume of rainwater is determined by the product of the surface area and the depth of the rainfall. Given the area A = 1.5 hectares and knowing that 1 hectare equals 10,000 square meters, the area in square meters is 1.5 × 10,000 = 15,000 m². The depth of the rainfall h is given as 5 cm, which is equivalent to 0.05 meters (5/100 m). Using the formula for volume V = A × h, the calculation is V = 15,000 × 0.05. This yields a total volume of 750 cubic meters (m³).

Explanation

The volume of water falling into the sea per minute is calculated using the formula for the volume of a rectangular prism, where the length of the water column is the distance the water travels in one minute.

The cross-sectional area of the river is the product of its width and depth: Area = 40 m × 3 m = 120 m².

The flow rate is 2 km/hr. To find the distance the water travels in one minute, we convert this speed into meters per minute: Speed = (2 × 1000 meters)/(60 minutes) = ¹⁰⁰⁄₃ meters/minute.

The volume of water flowing in one minute is: Volume in m³ = Area × Distance = 120 × ¹⁰⁰⁄₃ = 40 × 100 = 4000 m³.

To convert cubic meters to liters, we use the standard conversion factor where 1 m³ = 1000 liters: Total volume in liters = 4000 × 1000 = 4,000,000 liters.

Explanation

The distance D covered by a circular wheel is the product of the number of revolutions n and the circumference of the wheel 2π r. The relationship is expressed by the formula D = n × 2π r.

First, convert the total distance from kilometers to centimeters to match the required unit for the radius. Given D = 11 km, we have D = 11 × 1000 × 100 = 1,100,000 cm.

Substitute the known values n = 5000 and π ≈ ²²⁄₇ into the formula:

1,100,000 = 5000 × 2 × ²²⁄₇ × r

1,100,000 = (220,000)/(7) × r

To find r, rearrange the equation:

r = (1,100,000 × 7)/(220,000)

r = 5 × 7

r = 35 cm

Explanation

Let the three sides of the triangle be a, b, and c, where c = 17 cm is the largest side. The perimeter is given as a + b + c = 40 cm, so a + b + 17 = 40, which simplifies to a + b = 23 cm. The semi-perimeter s is half the perimeter, so s = 40 / 2 = 20 cm.

Using Heron's formula for the area, Area = √(s(s-a)(s-b)(s-c)), we substitute the known values: 60 = √(20(20-a)(20-b)(20-17)) 60 = √(60(20-a)(20-b))

Squaring both sides yields: 3600 = 60(20-a)(20-b) 60 = (20-a)(20-b) 60 = 400 - 20(a+b) + ab

Substituting a+b = 23 into the equation: 60 = 400 - 20(23) + ab 60 = 400 - 460 + ab ab = 120

The sides a and b are values that sum to 23 and multiply to 120. They are the roots of the quadratic equation x² - 23x + 120 = 0. Factoring this gives (x - 8)(x - 15) = 0, resulting in x = 8 and x = 15. The three sides of the triangle are therefore 8 cm, 15 cm, and 17 cm. Comparing these lengths, the smallest side is 8 cm.

Explanation

The single equivalent discount for two successive discounts d₁% and d₂% is given by the formula D = (d₁ + d₂ - d₁ × d₂{100}) %.

Given the successive discounts are d₁ = 20% and d₂ = 15%, substitute these values into the formula: D = 20 + 15 - (20 × 15)/(100) D = 35 - ³⁰⁰⁄₁₀₀ D = 35 - 3 D = 32%

This can also be calculated by considering an initial price of 100. After the first discount of 20%, the price becomes 100 - 20 = 80. The second discount of 15% is applied to this new price: 15% of 80 = ¹⁵⁄₁₀₀ × 80 = 12. The total discount is 20 + 12 = 32, which represents a single discount of 32% on the original price of 100.

Explanation

The given ratio is A:B = 1:3, B:C = 1:3, and C:D = 1:3. To find the continuous ratio A:B:C:D, we express all shares in terms of A. Since A/B = 1/3, we have B = 3A. Similarly, B/C = 1/3 implies C = 3B = 3(3A) = 9A. Finally, C/D = 1/3 implies D = 3C = 3(9A) = 27A.

The ratio of the shares of A, B, C, and D is 1:3:9:27. The sum of the ratio parts is 1 + 3 + 9 + 27 = 40. Given the total profit is 4,00,000, the value of one unit in the ratio is 4,00,000 / 40 = 10,000. C’s share corresponds to 9 units in the ratio, so C's share is 9 × 10,000 = 90,000.

Explanation

Let the total volume of the can be V litres. Since 9 litres of mixture are removed and replaced by 9 litres of liquid B, the total volume of the liquid in the can remains V at the end of the process. Initially, the ratio of A to B is 7:5, meaning liquid A constitutes ⁷⁄₁₂ of the mixture.

When 9 litres of the mixture are drawn off, the remaining volume is V - 9. The quantity of liquid A in this remaining mixture is ⁷⁄₁₂(V - 9). Because the can is سپس filled only with liquid B, the quantity of liquid A does not change during the refilling step. Thus, the quantity of A in the final mixture is still ⁷⁄₁₂(V - 9).

According to the new ratio 7:9, the quantity of liquid A in the final V litres is ⁷⁄₇₊₉V = ⁷⁄₁₆V. Equating these two expressions for the quantity of A:

⁷⁄₁₂(V - 9) = ⁷⁄₁₆V

Dividing both sides by 7:

(V - 9)/(12) = (V)/(16)

Multiplying both sides by 48 (the least common multiple of 12 and 16):

4(V - 9) = 3V

4V - 36 = 3V

V = 36

The initial quantity of liquid A was ⁷⁄₁₂ × 36 = 21 litres.

Explanation

Let the number to be added to or subtracted from each term of the ratio be x. According to the problem, the resulting ratio is 1:2. This can be expressed as:

(17 + x)/(24 + x) = ½

To find x, cross-multiply the terms of the equation:

2(17 + x) = 1(24 + x) 34 + 2x = 24 + x

Rearrange the terms to solve for x:

2x - x = 24 - 34 x = -10

The negative result indicates that the magnitude of 10 must be subtracted from each term. When 10 is subtracted from 17 and 24, the terms become 7 and 14 respectively, which simplifies to the ratio 7:14 or 1:2.

Explanation

Let the weekly incomes of A and B be 9x and 7x, and their weekly expenditures be 4y and 3y. Since Savings = Income - Expenditure, and both save 200 per week, we can set up the following equations:

9x - 4y = 200 7x - 3y = 200

By equating the two expressions (since both equal 200): 9x - 4y = 7x - 3y 2x = y

Substituting y = 2x into the first equation: 9x - 4(2x) = 200 9x - 8x = 200 x = 200

The sum of their weekly incomes is 9x + 7x = 16x. Substituting the value of x gives: 16 × 200 = 3200

Explanation

Let the three numbers be x, y, and z. Based on the problem, the first number x is related to the others as x = 2y and x = 3z. This implies y = ˣ⁄₂ and z = ˣ⁄₃.

The average of the three numbers is given as 49.5, so their sum is 49.5 × 3 = 148.5. Substituting the values of y and z in terms of x into the sum equation:

x + ˣ⁄₂ + ˣ⁄₃ = 148.5

To solve for x, find a common denominator:

(6x + 3x + 2x)/(6) = 148.5 ¹¹ˣ⁄₆ = 148.5 11x = 891 x = 81

The first number is 81. The third number is z = ˣ⁄₃ = ⁸¹⁄₃ = 27. The difference between the first and the third number is x - z = 81 - 27 = 54.

Explanation

The correct mean can be calculated by adjusting the total sum of the numbers based on the errors found. The initial sum of the 50 numbers is calculated as the mean multiplied by the total count: Sum = 30 × 50 = 1500.

To find the correct sum, we subtract the incorrect entries and add the correct ones: Correct Sum = 1500 - (82 + 13) + (28 + 31) Correct Sum = 1500 - 95 + 59 Correct Sum = 1405 + 59 = 1464

The correct mean is then the correct sum divided by the number of entries: Correct Mean = ¹⁴⁶⁴⁄₅₀ = 29.28

Alternatively, the change in the mean can be calculated directly from the difference in the entries: Change in Sum = (28 + 31) - (82 + 13) = 59 - 95 = -36 Change in Mean = ⁻³⁶⁄₅₀ = -0.72 Correct Mean = 30 - 0.72 = 29.28

Explanation

Let the cost price (CP) of one article be x. According to the problem, the selling price (SP) of 10 articles is equal to the cost price of 11 articles. This can be expressed as 10 × SP = 11 × CP.

The cost price of 10 articles is 10x, and the selling price of these same 10 articles is 11x. The profit made on the sale of 10 articles is the difference between their selling price and their cost price: Profit = 11x - 10x = x.

The gain percentage is calculated using the formula: Gain % = (Total Profit)/(Total Cost Price) × 100

Substituting the values: Gain % = ˣ⁄₁₀ₓ × 100 = ⅒ × 100 = 10%

Alternatively, when the selling price of n articles equals the cost price of m articles, the gain percent is determined by the relation (m - n)/(n) × 100. Here, m = 11 and n = 10. Applying the formula directly: Gain % = (11 - 10)/(10) × 100 = 10%

Explanation

The reduction in profit is directly caused by the increase in the discount offered on the marked price. The difference between the two discount rates is 6% - 5% = 1%. Since the cost price of the watch remains the same, this 1% difference in the discount amount is exactly equal to the 15 reduction in profit.

Let the marked price of the watch be M. The relationship can be expressed as: 1% of M = 15 ¹⁄₁₀₀ × M = 15 M = 15 × 100 = 1,500

Explanation

Since Krishna buys an equal number of articles of both types, let the number of articles of each type be 1.

The Cost Price (CP) of the first article is 10 and the Cost Price (CP) of the second article is 14. The total Cost Price for these 2 articles is 10 + 14 = 24.

He sells all the articles at 13 each. Since he has 2 articles in total, the total Selling Price (SP) is 13 × 2 = 26.

Since the total Selling Price is greater than the total Cost Price, there is a gain. The gain amount is 26 - 24 = 2.

The gain percentage is calculated using the formula: Gain% = (Total Gain)/(Total CP) × 100

Substituting the values: Gain% = ²⁄₂₄ × 100 Gain% = ¹⁄₁₂ × 100 Gain% = ²⁵⁄₃ Gain% = 8 ⅓%

Explanation

Let the cost prices of the two horses be CP₁ and CP₂ . The problem states that the total cost price is CP₁ + CP₂ = 19,500 . The first horse is sold at a 20% loss, so its selling price is 0.80 CP₁ . The second horse is sold at a 15% profit, so its selling price is 1.15 CP₂ .

Since the selling price of each horse is the same, we equate them: 0.80 CP₁ = 1.15 CP₂ CP₁{CP₂} = (1.15)/(0.80) = ¹¹⁵⁄₈₀ = ²³⁄₁₆

The sum of the ratio parts is 23 + 16 = 39 . To find the value of one part, we divide the total cost price by the sum of the ratio parts: Value per part = (19,500)/(39) = 500

Now, calculate the individual cost prices: CP₁ = 23 × 500 = 11,500 CP₂ = 16 × 500 = 8,000

Explanation

Let the selling price (SP) be x. The cost price (CP) is given as 40% of the selling price, which can be expressed as CP = 0.4x. To find what percent of the cost price the selling price is, we use the ratio (SP)/(CP) × 100.

Substituting the values into the formula: (x)/(0.4x) × 100 = (1)/(0.4) × 100

Simplifying the calculation: ¹⁰⁄₄ × 100 = 2.5 × 100 = 250%

Alternatively, if we assume SP = 100, then CP = 40. The required percentage is ¹⁰⁰⁄₄₀ × 100 = ⁵⁄₂ × 100 = 250%.

Explanation

The first condition states that 90% of A = 30% of B. This can be expressed as: ⁹⁰⁄₁₀₀A = ³⁰⁄₁₀₀B 90A = 30B B = 3A

The second condition states that B = (2x)% of A, which is: B = ²ˣ⁄₁₀₀A

By substituting the value of B from the first equation into the second equation, we get: 3A = ²ˣ⁄₁₀₀A 3 = ²ˣ⁄₁₀₀ 300 = 2x x = 150

Explanation

When the total expenditure remains constant, the price and quantity are inversely proportional. The reduction in price is 10%, making the ratio of the original price to the new price 100 : 90, or 10 : 9. Since quantity is inversely proportional to price, the ratio of the original quantity to the new quantity is 9 : 10. The difference in the ratio units is 10 - 9 = 1 unit. According to the problem, this 1 unit corresponds to an increase of 1 kg. Therefore, the original quantity purchased was 9 units, which equals 9 kg. The original price per kg is determined by dividing the total expenditure by the original quantity: Original Price = ²⁷⁰⁄₉ = 30

Explanation

Total expenditure is the product of price and consumption. When expenditure remains constant, the percentage reduction in consumption required due to a price increase of r% is given by the formula:

Percentage reduction in consumption = ( (r)/(100 + r) ) × 100%

Substituting the given price increase of r = 25:

Reduction = ( (25)/(100 + 25) ) × 100% Reduction = ( ²⁵⁄₁₂₅ ) × 100% Reduction = ⅕ × 100 = 20%

Alternatively, using the ratio method, a 25% increase means the price ratio changes from 100 to 125, which simplifies to a ratio of 4:5. Since expenditure is constant, consumption must be inversely proportional to price. Therefore, the consumption ratio must change from 5:4. The reduction in consumption is 1 unit out of the original 5 units. Expressed as a percentage, this is ( ⅕ ) × 100 = 20%.

Explanation

The problem states that 30% of A added to 40% of B equals 80% of B. This can be written as the linear equation: 0.3A + 0.4B = 0.8B

To solve for the relationship between A and B, subtract 0.4B from both sides of the equation: 0.3A = 0.8B - 0.4B 0.3A = 0.4B

The question asks for the percentage of A that B represents, which is calculated using the ratio (B)/(A) × 100. Rearranging the simplified equation gives: (B)/(A) = (0.3)/(0.4) = ¾

To find the percentage: ¾ × 100 = 75%

Explanation

The downstream speed D is 6 + 2 = 8 km/h and the upstream speed U is 6 - 2 = 4 km/h. Let d be the distance. According to the problem, the difference between the time taken to travel upstream and downstream is 3 hours. Using the relation Time = Distance / Speed, we set up the equation d/4 - d/8 = 3. To solve for d, we find a common denominator: (2d - d) / 8 = 3. This simplifies to d / 8 = 3, which gives d = 24 km.

Explanation

To find the distance D when two different speeds and their corresponding time offsets are given, the formula D = s₁ × s₂{|s₁ - s₂|} × Δ T is used, where s₁ and s₂ are the speeds and Δ T is the total time difference in hours. Given s₁ = 2.5 km/h and s₂ = 3 km/h, the difference in speed is 0.5 km/h. The student is 6 minutes late in the first scenario and 10 minutes early in the second, so the total time difference is 6 + 10 = 16 minutes. Converting this to hours, Δ T = ¹⁶⁄₆₀ hours. Substituting these values into the formula results in D = (2.5 × 3)/(3 - 2.5) × ¹⁶⁄₆₀ = (7.5)/(0.5) × ¹⁶⁄₆₀. Since 7.5 divided by 0.5 is 15, the equation becomes D = 15 × ¹⁶⁄₆₀ = ¹⁶⁄₄ = 4 km.

Explanation

In compound interest, if a sum becomes n times in T years, it becomes nᵐ times in m × T years. Here, the sum doubles itself (n = 2) in T = 4 years. We need to find the time taken for it to become four times itself. Since 4 = 2², the exponent m is 2. Therefore, the total time required is m × T = 2 × 4 = 8 years.

This can also be understood through the standard formula A = P(1 + r/100)ᵗ. From the first condition, 2P = P(1 + r/100)⁴, which simplifies to 2 = (1 + r/100)⁴. For the sum to become four times, we need 4 = (1 + r/100)ᵗ. Squaring the first equation gives 2² = [(1 + r/100)⁴]², which results in 4 = (1 + r/100)⁸. Comparing the two equations for the amount to be four times, we find t = 8.

Explanation

The simple interest SI is calculated using the formula 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. The problem states that the simple interest for 5 years is one-fourth of the sum, which gives the relation SI = ¼P and T = 5.

Substituting these values into the formula:

¼P = (P × R × 5)/(100)

By cancelling P from both sides, the equation becomes:

¼ = (5R)/(100)

Simplifying the right side:

¼ = (R)/(20)

Multiplying both sides by 20 to solve for R:

R = ²⁰⁄₄ = 5

The rate of interest per annum is 5%.

Explanation

The difference D between Compound Interest and Simple Interest for a duration of 2 years is determined by the formula D = P((R)/(100))², where P is the principal amount and R is the annual rate of interest. Substituting the given values P = 10,000 and D = 25 into the formula:

25 = 10,000 × ((R)/(100))²

25 = 10,000 × (R²)/(10,000)

By canceling 10,000 from both the numerator and the denominator, the equation simplifies to R² = 25. Taking the square root of both sides gives R = 5.

Explanation

Let the radius of the base for both the hemisphere and the cone be r. For a hemisphere, the height is always equal to its radius, so h = r. Since the problem states the heights are equal, the height of the cone is also r.

The curved surface area of a hemisphere is given by the formula 2π r². The curved surface area of a cone is given by π rl, where l is the slant height. The slant height l of the cone is calculated using the Pythagorean theorem as l = √(r²⁾⁺ʰ^^^². Substituting h = r, we get l = √(r²⁾⁺ʳ^^^² = √(2r²⁾ = r√2.

The curved surface area of the cone is therefore π · r · r√2 = √2π r². To find the ratio of the curved surface of the hemisphere to that of the cone, we divide the two areas: 2π r² : √2π r². This simplifies to 2 : √2. By rationalizing the ratio, we get √2 : 1.

Explanation

First, rationalize the expression inside the square root for x by multiplying the numerator and denominator by the conjugate of the denominator, √5 + 1: x = √({5) + 1{√5 - 1} × √5 + 1{√5 + 1}} = √(({5) + 1)²{5 - 1}} = √(({5) + 1)²{4}} = √5 + 1{2}

Rearranging the equation x = √5 + 1{2} gives 2x - 1 = √5. Squaring both sides of this equation yields (2x - 1)² = (√5)², which simplifies to 4x² - 4x + 1 = 5. Subtracting 5 from both sides gives 4x² - 4x - 4 = 0. Dividing the entire equation by 4 results in the simplified quadratic identity x² - x - 1 = 0, or x² - x = 1.

The expression to be evaluated is 5x² - 5x - 1. By factoring out 5 from the first two terms, we can rewrite the expression as 5(x² - x) - 1. Substituting the previously derived value x² - x = 1 into this expression gives 5(1) - 1 = 4.

Explanation

The remainder when a number is divided by 10 is always equal to the unit digit of that number. To find the unit digit of 2³³, we use the concept of cyclicity. The unit digits of the powers of 2 follow a repeating cycle of four: 2¹=2, 2²=4, 2³=8, and 2⁴=16 (unit digit 6).

The cycle of unit digits is {2, 4, 8, 6}. To determine which digit in the cycle corresponds to 2³³, we divide the exponent 33 by the cycle length 4. Using the division algorithm, 33 = (4 × 8) + 1. Since the remainder is 1, the unit digit of 2³³ is the same as the unit digit of 2¹, which is 2. Consequently, when 2³³ is divided by 10, the remainder is 2.

Explanation

To solve the continued fraction, simplify the expression starting from the bottom denominator. First, calculate the lowest part: 3 + ⅓ = ¹⁰⁄₃.

Next, substitute this value into the next level of the fraction: 3 + (1)/(10){3} = 3 + ³⁄₁₀ = ³³⁄₁₀.

Now, substitute this result back into the main expression: 3 + (3)/(33){10}. This simplifies to 3 + 3 × ¹⁰⁄₃₃, which is 3 + ³⁰⁄₃₃. By dividing the numerator and denominator of the fraction ³⁰⁄₃₃ by 3, it reduces to ¹⁰⁄₁₁.

Finally, calculate the sum: 3 + ¹⁰⁄₁₁ = (33 + 10)/(11) = ⁴³⁄₁₁.

Explanation

The unit digit of an exponential expression depends only on the unit digit of the base. For the first term (4387)²⁴⁵, the unit digit is determined by 7²⁴⁵. The unit digits of powers of 7 follow a cyclic pattern of length 4: 7¹ → 7, 7² → 9, 7³ → 3, and 7⁴ → 1. To find the position in the cycle, we divide the exponent 245 by 4. Since 245 = (61 × 4) + 1, the remainder is 1. Therefore, the unit digit of (4387)²⁴⁵ is the same as 7¹, which is 7.

For the second term (621)⁷², the unit digit is determined by 1⁷². Any power of a number ending in 1 will always have 1 as its unit digit. Thus, the unit digit of (621)⁷² is 1. Multiplying the unit digits of the two terms gives 7 × 1 = 7, which is the unit digit of the final product.

Explanation

The given expression is 2√2 + √2 + (1)/(2 + √(2)) + (1)/(√(2) + 1).

To solve this, first simplify the whole number terms and then rationalize the denominators of the fractional terms.

The sum of the first two terms is: 2√2 + √2 = 3√2

Rationalizing the third term (1)/(2 + √(2)) by multiplying the numerator and denominator by the conjugate 2 - √2: 1(2 - √2){(2 + √2)(2 - √2)} = 2 - √2{4 - 2} = 2 - √2{2} = 1 - √2{2}

Rationalizing the fourth term (1)/(√(2) + 1) by multiplying the numerator and denominator by the conjugate √2 - 1: 1(√2 - 1){(√2 + 1)(√2 - 1)} = √2 - 1{2 - 1} = √2 - 1

Now, combine all simplified parts: 3√2 + (1 - √2{2}) + (√2 - 1) 3√2 + √2 - √2{2} + 1 - 1 4√2 - 0.5√2 = 3.5√2

Substitute the value √2 = 1.4142: 3.5 × 1.4142 = 4.9497

Explanation

The operation is defined by the formula XY = X² + Y² - XY. To find the value of 1113, substitute X = 11 and Y = 13 into the given expression.

11^*13 = 11² + 13² - (11 × 13)

Calculating each term: 11² = 121 13² = 169 11 × 13 = 143

Substituting these values back into the expression: 11^*13 = 121 + 169 - 143 11^*13 = 290 - 143 = 147

Explanation

The given expression follows the algebraic form (a³ - b³)/(a² + ab + b²) where a = 3.2 and b = 0.2. This is evident because 0.008 = (0.2)³, 0.64 = 3.2 × 0.2, and 0.04 = (0.2)².

Using the algebraic identity a³ - b³ = (a - b)(a² + ab + b²), the expression can be rewritten as:

((3.2 - 0.2)((3.2)² + (3.2 × 0.2) + (0.2)²))/((3.2)² + 0.64 + 0.04)

Since the term (3.2)² + 0.64 + 0.04 is common to both the numerator and the denominator, they cancel each other out:

3.2 - 0.2 = 3

Explanation

Given the equation a¹^/³ = 11, we determine the value of a by cubing both sides of the equation: a = 11³ = 1331. To find the value of a² - 331a, we factor the expression as a(a - 331). Substituting the value of a into this simplified expression gives 1331(1331 - 331), which simplifies further to 1331 × 1000. Multiplying these values results in 1331000.

Explanation

Given the equation √(1 + ˣ⁄₉₆₁) = ³²⁄₃₁, we square both sides to eliminate the square root:

1 + ˣ⁄₉₆₁ = (³²⁄₃₁)²

Since 32² = 1024 and 31² = 961, the equation becomes:

1 + ˣ⁄₉₆₁ = ¹⁰²⁴⁄₉₆₁

Subtracting 1 from both sides:

ˣ⁄₉₆₁ = ¹⁰²⁴⁄₉₆₁ - 1

ˣ⁄₉₆₁ = (1024 - 961)/(961)

ˣ⁄₉₆₁ = ⁶³⁄₉₆₁

By comparing the numerators, we find x = 63.

A useful shortcut for equations of the form √(1 + (x)/(n²)) = ⁿ⁺¹⁄ₙ is that x will always equal the sum of the numerator and the denominator of the fraction on the right side: x = (n+1) + n. In this case, x = 32 + 31 = 63.

Explanation

Let a and b be odd numbers. According to the properties of parity, the sum of two odd numbers is always even, which means a + b is an even number. Additionally, any integer multiplied by 2 results in an even number, so the term 2ab is also even. The sum of two even numbers is always even, therefore the expression (a + b) + 2ab results in an even value. For example, if a = 1 and b = 3, then 1 + 3 + 2(1 × 3) = 4 + 6 = 10, which is an even number.

Explanation

The expression 2¹⁶ - 1 can be simplified using the algebraic identity a² - b² = (a - b)(a + b). By repeatedly applying this identity, we get: 2¹⁶ - 1 = (2⁸ - 1)(2⁸ + 1) 2⁸ - 1 = (2⁴ - 1)(2⁴ + 1) = (16 - 1)(16 + 1) = 15 × 17

Substituting this back into the first equation, we find that 2¹⁶ - 1 = 15 × 17 × (2⁸ + 1). Since 17 is a prime factor of the product, the original expression is divisible by 17. Alternatively, according to Fermat's Little Theorem, aᵖ⁻¹ ≡ 1 p for any prime p that does not divide a. For a = 2 and p = 17, the theorem gives 2¹⁷⁻¹ = 2¹⁶ ≡ 1 17, confirming that 2¹⁶ - 1 is divisible by 17.

Explanation

The sum of a set of numbers is calculated as the product of their average and the total count of numbers (Sum = Average × n).

First, calculate the total sum of all five numbers: Sumₜₒₜₐₗ = 34.4 × 5 = 172

Next, find the sum of the first two numbers: Sum₁_,₂ = 46.5 × 2 = 93

Then, find the sum of the fourth and fifth numbers: Sum₄_,₅ = 18 × 2 = 36

To find the third number, subtract the sum of the first, second, fourth, and fifth numbers from the total sum of all five numbers: Third Number = 172 - (93 + 36) Third Number = 172 - 129 = 43

Explanation

The least number that leaves a remainder of 3 when divided by 5, 6, 7, and 8 must be of the form LCM(5, 6, 7, 8) × k + 3. To find the LCM(5, 6, 7, 8), we identify the highest powers of prime factors involved: 5 = 5¹, 6 = 2¹ × 3¹, 7 = 7¹, and 8 = 2³. Thus, LCM = 2³ × 3 × 5 × 7 = 840.

The number is expressed as 840k + 3. We need to find the smallest integer k such that this expression is divisible by 9. Testing values for k: If k = 1, 840(1) + 3 = 843 (sum of digits 8+4+3=15, not divisible by 9). If k = 2, 840(2) + 3 = 1683 (sum of digits 1+6+8+3=18, which is divisible by 9).

The number 1683 fulfills all conditions. It leaves a remainder of 3 when divided by 5, 6, 7, and 8 because it is exactly 3 more than their least common multiple (multiplied by 2). Additionally, the sum of its digits is 18, confirming its divisibility by 9.

Explanation

Let the three numbers be 3x, 4x, and 5x according to the given ratio 3:4:5. In this sequence, 3x is the smallest number, 4x is the second number, and 5x is the largest number.

The problem states that the sum of the largest and smallest numbers is equal to the sum of the second number and 52. This relationship is expressed as: 3x + 5x = 4x + 52

Simplifying the equation by combining terms on the left side: 8x = 4x + 52

Subtracting 4x from both sides of the equation: 4x = 52

Dividing by 4 gives the value of the common ratio factor: x = 13

The smallest number is 3x. By substituting the value of x: 3 × 13 = 39

Explanation

The area of a circle is calculated using the formula A = π r², where r represents the radius. When a quantity is proportional to the square of a variable, the net percentage change resulting from an increase in that variable can be determined using the successive percentage change formula x + y + ˣʸ⁄₁₀₀.

Applying this to the radius increase of 50%, we let x = 50 and y = 50. The percentage increase in area is 50 + 50 + (50 × 50)/(100) = 100 + 25 = 125%.

Alternatively, if the initial radius is r, the initial area is π r². A 50% increase makes the new radius 1.5r. The new area is π (1.5r)² = 2.25π r². The increase in area is 2.25π r² - π r² = 1.25π r². To find the percentage increase, we calculate 1.25π r²{π r²} × 100 = 125%.

Explanation

Let the total work be the Least Common Multiple (LCM) of the time taken by A and B. Total Work = LCM(9, 12) = 36 units.

The efficiency of A is 36 / 9 = 4 units per day, and the efficiency of B is 36 / 12 = 3 units per day.

Since they work on alternate days starting with A, we consider a cycle of 2 days. Work done in 1 cycle (2 days) = 4 + 3 = 7 units.

To find the number of completed cycles, we divide the total work by the work done per cycle: 36 / 7 = 5 cycles with a remainder of 1 unit. Time taken for 5 cycles = 5 × 2 = 10 days. Work completed in 10 days = 5 × 7 = 35 units.

Remaining work = 36 - 35 = 1 unit. On the 11th day, it is A's turn to work. Since A can do 4 units in a full day, the time taken by A to complete the remaining 1 unit is 1 / 4 day.

Total time = 10 + 1/4 = 10 ¼ days.

Explanation

Let the total work be the Least Common Multiple (LCM) of the given time periods 10, 6, and 12, which is 60 units.

The efficiency (work done per day) of the pairs is calculated as: Efficiency of A + B = 60 / 10 = 6 units/day Efficiency of B + C = 60 / 6 = 10 units/day Efficiency of A + C = 60 / 12 = 5 units/day

Summing these efficiencies gives: 2(A + B + C) = 6 + 10 + 5 = 21 units/day Efficiency of A + B + C = 21 / 2 = 10.5 units/day

The time taken by A, B, and C together to complete the work is: Time = Total Work / Efficiency Time = 60 / 10.5 = 600 / 105 = 40 / 7 days

Converting the improper fraction 40 / 7 into a mixed fraction: 40 / 7 = 5 ⁵⁄₇ days

Explanation

The efficiency or rate of work R is defined as Work / Time. Let R_A and R_B be the rates of A and B respectively. According to the problem, A performs half the work of B in three-fourths of the time. If B performs work W in time T, then A performs W/2 work in 3/4 T time.

R_A = (W/2)/(3T/4) = ½ × ⁴⁄₃ × (W)/(T) = ⅔ R_B

The ratio of their efficiencies R_A : R_B is 2:3. The combined efficiency of A and B working together is the sum of their individual efficiencies:

Rₜₒₜₐₗ = R_A + R_B = ⅔ R_B + R_B = ⁵⁄₃ R_B

Since they complete the work together in 18 days, the total work can be expressed as:

Total Work = Rₜₒₜₐₗ × 18 = ⁵⁄₃ R_B × 18 = 30 R_B

The time taken by B alone to complete the total work is calculated by dividing the total work by B's efficiency:

Time_B = (30 R_B)/(R_B) = 30 days

Explanation

The area of the square is 81 cm², so the length of its side s is √(81) = 9 cm. The total length of the wire is equal to the perimeter of the square, calculated as 4s = 4 × 9 = 36 cm.

When the wire is rebent into a semicircular shape, the length of the wire corresponds to the perimeter of the semicircle, which consists of the curved arc π r and the straight diameter 2r. Setting the total length to 36 cm: r(π + 2) = 36 Substituting π = ²²⁄₇: r(²²⁄₇ + 2) = 36 r(³⁶⁄₇) = 36 r = 7 cm

The area of the semicircle is defined by the formula ½ π r². Substituting the radius: Area = ½ × ²²⁄₇ × 7 × 7 Area = 11 × 7 = 77 cm²

Explanation

The volume of water displaced in the cylindrical beaker is equal to the total volume of the spherical marbles submerged in it. Let n be the number of marbles dropped. The volume of the displaced water is calculated using the formula for the volume of a cylinder, V_c_yₗ = π R² h, where R is the radius of the beaker and h is the rise in the water level. The volume of a single spherical marble is Vₛₚₕ = ⁴⁄₃ π r³, where r is the radius of the marble.

The radius of the beaker R is 3.5 cm (half of 7 cm) and the rise in water level h is 5.6 cm. The radius of each marble r is 0.7 cm (half of 1.4 cm). Setting the total volume of n marbles equal to the volume of the displaced water:

n × ⁴⁄₃ π r³ = π R² h

n = (3 R² h)/(4 r³)

Substituting the given values:

n = (3 × (3.5)² × 5.6)/(4 × (0.7)³)

n = (3 × 12.25 × 5.6)/(4 × 0.343)

n = (205.8)/(1.372)

n = 150

Explanation

The volume of water required to fill the cistern is calculated using the formula for the volume of a cylinder, V = π R² H. Given the cistern diameter is 10m, the radius R is 5m, and the depth H is 2m. Thus, the total volume to be filled is V_cᵢₛₜₑᵣₙ = π × 5² × 2 = 50π m³.

The rate of water flow through the circular pipe depends on its cross-sectional area and the speed of the water. The internal diameter of the pipe is 20cm, which equals 0.2m, so the radius r is 0.1m. The speed of water is 3km/hr, which is 3000m/hr. The volume of water flowing out of the pipe per hour is V_fₗₒ_w = π r² × speed = π × (0.1)² × 3000 = π × 0.01 × 3000 = 30π m³/hr.

The time required to fill the cistern is the total volume divided by the flow rate: Time = (50π)/(30π) = ⁵⁄₃ hours. Converting this into hours and minutes, ⁵⁄₃ hours is equal to 1 hour and ⅔ of an hour. Since ⅔ × 60 = 40 minutes, the total time taken is 1 hour 40 minutes.

Explanation

The total distance covered by a wheel is given by the formula D = n × C, where n is the number of revolutions and C is the circumference of the wheel. The circumference is calculated as C = π d, where d represents the diameter.

First, convert the total distance from kilometers to centimeters to match the required unit: 11 km = 11 × 1000 × 100 = 1,100,000 cm.

Using the formula D = n × π d: 1,100,000 = 5000 × ²²⁄₇ × d

Divide both sides by 5000: 220 = ²²⁄₇ × d

Solving for d: d = (220 × 7)/(22) d = 10 × 7 d = 70 cm