SSC CGL Tier 1 2017 Quantitative Aptitude Questions with Solutions

Explanation

A number is divisible by 80 if it is divisible by both 10 and 8. For the number 347XY to be divisible by 10, the last digit Y must be 0. With Y = 0, the number becomes 347X0.

For 347X0 to be divisible by 80, the expression 347X0 / 80 must result in an integer. This is equivalent to 347X being divisible by 8. According to the divisibility rule for 8, the last three digits of the number, which are 47X, must be divisible by 8.

We can determine X by dividing 470 by 8 and looking at the remainder. 470 = (8 × 58) + 6. To make the number 47X divisible by 8, we need to add a value to the remainder 6 to reach the next multiple of 8, which is 64. Specifically, since 472 = 8 × 59, the digit X must be 2. If we check the next possibility, 472 + 8 = 480, which would change the hundreds digit. Thus, X = 2 is the only single-digit solution.

With X = 2 and Y = 0, the value of X + Y = 2 + 0 = 2.

Explanation

Let the total work be the Least Common Multiple (LCM) of the individual times taken by A, B, and C (20, 24, and 30), which is 120 units. The daily work efficiencies of A, B, and C are calculated as 120/20 = 6 units/day, 120/24 = 5 units/day, and 120/30 = 4 units/day respectively.

Let the total duration of the work be T days. According to the problem, A worked for 4 days, C worked for the entire duration of T days, and B worked for T - 6 days because B left 6 days before completion. The sum of the work done by each individual must equal the total work of 120 units:

Work done by A + Work done by B + Work done by C = 120 (6 × 4) + 5(T - 6) + 4T = 120 24 + 5T - 30 + 4T = 120 9T - 6 = 120 9T = 126 T = 14

Alternatively, we can adjust the total work based on the departures. Starting with 120 units, we subtract the 4 days of work completed by A (6 × 4 = 24 units), leaving 96 units. Since B left 6 days early, we can imagine B stayed until the end by adding the work B would have done in those 6 days (5 × 6 = 30 units) to the remaining total. The adjusted work for B and C is 96 + 30 = 126 units. The total time T is then this adjusted work divided by the combined efficiency of B and C: 126 / (5 + 4) = 14 days.

Explanation

When a solid sphere is cut into two equal halves (hemispheres), two new circular cross-sections are created at the base of each hemisphere. The increase in the total surface area is equal to the sum of the areas of these two new circular faces. Since the area of one circular face is π r², the total increase is 2π r².

Given the diameter d = 17.5 cm, the radius r = (17.5)/(2) cm. Substituting these values into the formula for the increase:

Increase in surface area = 2 × π × r² Increase = 2 × ²²⁄₇ × ((17.5)/(2)) × ((17.5)/(2)) Increase = ¹¹⁄₇ × 17.5 × 8.75 Increase = 11 × 2.5 × 8.75 Increase = 27.5 × 8.75 = 481.25

The resulting increase in the total surface area is 481.25 cm².

Explanation

The relationship between the Marked Price (MP) and the Selling Price (SP) after a discount is given by the formula SP = MP × (1 - (Discount %)/(100)).

Given that the discount is 34% and the article is sold for Rs 3168, we can substitute the values into the formula: 3168 = MP × (1 - ³⁴⁄₁₀₀) 3168 = MP × ⁶⁶⁄₁₀₀

To find the Marked Price, rearrange the equation: MP = (3168 × 100)/(66)

Dividing 3168 by 66: 3168 ÷ 6 = 528 66 ÷ 6 = 11 528 ÷ 11 = 48

Therefore: MP = 48 × 100 = 4800

By using the percentage method, since the discount is 34%, the Selling Price represents 66% (100% - 34%) of the Marked Price. If 66% corresponds to Rs 3168, then 1% corresponds to 3168 / 66 = 48. Thus, the total 100% Marked Price is 48 × 100 = 4800.

Explanation

To find the ratio of P and Q, we start with the given equation: ³⁄₇P = ⁴⁄₁₁Q We isolate the ratio (P)/(Q) by moving Q to the denominator of the left side and moving the coefficient of P to the right side using its reciprocal: (P)/(Q) = ⁴⁄₁₁ ÷ ³⁄₇ In fraction division, we multiply the first fraction by the reciprocal of the second: (P)/(Q) = ⁴⁄₁₁ × ⁷⁄₃ Multiplying the numerators together and the denominators together gives: (P)/(Q) = ²⁸⁄₃₃ This results in the ratio P : Q = 28 : 33.

Explanation

The sum of all 17 results is calculated as 17 × 60 = 1020. The sum of the first 9 results is 9 × 57 = 513, and the sum of the last 9 results is 9 × 65 = 585. When these two subsets are added together, the 9th result is counted twice. Therefore, the value of the 9th result is the sum of these two subsets minus the total sum of all 17 results: (513 + 585) - 1020 = 1098 - 1020 = 78. Alternatively, using the deviation method, the first 9 results deviate from the mean by -3 each (9 × -3 = -27) and the last 9 results deviate by +5 each (9 × 5 = +45). The 9th result is the mean plus the net deviation: 60 + (-27 + 45) = 60 + 18 = 78.

Explanation

Let the initial cost price be CP = 100. Since the profit is 170% of the cost price, the profit is 170. The selling price is calculated as SP = CP + Profit, which gives SP = 100 + 170 = 270.

When the cost price increases by 20%, the new cost price becomes CPₙₑ_w = 100 + 20 = 120. The selling price remains constant at 270. The new profit is determined by the difference between the selling price and the new cost price: New Profit = 270 - 120 = 150.

The new profit percentage is calculated on the new cost price using the formula: New Profit Percentage = (New Profit)/(CPₙₑ_w₎ × 100 New Profit Percentage = ¹⁵⁰⁄₁₂₀ × 100 New Profit Percentage = ⁵⁄₄ × 100 = 125%

Explanation

Let the unknown number be denoted as x. The problem states that the difference between 32% of this number and 17% of the same number is 120. Since both percentages are calculated on the same base value x, we can directly subtract the percentages to find the relative difference.

Percentage difference = 32% - 17% = 15%

This 15% of the number is equivalent to the absolute value 120. Using the basic percentage formula:

¹⁵⁄₁₀₀ × x = 120

To solve for x, rearrange the equation:

x = (120 × 100)/(15)

Dividing 120 by 15 gives 8. Therefore:

x = 8 × 100 = 800

Explanation

Let the speed of the boat in still water be u km/hr and the speed of the stream be v km/hr. The upstream speed is x = u - v and the downstream speed is y = u + v.

The first condition gives: 15/x + 10.5/y = 3.25 Multiplying by 4 to clear decimals: 60/x + 42/y = 13 --- (i)

The second condition gives: 12/x + 14/y = 3 --- (ii)

To eliminate y, multiply equation (ii) by 3: 36/x + 42/y = 9 --- (iii)

Subtracting equation (iii) from equation (i): 24/x = 4 ⟹ x = 6

Substituting x = 6 into equation (ii): 12/6 + 14/y = 3 ⟹ 2 + 14/y = 3 ⟹ 14/y = 1 ⟹ y = 14

The speed of the boat in still water is the average of the downstream and upstream speeds: u = (x + y)/2 = (6 + 14)/2 = 20/2 = 10 km/hr

Explanation

Let the sum lent be P. The simple interest is calculated using the formula SI = (P × R × T)/(100), where R is the rate of interest per annum and T is the time in years. Substituting the given values R = 5 and T = 15, the interest is SI = (P × 5 × 15)/(100) = (75P)/(100) = 0.75P.

According to the question, the interest is Rs 250 less than the sum lent, which can be expressed as SI = P - 250. Equating the two expressions for SI gives: 0.75P = P - 250 P - 0.75P = 250 0.25P = 250 P = (250)/(0.25) = 1000

Explanation

Given x = 2 + √3, first calculate 2x = 2(2 + √3) = 4 + 2√3. To find √(2x), express 4 + 2√3 as a perfect square using the identity (a + b)² = a² + b² + 2ab. By observation, 4 + 2√3 = 3 + 1 + 2√3 = (√3)² + 1² + 2(√3)(1) = (√3 + 1)². Therefore, √(2x) = √3 + 1.

Next, calculate the reciprocal (1)/(√(2x)) by rationalizing the denominator: (1)/(√(3) + 1) = (1)/(√(3) + 1) × √3 - 1{√3 - 1} = √3 - 1{3 - 1} = √3 - 1{2}.

Now, find the sum of the two terms: √(2x) + (1)/(√(2x)) = (√3 + 1) + √3 - 1{2} = 2(√3 + 1) + (√3 - 1){2} = 2√3 + 2 + √3 - 1{2} = 3√3 + 1{2}.

Explanation

Given x + ¹⁄ₓ = k, we use the algebraic relation x³ + (1)/(x³) = k³ - 3k. Substituting k = 4: x³ + (1)/(x³) = 4³ - 3(4) = 64 - 12 = 52

To reach the sixth power, we apply the squaring identity a² + (1)/(a²) = (a + ¹⁄ₐ)² - 2. Letting a = x³, we substitute the value found above: x⁶ + (1)/(x⁶) = (x³ + (1)/(x³))² - 2 x⁶ + (1)/(x⁶) = 52² - 2 = 2704 - 2 = 2702

This can also be verified by first squaring x + ¹⁄ₓ to get x² + (1)/(x²) = 4² - 2 = 14, and then cubing that result: 14³ - 3(14) = 2744 - 42 = 2702.

Explanation

The given expression can be simplified by identifying a common denominator for the two terms. Since y² - 1 = (y + 1)(y - 1), the expression is rewritten as: (1)/(y+1) + (2y+1)/((y+1)(y-1)) = ((y-1) + (2y+1))/((y+1)(y-1)) = (3y)/(y²-1)

To evaluate this in terms of x, substitute y = ²⁻ˣ⁄₁₊ₓ into the simplified form. The numerator 3y becomes: 3y = ³⁽²⁻ˣ⁾⁄₁₊ₓ

The denominator y²-1 is calculated by substituting y: y²-1 = (²⁻ˣ⁄₁₊ₓ)² - 1 = ((2-x)² - (1+x)²)/((1+x)²) Expanding the squares in the numerator gives (4 - 4x + x²) - (1 + 2x + x²) = 3 - 6x, which simplifies to 3(1-2x). Thus, the denominator is: y²-1 = (3(1-2x))/((1+x)²)

Combining the numerator and denominator: ³⁽²⁻ˣ⁾⁄₁₊ₓ ÷ (3(1-2x))/((1+x)²) = ³⁽²⁻ˣ⁾⁄₁₊ₓ · ((1+x)²)/(3(1-2x)) After canceling the common factor of 3 and one factor of (1+x), the resulting expression is: ⁽²⁻ˣ⁾⁽¹⁺ˣ⁾⁄₁₋₂ₓ

This result matches the numerator and denominator pair provided in option c.

Explanation

The point P is the incenter of the triangle ABC because it is the intersection of the internal angle bisectors of ∠ ABC and ∠ ACB. In any triangle, the angle formed at the incenter by the bisectors of two angles is given by the formula ∠ BPC = 90° + ½ ∠ BAC.

Given that ∠ BAC = 50°, we substitute this value into the formula: ∠ BPC = 90° + ½ (50°) ∠ BPC = 90° + 25° ∠ BPC = 115°

Alternatively, in triangle ABC, the sum of angles is 180°, so ∠ ABC + ∠ ACB = 180° - 50° = 130°. Since BP and CP are bisectors, ∠ PBC + ∠ PCB = ½ (∠ ABC + ∠ ACB) = ½ (130°) = 65°. In triangle BPC, the sum of angles must also be 180°, therefore ∠ BPC = 180° - (∠ PBC + ∠ PCB) = 180° - 65° = 115°.

Explanation

In two intersecting circles of equal radius r, the line segment connecting the centers C₁C₂ and the common chord PQ are perpendicular bisectors of each other. Let M be their point of intersection. The distance from a center to this intersection point is C₁M = C₁C₂/2 = ⁴⁰⁄₂ = 20 cm. The distance from the intersection point to an endpoint of the chord is PM = (PQ)/(2) = ³⁰⁄₂ = 15 cm.

The radius r forms the hypotenuse of the right-angled triangle △ C₁MP. According to the Pythagorean theorem, r² = C₁M² + PM². Substituting the values, r² = 20² + 15², which gives r² = 400 + 225 = 625. Taking the square root of both sides, r = √(625) = 25 cm. This follows the 3:4:5 Pythagorean triplet ratio scaled by a factor of 5, where 15:20:25 identifies the radius directly.

Explanation

Since ST is the perpendicular bisector of side PQ, every point on ST is equidistant from the endpoints P and Q. Point T lies on this bisector, so PT = QT. In △ PTQ, since PT = QT, the base angles are equal, meaning ∠ TPQ = ∠ TQP.

Let ∠ PQR (which is the same as ∠ TQP) be x. Then ∠ TPQ is also x. The total angle at vertex P in △ PQR is the sum of ∠ TPR and ∠ TPQ, so ∠ P = 46° + x.

In △ PQR, the sum of the interior angles is 180°. Substituting the known values: ∠ P + ∠ Q + ∠ R = 180° (46° + x) + x + 54° = 180° 2x + 100° = 180° 2x = 80° x = 40°

Explanation

The expression (cot A + tan B)/(cot B + tan A) can be simplified by substituting cot A = (1)/(tan A) and cot B = (1)/(tan B) into the formula. This transformation allows all terms to be expressed in terms of the tangent function:

(1)/(tan A) + tan B{(1)/(tan B) + tan A}

Simplifying the numerator and the denominator separately by taking common denominators:

(1 + tan A tan B)/(tan A){(1 + tan A tan B)/(tan B)}

By dividing the two fractions, the term (1 + tan A tan B) cancels out:

(1 + tan A tan B)/(tan A) × (tan B)/(1 + tan A tan B) = (tan B)/(tan A)

Using the trigonometric relation (1)/(tan A) = cot A, the expression simplifies directly to tan B cot A.

Explanation

Using the trigonometric identity cos² θ = 1 - sin² θ, the given equation cos² θ - sin θ = 1/4 can be expressed entirely in terms of sin θ as: 1 - sin² θ - sin θ = 1/4

To eliminate the fraction, multiply the entire equation by 4: 4 - 4sin² θ - 4sin θ = 1

Rearranging the terms into a standard quadratic equation form ax² + bx + c = 0 where x = sin θ: 4sin² θ + 4sin θ - 3 = 0

Solving this quadratic equation using the factorization method: 4sin² θ + 6sin θ - 2sin θ - 3 = 0 2sin θ(2sin θ + 3) - 1(2sin θ + 3) = 0 (2sin θ - 1)(2sin θ + 3) = 0

Setting each factor to zero gives sin θ = 1/2 or sin θ = -3/2. Since the range of the sine function is restricted to the interval [-1, 1], the value -3/2 is mathematically impossible for a real angle θ. Therefore, the only valid solution is sin θ = 1/2.

Explanation

The number of students studying Commerce in the year 2012 was 148, and the number of students studying Commerce in the year 2016 was 168. To calculate the percentage increase, the standard formula is applied:

Percentage Increase = (Final Value - Initial Value)/(Initial Value) × 100

Substituting the given values: Increase = 168 - 148 = 20 Percentage Increase = ²⁰⁄₁₄₈ × 100

By simplifying the fraction: ²⁰⁄₁₄₈ × 100 = ⁵⁄₃₇ × 100 = ⁵⁰⁰⁄₃₇

Performing the division: 500 ÷ 37 ≈ 13.5135

Rounding to two decimal places, the percentage increase in the number of Commerce students from 2012 to 2016 is 13.51.

Explanation

The simple annual growth rate (SAGR) is calculated by dividing the total percentage growth over a period by the number of years in that period. The formula is: SAGR = V_fᵢₙₐₗ - Vᵢₙᵢₜᵢₐₗ{Vᵢₙᵢₜᵢₐₗ × n} × 100

From the given table, the number of students in Business for the year 2012 (Vᵢₙᵢₜᵢₐₗ) is 32 and for the year 2016 (V_fᵢₙₐₗ) is 48. The number of years (n) from 2012 to 2016 is 2016 - 2012 = 4.

First, calculate the total growth percentage over the four-year period: Total Growth Percentage = (48 - 32)/(32) × 100 = ¹⁶⁄₃₂ × 100 = 50%

Now, divide the total growth percentage by the number of years to find the simple annual growth rate: SAGR = (50%)/(4) = 12.5%

Explanation

The ratio of the average number of students in two categories over the same period of time is equivalent to the ratio of their total sums. To find the ratio of students in Arts to those in Science, we first sum the values for each category from 2012 to 2016.

Total number of students in Arts = 48 + 56 + 64 + 78 + 92 = 338 Total number of students in Science = 105 + 123 + 125 + 148 + 161 = 662

The average number of students is calculated by dividing the total sum by the number of years (n = 5). The required ratio is:

Ratio = (Sum of Arts / 5)/(Sum of Science / 5) = ³³⁸⁄₆₆₂

Simplifying the fraction by dividing both the numerator and the denominator by their common factor of 2:

Ratio = ¹⁶⁹⁄₃₃₁

Explanation

The total number of students for each year is calculated by summing the students in Arts, Science, Commerce, and Business. For 2012, the total is 48 + 105 + 148 + 32 = 333. For 2013, the total is 56 + 123 + 136 + 30 = 345. For 2014, the total is 64 + 125 + 144 + 36 = 369. For 2015, the total is 78 + 148 + 156 + 36 = 418. For 2016, the total is 92 + 161 + 168 + 48 = 469.

To find the percentage increase over the previous year, use the formula Percentage Increase = Total_cᵤᵣᵣₑₙₜ - Totalₚᵣₑᵥᵢₒᵤₛ{Totalₚᵣₑᵥᵢₒᵤₛ} × 100.

The increases are as follows: For 2013: (345 - 333)/(333) × 100 = ¹²⁄₃₃₃ × 100 ≈ 3.61% For 2014: (369 - 345)/(345) × 100 = ²⁴⁄₃₄₅ × 100 ≈ 6.96% For 2015: (418 - 369)/(369) × 100 = ⁴⁹⁄₃₆₉ × 100 ≈ 13.28% For 2016: (469 - 418)/(418) × 100 = ⁵¹⁄₄₁₈ × 100 ≈ 12.20%

Comparing these values, the year 2015 shows the maximum percentage increase of approximately 13.28%.

Explanation

A number that is both a perfect square and a perfect cube must be a perfect sixth power. If n = x² and n = y³, the exponents in the prime factorization of n must be multiples of both 2 and 3, meaning they must be multiples of lcm(2, 3) = 6. Therefore, we are looking for integers k such that 2000 ≤ k⁶ ≤ 7000.

Evaluating consecutive sixth powers: 1⁶ = 1 2⁶ = 64 3⁶ = 729 4⁶ = (2²)⁶ = 2¹² = 4096 5⁶ = 15625

Comparing these values to the given range of 2000 to 7000, the only value that fits is 4096. Consequently, there is exactly one such number.

Explanation

For work problems involving or and and conditions, the time taken by the combined group can be solved using the ratio of efficiencies or a direct formula. Given that 3 men or 4 women can complete the job in 120 days, we can establish the efficiency relationship as 3M = 4W, which means 1M = ⁴⁄₃W.

To find the time taken by 12 men and 16 women working together, we first convert the entire group into a single unit (women). The combined strength is: 12M + 16W = 12(⁴⁄₃W) + 16W = 16W + 16W = 32W

Using the work-rate formula M₁ D₁ = M₂ D₂, where M₁ is 4 women and D₁ is 120 days: 4 × 120 = 32 × D₂ 480 = 32 × D₂ D₂ = ⁴⁸⁰⁄₃₂ = 15 days

Alternatively, using the shortcut formula for such problems: Time = (Given Days)/(And Men){Or Men + (And Women)/(Or Women)} Time = (120)/(12){3 + ¹⁶⁄₄} Time = (120)/(4 + 4) = ¹²⁰⁄₈ = 15 days

Explanation

The volume V of a hemisphere is calculated using the formula V = ⅔ π r³, where r is the radius of the hemisphere. Given the diameter is 21 cm, the radius r = ²¹⁄₂ = 10.5 cm. Substituting the values into the formula:

V = ⅔ × ²²⁄₇ × (²¹⁄₂)³

V = ⅔ × ²²⁄₇ × ²¹⁄₂ × ²¹⁄₂ × ²¹⁄₂

To simplify the calculation, cancel the common factors in the numerator and denominator. The terms 3 × 7 in the denominator cancel out one 21 in the numerator. One factor of 2 in the denominator cancels the 2 from the ⅔ coefficient, and another factor of 2 reduces 22 to 11. This leaves:

V = (11 × 21 × 21)/(2)

V = (11 × 441)/(2)

V = ⁴⁸⁵¹⁄₂ = 2425.5

The volume of the hemisphere is 2425.5 cm³.

Explanation

The marked price of the article can be calculated using the successive discount formula. Let the marked price be M. The selling price S after two successive discounts d₁ and d₂ is given by the relation S = M × (1 - (d₁)/(100)) × (1 - (d₂)/(100)).

Given S = 16896, d₁ = 20%, and d₂ = 12%, we substitute the values into the formula:

16896 = M × (1 - ²⁰⁄₁₀₀) × (1 - ¹²⁄₁₀₀) 16896 = M × 0.8 × 0.88 16896 = M × 0.704

To find M, divide the selling price by the combined multiplier:

M = (16896)/(0.704) M = 24000

Alternatively, the effective single discount for two successive discounts x% and y% is x + y - ˣʸ⁄₁₀₀. Substituting the given rates, the effective discount is 20 + 12 - (20 × 12)/(100) = 32 - 2.4 = 29.6%. This means the selling price is 100% - 29.6% = 70.4% of the marked price. Setting 70.4% of M = 16896 yields M = (16896)/(70.4) × 100 = 24000.

Explanation

For a constant distance, speed is inversely proportional to the time taken, as derived from the formula D = S × T. Given the ratio of speeds S₁ : S₂ : S₃ is 3 : 4 : 6, the ratio of the time taken T₁ : T₂ : T₃ is represented by the reciprocals of the speeds: ⅓ : ¼ : ⅙. To convert these fractions into a whole number ratio, multiply each term by the Least Common Multiple (LCM) of the denominators 3, 4, and 6, which is 12. The calculation follows as (⅓ × 12) : (¼ × 12) : (⅙ × 12), which simplifies to 4 : 3 : 2.

Explanation

The total runs scored by all 7 players is determined by the formula Total = Average × Number of observations. Given the average is 53 for 7 players, the total runs scored are 53 × 7 = 371.

The sum of the runs scored by the first 6 players is 121 + 40 + 26 + 56 + 37 + 48 = 328. To find the runs scored by the 7th player, subtract the sum of the runs of these 6 players from the total runs of all 7 players: 371 - 328 = 43.

Explanation

The total cost price (CP) of the bicycle is the sum of the purchase price and the repair expenses. Therefore, CP = 2700 + 500 = 3200. The selling price (SP) is 3520.

Profit is determined by the formula SP - CP, which gives 3520 - 3200 = 320.

The profit percentage is calculated using the formula: Profit Percentage = (Profit)/(CP) × 100

Substituting the values: Profit Percentage = ³²⁰⁄₃₂₀₀ × 100 = ⅒ × 100 = 10%

Explanation

When the expenditure on a commodity is to remain constant, the consumption must be inversely proportional to the price. This relationship is defined by the formula Expenditure = Price × Consumption. To find the percentage reduction in consumption required after a price increase, the formula is:

Percentage reduction = (New Price - Original Price)/(New Price) × 100

In this case, the original price is Rs 24/kg and the new price is Rs 36/kg. The absolute increase in price is 36 - 24 = 12. Substituting these values into the formula:

Percentage reduction = ¹²⁄₃₆ × 100

Percentage reduction = ⅓ × 100

The calculation results in 33.33%. This means that the consumption must be reduced by one-third of its original level to offset the 50% increase in price, thereby keeping the total budget for onions unchanged. This can be verified by noting that if a household originally bought 3 kg for Rs 72 (24 × 3), they must now buy 2 kg for the same Rs 72 (36 × 2), representing a reduction of 1 kg out of 3 kg.

Explanation

The ratio of the speed of the train to the speed of the car is 1.2 : 1, which simplifies to 6 : 5. Since speed and time are inversely proportional for a fixed distance, the ratio of the time taken by the train (moving time) to the time taken by the car is 5 : 6.

The difference in these time ratios is 6 - 5 = 1 unit. This 1 unit represents the 30 minutes, or 0.5 hours, that the train spent at various stations. Because the train stopped for this duration, it arrived at the same time as the slower-moving car despite being faster.

Using the value of the time unit: 1 unit = 0.5 hours Moving time of the train = 5 × 0.5 = 2.5 hours

The speed of the train is calculated using the formula: Speed = (Distance)/(Time) Speed = (180 km)/(2.5 hours) = 72 km/hr

Explanation

The equation (x/y)ᵃ⁻⁴ = (y/x)²ᵃ⁻⁵ can be simplified by expressing both sides with the same base. Since (y/x) = (x/y)⁻¹, the expression becomes (x/y)ᵃ⁻⁴ = ((x/y)⁻¹)²ᵃ⁻⁵, which simplifies to (x/y)ᵃ⁻⁴ = (x/y)⁵⁻²ᵃ.

For an equation of the form uᵐ = uⁿ to hold, there are multiple possible mathematical conditions. One condition is that the exponents are equal, m = n. Setting the exponents equal, a - 4 = 5 - 2a, results in 3a = 9, which gives a = 3. If the constant a equals 3, the original equation becomes (x/y)⁻¹ = (y/x)¹, which simplifies to y/x = y/x. This is an identity that is true for any non-zero values of x and y, allowing for any relationship where x is greater than, less than, or equal to y.

Another condition is that the base x/y equals 1, which implies x = y. If x = y, the equation 1ᵃ⁻⁴ = 1²ᵃ⁻⁵ is true for any value of a. Because the problem does not specify the value of a, it is not possible to determine whether x must equal y or if x and y can be any non-zero numbers. The lack of constraints on the constant a means the relationship between x and y remains undefined.

Explanation

The given equation is 3a - ³⁄ₐ - 3 = 0. Dividing the entire equation by 3, we get a - ¹⁄ₐ - 1 = 0, which simplifies to a - ¹⁄ₐ = 1.

To find the value of a³ - (1)/(a³), we use the algebraic identity (x - y)³ = x³ - y³ - 3xy(x - y). Substituting x = a and y = ¹⁄ₐ, we get: (a - ¹⁄ₐ)³ = a³ - (1)/(a³) - 3(a)(¹⁄ₐ)(a - ¹⁄ₐ) (a - ¹⁄ₐ)³ = a³ - (1)/(a³) - 3(a - ¹⁄ₐ)

Now, substitute the value a - ¹⁄ₐ = 1 into this equation: 1³ = a³ - (1)/(a³) - 3(1) 1 = a³ - (1)/(a³) - 3 a³ - (1)/(a³) = 1 + 3 = 4

The question asks for the value of a³ - (1)/(a³) + 2. Substituting the calculated value: 4 + 2 = 6

Explanation

In △ ABC, AD is the median to side BC, which means BD = DC = ½ BC. The problem states that AD = ½ BC, so AD = BD = DC. Since the distance from point D to all three vertices A, B, and C is equal, D is the circumcenter of △ ABC, and BC is the diameter of the circumcircle. By Thales's theorem, the angle subtended by the diameter at the circumference is 90°, thus ∠ BAC = 90°. In △ ADC, since AD = DC, the triangle is isosceles and ∠ DAC = ∠ ACD = 40°. Therefore, ∠ DAB = ∠ BAC - ∠ DAC = 90° - 40° = 50°.

Explanation

In △ ABC, the sum of all internal angles is 180°. Given ∠ BAC = 75° and ∠ BCA = 80°, we find the third angle ∠ ABC = 180° - (75° + 80°) = 25°.

The circum-centre O of a triangle is the center of the circle passing through all vertices A, B, and C. According to the central angle theorem, the angle subtended by an arc at the center is twice the angle subtended at the circumference. Therefore, the angle subtended by chord AC at the center O is ∠ AOC = 2 × ∠ ABC = 2 × 25° = 50°.

In △ OAC, the sides OA and OC are equal as they are radii of the same circum-circle. This makes △ OAC an isosceles triangle where the base angles ∠ OAC and ∠ OCA are equal. Using the angle sum property for △ OAC: ∠ OAC + ∠ OCA + ∠ AOC = 180° 2 × ∠ OAC + 50° = 180° 2 × ∠ OAC = 130° ∠ OAC = 65°

Explanation

In a rhombus where the smaller diagonal is equal to the side length, the rhombus is composed of two equilateral triangles joined together. If the side length is s = 4 cm, the smaller diagonal d₁ is also 4 cm. The larger diagonal d₂ of such a rhombus is given by the formula d₂ = √3 × s. Substituting the value of s, we get d₂ = 4√3 cm.

The side of the required equilateral triangle is equal to this bigger diagonal, so a = 4√3 cm. The area of an equilateral triangle is calculated using the formula Area = √3{4} a². Substituting the value of a:

Area = √3{4} × (4√3)² Area = √3{4} × 48 Area = 12√3 cm²

Explanation

To simplify the expression (cos A + sin A)(cot A + tan A), we utilize the standard trigonometric identity cot A + tan A = (1)/(sin A cos A). This identity is derived by expressing cot A as (cos A)/(sin A) and tan A as (sin A)/(cos A), which gives (cos² A + sin² A)/(sin A cos A) = (1)/(sin A cos A).

Substituting this into the original expression, we get: (cos A + sin A) × (1)/(sin A cos A)

Distributing the terms yields: (cos A)/(sin A cos A) + (sin A)/(sin A cos A)

By canceling the common terms in the numerator and denominator, we arrive at: (1)/(sin A) + (1)/(cos A)

Using the reciprocal identities (1)/(sin A) = csc A and (1)/(cos A) = sec A, the expression simplifies to csc A + sec A, which is the same as sec A + csc A.

Explanation

To find the value of cos θ, we rearrange the given equation 2 cos θ = 2 - sin θ to isolate the sine term as sin θ = 2 - 2 cos θ. Squaring both sides of the equation gives sin² θ = (2 - 2 cos θ)², which expands to sin² θ = 4 - 8 cos θ + 4 cos² θ. Using the fundamental identity sin² θ = 1 - cos² θ, we substitute to get 1 - cos² θ = 4 - 8 cos θ + 4 cos² θ.

Combining like terms results in the quadratic equation 5 cos² θ - 8 cos θ + 3 = 0. This can be factored as (5 cos θ - 3)(cos θ - 1) = 0. Setting each factor to zero provides the possible values for cos θ. From 5 cos θ - 3 = 0, we get cos θ = 3/5, and from cos θ - 1 = 0, we get cos θ = 1. Both values are valid as they fall within the range of the cosine function and satisfy the original relationship.

Explanation

The difference between the average exports and average imports is determined by calculating the average of the annual trade balances (Export minus Import). Based on the data associated with this total trade series, the export and import values for each of the five years are: 1850 and 1950 for the 3800 total, 1750 and 1850 for the 3600 total, 1350 and 1450 for the 2800 total, 2100 and 1800 for the 3900 total, and 2650 and 2350 for the 5000 total.

First, calculate the sum of exports: 1850 + 1750 + 1350 + 2100 + 2650 = 9700 crore dollars. Next, calculate the sum of imports: 1950 + 1850 + 1450 + 1800 + 2350 = 9400 crore dollars. The difference in the averages is then found using the following relation:

Difference in averages = (Σ Exports - Σ Imports)/(n)

Difference in averages = (9700 - 9400)/(5) = ³⁰⁰⁄₅ = 60

Explanation

To find the number of integers between 1 and 200 that are divisible by 3 but not by 7, we use the principle of exclusion. We first calculate the total number of multiples of 3 in the range and then subtract those that are also multiples of 7. A number is divisible by both 3 and 7 if and only if it is divisible by their least common multiple, which is LCM(3, 7) = 21.

The number of multiples of n up to x is given by the floor function ⌊ ˣ⁄ₙ ⌋.

Number of multiples of 3: ⌊ ²⁰⁰⁄₃ ⌋ = 66

Number of multiples of 21: ⌊ ²⁰⁰⁄₂₁ ⌋ = 9

The count of numbers divisible by 3 but not by 7 is 66 - 9 = 57.

Explanation

The total work done is the product of the number of individuals, the time taken, and their respective efficiencies. Let the efficiency of a woman be W, a man be M, and a child be C. Since the work done is the same in all three cases, we can equate the total work:

10 × 6 × W = 6 × 5 × M = 8 × 10 × C

60W = 30M = 80C

Dividing the entire equation by 10, we get:

6W = 3M = 8C

To find the ratio W : M : C, we divide each term by the Least Common Multiple (LCM) of the coefficients 6, 3, and 8. The LCM of 6, 3, and 8 is 24.

(6W)/(24) = (3M)/(24) = (8C)/(24)

(W)/(4) = (M)/(8) = (C)/(3)

Thus, the ratio of the efficiency of a woman, a man, and a child is 4 : 8 : 3.

Explanation

The volume of a cylinder is given by the formula V = π r² h = A × h, where A is the base area and h is the height. Let the volumes, base areas, and heights of the two cylinders be V₁, V₂, A₁, A₂ and h₁, h₂ respectively.

The ratio of the volumes is: (V₁)/(V₂) = (A₁ × h₁)/(A₂ × h₂)

Substituting the given values (V₁)/(V₂) = ⁷⁄₃, (h₁)/(h₂) = ⁷⁄₉, and A₂ = 154 cm²: ⁷⁄₃ = (A₁)/(154) × ⁷⁄₉

Solving for A₁: A₁ = ⁷⁄₃ × ⁹⁄₇ × 154 = 3 × 154 = 462 cm²

Since the base area A₁ = π r₁², where r₁ is the radius of the first cylinder: ²²⁄₇ × r₁² = 462 r₁² = (462 × 7)/(22) r₁² = 21 × 7 = 147 r₁ = √(147) = √(49 × 3) = 7√3 cm

Explanation

Let the Marked Price (MP) of the clock be 100. Kanchan purchased the clock at a 25% discount, so her Cost Price (CP) is 100 - (0.25 × 100) = 75. She then sold the clock at a 75% profit on her purchase price. Her Selling Price (SP) is 75 + (0.75 × 75) = 75 + 56.25 = 131.25.

The question asks for the profit percentage relative to the original Marked Price. The profit earned over the Marked Price is SP - MP = 131.25 - 100 = 31.25. The percentage profit on the Marked Price is calculated as ((SP - MP)/(MP)) × 100. Substituting the values, we get ((31.25)/(100)) × 100 = 31.25%.

Explanation

The total amount of Rs 8400 is divided in the ratio 6 : 8 : 7. The sum of the ratios is 6 + 8 + 7 = 21. The value of one unit for income is 8400 / 21 = 400. Thus, the incomes of A, B, and C are 6 × 400 = 2400, 8 × 400 = 3200, and 7 × 400 = 2800 respectively.

The savings are in the ratio 3 : 2 : 4. Given that B's saving is Rs 400, which corresponds to 2 units in the ratio, the value of one unit for savings is 400 / 2 = 200. Consequently, the savings of A, B, and C are 3 × 200 = 600, 2 × 200 = 400, and 4 × 200 = 800 respectively.

Expenditure is calculated as Income minus Savings. Expenditure of A = 2400 - 600 = 1800 Expenditure of B = 3200 - 400 = 2800 Expenditure of C = 2800 - 800 = 2000

The ratio of their expenditures is 1800 : 2800 : 2000. Dividing each term by 200, we get the simplified ratio 9 : 14 : 10.

Explanation

The correct average can be calculated using the adjustment formula: Correct Average = Incorrect Average + (Actual Value - Incorrect Value)/(n). Here, the number of students n = 24, the initial average is 12 years, the actual age of the student is 8 years, and the incorrectly recorded age is 14 years.

Substituting these values: Correct Average = 12 + (8 - 14)/(24) Correct Average = 12 - ⁶⁄₂₄ Correct Average = 12 - 0.25 Correct Average = 11.75

The error resulted in the total sum being overestimated by 14 - 8 = 6 years. To find the correct average, this excess sum must be distributed across all 24 students. The average was therefore inflated by 6 / 24 = 0.25 years, which is subtracted from the incorrect average of 12 to find the true value.

Explanation

According to the given condition, 70% of the cost price (CP) is equal to 40% of the selling price (SP). This relationship can be written as: 0.70 × CP = 0.40 × SP

Rearranging the terms to find the ratio of the selling price to the cost price: (SP)/(CP) = ⁷⁰⁄₄₀ = ⁷⁄₄

This implies that if the cost price is 4 units, the selling price is 7 units. Since the selling price is greater than the cost price, there is a profit. The profit is calculated as: Profit = SP - CP = 7 - 4 = 3 units

The profit percentage is determined using the formula: Profit % = ( (Profit)/(CP) ) × 100 Profit % = ( ¾ ) × 100 = 75%

Explanation

The percentage formula states that x% of y = (x × y)/(100). Applying this to the given expression: a% of b = (a × b)/(100) = ᵃᵇ⁄₁₀₀ b% of a = (b × a)/(100) = ᵇᵃ⁄₁₀₀ Since multiplication is commutative (ab = ba), the sum is: ᵃᵇ⁄₁₀₀ + ᵃᵇ⁄₁₀₀ = ²ᵃᵇ⁄₁₀₀ This result can be rearranged as ²ᵃ⁄₁₀₀ × b, which is the mathematical definition of 2a% of b.

Explanation

Since the distance to the office remains constant, speed and time are inversely proportional (S ∝ 1/T). If the speed increases to 7/6 of the usual speed, the time taken will be the reciprocal of this ratio, which is 6/7 of the usual time.

Let the usual time be T. The new time is ⁶⁄₇T. The difference between the usual time and the new time is given as 15 minutes:

T - ⁶⁄₇T = 15

⅐T = 15

T = 15 × 7 = 105

Alternatively, using the ratio method, if the speed ratio is 6:7, the time ratio is 7:6. The difference of 1 unit in the ratio corresponds to 15 minutes. Therefore, the usual time of 7 units is 7 × 15 = 105 minutes.

Explanation

The total simple interest received from both B and C is the sum of the individual interests calculated using the formula SI = (P · R · T)/(100).

For B, the interest is (10000 · R · 3)/(100) = 300R. For C, the interest is (6000 · R · 4)/(100) = 240R.

The total interest received is given as Rs 5400. Equating the sum of the individual interests to the total: 300R + 240R = 5400 540R = 5400 R = ⁵⁴⁰⁰⁄₅₄₀ = 10

The rate of interest is 10% per annum.

Explanation

According to the Factor Theorem, if a polynomial P(x) is divisible by (x - a), then P(a) = 0. In this case, the polynomial P(x) = x³ + 2x² - 5x + k is divisible by (x + 1). Therefore, substituting x = -1 into the polynomial must result in zero.

P(-1) = (-1)³ + 2(-1)² - 5(-1) + k = 0

-1 + 2(1) + 5 + k = 0

-1 + 2 + 5 + k = 0

6 + k = 0

k = -6