SSC CGL Tier 1 2011 Quantitative Aptitude Questions with Solutions

Explanation

The relationship between the number of men and the time taken to complete a constant amount of work is governed by the formula M₁ D₁ = M₂ D₂, where M is the number of men and D is the number of days.

According to the problem: Initial number of men M₁ = x Initial number of days D₁ = 30 New number of men M₂ = x + 6 New number of days D₂ = 30 - 10 = 20

Substituting these values into the formula: x × 30 = (x + 6) × 20 30x = 20x + 120 30x - 20x = 120 10x = 120 x = 12

Explanation

Let the total work be the Least Common Multiple (LCM) of the given durations: 12, 15, and 20 days. LCM(12, 15, 20) = 60 units.

The daily efficiency (work done per day) for each pair is calculated as: Efficiency of (P + Q) = ⁶⁰⁄₁₂ = 5 units/day Efficiency of (Q + R) = ⁶⁰⁄₁₅ = 4 units/day Efficiency of (R + P) = ⁶⁰⁄₂₀ = 3 units/day

Summing these efficiencies gives: 2(P + Q + R) = 5 + 4 + 3 = 12 units/day P + Q + R = ¹²⁄₂ = 6 units/day

To find the individual efficiency of P, subtract the efficiency of the pair (Q + R) from the combined efficiency of (P + Q + R): Efficiency of P = (P + Q + R) - (Q + R) = 6 - 4 = 2 units/day

The time required for P to finish the work alone is: Time = (Total Work)/(Efficiency of P) = ⁶⁰⁄₂ = 30 days

Explanation

The efficiency ratio of A to B is 3:1. Since efficiency is inversely proportional to the time taken to complete a task, the ratio of time taken by A to B is 1:3. Let the time taken by A be x days and the time taken by B be 3x days.

The difference in time is given as 3x - x = 60 days, which simplifies to 2x = 60, or x = 30 days. This means A completes the work in 30 days and B completes it in 90 days.

To find the time taken when working together, we use the combined work formula T = (D1 × D2)/(D1 + D2), where D1 and D2 are the individual days taken by A and B respectively. Substituting the values:

T = (30 × 90)/(30 + 90) T = ²⁷⁰⁰⁄₁₂₀ T = ²⁷⁰⁄₁₂ T = ⁴⁵⁄₂ days.

Explanation

The area of the square is given as 121 cm². Since the area of a square is s², the side s of the square is √(121) = 11 cm. The total length of the copper wire is equal to the perimeter of this square, which is calculated as 4 × s = 4 × 11 = 44 cm.

When the same wire is bent into a circle, the circumference of the circle 2π r must equal the length of the wire. Setting the circumference to 44 cm and using the approximation π = ²²⁄₇, we get 2 × ²²⁄₇ × r = 44. Simplifying this equation, ⁴⁴⁄₇ × r = 44, which results in r = 7 cm.

Explanation

Let the dimensions of the cuboid be length l, width w, and height h. The areas of the three consecutive faces are given by the products of these dimensions: A₁ = l × w, A₂ = w × h, and A₃ = h × l. The volume V of a cuboid is calculated as V = l × w × h.

Multiplying the areas of the three faces results in: A₁ × A₂ × A₃ = (l × w) × (w × h) × (h × l) = l² × w² × h² = (l × w × h)²

This establishes the direct relationship V² = A₁ × A₂ × A₃, which means V = √(A₁ × A₂ × A₃). Substituting the given values: V = √(12 × 20 × 15) V = √(3600) V = 60

The volume of the cuboid is 60 cm³.

Explanation

The volume of water required to raise the level in the rectangular tank is given by V = length × width × height. Substituting the given dimensions, V = 50 × 44 × 0.07 = 154 m³. The pipe has a circular cross-section with a radius r = 7 cm = 0.07 m. The speed of water flow is 5 km/h, which is equivalent to 5000 m/h.

The volume of water flowing through the pipe in t hours is calculated as Area of cross-section × Speed × Time. This gives V = π r² × 5000 × t = ²²⁄₇ × 0.07 × 0.07 × 5000 × t. Simplifying this expression, we get 22 × 0.01 × 0.07 × 5000 × t = 77t m³. Equating the volume of water from the pipe to the required volume in the tank, we have 77t = 154, which yields t = 2 hours.

Explanation

The distance covered by a wheel in a single revolution is equal to its circumference, defined as π d, where d represents the diameter. For n revolutions, the total distance D is expressed by the formula D = n π d. Given that the car moves 440 meters in 1000 revolutions, we substitute these values into the equation using π ≈ ²²⁄₇:

440 = 1000 × ²²⁄₇ × d

Rearranging the formula to solve for the diameter d:

d = (440 × 7)/(1000 × 22)

Simplifying the calculation:

d = (20 × 7)/(1000) = ¹⁴⁰⁄₁₀₀₀ = 0.14 meters

Explanation

Let the lengths of the sides of the triangle be 2x, 3x, and 4x. Since the perimeter is the sum of all sides, we have 2x + 3x + 4x = 18, which simplifies to 9x = 18, giving x = 2. Therefore, the sides of the triangle are a = 4 cm, b = 6 cm, and c = 8 cm.

To find the area of the triangle, we use Heron's formula. First, calculate the semi-perimeter s: s = (a + b + c)/(2) = (4 + 6 + 8)/(2) = ¹⁸⁄₂ = 9 cm

The area A is given by: A = √(s(s - a)(s - b)(s - c)) A = √(9(9 - 4)(9 - 6)(9 - 8)) A = √(9 × 5 × 3 × 1) A = √(135) A = √(9 × 15) A = 3√(15) cm²

Explanation

The relationship between Cost Price (CP) and Marked Price (MP) can be directly expressed using the formula CP/MP = (100 - Discount%)/(100 + Profit%). Given that the Marked Price is Rs. 275, the discount is 5%, and the profit is 4.5%, we substitute these values into the ratio.

CP / 275 = (100 - 5) / (100 + 4.5) CP / 275 = 95 / 104.5

To find the CP, we rearrange the equation: CP = (95 / 104.5) × 275

Multiplying both the numerator and denominator of the fraction by 10 to remove the decimal gives: CP = (950 / 1045) × 275

Simplifying the calculation: CP = (190 / 209) × 275 CP = (190 × 275) / 209 CP = 52250 / 209 CP = 250

Explanation

The equivalent single discount for two successive discounts of a% and b% is calculated using the formula a + b - ᵃᵇ⁄₁₀₀. For successive discounts of 36% and 4%, the equivalent discount rate is 36 + 4 - (36 × 4)/(100) = 40 - 1.44 = 38.56%.

The difference between the single discount of 40% and the successive discounts is 40% - 38.56% = 1.44%. To find the monetary value of this difference on Rs. 500, calculate 1.44% of the total amount: Difference = (1.44)/(100) × 500 Difference = 1.44 × 5 = 7.20

Alternatively, when the sum of two successive discounts (36 + 4 = 40) equals the single discount (40), the difference is always ᵃᵇ⁄₁₀₀% of the principal. Thus, (36 × 4)/(100)% of 500 = 1.44% of 500 = 7.20.

Explanation

Given the ratio x:y = 4:5, we can substitute x = 4k and y = 5k into the expression (3x + y):(5x + 3y). Since the expression is a ratio of homogeneous linear polynomials, the constant k will cancel out, allowing us to directly substitute x = 4 and y = 5.

For the first part of the ratio: 3x + y = 3(4) + 5 = 12 + 5 = 17

For the second part of the ratio: 5x + 3y = 5(4) + 3(5) = 20 + 15 = 35

Combining these results, the ratio (3x + y):(5x + 3y) is 17:35.

Explanation

To keep the quantity of water constant across both states of the mixture, we normalize the ratios of Acid to Water. Initially, the ratio is 1:3. After adding acid, the ratio becomes 1:2. Multiplying the initial ratio by 2 and the new ratio by 3 gives: Initial Ratio = 2:6 New Ratio = 3:6

The water remains at 6 units in both cases. The increase in the acid component from 2 units to 3 units represents a difference of 1 unit. Since 5 litres of acid were added, 1 unit is equal to 5 litres.

The new mixture consists of 3 units of acid and 6 units of water, totaling 3 + 6 = 9 units. Total quantity = 9 × 5 = 45 litres.

Explanation

Let the two numbers be 2x and 3x. According to the problem, when each number is increased by 4, the new ratio becomes 5:7. This can be expressed as the equation:

(2x + 4)/(3x + 4) = ⁵⁄₇

To solve for x, cross-multiply the terms:

7(2x + 4) = 5(3x + 4)

14x + 28 = 15x + 20

Subtracting 14x from both sides and subtracting 20 from both sides gives:

x = 8

The two numbers are 2x = 2 × 8 = 16 and 3x = 3 × 8 = 24. The difference between the numbers is 3x - 2x = x. Since x = 8, the difference is 8.

Explanation

The savings of an individual is calculated as Income - Expenditure = Savings. Let the monthly incomes of A and B be 4x and 3x, and their monthly expenses be 3y and 2y respectively. Given that both save Rs. 6,000, we can use the ratio difference method.

In the income ratio 4:3 and the expense ratio 3:2, the numerical difference between the income and expense parts for A is 4 - 3 = 1 unit, and for B is 3 - 2 = 1 unit. Since both individuals save the same amount, this common difference of 1 unit corresponds directly to the savings of Rs. 6,000.

1 unit = 6,000

The income of A is represented by 4 units in the income ratio. Therefore, the monthly income of A is calculated as: 4 × 6,000 = 24,000

Explanation

Let the three consecutive odd numbers be x, x+2, and x+4. The average of any three consecutive numbers in an arithmetic progression is simply the middle term. Therefore, the average is x+2.

According to the given condition, this average is 12 more than one third of the first number x. This relationship can be expressed by the following equation: x + 2 = ⅓x + 12

To solve for x, rearrange the terms to group the variables on one side: x - ⅓x = 12 - 2 ⅔x = 10 x = (10 × 3)/(2) x = 15

Since x = 15, the three consecutive odd numbers are 15, 17, and 19. The question asks for the last of the three numbers, which is x + 4 = 15 + 4 = 19.

Explanation

The correct average is determined by calculating the net adjustment required for the total sum. The sum of the values correctly intended for entry is 64 + 28 = 92, while the sum of the values actually entered incorrectly is 46 + 82 = 128. The net change in the total sum is 92 - 128 = -36. To find the correction needed for the average, this net change is divided by the total number of observations: ⁻³⁶⁄₁₈ = -2. Applying this correction to the initially recorded average, the correct average is 124 - 2 = 122.

Explanation

Let the cost price of one article be C and the selling price of one article be S. Based on the problem, the cost price of 15 articles is equal to the selling price of 12 articles, which gives the equation 15C = 12S. By simplifying this, we find the selling price of one article relative to its cost price: S = ¹⁵⁄₁₂C = ⁵⁄₄C = 1.25C.

The gain is the difference between the selling price and the cost price: 1.25C - C = 0.25C. To find the gain percentage, we use the formula Gain % = (Gain)/(C) × 100. Substituting the values gives (0.25C)/(C) × 100 = 25%.

This can also be solved using the shortcut formula for situations where the cost price of x items equals the selling price of y items: Gain % = (x - y)/(y) × 100. Plugging in x = 15 and y = 12, we get (15 - 12)/(12) × 100 = ³⁄₁₂ × 100 = ¼ × 100 = 25%.

Explanation

Let the marked price (MP) of the article be 100. The cost price (CP) is given as 64% of the marked price, so CP = 64.

A discount of 12% is allowed on the marked price to determine the selling price (SP). SP = MP - Discount = 100 - 12 = 88

The profit or gain is the difference between the selling price and the cost price: Gain = SP - CP = 88 - 64 = 24

The gain percentage is calculated on the cost price using the formula: Gain % = (Gain)/(CP) × 100 Gain % = ²⁴⁄₆₄ × 100

Simplifying the fraction ²⁴⁄₆₄ by dividing both the numerator and denominator by 8 gives ⅜. Gain % = ⅜ × 100 = 37.5%

Explanation

The cost price (CP) of 3 eggs is Rs. 5, so the CP of 1 egg is Rs. ⁵⁄₃. The selling price (SP) of 5 eggs is Rs. 12, so the SP of 1 egg is Rs. ¹²⁄₅.

To simplify the calculation, we consider the Least Common Multiple (LCM) of the quantities 3 and 5, which is 15. CP of 15 eggs = 15 × ⁵⁄₃ = 25 Rs. SP of 15 eggs = 15 × ¹²⁄₅ = 36 Rs. Gain on 15 eggs = 36 - 25 = 11 Rs.

Since a gain of Rs. 11 is realized on 15 eggs, the number of eggs required to realize a total gain of Rs. 143 is determined by the ratio of the total gain to the gain per unit of 15 eggs: Number of eggs = ¹⁴³⁄₁₁ × 15 = 13 × 15 = 195.

Explanation

The selling price of each article is 4,000, making the total selling price 8,000. For the first article sold at a 25% profit, the cost price is 4,000 / 1.25 = 3,200, resulting in an 800 profit. To achieve no net gain or loss, the second article must have a loss of 800, meaning its cost price was 4,000 + 800 = 4,800, and the loss percentage is (800 / 4,800) * 100 = 50/3%.

Explanation

The cost price CP is related to the selling price SP and loss L by the formula CP = SP + L. Given that the loss is 20% of the selling price, we define L = 0.20 × SP. Substituting this into the cost price equation gives CP = SP + 0.20 × SP = 1.20 × SP. The loss percentage on the cost price is calculated using the formula:

Loss % on CP = (L)/(CP) × 100

By substituting the known expressions for L and CP:

Loss % on CP = (0.20 × SP)/(1.20 × SP) × 100

Loss % on CP = (0.20)/(1.20) × 100 = ⅙ × 100 = ⁵⁰⁄₃%

Explanation

The problem provides the relationship ⁹⁰⁄₁₀₀ × A = ³⁰⁄₁₀₀ × B. This simplifies to 90A = 30B, which further reduces to B = 3A. The second condition states that B = x% of A, which is expressed mathematically as B = ˣ⁄₁₀₀ × A. By equating these two expressions for B, we obtain the equation 3A = ˣ⁄₁₀₀ × A. Dividing both sides by A gives 3 = ˣ⁄₁₀₀. Multiplying both sides by 100 yields x = 300.

Explanation

Expenditure is defined as the product of price (P) and consumption (C). When expenditure is kept constant, price and consumption are inversely proportional, represented by the relation P₁ C₁ = P₂ C₂. To find the percentage reduction in consumption when the price increases, the formula is: Percentage Reduction = (P₂ - P₁)/(P₂) × 100 Given the initial price P₁ = 6 and the new price P₂ = 7.50, the increase in price is 7.50 - 6 = 1.50. Substituting these values into the equation: Percentage Reduction = (1.50)/(7.50) × 100 Percentage Reduction = ¹⁵⁄₇₅ × 100 = ⅕ × 100 = 20%

Explanation

Let the third number be 100. Based on the given conditions, the first number is 30% less than the third, making it 100 - 30 = 70. Similarly, the second number is 37% less than the third, making it 100 - 37 = 63.

To find the percentage by which the second number is less than the first number, calculate the difference between the two and divide by the first number. The difference is 70 - 63 = 7. The required percentage is calculated as:

Percentage = (Difference)/(First Number) × 100

Percentage = ⁷⁄₇₀ × 100 = ⅒ × 100 = 10%

Explanation

For a constant distance, speed is inversely proportional to time. If the man walks at ⁶⁄₇ of his usual speed, the time taken will be the reciprocal of the speed ratio, which is ⁷⁄₆ of his usual time.

Let his usual time be T. The new time is ⁷⁄₆T. The difference between the new time and the usual time is given as 25 minutes.

⁷⁄₆T - T = 25

⅙T = 25

T = 25 × 6 = 150 minutes

To convert minutes into hours, divide by 60: 150 minutes = ¹⁵⁰⁄₆₀ hours = 2.5 hours

2.5 hours is equivalent to 2 hours and 30 minutes.

Explanation

To find the distance between the school and the house, we use the formula for distance when two different speeds result in two different arrival times relative to a fixed schedule:

Distance = (S₁ × S₂)/(|S₁ - S₂|) × Δ T

Here, the speeds are S₁ = 5 km/hr and S₂ = 3 km/hr. The student is 15 minutes early in the first case and 9 minutes late in the second. The total time difference (Δ T) between these two scenarios is the sum of the early and late durations:

Δ T = 15 + 9 = 24 minutes

Since speeds are in km/hr, we convert the time difference into hours:

Δ T = ²⁴⁄₆₀ = ⅖ hours

Substituting these values into the formula:

Distance = (5 × 3)/(5 - 3) × ⅖

Distance = ¹⁵⁄₂ × ⅖

Distance = ³⁰⁄₁₀ = 3 km

Explanation

Under compound interest, if a principal becomes n times its value in t years, it will become nᵏ times its value in k × t years. In this problem, the sum doubles (n = 2) in t = 5 years. The total duration given is 20 years.

To find the number of doubling periods (k), we divide the total time by the time taken to double: k = ²⁰⁄₅ = 4

The total growth factor after 20 years is 2⁴ = 16. The final amount is calculated by multiplying the initial principal by this growth factor: Amount = 12,000 × 16 = 1,92,000

This follows the standard compound interest logic where the amount increases geometrically over equal intervals of time. Since there are four 5-year intervals in 20 years, the money doubles four times sequentially.

Explanation

The simple interest formula is SI = (P × R × T)/(100), where P is the principal sum, R is the rate of interest per annum, and T is the time in years. According to the problem, the simple interest for 6 years is ⁹⁄₂₅ of the principal sum.

Substituting the given values into the formula: ⁹⁄₂₅ P = (P × R × 6)/(100)

Dividing both sides by P and rearranging the equation to solve for R: ⁹⁄₂₅ = (6R)/(100) R = (9 × 100)/(25 × 6) R = (9 × 4)/(6) R = ³⁶⁄₆ R = 6%

Explanation

For a two-year period, the difference D between compound interest and simple interest is determined by the formula D = P(ʳ⁄₁₀₀)², where P represents the principal and r represents the annual interest rate.

Substitute the given principal P = 5000 and the difference D = 32 into the formula: 32 = 5000 × (ʳ⁄₁₀₀)² 32 = 5000 × (r²)/(10000) 32 = (r²)/(2) r² = 64 r = √(64) = 8

Explanation

The greatest 4-digit number is 9999. To find the least number that must be added to it to make the sum divisible by 345, we first divide 9999 by 345 to find the remainder.

Using the division algorithm: 9999 = (345 × 28) + 339

The remainder is 339. The least number to be added to the dividend to make it perfectly divisible by the divisor is calculated as: Required Number = Divisor - Remainder Required Number = 345 - 339 = 6

When 6 is added to 9999, the sum is 10005, which is exactly divisible by 345 (345 × 29 = 10005).

Explanation

To find the remainder when 17²⁰⁰ is divided by 18, we apply the property of modular arithmetic. We observe that 17 can be expressed as 18 - 1, which means 17 ≡ -1 18. According to the power rule in modular arithmetic, aⁿ m ≡ (a m)ⁿ m. Therefore, 17²⁰⁰ 18 ≡ (-1)²⁰⁰ 18. Since the exponent 200 is an even number, (-1)²⁰⁰ simplifies to 1. Consequently, the remainder is 1.

Explanation

The unit digit of a number raised to any power depends only on the unit digit of the base. For the base 124, the unit digit is 4. The powers of 4 follow a cycle of 2 regarding their unit digits: 4¹ = 4, 4² = 16, 4³ = 64, 4⁴ = 256, and so on. This pattern shows that 4ⁿ results in a unit digit of 6 when n is even and 4 when n is odd.

In the first term (124)³⁷², the exponent 372 is an even number, so the unit digit is 6. In the second term (124)³⁷³, the exponent 373 is an odd number, so the unit digit is 4. Adding the unit digits of both terms gives 6 + 4 = 10. The unit digit of this sum is 0.

Explanation

The operation a^* b is defined by the rule aᵇ, where the first operand is the base and the second operand is the exponent. Given the values a = 5 and b = 3, we substitute them into the formula to get 5³. Calculating the power, 5 × 5 × 5 = 125.

Explanation

From the given equation 1.5x = 0.04y, we find the ratio of y to x: ʸ⁄ₓ = (1.5)/(0.04) = ¹⁵⁰⁄₄ = ⁷⁵⁄₂

The expression to be evaluated is (y² - x²)/(y² - 2xy + x²). By applying the algebraic identities a² - b² = (a - b)(a + b) and (a - b)² = a² - 2ab + b², the expression simplifies as follows: (y² - x²)/(y² - 2xy + x²) = ((y - x)(y + x))/((y - x)²) = (y + x)/(y - x)

Dividing both the numerator and the denominator by x allows for the direct substitution of the previously derived ratio: (y + x)/(y - x) = (y)/(x) + 1{ʸ⁄ₓ - 1}

Substituting ʸ⁄ₓ = ⁷⁵⁄₂ into the equation: (75)/(2) + 1{⁷⁵⁄₂ - 1} = (75 + 2)/(2){(75 - 2)/(2)} = (77/2)/(73/2) = ⁷⁷⁄₇₃

Explanation

To find the value of X = 0.3 × 0.3, multiply the numbers as if they were whole numbers and then account for the decimal places. Since 3 × 3 = 9 and each factor has one decimal place, the product must have a total of two decimal places. Moving the decimal point two places to the left from the digit 9 results in 0.09. Alternatively, converting the decimals to fractions gives ³⁄₁₀ × ³⁄₁₀ = ⁹⁄₁₀₀, which equals 0.09.

Explanation

To find the remainder of 3²¹ divided by 5, we apply Fermat's Little Theorem, which states that if p is a prime number, then aᵖ⁻¹ ≡ 1 p for any integer a not divisible by p. Here, a = 3 and p = 5.

According to the theorem, 3⁵⁻¹ = 3⁴ ≡ 1 5. We can break down the exponent 21 based on this periodicity: 21 = (4 × 5) + 1. Using the properties of indices and modular arithmetic:

3²¹ = (3⁴)⁵ × 3¹

Substituting the congruence 3⁴ ≡ 1 5:

3²¹ ≡ (1)⁵ × 3¹ 5

3²¹ ≡ 1 × 3 ≡ 3 5

Explanation

To find the value of the expression, the numerator and the denominator are simplified separately.

The numerator is 2⅓ - 1²⁄₁₁. Converting these mixed fractions to improper fractions results in ⁷⁄₃ - ¹³⁄₁₁. By finding a common denominator of 33, the expression becomes (77 - 39)/(33) = ³⁸⁄₃₃.

The denominator is a continued fraction: 3 + (1)/(3 + 1){3 + ⅓}. Solving from the bottom up, the lowest term is 3 + ⅓ = ¹⁰⁄₃. Substituting this back gives 3 + (1)/(3 + 1){10/3} = 3 + (1)/(3 + 3){10}. Simplifying the next layer yields 3 + ³⁄₁₀ = ³³⁄₁₀. Substituting this back gives 3 + (1)/(33/10) = 3 + ¹⁰⁄₃₃. The final value of the denominator is (99 + 10)/(33) = ¹⁰⁹⁄₃₃.

By dividing the simplified numerator by the simplified denominator, the final value is calculated as (38/33)/(109/33) = ³⁸⁄₃₃ × ³³⁄₁₀₉ = ³⁸⁄₁₀₉.

Explanation

To find the last digit of the expression (1001)²⁰⁰⁸ + 1002, we focus solely on the units digit of each component. The last digit of any power aⁿ depends only on the last digit of the base a. For the term (1001)²⁰⁰⁸, the base 1001 ends in 1. Since 1 raised to any positive integer power always results in a number ending in 1 (1ⁿ = 1), the last digit of (1001)²⁰⁰⁸ is 1. The second term in the expression is 1002, which has a units digit of 2. By adding the units digits of both terms, we get 1 + 2 = 3.

Explanation

To solve this expression, we rationalize the denominator of each term individually by multiplying the numerator and denominator by the conjugate of the denominator.

For the first term, we multiply by √6 - √3: 3√2{√6 + √3} × √6 - √3{√6 - √3} = 3√(12) - 3√6{6 - 3} = 3(2√3 - √6){3} = 2√3 - √6

For the second term, we multiply by √6 - √2: 4√3{√6 + √2} × √6 - √2{√6 - √2} = 4√(18) - 4√6{6 - 2} = 4(3√2 - √6){4} = 3√2 - √6

For the third term, we multiply by √3 - √2: √6{√3 + √2} × √3 - √2{√3 - √2} = √(18) - √(12){3 - 2} = 3√2 - 2√3{1} = 3√2 - 2√3

Now, substituting these simplified values back into the original expression: (2√3 - √6) - (3√2 - √6) + (3√2 - 2√3) 2√3 - √6 - 3√2 + √6 + 3√2 - 2√3

Grouping the like terms together: (2√3 - 2√3) + (-√6 + √6) + (-3√2 + 3√2) = 0 + 0 + 0 = 0

Explanation

The operation x^*y is defined as (x + 3)²(y - 1). By substituting x = 5 and y = 4 into this formula, we calculate the result as follows:

5^*4 = (5 + 3)²(4 - 1) 5^*4 = 8² × 3 5^*4 = 64 × 3 5^*4 = 192

The calculation follows the order of operations by first solving the expressions within the parentheses, applying the exponent to the first term, and then performing the multiplication.

Explanation

Each term in the numerator is exactly ten times the corresponding term in the denominator. Specifically, 0.05 = 10 × 0.005, 0.41 = 10 × 0.41, and 0.073 = 10 × 0.0073. By substituting these relationships into the expression, we let a = 0.005, b = 0.041, and c = 0.0073. The expression becomes:

(10a)² + (10b)² + (10c)²{a² + b² + c²}

Applying the exponent to the product inside the parentheses, (10k)² = 100k², the numerator transforms into 100a² + 100b² + 100c². Factoring out the common multiplier of 100 from the numerator gives:

100(a² + b² + c²){a² + b² + c²}

The term (a² + b² + c²) in the numerator and denominator cancels out completely, leaving the constant value of 100.

Explanation

To determine the value of x, we compare the algebraic terms on both sides of the equation. The given equation is: 9 - ʸ⁄ₓ = -ʸ⁄₁₂ + constant

For the expression to be consistent across the terms involving y, the coefficient of y on the left-hand side must be reproducible on the right-hand side using the common denominator of 12. Using the value x = 3 from the options, the term ʸ⁄ₓ becomes: ʸ⁄₃ = ⁴ʸ⁄₁₂

Substituting this into the left-hand side of the equation, we have: 9 - ⁴ʸ⁄₁₂

We can decompose the term -⁴ʸ⁄₁₂ to match the structure of the right-hand side: 9 - ( ʸ⁄₁₂ + ³ʸ⁄₁₂ ) = -ʸ⁄₁₂ + 9 - ³ʸ⁄₁₂

By comparing the resulting expression -ʸ⁄₁₂ + (9 - ʸ⁄₄) with the right-hand side of the given equation -ʸ⁄₁₂ + constant, we see that the term ʸ⁄ₓ corresponds to a coefficient that is a multiple of ¹⁄₁₂. Specifically: ¹⁄ₓ = ⁴⁄₁₂ ¹⁄ₓ = ⅓ x = 3

Explanation

Let the two numbers be x and y. According to the given conditions, the product of their sum and the first number is x(x + y) = 247, and the product of their sum and the second number is y(x + y) = 114.

By adding these two equations together, we get: x(x + y) + y(x + y) = 247 + 114 (x + y)(x + y) = 361 (x + y)² = 361

Taking the square root of both sides gives the sum of the numbers: x + y = √(361) = 19

Explanation

Let the required number be x. According to the problem, one-seventh of the number exceeds its eleventh part by 100. This relationship can be expressed by the linear equation:

ˣ⁄₇ - ˣ⁄₁₁ = 100

To solve for x, find a common denominator for the fractions, which is the product of 7 and 11, namely 77. Rewrite the terms using this common denominator:

(11x - 7x)/(77) = 100

Simplify the numerator:

⁴ˣ⁄₇₇ = 100

Isolate x by multiplying both sides by 77 and then dividing by 4:

4x = 100 × 77

4x = 7700

x = ⁷⁷⁰⁰⁄₄ = 1925

Explanation

The Highest Common Factor (HCF) of any set of numbers must always be a divisor of their Least Common Multiple (LCM). This relationship exists because the LCM is composed of all the prime factors of the numbers involved, including those that make up the HCF. Consequently, the division of the LCM by the HCF must result in a whole number. Given that the LCM is 120, 35 cannot be the HCF because 120 is not exactly divisible by 35. The calculation 120 ÷ 35 results in a remainder of 15, which violates the fundamental property that the HCF must be a factor of the LCM.

Explanation

Let the number be N. According to the dividend-divisor relationship, N = 49k + 32, where k is an integer quotient. To find the remainder when N is divided by 7, we calculate N 7. Since the first divisor 49 is a multiple of the second divisor 7, the term 49k is divisible by 7 without any remainder. Therefore, the required remainder is found by dividing the original remainder 32 by 7. Expressing 32 in terms of 7, we get 32 = 7 × 4 + 4. The remainder is 4.

Explanation

Let x be the number of correct answers. Since the student attempts all 75 questions, the number of wrong answers is 75 - x. Based on the marking scheme of 4 marks for a correct answer and a penalty of 1 mark for a wrong answer, the total marks can be represented by the equation:

4x - 1(75 - x) = 125

Solving the equation: 4x - 75 + x = 125 5x = 125 + 75 5x = 200 x = 40

Alternatively, this can be solved by considering the total potential marks. If all 75 questions were answered correctly, the student would have earned 75 × 4 = 300 marks. The actual score of 125 marks indicates a total loss of 300 - 125 = 175 marks. Every incorrect answer costs the student 5 marks (4 marks for the missed correct answer plus 1 mark for the penalty). Dividing the total loss by the loss per incorrect answer gives 175 / 5 = 35 wrong answers. Thus, the number of correct answers is 75 - 35 = 40.

Explanation

To determine the next time the traffic lights will change simultaneously, we find the Least Common Multiple (LCM) of the given intervals: 24, 36, and 54 seconds. The prime factorization of these numbers is: 24 = 2³ × 3 36 = 2² × 3² 54 = 2 × 3³

The LCM is calculated by taking the highest power of each prime factor present in the numbers: LCM = 2³ × 3³ = 8 × 27 = 216 seconds

Converting 216 seconds into minutes and seconds: 216 / 60 = 3 minutes and 36 seconds

Adding this duration to the initial simultaneous change time of 10:15:00 AM: 10:15:00 AM + 3 minutes 36 seconds = 10:18:36 AM

Explanation

A's daily work rate is 1/12 because he can complete the entire task in 12 days. In the first 3 days, A completes 3 × (1/12) = 1/4 of the work. The remaining work to be completed is 1 - 1/4 = 3/4.

A and B work together to finish this remaining 3/4 of the work in 3 days. Therefore, their combined daily work rate is (3/4) / 3 = 1/4. If we denote B's daily work rate as 1/B, the sum of their individual rates equals their combined rate: 1/12 + 1/B = 1/4.

To find B's rate, subtract A's rate from the combined rate: 1/B = 1/4 - 1/12. Using a common denominator of 12, this becomes 1/B = 3/12 - 1/12 = 2/12 = 1/6. Since B's work rate is 1/6 per day, B alone can finish the work in 6 days.

Explanation

The combined efficiency of two people working together is the reciprocal of the time taken to complete the work. Let the daily work rates of A, B, and C be denoted by their respective letters. According to the problem: A + B = ⅛ B + C = ¹⁄₁₂ C + A = ⅛

Adding these equations yields 2(A + B + C) = ⅛ + ¹⁄₁₂ + ⅛. To sum these fractions, use the common denominator of 24: 2(A + B + C) = (3 + 2 + 3)/(24) = ⁸⁄₂₄ = ⅓

Dividing by 2 gives the joint efficiency of all three individuals: A + B + C = ⅙

Since the time required to complete the work is the reciprocal of the total efficiency, A, B, and C working together will complete the task in 6 days.