SSC CGL Tier 1 2013 Quantitative Aptitude Questions with Solutions

Explanation

Let the fraction be (x)/(y). According to the first condition, when 2 is added to both the numerator and the denominator, the fraction becomes ⁹⁄₁₁. Testing the fraction ⁷⁄₉, we see that (7 + 2)/(9 + 2) = ⁹⁄₁₁, which satisfies the first condition.

According to the second condition, when 3 is added to both the numerator and the denominator, the fraction becomes ⅚. Testing ⁷⁄₉ with this condition gives (7 + 3)/(9 + 3) = ¹⁰⁄₁₂. Simplifying ¹⁰⁄₁₂ by dividing both terms by 2 yields ⅚, which satisfies the second condition.

Alternatively, this can be solved using simultaneous equations. The conditions provide 11(x + 2) = 9(y + 2) and 6(x + 3) = 5(y + 3). Simplifying these leads to 11x - 9y = -4 and 6x - 5y = -3. Solving for x and y gives x = 7 and y = 9.

Explanation

The given equation (a)/(b) + ᵇ⁄ₐ - 1 = 0 can be simplified by multiplying the entire expression by the common denominator ab. This yields a² + b² - ab = 0, which is equivalent to a² - ab + b² = 0.

The algebraic identity for the sum of two cubes is a³ + b³ = (a + b)(a² - ab + b²). By substituting the value of the quadratic factor derived from the initial condition (a² - ab + b² = 0) into this identity, we obtain a³ + b³ = (a + b) × 0 = 0.

Explanation

The given equation x + ¹⁄ₓ = 99 can be rewritten by multiplying both sides by x, which results in x² + 1 = 99x. The expression to be evaluated is (100x)/(2x² + 102x + 2). By factoring out a 2 from the first and last terms of the denominator, the expression becomes (100x)/(2(x² + 1) + 102x). Substituting the value x² + 1 = 99x into the denominator yields (100x)/(2(99x) + 102x). This simplifies to (100x)/(198x + 102x) = ¹⁰⁰ˣ⁄₃₀₀ₓ. Dividing the numerator and denominator by 100x gives the final value of ⅓.

Explanation

Given that x = 3 + 2√2, we first find the value of ¹⁄ₓ by rationalizing the denominator. Multiplying the numerator and denominator by the conjugate 3 - 2√2, we get: ¹⁄ₓ = (1)/(3 + 2√(2)) × 3 - 2√2{3 - 2√2} = 3 - 2√2{3² - (2√2)²} = 3 - 2√2{9 - 8} = 3 - 2√2

Next, we calculate the sum of x and ¹⁄ₓ: x + ¹⁄ₓ = (3 + 2√2) + (3 - 2√2) = 6

To find the value of x² + (1)/(x²), we use the algebraic identity (a + b)² = a² + b² + 2ab. Setting a = x and b = ¹⁄ₓ, we have: (x + ¹⁄ₓ)² = x² + (1)/(x²) + 2(x)(¹⁄ₓ) 6² = x² + (1)/(x²) + 2 36 = x² + (1)/(x²) + 2 x² + (1)/(x²) = 36 - 2 = 34

Explanation

The given expression can be simplified by finding a common denominator for the three terms. The least common multiple (LCM) of bc, ca, and ab is abc. Rewriting the expression with this common denominator gives:

a²{bc} + b²{ca} + c²{ab} = a³ + b³ + c³{abc}

In algebra, there is a standard identity: a³ + b³ + c³ - 3abc = (a + b + c)(a² + b² + c² - ab - bc - ca). When it is given that a + b + c = 0, the entire right side of this identity becomes zero, which leads to the relation:

a³ + b³ + c³ = 3abc

By substituting 3abc for the numerator in our simplified expression, we get:

(3abc)/(abc) = 3

Explanation

For any two positive real numbers a and b, the Arithmetic Mean-Geometric Mean (AM-GM) inequality states that (a + b)/(2) ≥ √(ab).

In this problem, we seek the minimum value of 2x + y given that x and y are positive real numbers and xy = 8. Let a = 2x and b = y. Applying the AM-GM inequality:

(2x + y)/(2) ≥ √((2x)(y))

Substitute the given value xy = 8 into the inequality:

(2x + y)/(2) ≥ √(2(xy)) (2x + y)/(2) ≥ √(2(8)) (2x + y)/(2) ≥ √(16) (2x + y)/(2) ≥ 4

Multiplying both sides of the inequality by 2 gives:

2x + y ≥ 8

The minimum value of the expression is reached when 2x = y. To verify this consistency with the constraint xy = 8, substituting y = 2x yields x(2x) = 8, which simplifies to 2x² = 8 or x² = 4. Since x is a positive real number, x = 2 and y = 4. Substituting these values back into the objective function gives 2(2) + 4 = 8.

Explanation

To find the point of intersection of the two lines, we solve the given equations for x and y independently.

The first equation is 2x + 1 = 0. Subtracting 1 from both sides gives 2x = -1. Dividing by 2, we find x = -1/2. This equation represents a vertical line where every point has an x-coordinate of -1/2.

The second equation is 3y - 9 = 0. Adding 9 to both sides gives 3y = 9. Dividing by 3, we find y = 3. This equation represents a horizontal line where every point has a y-coordinate of 3.

The intersection of a vertical line x = a and a horizontal line y = b is the unique point (a, b). Substituting the calculated values, the intersection point is (-1/2, 3).

Explanation

According to the exterior angle theorem, the exterior angle of a triangle is equal to the sum of the two interior opposite angles. Since BC is produced to D, ∠ ACD is the exterior angle at vertex C.

x = ∠ BAC + ∠ ABC = 75° + 45° = 120°

The question asks for (x / 3)% of 60°. Substituting the value x = 120:

(120 / 3)% of 60° = 40% of 60°

Applying the percentage formula:

(40 / 100) × 60 = 24°

Explanation

Let the four numbers be n₁, n₂, n₃, and n₄. The sum of the first three numbers is 3 × 15 = 45, which can be expressed as n₁ + n₂ + n₃ = 45. The sum of the last three numbers is 3 × 16 = 48, which can be expressed as n₂ + n₃ + n₄ = 48. Subtracting the first equation from the second gives (n₂ + n₃ + n₄) - (n₁ + n₂ + n₃) = 48 - 45, simplifying to n₄ - n₁ = 3. Since the last number n₄ is 19, substituting this value gives 19 - n₁ = 3, which results in n₁ = 16.

Explanation

The total age of the family of 5 members at present is calculated as 33 × 5 = 165 years. To determine the average age just before the birth of the youngest member, we calculate the state of the family 9 years ago. During this 9-year interval, the sum of the ages of all 5 members decreased by 5 × 9 = 45 years. Thus, the total age of the family 9 years ago was 165 - 45 = 120 years. At that specific point in time, just before the birth, the family consisted of only 4 members. The average age is therefore the total age divided by the number of members: 120 / 4 = 30 years.

Explanation

In cases where a trader sells goods at the cost price but uses a false weight, the gain percentage is calculated using the standard relation:

Gain % = ( (True Weight - False Weight)/(False Weight) ) × 100

Given that the true weight is 1000 g (1 kg) and the false weight used by the dealer is 850 g, the calculation is as follows:

Gain % = ( (1000 - 850)/(850) ) × 100

Gain % = ( ¹⁵⁰⁄₈₅₀ ) × 100

Simplifying the fraction by dividing the numerator and the denominator by 50:

Gain % = ( ³⁄₁₇ ) × 100 = ³⁰⁰⁄₁₇%

By dividing 300 by 17, we get a quotient of 17 and a remainder of 11. Expressing this as a mixed fraction results in 17 ¹¹⁄₁₇%.

Explanation

Let the total number of students who applied for the groups be x. According to the problem, 40% of the students were allotted group A, leaving 60% of the total. The remaining students are calculated as 0.60x.

Out of this remainder, 75% were given group B. Therefore, the number of students in group B is 0.75 × 0.60x = 0.45x.

The students assigned to group C represent the final remaining portion of the total. This can be found by subtracting the students in groups A and B from the total: x - (0.40x + 0.45x) = x - 0.85x = 0.15x

Given that the number of students in group C is 12, we set up the equation: 0.15x = 12 x = (12)/(0.15) x = ¹²⁰⁰⁄₁₅ x = 80

Alternatively, using the remaining fraction method, if 40% are in group A, 60% remain. If 75% of those go to group B, then 25% of the 60% remain for group C. 0.25 × 0.60 × x = 12 0.15x = 12 x = 80

Explanation

The speed of the first train v₁ is determined by the time it takes to cross a stationary milestone. Using the formula for speed v = d/t, where d is the length of the train: v₁ = 150/15 = 10 m/s.

When two trains of lengths L₁ and L₂ cross each other while traveling in opposite directions, the total distance covered is the sum of their lengths, L₁ + L₂. The relative speed is the sum of their individual speeds, v₁ + v₂. According to the problem, the second train has the same length as the first (L₂ = 150 m) and they cross in 12 seconds.

The relative speed is calculated as: v₁ + v₂ = (150 + 150) / 12 = 300 / 12 = 25 m/s.

By substituting the speed of the first train into the equation: 10 + v₂ = 25 v₂ = 15 m/s.

To convert the speed of the second train from m/s to km/hr, multiply by the conversion factor 18/5: 15 × (18/5) = 3 × 18 = 54 km/hr.

Explanation

The upstream speed u is calculated by dividing the distance by the time taken in hours: u = (1 km)/(10/60 h) = 6 km/h. Similarly, the downstream speed v is calculated as v = (1 km)/(4/60 h) = 15 km/h. The speed of the stream is derived using the standard relation Stream Speed = (v - u)/(2). Substituting the obtained values into this formula, we get (15 - 6)/(2) = ⁹⁄₂ = 4.5 km/h.

Explanation

For a sum of money to double itself under simple interest, the interest earned must be equal to the principal amount (P). Let the rate of interest be R and the time be T. The formula for simple interest is SI = (P × R × T)/(100). Since the amount doubles, the interest SI is equal to P. Substituting the given time T = 15 years into the formula, we get P = (P × R × 15)/(100). By cancelling P from both sides, the equation becomes 1 = (15R)/(100). Solving for R gives R = ¹⁰⁰⁄₁₅. Simplifying the fraction by dividing both the numerator and the denominator by 5 results in R = ²⁰⁄₃, which is equivalent to 6 ⅔ percent per annum.

Explanation

In an isosceles triangle, let the two equal angles be y and the third, unequal angle be x. According to the problem, the unequal angle is twice the sum of the equal angles, which gives the relation x = 2(y + y), or x = 4y.

The sum of all interior angles in any triangle is 180°. Therefore, x + y + y = 180°. Substituting the value of x from the first relation into this equation, we get 4y + 2y = 180°, which simplifies to 6y = 180°. Dividing by 6, we find y = 30°. Thus, each of the equal angles measures 30°.

Explanation

Let h be the height of the pole and L be the length of the shadow cast by the pole on the ground. The relationship between the height, the shadow length, and the angle of elevation θ of the Sun is given by the trigonometric ratio: tan θ = (Height of the pole)/(Length of the shadow)

According to the given condition, the length of the shadow is √3 times the height of the pole, which can be expressed as L = √3h. Substituting this value into the formula: tan θ = (h)/(√(3)h) tan θ = (1)/(√(3))

In trigonometry, the tangent of an angle equals (1)/(√(3)) when the angle is 30°. Therefore, θ = 30°.

Explanation

Using the trigonometric identity sec²Θ = 1 + tan²Θ, we substitute this into the given equation 3(sec²Θ + tan²Θ) = 5:

3(1 + tan²Θ + tan²Θ) = 5 3(1 + 2tan²Θ) = 5 3 + 6tan²Θ = 5 6tan²Θ = 2 tan²Θ = ²⁄₆ = ⅓

Given that Θ is a positive acute angle, we take the positive square root: tanΘ = (1)/(√(3)) This corresponds to Θ = 30°.

Now, we evaluate the expressions in the correct relation: cos 2Θ = cos(2 × 30°) = cos 60° = ½ sin Θ = sin 30° = ½

Since both sides equal ½, the equality cos 2Θ = sin Θ holds true.

Explanation

When two circles touch each other externally, the distance between their centers is equal to the sum of their radii. This relationship is expressed by the formula d = r₁ + r₂, where d is the distance between the centers and r₁ and r₂ are the radii of the two circles. Substituting the given values into this equation, we have 7 = 4 + r₂. By subtracting 4 from both sides, we find that r₂ = 7 - 4, which equals 3 cm.

Explanation

In Δ ABC, since AB = AC, the angles opposite to these sides are equal, so ∠ ABC = ∠ ACB. Let ∠ ABC = ∠ ACB = x. Since AC = AD and AB = AC, it follows that AB = AD, making A the midpoint of the segment BD. In Δ ACD, since AC = AD, the angles opposite to these sides are also equal, so ∠ ACD = ∠ ADC. Let ∠ ACD = ∠ ADC = y.

In the large triangle BCD, the sum of the interior angles must be 180°. Therefore, ∠ CBD + ∠ BDC + ∠ BCD = 180°. Substituting the respective values, we have x + y + (x + y) = 180°, which simplifies to 2(x + y) = 180°, leading to x + y = 90°. Since ∠ BCD = ∠ ACB + ∠ ACD = x + y, the measure of ∠ BCD is 90°. Geometrically, because AB = AC = AD, point A is equidistant from B, C, and D, making A the circumcenter of Δ BCD. Since A lies on the line segment BD, BD is the diameter of the circumcircle, and the angle ∠ BCD subtended by the diameter is always 90°.

Explanation

In any right-angled triangle, the circumcentre lies at the midpoint of the hypotenuse, which means the circumradius (R) is equal to half the length of the hypotenuse. Given ∠ ABC = 90°, the side AC is the hypotenuse. According to the Pythagorean theorem:

AC = √(AB² + BC²) AC = √(5² + 12²) = √(25 + 144) = √(169) = 13 cm

The radius of the circumcircle is therefore half of the hypotenuse: R = (AC)/(2) = ¹³⁄₂ = 6.5 cm

Explanation

In an equilateral triangle, the circumcenter and the centroid coincide. The centroid divides the median, which also serves as the height or altitude in an equilateral triangle, in the ratio 2:1. The distance from the vertex to the centroid is the circumradius R, while the distance from the centroid to the midpoint of the opposite side is the inradius r.

The height h of the triangle is the sum of these two segments: h = R + r. In an equilateral triangle, the circumradius is twice the inradius, such that R = 2r. By substituting r = (R)/(2) into the height equation, we get the direct relationship h = ³⁄₂R.

Given the circumradius R = 8 cm, the height is calculated as: h = ³⁄₂ × 8 = 12 cm.

Explanation

According to the power of a point theorem, the square of the length of a tangent segment from an external point to a circle is equal to the product of the lengths of the segments of any secant line passing through that point and the circle. In this configuration, P is a point on the line BA produced, making PAB a common secant for both intersecting circles. For the first circle, where PT is the tangent, the relation is PT² = PA · PB. For the second circle, where PQ is the tangent, the relation is PQ² = PA · PB. By comparing these two equations, it is evident that PT² = PQ². Taking the square root of both sides, we find that the lengths of the tangents are equal, PT = PQ.

Explanation

The circumcentre O is the centre of the circle passing through the vertices A, B, and C of the triangle. By the circle theorem, the angle subtended by an arc BC at the centre is double the angle subtended by it at any point on the remaining part of the circle. Therefore, ∠BOC = 2∠BAC, which can be written as ∠BOC = 2∠A.

In △BOC, the sides OB and OC are equal as they are both radii of the circumcircle. This makes △BOC an isosceles triangle. Given that OD ⊥ BC, the altitude OD from the vertex O to the base BC of an isosceles triangle also bisects the vertex angle ∠BOC. Consequently, ∠BOD = ½∠BOC. Substituting the relationship derived from the circle theorem, ∠BOD = ½(2∠A) = ∠A.

Explanation

The expression is simplified using the fundamental trigonometric identities 1 + cot²θ = csc²θ and 1 + tan²θ = sec²θ. Substituting these into the terms gives:

(1)/(csc²θ) + (3)/(sec²θ) + 2sin²θ

By applying the reciprocal identities where sinθ = (1)/(cscθ) and cosθ = (1)/(secθ), the expression transforms into:

sin²θ + 3cos²θ + 2sin²θ

Combining the sin²θ terms results in:

3sin²θ + 3cos²θ

Factoring out the common coefficient 3, we use the Pythagorean identity sin²θ + cos²θ = 1:

3(sin²θ + cos²θ) = 3(1) = 3

Explanation

Using the trigonometric identities 1 + tan²θ = sec²θ and 1 + cot²θ = csc²θ, the given expression can be simplified as follows:

The first term is (4)/(1 + tan²θ) = (4)/(sec²θ) = 4cos²θ. The second term is (1)/(1 + cot²θ) = (1)/(csc²θ) = sin²θ.

Substituting these into the original equation, we have: 4cos²θ + sin²θ + 3sin²θ

Combining the sin²θ terms: 4cos²θ + 4sin²θ

Factoring out the common coefficient: 4(cos²θ + sin²θ)

Applying the fundamental Pythagorean identity sin²θ + cos²θ = 1: 4(1) = 4

Explanation

Since the given expression is a trigonometric identity whose value is independent of x, the most efficient approach for competition exams is the substitution method. By substituting a convenient value such as x = 0, the calculation becomes straightforward because sin 0 = 0 and cos 0 = 1.

Substituting x = 0 into the expression: 3(sin 0 - cos 0)⁴ + 6(sin 0 + cos 0)² + 4(sin⁶ 0 + cos⁶ 0) = 3(0 - 1)⁴ + 6(0 + 1)² + 4(0⁶ + 1⁶) = 3(-1)⁴ + 6(1)² + 4(1) = 3(1) + 6(1) + 4 = 13

Alternatively, using algebraic expansion and trigonometric identities, the expression can be simplified as follows: 3((sin x - cos x)²)² + 6(1 + 2 sin x cos x) + 4((sin² x + cos² x)³ - 3 sin² x cos² x (sin² x + cos² x)) = 3(1 - 2 sin x cos x)² + 6(1 + 2 sin x cos x) + 4(1 - 3 sin² x cos² x) = 3(1 - 4 sin x cos x + 4 sin² x cos² x) + 6 + 12 sin x cos x + 4 - 12 sin² x cos² x = 3 - 12 sin x cos x + 12 sin² x cos² x + 6 + 12 sin x cos x + 4 - 12 sin² x cos² x Summing the constant terms 3 + 6 + 4 yields 13, while the terms involving sin x cos x and sin² x cos² x cancel each other out.

Explanation

The expression within the parentheses can be simplified by identifying the common denominator and using trigonometric identities. Given the structure, we combine the terms:

(1 + sin θ)/(cos θ) - (cos θ)/(1 + sin θ) = ((1 + sin θ)² - cos² θ)/(cos θ (1 + sin θ))

Expanding the numerator: 1 + 2sin θ + sin² θ - cos² θ

Using the identity 1 - cos² θ = sin² θ, the numerator becomes: sin² θ + 2sin θ + sin² θ = 2sin² θ + 2sin θ = 2sin θ (1 + sin θ)

Substituting this back into the fraction: (2sin θ (1 + sin θ))/(cos θ (1 + sin θ)) = (2sin θ)/(cos θ) = 2tan θ

Now, multiply this by the external term cosec θ sec θ: ((1)/(sin θ) · (1)/(cos θ)) · 2tan θ = (1)/(sin θ cos θ) · (2sin θ)/(cos θ) = (2)/(cos² θ) = 2sec² θ

Finally, subtract the remaining term 2tan² θ: 2sec² θ - 2tan² θ = 2(sec² θ - tan² θ)

Since sec² θ - tan² θ = 1, the expression evaluates to: 2(1) = 2

Explanation

Let sin Θ = x. Since csc Θ = (1)/(sin Θ), the given equation sin Θ + csc Θ = 2 can be written as x + ¹⁄ₓ = 2. Multiplying the entire equation by x yields the quadratic equation x² + 1 = 2x, which simplifies to x² - 2x + 1 = 0. This expression is a perfect square, (x - 1)² = 0, which results in the unique solution x = 1.

Consequently, we find that sin Θ = 1 and csc Θ = ¹⁄₁ = 1. Substituting these values into the required expression, we get sin⁵ Θ + csc⁵ Θ = 1⁵ + 1⁵ = 1 + 1 = 2. This follows the general algebraic property where if x + ¹⁄ₓ = 2 for a real variable x, then xⁿ + (1)/(xⁿ) = 2 for any positive integer power n.

Explanation

To find the fourth root of 24010000, we can decompose the number into its powers of 10 and its coefficient: 24010000 = 2401 × 10⁴. Applying the property of exponents ⁴√(a × b) = ⁴√a × ⁴√b, we calculate the fourth root of each factor individually.

The number 2401 is the fourth power of 7, as 7 × 7 × 7 × 7 = 49 × 49 = 2401. Therefore, ⁴√(2401) = 7. For the second factor, the fourth root of 10⁴ is simply 10. Multiplying these results gives 7 × 10 = 70. This can be verified by raising 70 to the fourth power: 70⁴ = (7 × 10)⁴ = 7⁴ × 10⁴ = 2401 × 10000 = 24010000.

Explanation

The largest 4-digit number is 9999. Since 100² = 10000 is the smallest 5-digit number, the largest 4-digit perfect square must be the square of the integer immediately preceding 100, which is 99.

To calculate 99² quickly, use the identity (a - b)² = a² - 2ab + b²: 99² = (100 - 1)² 99² = 100² - 2(100)(1) + 1² 99² = 10000 - 200 + 1 99² = 9801

Alternatively, using the long division method for finding the square root of 9999, we find that 99 is the quotient and 198 is the remainder. Subtracting the remainder from the original number yields the largest perfect square: 9999 - 198 = 9801.

Explanation

Let the total work be the Least Common Multiple (LCM) of 12, 15, and 20, which is 60 units.

The daily work efficiency of Ram and Shyam together is 60 / 12 = 5 units per day. The daily work efficiency of Shyam and Hari together is 60 / 15 = 4 units per day. The daily work efficiency of Hari and Ram together is 60 / 20 = 3 units per day.

Summing these efficiencies, we get twice the combined efficiency of Ram, Shyam, and Hari: 2 × (Ram + Shyam + Hari) = 5 + 4 + 3 = 12 units per day. Therefore, the combined efficiency of Ram, Shyam, and Hari is 12 / 2 = 6 units per day.

To find the efficiency of Ram alone, subtract the combined efficiency of Shyam and Hari from the total combined efficiency: Ram's efficiency = (Ram + Shyam + Hari) - (Shyam + Hari) = 6 - 4 = 2 units per day.

The time taken by Ram to complete the work alone is: Total work / Ram's efficiency = 60 / 2 = 30 days.

Explanation

The most efficient way to solve this type of problem is by using the combined efficiency formula for and/or scenarios: Required Days = (D)/(M₂₎{M₁ + W₂{W₁}} In this formula, D is the number of days given in the initial condition (12), M₁ and W₁ are the counts in the or condition (3 men and 5 women), and M₂ and W₂ are the counts in the and condition (6 men and 5 women). Substituting the given values into the formula: Required Days = (12)/(6){3 + ⁵⁄₅} Required Days = (12)/(2 + 1) Required Days = ¹²⁄₃ = 4 days.

Explanation

Let the total work be the LCM of 12, 15, and 20, which is 60 units. The daily work done by A and B together is 60 / 12 = 5 units. The daily work done by B and C together is 60 / 15 = 4 units, and the daily work done by C and A together is 60 / 20 = 3 units. By adding these efficiencies, we get 2(A + B + C) = 5 + 4 + 3 = 12 units per day. Thus, the combined efficiency of A, B, and C is 12 / 2 = 6 units per day. To isolate the efficiency of A, we subtract the combined efficiency of B and C from the total: 6 - 4 = 2 units per day. The time taken by A to complete the work alone is the total work divided by A's efficiency, which is 60 / 2 = 30 days.

Explanation

The perimeter (circumference) of a circle is calculated using the formula C = π d, where d represents the diameter. According to the problem, the difference between the perimeter and the diameter is given as X. This relationship is expressed by the equation π d - d = X. By factoring out the diameter d from the left side, the equation becomes d(π - 1) = X. To isolate the diameter d, both sides of the equation are divided by (π - 1), yielding d = (X)/(π - 1). Thus, the diameter of the circle is X/(π - 1) units.

Explanation

The base of a right circular cylinder is a circle. The perimeter of the base, given as a, is equal to the circumference of the circle, 2π r, where r is the radius. From this, the radius can be expressed as r = a / (2π).

The volume V of a right circular cylinder is given by the formula V = π r² h, where h is the height. Substituting the expression for r into the volume formula yields V = π (a / (2π))² h. Simplifying this expression, V = π (a² / (4π²)) h, which reduces to V = (a² / (4π)) h. To find the height, the equation is rearranged as h = 4π V / a².

Explanation

According to Archimedes' principle, the volume of the water displaced by a completely submerged object is equal to the volume of the object itself. Let the rise in the water level in the cylindrical vessel be h.

The radius of the sphere is rₛ = ⁶⁄₂ = 3 cm, and the radius of the cylindrical vessel is r_c = ¹²⁄₂ = 6 cm.

The volume of the sphere is given by the formula Vₛ = ⁴⁄₃ π rₛ³. Substituting the radius, we get Vₛ = ⁴⁄₃ π (3)³ = ⁴⁄₃ π × 27 = 36π cm³.

The volume of the displaced water in the cylindrical vessel forms a cylinder of height h and radius r_c. This volume is V_w = π r_c² h = π (6)² h = 36π h cm³.

Equating the volume of the sphere to the volume of the displaced water: 36π h = 36π h = 1 cm

Explanation

The net gain percentage can be calculated using the successive percentage change formula x + y + ˣʸ⁄₁₀₀, where x represents the markup percentage and y represents the discount percentage. Given a markup of 20% (x = 20) and a discount of 15% (y = -15), the calculation is:

Net Gain % = 20 - 15 + (20 × (-15))/(100) Net Gain % = 5 - ³⁰⁰⁄₁₀₀ Net Gain % = 5 - 3 = 2%

Alternatively, if we assume the Cost Price is 100, the Marked Price becomes 120 (20% above cost). The discount is applied to the Marked Price, so 15% of 120 is calculated as 0.15 × 120 = 18. The Selling Price is the Marked Price minus the discount: 120 - 18 = 102. A Selling Price of 102 compared to a Cost Price of 100 yields a profit of 2, which is exactly 2%.

Explanation

The relationship between the Cost Price (CP) and the Marked Price (MP) can be expressed using the formula: (CP)/(MP) = (100 - Discount percentage)/(100 + Profit percentage)

Given a discount of 10% and a profit of 12%, substituting these values into the formula gives: (CP)/(MP) = (100 - 10)/(100 + 12) = ⁹⁰⁄₁₁₂

To find the simplest form of the ratio, both the numerator and the denominator are divided by their greatest common divisor, which is 2: (90 ÷ 2)/(112 ÷ 2) = ⁴⁵⁄₅₆

This calculation provides the ratio of the cost price to the printed price.

Explanation

The list price of the article is 160. After the first discount of 10% is applied, the price reduces to: 160 - (10% of 160) = 160 - 16 = 144

The customer eventually buys the article for 122.40 after a second successive discount. This second discount is applied to the reduced price of 144. The absolute value of the second discount is: 144 - 122.40 = 21.60

To find the second discount percentage, we use the formula: Second Discount % = (Second Discount Amount)/(Price after first discount) × 100 Second Discount % = (21.60)/(144) × 100 = ²¹⁶⁰⁄₁₄₄

Performing the division: 2160 / 144 = 15%

Explanation

Let the number of boys be B and the number of girls be G. According to the problem, 10% of the number of girls is equal to 1/20 of the number of boys. Expressing the percentage as a fraction, 10% = 10/100 = 1/10. The given condition can be written as the following equation:

(1/10) × G = (1/20) × B

To find the ratio of the number of boys to the number of girls (B:G), rearrange the terms to isolate B/G:

B / G = (1/10) / (1/20) B / G = (1/10) × (20/1) B / G = 2 / 1

This simplifies to B:G = 2:1.

Explanation

The expression relates to the ratio of exponents where the terms are 25²^.⁵ : 5³. To simplify this, we express the base 25 as a power of 5 using the identity 25 = 5².

Substituting this into the expression and applying the power rule (aᵐ)ⁿ = aᵐ^×ⁿ: 25²^.⁵ = (5²)²^.⁵ = 5²^ײ^.⁵ = 5⁵

The ratio then becomes: 5⁵ : 5³

Using the law of exponents for division aᵐ / aⁿ = aᵐ⁻ⁿ: (5⁵)/(5³) = 5⁵⁻³ = 5² = 25

This results in the ratio 25:1.

Explanation

In an Arithmetic Progression with an odd number of terms, the average is always equal to the middle term. Since a, b, c, d, e are five consecutive odd numbers, they form an Arithmetic Progression with a common difference of 2. We can express each term in relation to the first term a as follows: b = a + 2, c = a + 4, d = a + 6, and e = a + 8.

The average is calculated by dividing the sum of the terms by the total number of terms: Average = (a + b + c + d + e)/(5) Average = (a + (a + 2) + (a + 4) + (a + 6) + (a + 8))/(5) Average = (5a + 20)/(5) Average = a + 4

Since there are five terms, the middle term is the third term, which is c = a + 4. Substituting the values into the average formula directly yields a + 4.

Explanation

The total sum of the 20 numbers is found by multiplying the count by the average: 20 × 15 = 300. Similarly, the sum of the first five numbers is 5 × 12 = 60. The sum of the remaining 15 numbers is the difference between the total sum and the sum of the first five: 300 - 60 = 240. The average of these remaining numbers is ²⁴⁰⁄₁₅ = 16.

Alternatively, using the deviation method, the average of the first five numbers is 3 units less than the overall average (15 - 12 = 3). This creates a total deficit of 5 × 3 = 15 units. To maintain the overall average of 15, the remaining 15 numbers must collectively provide an extra 15 units. Distributing this surplus equally among the 15 remaining numbers adds ¹⁵⁄₁₅ = 1 unit to the overall average. Consequently, the average of the rest is 15 + 1 = 16.

Explanation

The difference in the percentage values between a loss of 20% and a gain of 5% represents the total change in the selling price. Since the transition is from a 20% loss to a 5% gain, the total percentage increase is:

5% - (-20%) = 25%

This 25% increase in value corresponds to the absolute increase in the selling price, which is given as 100. Let the Cost Price be CP.

25% of CP = 100

²⁵⁄₁₀₀ × CP = 100

CP = (100 × 100)/(25) = 400

Alternatively, if the Cost Price is taken as 100 units, a 20% loss results in a selling price of 80 units, and a 5% gain results in a selling price of 105 units. The difference between these two selling prices is 25 units. Given that 25 units is equal to 100, then 1 unit is equal to 4. Therefore, the Cost Price (100 units) is 100 × 4 = 400.

Explanation

Let the original price of the article be x. The problem states that the price is first decreased by 20% and then increased by 30%, resulting in a final price of 416.

The relationship between the original price and the final price can be expressed using decimal multipliers for successive percentage changes: x × (1 - 0.20) × (1 + 0.30) = 416 x × 0.8 × 1.3 = 416

Multiplying the decimal factors gives the net multiplier: x × 1.04 = 416

Solving for x: x = (416)/(1.04) x = ⁴¹⁶⁰⁰⁄₁₀₄ x = 400

This can also be calculated using the net percentage change formula a + b + ᵃᵇ⁄₁₀₀, where a = -20 and b = 30. The net change is -20 + 30 + ⁽⁻²⁰⁾⁽³⁰⁾⁄₁₀₀ = 10 - 6 = 4%. An increase of 4% on the original price yields the final price: 104% of x = 416 1.04x = 416 x = 400

Explanation

To find the total journey distance x, we sum the fractions of the journey covered by different modes of transport. The fraction of the journey covered by train is ²⁄₁₅ and by bus is ⁹⁄₂₀.

The total fraction covered by train and bus is: ²⁄₁₅ + ⁹⁄₂₀ = (8 + 27)/(60) = ³⁵⁄₆₀ = ⁷⁄₁₂

The remaining portion of the journey covered on foot represents the difference between the whole journey and the parts covered by vehicle: 1 - ⁷⁄₁₂ = ⁵⁄₁₂

Since the distance covered on foot is given as 10 km, we set up the equation: ⁵⁄₁₂ × x = 10 x = (10 × 12)/(5) x = 2 × 12 = 24 km

Explanation

For a constant distance, speed is inversely proportional to time. If the man walks at ¾ of his usual speed, the ratio of his usual speed to his new speed is 4 : 3. Consequently, the ratio of his usual time to his new time is 3 : 4.

The difference between the new time and the usual time is 4 - 3 = 1 unit. According to the problem, this 1 unit difference corresponds to a delay of 20 minutes. Since the usual time is represented by 3 units in the ratio, the calculation is 3 × 20 = 60 minutes.

Explanation

The relationship between compound interest (CI) and simple interest (SI) for a period of two years at a rate of interest R is given by the formula CI = SI × (1 + (R)/(200)).

Substituting the given values into the formula: 2,544 = SI × (1 + ¹²⁄₂₀₀) 2,544 = SI × (1 + ³⁄₅₀) 2,544 = SI × ⁵³⁄₅₀

Solving for simple interest: SI = (2,544 × 50)/(53) SI = 48 × 50 SI = 2,400

Alternatively, the effective rate of simple interest for 2 years is 12% + 12% = 24%. The effective rate of compound interest for 2 years is calculated using the successive percentage formula x + y + ˣʸ⁄₁₀₀, which gives 12 + 12 + (12 × 12)/(100) = 25.44%. Given that 25.44% of the sum is 2,544, the simple interest representing 24% of the sum is (2,544)/(25.44) × 24 = 100 × 24 = 2,400.

Explanation

Let b be the cost of a bucket and m be the cost of a mug. According to the problem, we can establish the following system of equations:

8b + 5m = 92 5b + 8m = 77

Summing the two equations yields 13b + 13m = 169. Dividing this entire equation by 13 gives b + m = 13. Subtracting the second equation from the first yields 3b - 3m = 15. Dividing this equation by 3 gives b - m = 5.

To find the individual costs, add the two simplified equations: (b + m) + (b - m) = 13 + 5, which simplifies to 2b = 18, resulting in b = 9. Substituting b = 9 into b + m = 13 gives 9 + m = 13, resulting in m = 4.

The cost of 2 mugs and 3 buckets is calculated as 2m + 3b = 2(4) + 3(9) = 8 + 27 = 35.