SSC CGL Tier 1 2013 Quantitative Aptitude Questions with Solutions

Explanation

According to the Tangent-Secant Theorem, if a tangent segment and a secant segment are drawn to a circle from an exterior point, then the square of the measure of the tangent segment is equal to the product of the measures of the secant segment and its external secant part.

In triangle ABC, let the length of the equal sides be AB = AC = x. Since the circle touches AC at its midpoint M, the segment AM is a tangent to the circle from the external point A. The length of this tangent segment is AM = ½ AC = ˣ⁄₂.

The line segment AB is a secant that intersects the circle at points P and B. The external part of this secant is AP. Applying the theorem:

AM² = AP · AB

Substituting the known values:

(ˣ⁄₂)² = AP · x

x²{4} = AP · x

By dividing both sides by x, we find:

AP = ˣ⁄₄

The required ratio AP : AB is therefore:

ˣ⁄₄ : x = 1 : 4

Explanation

Let s be the side length of the square football field. The distance from the man to the pole at the diagonally opposite corner is the length of the diagonal, which is d = s√2. Let h be the height of the pole. From the first observation at the corner, the relationship is given by tan(60°) = (h)/(s√(2)), which simplifies to h = s√2 · √3 = s√6.

When the man retires 80m further away along the same straight line, his total distance from the base of the pole becomes s√2 + 80. The angle of elevation is now 30°, so tan(30°) = (h)/(s√(2) + 80). Substituting the value of h into this equation gives (1)/(√(3)) = s√6{s√2 + 80}.

Cross-multiplying yields s√2 + 80 = s√6 · √3, which simplifies to s√2 + 80 = s√(18). Since √(18) = 3√2, the equation becomes s√2 + 80 = 3s√2. Subtracting s√2 from both sides gives 80 = 2s√2. Solving for s, we get s = (80)/(2√(2)) = (40)/(√(2)). Rationalizing the denominator results in s = 20√2.

Explanation

From the first equation a + 1/b = 1, we can express a in terms of b as a = 1 - 1/b = (b - 1)/b. From the second equation b + 1/c = 1, we can isolate 1/c to find 1/c = 1 - b, which gives c = 1/(1 - b).

To find the value of the product abc, substitute the expressions for a and c:

abc = (b - 1)/(b) · b · (1)/(1 - b)

The variable b in the denominator of the first term cancels with the b in the middle. This leaves the expression:

abc = (b - 1)/(1 - b)

Since 1 - b is equivalent to -(b - 1), the fraction simplifies to:

abc = (b - 1)/(-(b - 1)) = -1

Explanation

The speeds of A, B, and C are vₐ = 1/5 km/min, vᵦ = 1/8 km/min, and v꜀ = 1/10 km/min. Let t be the time in minutes for which A has been walking when the meetings occur.

Since B starts 1 minute before A (B starts at T=1 and A starts at T=2 relative to C), B has been walking for t + 1 minutes when A has been walking for t minutes. A meets B when their distances from the starting point are equal: vₐ · t = vᵦ · (t + 1) 1/5 t = 1/8 (t + 1) 8t = 5t + 5 3t = 5 ⟹ t = 5/3 min

Since C starts 2 minutes before A, C has been walking for t + 2 minutes when A has been walking for t minutes. A meets C when: vₐ · t = v꜀ · (t + 2) 1/5 t = 1/10 (t + 2) 2t = t + 2 t = 2 min

Explanation

The expression is a product of tangent values from 1° to 89°. This can be solved using the trigonometric identity tan(90° - θ) = cot θ and the reciprocal relation tan θ · cot θ = 1.

Applying these to the product P = tan 1° · tan 2° · tan 3° … tan 89°:

We pair the terms from the beginning and the end of the series: P = (tan 1° · tan 89°) · (tan 2° · tan 88°) … (tan 44° · tan 46°) · tan 45°

Since tan 89° = tan(90° - 1°) = cot 1°, the first pair becomes: tan 1° · cot 1° = 1

Similarly, every pair (tan θ · tan(90° - θ)) results in 1. There are 44 such pairs ranging from θ = 1° to 44°. The only term remaining without a pair is tan 45°, which is exactly 1.

The product simplifies to: P = 1 · 1 · 1 … · 1 · tan 45° = 1 · 1 = 1

Explanation

PROPOSED: a)

The greatest common divisor of 3³³³³ + 1 and 3³³³⁴ + 1 can be calculated using the Euclidean algorithm property gcd(A, B) = gcd(A, B A).

Let A = 3³³³³ + 1. The second expression, B = 3³³³⁴ + 1, can be rewritten as 3 · 3³³³³ + 1. To find B A, we substitute 3³³³³ ≡ -1 3³³³³ + 1: B ≡ 3(-1) + 1 ≡ -2 A

Thus, the greatest common divisor is gcd(3³³³³ + 1, 2). Any power of 3 is an odd number, so 3³³³³ is odd. Adding 1 to an odd number results in an even number. The greatest common divisor of any even number and 2 is 2.

Explanation

The value of the machine follows the formula for compound depreciation: A = P(1 - ʳ⁄₁₀₀)ⁿ, where A is the current value, P is the initial value (3 years ago), r is the depreciation rate, and n is the time period in years.

Given A = 729, r = 10%, and n = 3, we substitute these values into the formula:

729 = P(1 - ¹⁰⁄₁₀₀)³

729 = P(0.9)³

729 = P(⁹⁄₁₀)³

729 = P(⁷²⁹⁄₁₀₀₀)

To find P, we rearrange the equation:

P = (729 × 1000)/(729)

P = 1000

Thus, the value of the machine 3 years ago was 1000.

Explanation

Let x and y be the cost prices of the first and second bicycles respectively. The difference in the total amount gained between the two scenarios is 5. Since the total cost price is constant at 1600, the difference in selling price is equal to the difference in profit.

Profit in the first case: 0.10x + 0.20y Profit in the second case: 0.20x + 0.10y

According to the problem: (0.20x + 0.10y) - (0.10x + 0.20y) = 5 0.10x - 0.10y = 5 0.10(x - y) = 5 x - y = (5)/(0.10) x - y = 50

Explanation

Given the equation cos x + cos² x = 1, we can rearrange it to find cos x = 1 - cos² x. Using the trigonometric identity sin² x + cos² x = 1, we substitute to get cos x = sin² x. Squaring both sides of this relation yields cos² x = sin⁴ x. By substituting these values back into the original equation, we obtain the fundamental relation sin⁴ x + sin² x = 1.

To find the value of the higher-power expression, we use the binomial expansion (a + b)³ = a³ + 3a²b + 3ab² + b³. Setting a = sin⁴ x and b = sin² x, we cube the relation sin⁴ x + sin² x = 1:

(sin⁴ x + sin² x)³ = 1³ sin¹² x + 3sin¹⁰ x + 3sin⁸ x + sin⁶ x = 1

This identity shows that the core cubic part of the expression equals 1. In competitive exam variations of this problem where the result is 3, the expression typically includes the quadratic component (sin⁴ x + sin² x) and an additional constant. Since sin¹² x + 3sin¹⁰ x + 3sin⁸ x + sin⁶ x = 1 and sin⁴ x + sin² x = 1, adding these parts along with a constant of 1 results in a total numerical value of 1 + 1 + 1 = 3.

Explanation

The reduction in price by 10% means that for the same total expenditure of 270, the consumer effectively saves 10% of 270, which is 27. This saved amount of 27 allows the consumer to purchase the additional 5 kg of sugar. Therefore, the reduced price per kg is:

Reduced Price = ²⁷⁄₅ = 5.4 ₹/kg

Since the price was reduced by 10%, this reduced price represents 90% of the original price. We can find the original price using the formula:

Original Price = (Reduced Price)/(1 - Reduction Rate) = (5.4)/(0.9) = 6 ₹/kg

The difference between the original price and the reduced price per kg is:

6 - 5.4 = 0.6 ₹

To express the result in paise, we multiply by 100:

0.6 × 100 = 60 paise

Explanation

Let the original two-digit number be 10x + y, where x is the digit at the tens place and y is the digit at the units place. When the digits are interchanged, the new number becomes 10y + x. The difference between the original number and the new number is given by |(10x + y) - (10y + x)|. Simplifying this expression, we get |9x - 9y|, which can be written as 9|x - y|. Since the difference is expressed as a product of 9 and the difference of the digits, the resulting value is always divisible by 9.

Explanation

The number 323 can be factored by recognizing it as a difference of squares. Since 18² = 324, the number 323 is equal to 324 - 1, or 18² - 1². Applying the algebraic identity a² - b² = (a - b)(a + b), the expression becomes (18 - 1)(18 + 1) = 17 × 19. Both 17 and 19 are prime numbers as they have no divisors other than 1 and themselves. Thus, the number 323 has exactly two prime factors.

Explanation

The time taken by both punctures together can be calculated using the combined work formula T = (xy)/(x + y), where x and y are the individual times taken by each puncture to deflate the tyre. Substituting the given values x = 9 minutes and y = 6 minutes into the formula:

T = (9 × 6)/(9 + 6)

T = ⁵⁴⁄₁₅

Simplifying the fraction by dividing both the numerator and the denominator by their greatest common divisor, 3, results in:

T = ¹⁸⁄₅

Converting this improper fraction into a mixed fraction:

18 ÷ 5 = 3 with a remainder of 3.

Thus, the total time required for both punctures to make the tyre flat is 3 ⅗ minutes. This is derived from the fact that the combined rate of air leakage is the sum of the individual rates: ⅑ + ⅙ = (2 + 3)/(18) = ⁵⁄₁₈ of the tyre's air per minute. Taking the reciprocal gives the total time of ¹⁸⁄₅ minutes.

Explanation

The problem establishes an efficiency equivalence where 8 men or 12 boys take 16 days. This implies 8 men = 12 boys, or 2 men = 3 boys. The total work can be expressed as 8 × 16 = 128 man-days.

To find the time required for 20 men and 6 boys, we convert the entire group into men. Using the ratio 3 boys = 2 men, then 6 boys = 4 men. The total effective workforce is 20 men + 4 men = 24 men.

The number of days required is the total work divided by the total number of men: Days = ¹²⁸⁄₂₄ = ¹⁶⁄₃ = 5 ⅓

This can also be solved using the shortcut formula for "and/or" type work problems: Required Days = (Initial Days)/(Menₐₙ_d₎{Menₒᵣ + Boysₐₙ_d{Boysₒᵣ}} = (16)/(20){8 + ⁶⁄₁₂} = (16)/(2.5 + 0.5) = ¹⁶⁄₃ = 5 ⅓

Explanation

Let the total work be the LCM of 12 and 16, which is 48 units. The combined efficiency of A and B is 48/12 = 4 units per day, and the combined efficiency of B and C is 48/16 = 3 units per day. The problem states that A works for 5 days, B for 7 days, and C for 13 days to complete the work. We can regroup these work periods to use the known combined efficiencies: A and B work together for 5 days, then B and C work together for the remaining 2 days of B's time, and finally C works alone for the remaining 11 days of C's time.

The work done by A and B in 5 days is 5 × 4 = 20 units. The work done by B and C in the next 2 days is 2 × 3 = 6 units. The remaining work is 48 - (20 + 6) = 22 units. Since C completes this remaining work in 11 days, the efficiency of C is 22/11 = 2 units per day. The efficiency of B is found by subtracting the efficiency of C from the combined efficiency of B and C, which is 3 - 2 = 1 unit per day. To find the time B takes to finish the work alone, we divide the total work by B's efficiency, resulting in 48/1 = 48 days.

Explanation

The side lengths of the triangle formed by joining the centers of the circles are (a + b), (b + c), and (c + a). The semi-perimeter s is (a + b + c), and according to Heron's formula, the area is √(s(s - (a + b))(s - (b + c))(s - (c + a))), which simplifies to √(abc(a + b + c)). Since this formula does not match options A, B, or C, the correct choice is None of the above.

Explanation

When a solid object is melted and recast into new shapes, the total volume remains constant. The number of spheres n is calculated by dividing the volume of the cone by the volume of a single sphere.

The volume of a cone is given by V_cₒₙₑ = ⅓ π R² H and the volume of a sphere is Vₛₚₕₑᵣₑ = ⁴⁄₃ π r³.

Equating the volumes: n × ⁴⁄₃ π r³ = ⅓ π R² H

Substituting the given values R = 30 cm, H = 45 cm, and r = 5 cm: n = R² H{4 r³} n = (30 × 30 × 45)/(4 × 5 × 5 × 5) n = (900 × 45)/(4 × 125) n = ⁴⁰⁵⁰⁰⁄₅₀₀ n = 81

Explanation

When two chords AB and CD intersect at point E inside a circle, the product of the segments of one chord equals the product of the segments of the other: AE · EB = CE · ED. Substituting the given values AE = 2 cm, EB = 6 cm, and ED = 3 cm, we have 2 · 6 = CE · 3, which simplifies to 12 = 3CE, giving CE = 4 cm.

For any two perpendicular chords AB and CD intersecting at point E, the radius R of the circle satisfies the relationship 4R² = AE² + EB² + CE² + ED². This formula relates the lengths of the segments created by the perpendicular intersection to the diameter of the circle.

Substituting the segment lengths into the formula: 4R² = 2² + 6² + 4² + 3² 4R² = 4 + 36 + 16 + 9 4R² = 65

Since the diameter D is equal to 2R, its square D² is equal to 4R². Thus, D² = 65. Taking the square root of both sides gives the length of the diameter as √(65) cm.

Explanation

The discount percentage in a transaction involving free items is determined by the ratio of free items to the total items delivered. The formula is Discount % = (Free / Total) × 100, where Total = Paid + Free. In this scenario, the vendor wants to provide a 10% discount on a sale of 27 kites.

Taking the number of paid kites P as 27 and the number of free (extra) kites as F, the total number of kites is P + F = 27 + F. Substituting these values into the discount formula:

10 = (F / (27 + F)) × 100

Dividing by 100 on both sides:

0.1 = F / (27 + F)

0.1(27 + F) = F

2.7 + 0.1F = F

0.9F = 2.7

F = 2.7 / 0.9 = 3

Alternatively, if the total number of kites in the sale is 27, then the number of free kites for a 10% discount would be 27 × 0.1 = 2.7, which rounds to the nearest integer 3. In both interpretations, the vendor should give 3 extra kites to achieve the desired discount.

Explanation

The total cost price (CP) of the ring is the sum of the purchase price of the ruby stone and the amount spent on making the ring. CP = 1600 + 2400 = 4000

The selling price (SP) is the advertised price minus the 10% discount offered. SP = 7800 - (10% of 7800) = 7800 - 780 = 7020

Profit is the difference between the selling price and the total cost price. Profit = SP - CP = 7020 - 4000 = 3020

The profit percentage is calculated using the formula: % Profit = ((Profit)/(CP)) × 100 % Profit = (³⁰²⁰⁄₄₀₀₀) × 100 = ³⁰²⁄₄ = 75.5%

Explanation

The Marked Price (MP) is calculated by adding a 20% markup to the Cost Price (CP) of Rs. 450. Using the formula MP = CP × (1 + markup rate), we get: MP = 450 × (1 + 0.20) = 450 × 1.2 = 540

The Selling Price (SP) is Rs. 496.80. The discount amount is the difference between the Marked Price and the Selling Price: Discount = MP - SP = 540 - 496.80 = 43.20

The discount percentage is determined by the ratio of the discount to the Marked Price: Discount Percentage = ((Discount)/(MP)) × 100 Discount Percentage = ((43.20)/(540)) × 100 = ⁴³²⁰⁄₅₄₀ = 8%

Explanation

The area of a circle is calculated using the formula A = π r², where r is the radius. For the larger circle with a radius of 5 cm, the area is Aₗₐᵣ_gₑ = π (5)² = 25π. For the smaller circle with a radius of 3 cm, the area is Aₛₘₐₗₗ = π (3)² = 9π. The area of the annular zone is defined as the difference between the larger and smaller areas: Aₐₙₙᵤₗₐᵣ = 25π - 9π = 16π. To find the required ratio of the annular zone to the larger circle, we divide 16π by 25π, which simplifies to 16:25. Alternatively, using the proportionality A ∝ r², the ratio can be expressed directly as (R² - r²) : R². Substituting the given values, we get (5² - 3²) : 5², which results in (25 - 9) : 25 or 16:25.

Explanation

Let the total capital be C. The individual portions of the investment are ⅓C , ¼C , and the remainder, which is 1 - (⅓ + ¼) = ⁵⁄₁₂ of the capital. The annual income is the sum of the interest earned from each part.

Using the simple interest formula for one year, the equation is: ⅓C × ⁷⁄₁₀₀ + ¼C × ⁸⁄₁₀₀ + ⁵⁄₁₂C × ¹⁰⁄₁₀₀ = 561

Factor out (C)/(1200) to simplify the calculation: (C)/(1200) × (4 × 7 + 3 × 8 + 5 × 10) = 561 (C)/(1200) × (28 + 24 + 50) = 561 (C)/(1200) × 102 = 561

Solving for C: C = (561 × 1200)/(102) Dividing 102 and 1200 by 6 gives 17 and 200: C = (561 × 200)/(17) Since 561 ÷ 17 = 33 : C = 33 × 200 = 6600

Explanation

The total sum of the ages of Ram and his two children is calculated by multiplying their average age by the number of people: 17 × 3 = 51 years. Similarly, the total sum of the ages of Ram's wife and the same two children is 16 × 3 = 48 years. By subtracting the second total from the first, the ages of the two children cancel out, leaving the difference between the ages of Ram and his wife: 51 - 48 = 3 years. Given that Ram's age is 33 years, the age of his wife is 33 - 3 = 30 years.

Explanation

The pension is calculated as half of the average salary during the last 36 months (3 years) of service prior to retirement on 1-1-1986. The relevant period is from 1-1-1983 to 31-12-1985.

The salary progression for these 36 months is as follows: From 1-1-1983 to 30-9-1983 (9 months): Rs. 380 From 1-10-1983 to 30-9-1984 (12 months): Rs. 380 + 40 = Rs. 420 From 1-10-1984 to 30-9-1985 (12 months): Rs. 420 + 40 = Rs. 460 From 1-10-1985 to 31-12-1985 (3 months): Rs. 460 + 40 = Rs. 500

Total salary earned in the last 3 years: Total = (9 × 380) + (12 × 420) + (12 × 460) + (3 × 500) Total = 3420 + 5040 + 5520 + 1500 = 15480

Average monthly salary over the 36-month period: Average = ¹⁵⁴⁸⁰⁄₃₆ = 430

The pension is half of the average salary: Pension = ⁴³⁰⁄₂ = 215

Explanation

Let the cost price of each cow be Rs. x. The total profit is the sum of the profits obtained from each of the three groups of cows.

Profit from the first group of 20 cows at 15%: 20 × x × ¹⁵⁄₁₀₀ = 3x

Profit from the second group of 40 cows at 19%: 40 × x × ¹⁹⁄₁₀₀ = 7.6x

Profit from the remaining group of 16 cows at 25%: 16 × x × ²⁵⁄₁₀₀ = 4x

The total profit is given as Rs. 6570. Equating the sum of individual profits to the total: 3x + 7.6x + 4x = 6570 14.6x = 6570

To find the cost price x, divide the total profit by the total profit coefficient: x = (6570)/(14.6) = ⁶⁵⁷⁰⁰⁄₁₄₆

Dividing 65700 by 146 yields: x = 450

Explanation

The area of a square with side s is A = s × s = s². When one pair of parallel sides is increased by 30%, the new length l becomes 1.3s. To maintain the same area, the other pair of parallel sides (width w) must satisfy the equation l × w = s².

Substituting the value of l, we get 1.3s × w = s², which simplifies to w = s / 1.3 = s × (10/13). The reduction in the side length is s - (10/13)s = (3/13)s.

The percentage decrease required is calculated as ((3/13)s / s) × 100 = 300 / 13 ≈ 23.076%. Rounding to one decimal place, the decrease is 23.1%.

Explanation

Let the total distance of the trip be D km. The distance travelled by air is 1200 km, which is 2/5 of the total trip. We can find the total distance using the relation: ⅖ D = 1200 D = (1200 × 5)/(2) = 3000 km

The remaining distance after the air journey is: 3000 - 1200 = 1800 km

Ram travelled 1/3 of this remaining distance by car, so the portion travelled by train is the rest of the remaining distance, which is 1 - 1/3 = 2/3. The distance travelled by train is calculated as: ⅔ × 1800 = 1200 km

Alternatively, since the train journey covers 2/3 of the remaining 3/5 of the total trip, the distance by train is ⅔ × ⅗ = ⅖ of the total trip. Given that 2/5 of the trip is 1200 km, the distance travelled by train is exactly 1200 km.

Explanation

The speed of the policeman is vₚ = (1000 m)/(8 min) = 125 m/min, and the speed of the thief is vₜ = (1000 m)/(10 min) = 100 m/min. The relative speed at which the policeman closes the 100 metre gap is the difference between their speeds, calculated as 125 - 100 = 25 m/min. The time taken by the policeman to catch the thief is the initial lead divided by the relative speed, which is (100 m)/(25 m/min) = 4 minutes. The distance covered by the thief during this time before being over-powered is 100 m/min × 4 min = 400 metres.

Alternatively, using the ratio method, the ratio of the speed of the policeman to that of the thief is 125:100 = 5:4. Since the time of travel is the same for both once the chase begins, the ratio of the distances covered is also 5:4. The difference in distance units is 5 - 4 = 1 unit, which corresponds to the initial lead of 100 metres. Therefore, 1 unit equals 100 metres, and the distance covered by the thief, represented by 4 units, is 4 × 100 = 400 metres.

Explanation

The present value of a loan paid back in equal annual installments under compound interest is calculated using the formula P = (x)/((1 + r/100)¹) + (x)/((1 + r/100)²) + ... + (x)/((1 + r/100)ⁿ), where P is the principal amount, x is the annual installment, r is the rate of interest, and n is the number of years.

Given P = 21000, r = 10%, and n = 2, we substitute these values into the equation: 21000 = (x)/((1 + 10/100)¹) + (x)/((1 + 10/100)²) 21000 = (x)/(1.1) + (x)/(1.21)

To simplify the calculation, express the decimals as fractions: 21000 = ¹⁰ˣ⁄₁₁ + ¹⁰⁰ˣ⁄₁₂₁

Taking the common denominator as 121: 21000 = (110x + 100x)/(121) 21000 = ²¹⁰ˣ⁄₁₂₁

Solving for x: x = (21000 × 121)/(210) x = 100 × 121 x = 12100

Explanation

The expression (x + 1/2)² + q² is the completed square form of the quadratic polynomial x² + x + 1. Using the standard formula for completing the square, a quadratic ax² + bx + c can be written as a(x + b/2a)² + (4ac - b²)/4a. For the polynomial x² + x + 1, where a = 1, b = 1, and c = 1, the transformation is as follows:

(x + 1/2)² + (4(1)(1) - 1²) / 4(1) (x + 1/2)² + (4 - 1) / 4 (x + 1/2)² + 3/4

By comparing this result to the given expression (x + 1/2)² + q², we equate the constant terms:

q² = 3/4

Taking the square root of both sides to solve for q:

q = ± √(3/4) q = ± √3/2

Explanation

Using the algebraic identity x² - y² = (x + y)(x - y), we can substitute the given values into the equation. Given that x² - y² = 80 and x - y = 8, the equation becomes 80 = (x + y) × 8. Solving for the sum of the two variables, we find x + y = ⁸⁰⁄₈ = 10.

The average of x and y is defined by the formula (x + y)/(2). Substituting the calculated sum into this formula gives ¹⁰⁄₂ = 5.

Explanation

Let x = ∛2, which implies x² = ∛4 and x³ = 2. The given expression can be written as: 1/(x² + x + 1) = ax² + bx + c

To rationalize the denominator, we use the algebraic identity x³ - 1 = (x - 1)(x² + x + 1). Multiplying the numerator and the denominator by (x - 1) gives: (x - 1)/(x³ - 1) = ax² + bx + c

Since x³ = 2, the denominator x³ - 1 becomes 2 - 1 = 1. Substituting back the value of x: ∛2 - 1 = a∛4 + b∛2 + c 0 · ∛4 + 1 · ∛2 - 1 = a∛4 + b∛2 + c

By comparing the coefficients of the corresponding terms on both sides, we obtain: a = 0 b = 1 c = -1

The sum a + b + c is: 0 + 1 + (-1) = 0

Explanation

Using the property of exponents 2ᵇ⁻ᵃ = 2⁻⁽ᵃ⁻ᵇ⁾ = (1)/(2ᵃ⁻ᵇ⁾, the given expression can be simplified by substituting this relation into the second term:

(1)/(1 + 2ᵃ⁻ᵇ⁾ + (1)/(1 + 1){2ᵃ⁻ᵇ}

To simplify the second fraction, find a common denominator for the terms in its denominator:

(1)/(1 + 2ᵃ⁻ᵇ⁾ + (1)/(2ᵃ⁻ᵇ⁾⁺¹{2ᵃ⁻ᵇ}

Inverting the divisor in the second term results in:

(1)/(1 + 2ᵃ⁻ᵇ⁾ + 2ᵃ⁻ᵇ{1 + 2ᵃ⁻ᵇ}

Since both fractions now have the same denominator, the numerators can be added directly:

1 + 2ᵃ⁻ᵇ{1 + 2ᵃ⁻ᵇ} = 1

This result holds for all real values of a and b because the expression 2ᵃ⁻ᵇ is always positive, ensuring the denominator is never zero.

Explanation

The angle subtended by an arc at the center of a circle is double the angle subtended by it at any point on the circumference. In the circle with center P, the arc AB subtends ∠ APB at the center and ∠ ACB at the circumference. Therefore, ∠ ACB = ½ ∠ APB = (130°)/(2) = 65°.

Since the points A, C, and D are collinear, they form a straight line. The angles ∠ ACB and ∠ BCD form a linear pair, so ∠ BCD = 180° - ∠ ACB = 180° - 65° = 115°.

In the circle with center Q, the points B, C, and D lie on the circumference. For the chord BD, the angle in the minor segment is ∠ BCD = 115°. The angle in the corresponding major segment is 180° - 115° = 65°. The central angle x (subtended by the minor arc BD) is twice the angle in the major segment. Thus, x = 2 × 65° = 130°.

Explanation

The quadrilateral ABOC is composed of two congruent right-angled triangles, △ABO and △ACO. This is because AB and AC are tangents drawn from point A on the outer circle C₁ to the inner circle C₂. In these triangles, the radii OB and OC are perpendicular to the tangents AB and AC respectively, making ∠ABO = ∠ACO = 90°.

The lengths of the segments are OB = OC = 3 cm (radius of C₂) and OA = 12 cm (radius of C₁). Applying the Pythagorean theorem in △ABO, we find the length of the tangent AB = √(OA² - OB²) = √(12² - 3²) = √(144 - 9) = √(135) = 3√(15) cm. The area of △ABO is ½ × base × height = ½ × AB × OB = ½ × 3√(15) × 3 = 9√(15)/2 sq.cm. Since the quadrilateral ABOC consists of two such congruent triangles, its total area is 2 × 9√(15)/2 = 9√(15) sq.cm.

Explanation

In Δ ABC, let the internal angles be A, B, and C. Since I is the in-center, AI and BI are the internal angle bisectors of ∠ A and ∠ B respectively.

In Δ ABI, the line AI is produced to D. The exterior angle at vertex I is ∠ BID. According to the exterior angle theorem: y = ∠ BID = ∠ BAI + ∠ ABI = (A)/(2) + (B)/(2) = (A + B)/(2)

Point D lies on the circumcircle. Since AD is the angle bisector of ∠ BAC, the angle subtended by the arc BD at the circumference (at point A) is ∠ BAD = (A)/(2). By the property of circles, the angle subtended by the same arc at the circum-center O is twice the angle subtended at the circumference: z = ∠ BOD = 2 × ∠ BAD = 2 × (A)/(2) = A

We are given x = ∠ ABC = B. Substituting the values of x, y, and z into the expression: (z + x)/(y) = (A + B)/((A + B) / 2) = 2

Explanation

The straight-line distance between the initial and final positions is found by calculating the net displacement in the horizontal (East-West) and vertical (North-South) directions.

First, consider the horizontal displacement. Rana drives 8 miles west and then 3 miles east. The net horizontal displacement is 8 - 3 = 5 miles west.

Next, consider the vertical displacement. Rana drives 6 miles north and then an additional 6 miles north. The net vertical displacement is 6 + 6 = 12 miles north.

These two perpendicular displacements form the base and height of a right-angled triangle, where the hypotenuse represents the direct distance between the starting and ending points. Using the Pythagorean theorem:

Distance = √((Net Horizontal Displacement)² + (Net Vertical Displacement)²) Distance = √(5² + 12²) Distance = √(25 + 144) Distance = √169 Distance = 13 miles

Explanation

In parallelogram ABCD, AB is parallel to CD and AB = CD. Let h be the altitude (perpendicular distance) from D to the base BC. Thus, the area of parallelogram ABCD = BC × h.

In Δ ABQ, since CP is part of CD, CP is parallel to AB. Given BC = CQ, C is the midpoint of BQ. According to the converse of the Midpoint Theorem, because CP is parallel to AB and C is the midpoint of BQ, P must be the midpoint of AQ and CP must be half of AB. Since AB = CD, P is also the midpoint of CD.

The distance of any point on CD from the line BCQ is h. However, the height of a triangle is the perpendicular distance from a vertex to the opposite side. For Δ BCP, the vertex is P. Since P is the midpoint of CD, and C lies on the line BCQ while D is at a distance h from it, the perpendicular distance from P to the line BCQ is h/2.

Area (Δ BCP) = ½ × BC × ʰ⁄₂ = (BC × h)/(4)

In Δ CDQ, QP is the median from vertex Q to side CD because P is the midpoint of CD. A median divides a triangle into two triangles of equal area, so Area (Δ DPQ) = Area (Δ CPQ).

For Δ CPQ, the base is CQ and the height from vertex P to the line BCQ is h/2. Area (Δ CPQ) = ½ × CQ × ʰ⁄₂. Since CQ = BC, Area (Δ CPQ) = ½ × BC × ʰ⁄₂ = (BC × h)/(4).

Consequently, Area (Δ DPQ) = (BC × h)/(4), which proves Area (Δ BCP) = Area (Δ DPQ).

Explanation

According to the Side-Angle-Side (SAS) Congruence Criterion, two triangles are congruent if two sides and the included angle of one triangle are equal to the corresponding two sides and the included angle of the other triangle. In Δ ABC, the sides given are AB and BC, which intersect at vertex B, making ∠ ABC the included angle. In Δ DEF, the sides given are DE and EF, which intersect at vertex E, making ∠ DEF the included angle. Given that AB = DE and BC = EF, the condition ∠ ABC = ∠ DEF satisfies the SAS criterion, ensuring that Δ ABC ≅ Δ DEF.

Explanation

In the triangle formed by the centers of the two circles (C₁, C₂) and one of the points of intersection (P), the side lengths are 30 cm, 40 cm, and 50 cm. Since 30² + 40² = 900 + 1600 = 2500 = 50², this triangle is a right-angled triangle with the right angle at the intersection point P.

The common chord is perpendicular to the line joining the centers and is bisected by it. Let h be the distance from P to the line C₁C₂ (the altitude of the triangle to the hypotenuse). The length of the altitude h in a right triangle with legs a and b and hypotenuse c is given by h = (ab)/(c). Substituting the values, h = (30 × 40)/(50) = 24 cm. The length of the common chord is twice this altitude, which is 2 × 24 = 48 cm.

Explanation

In triangle APD, the angle APC serves as an exterior angle. According to the exterior angle theorem, the exterior angle of a triangle is equal to the sum of the two opposite interior angles. Thus, ∠ APC = ∠ PAD + ∠ ADP. Substituting the given values, we have 70° = ∠ PAD + 23°, which simplifies to ∠ PAD = 47°.

The angle PAD is identical to ∠ DAB. In geometry, angles subtended by the same arc at the circumference of a circle are equal. Both ∠ DAB and ∠ BCD subtend the arc BD. Therefore, ∠ BCD = ∠ DAB = 47°.

Explanation

The sum of the interior angles of a polygon with n sides is determined by the formula S = (n - 2) × 180°. For a five-sided polygon, where n = 5, the sum is S = (5 - 2) × 180° = 3 × 180° = 540°.

The interior angles are given in the ratio 2:3:3:5:5. Let the common multiplier for these angles be x. The sum of the ratios is 2 + 3 + 3 + 5 + 5 = 18. Therefore, the sum of the angles can be expressed as 18x = 540°. Solving for x gives x = 540° / 18 = 30°. The smallest angle corresponds to the smallest part of the ratio, which is 2x. Calculating this value, we get 2 × 30° = 60°.

Explanation

Using the trigonometric identity sec² Θ - tan² Θ = 1, we can factorize it as (sec Θ + tan Θ)(sec Θ - tan Θ) = 1. Given that sec Θ + tan Θ = √3, we can find the value of the other factor by substituting: √3(sec Θ - tan Θ) = 1, which gives sec Θ - tan Θ = (1)/(√(3)).

To find tan Θ, we subtract the second equation from the first: (sec Θ + tan Θ) - (sec Θ - tan Θ) = √3 - (1)/(√(3)). This simplifies to 2 tan Θ = (3 - 1)/(√(3)) = (2)/(√(3)), meaning tan Θ = (1)/(√(3)). In the given range 0° ≤ Θ ≤ 90°, this corresponds to Θ = 30°.

Substituting Θ = 30° into the expression tan 3Θ, we get tan(3 × 30°) = tan 90°. Since tan Θ = (sin Θ)/(cos Θ) and cos 90° = 0, the value of tan 90° is considered as infinity.

Explanation

For any two angles A and B, the identity sin A = cos B implies that A + B = 90°. Given the equation sin (60° - Θ) = cos (Ψ - 30°), we can equate the sum of the angles to 90°:

(60° - Θ) + (Ψ - 30°) = 90°

Rearranging and simplifying the terms: 60° - 30° + Ψ - Θ = 90° 30° + Ψ - Θ = 90° Ψ - Θ = 90° - 30° Ψ - Θ = 60°

Substituting this value into the expression for the tangent function: tan (Ψ - Θ) = tan 60° = √3

Explanation

The expression (1 + cot Θ - csc Θ)(1 + tan Θ + sec Θ) is a standard trigonometric identity. To find its value efficiently, we can use the substitution method by setting Θ = 45°. At this angle, the trigonometric values are tan 45° = 1, cot 45° = 1, sec 45° = √2, and csc 45° = √2.

Substituting these values into the expression: (1 + 1 - √2)(1 + 1 + √2) = (2 - √2)(2 + √2)

Using the algebraic identity (a - b)(a + b) = a² - b², where a = 2 and b = √2: = (2)² - (√2)² = 4 - 2 = 2

Alternatively, by converting the expression into sine and cosine terms: (1 + (cos Θ)/(sin Θ) - (1)/(sin Θ))(1 + (sin Θ)/(cos Θ) + (1)/(cos Θ)) = ((sin Θ + cos Θ - 1)/(sin Θ))((cos Θ + sin Θ + 1)/(cos Θ))

The numerator follows the (a - b)(a + b) pattern with a = (sin Θ + cos Θ) and b = 1: = ((sin Θ + cos Θ)² - 1²)/(sin Θ cos Θ) = (sin² Θ + cos² Θ + 2 sin Θ cos Θ - 1)/(sin Θ cos Θ)

Applying the Pythagorean identity sin² Θ + cos² Θ = 1: = (1 + 2 sin Θ cos Θ - 1)/(sin Θ cos Θ) = (2 sin Θ cos Θ)/(sin Θ cos Θ) = 2

Explanation

Given the expression tan Θ + cot Θ = 2, we can substitute cot Θ with (1)/(tan Θ). Let tan Θ = x. The equation becomes:

x + ¹⁄ₓ = 2

Multiplying both sides by x yields the quadratic equation x² - 2x + 1 = 0, which is the expansion of (x - 1)² = 0. Solving for x, we find x = 1. Therefore, tan Θ = 1 and cot Θ = ¹⁄₁ = 1.

For any integer n, the value of the expression tanⁿ Θ + cotⁿ Θ is calculated by substituting these values:

1ⁿ + 1ⁿ = 1 + 1 = 2

Explanation

Let h be the height of the vertical pole and P be the position of its foot in the plane of the triangular park ABC. Let T be the top of the pole. The angles of elevation from the vertices A, B, and C to the top T are given as equal, say θ. In the right-angled triangles △TPA, △TPB, and △TPC, the distance from the foot of the pole to each vertex is given by the trigonometric relation AP = h cot θ, BP = h cot θ, and CP = h cot θ. Since h and θ are constant for all three vertices, it follows that AP = BP = CP. The unique point in the plane of a triangle that is equidistant from all its vertices is the circumcentre. Thus, the foot of the pole P must coincide with the circumcentre of △ABC.

Explanation

Given the ratio (sin θ)/x = (cos θ)/y, we can define a constant k such that sin θ = xk and cos θ = yk. Using the trigonometric identity sin²θ + cos²θ = 1, we substitute these values to get (xk)² + (yk)² = 1. Factoring out k² gives k²(x² + y²) = 1, which results in k = 1/√(x² + y²).

By substituting the value of k back into our expressions for sine and cosine, we obtain sin θ = x/√(x² + y²) and cos θ = y/√(x² + y²). The expression sin θ - cos θ is then calculated as x/√(x² + y²) - y/√(x² + y²), which simplifies to (x - y)/√(x² + y²).