SSC CGL Tier 1 2023 Quantitative Aptitude Questions with Solutions

Explanation

Anil reaches point B after traveling for 192 / 64 = 3 hours. Since he starts at 8:20 a.m., he reaches B and begins his return journey at 11:20 a.m.

At 11:20 a.m., Tirath has been driving for 1 hour and 30 minutes (from 9:50 a.m. to 11:20 a.m.), which is 1.5 hours. The distance covered by Tirath during this time is 96 × 1.5 = 144 km. At this moment, Anil is at point B (192 km from A) and Tirath is 144 km from A.

The distance between Anil and Tirath at 11:20 a.m. is 192 - 144 = 48 km. Since they are now moving towards each other (Anil from B to A and Tirath from A to B), their relative speed is the sum of their individual speeds: 64 + 96 = 160 km/h.

The time taken for them to meet after 11:20 a.m. is calculated as: Time = (Distance)/(Relative Speed) = ⁴⁸⁄₁₆₀ = 0.3 hours

Converting 0.3 hours into minutes: 0.3 × 60 = 18 minutes. Therefore, they meet at 11:20 a.m. + 18 minutes, which is 11:38 a.m.

Explanation

The arc length l of a circle is defined by the formula l = (θ)/(360) × 2π r, where θ is the angle subtended at the centre and r is the radius. Substituting the given values l = 19.25 cm and θ = 18°, we get 19.25 = ¹⁸⁄₃₆₀ × 2 × ²²⁄₇ × r. Simplifying this expression, 19.25 = ¹⁄₂₀ × ⁴⁴⁄₇ × r, which yields 19.25 = ¹¹⁄₃₅ × r. Solving for r, we find r = (19.25 × 35)/(11) = 1.75 × 35 = 61.25 cm.

The area of the circle A is calculated using the formula A = π r². Substituting the value of the radius, A = ²²⁄₇ × (61.25)² = ²²⁄₇ × 3751.5625. Dividing 3751.5625 by 7 gives 535.9375. Finally, A = 22 × 535.9375 = 11790.625 cm².

Explanation

Congruence in geometry signifies that two figures are identical in both shape and size. For two triangles to be congruent, every corresponding side must be of equal length and every corresponding angle must be of equal measure. If the size or shape of the triangles is not the same, they fail the criteria for congruence. The claim that triangles can be congruent regardless of whether their size and shape are equal is a direct contradiction of the fundamental mathematical definition of the term.

Explanation

In triangle PQR right-angled at Q, with PQ:QR = 3:4, the hypotenuse PR is 5 by the Pythagorean theorem. Therefore, sin P = QR/PR = 4/5, sin Q = sin 90 = 1, and sin R = PQ/PR = 3/5. Their sum is 4/5 + 1 + 3/5 = 12/5.

Explanation

The first shop offers a flat discount of 50%. For the second shop, the effective discount of two successive discounts of 30% and 30% is calculated using the formula x + y - ˣʸ⁄₁₀₀. Substituting the given values, the total discount is 30 + 30 - (30 × 30)/(100) = 60 - 9 = 51%.

The difference between the two discount rates is 51% - 50% = 1%. Since the final bill difference is ₹43, this 1% of the cost price (P) must equal ₹43.

1% of P = 43 ¹⁄₁₀₀ × P = 43 P = 43 × 100 = 4,300

Explanation

Let the selling price (SP) of one chair be x. According to the problem, total gain (4x) is equal to total SP (20x) minus total CP (18,000), leading to the equation 4x = 20x - 18,000. Solving for x gives 16x = 18,000, resulting in x = 1,125.

Explanation

The weighted average is calculated by dividing the total marks of both colleges (180 * 88 + 320 * 72 = 15,840 + 23,040 = 38,880) by the total number of students (180 + 320 = 500). Dividing 38,880 by 500 results in 77.76.

Explanation

Given x² + 1/x² = 18, we add 2 to both sides to find (x + 1/x)² = 20, which gives x + 1/x = √20 = 2√5. Using the identity x³ + 1/x³ = (x + 1/x)³ - 3(x + 1/x), we calculate (2√5)³ - 3(2√5) = 40√5 - 6√5 = 34√5.

Explanation

The average speed for a journey where two equal distances are covered at different speeds is calculated using the formula 2xy/(x+y). Plugging in the values 64 km/h and 80 km/h, the calculation is (2 * 64 * 80) / (64 + 80) = 10240 / 144, which equals 71.11 km/h.

Explanation

Each fraction in the expression can be simplified using the algebraic identity a² - b² = (a - b)(a + b).

Applying this identity to the first term, the numerator becomes (p - (q - r))(p + (q - r)) = (p - q + r)(p + q - r) and the denominator becomes ((p + r) - q)((p + r) + q) = (p + r - q)(p + r + q). Canceling the common factor (p - q + r), the first term simplifies to (p + q - r)/(p + q + r).

Applying the same logic to the second term, the numerator is (q - (p - r))(q + (p - r)) = (q - p + r)(q + p - r) and the denominator is ((p + q) - r)((p + q) + r) = (p + q - r)(p + q + r). Canceling the common factor (p + q - r), the second term simplifies to (q - p + r)/(p + q + r).

For the third term, the numerator is (r - (p - q))(r + (p - q)) = (r - p + q)(r + p - q) and the denominator is ((q + r) - p)((q + r) + p) = (q + r - p)(q + r + p). Canceling the common factor (r + q - p), the third term simplifies to (r + p - q)/(p + q + r).

Summing the simplified terms yields: (p + q - r)/(p + q + r) + (q - p + r)/(p + q + r) + (r + p - q)/(p + q + r) = (p + q - r + q - p + r + r + p - q)/(p + q + r) = (p + q + r)/(p + q + r) = 1

Alternatively, using a substitution shortcut for competitive exams, let p = 1, q = 1, r = 1. The expression becomes ⅓ + ⅓ + ⅓ = 1.

Explanation

The net profit or loss percentage when a price is first increased by a certain percentage and then decreased by a discount percentage can be calculated using the successive percentage change formula: x + y + ˣʸ⁄₁₀₀.

In this problem, the dealer marks the goods at 20% above the cost price, so x = +20. He then allows a discount of 15%, so y = -15. Substituting these values into the formula:

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 after a 20% markup. A 15% discount on the marked price of 120 is calculated as 120 × 0.15 = 18. The selling price is then 120 - 18 = 102. The gain percentage is the difference between the selling price and the cost price relative to the cost price: (102 - 100)/(100) × 100 = 2%.

Explanation

The volume of water a pipe can hold is the volume of its inner cylinder, given by V_wₐₜₑᵣ = π r² h, where r is the inner radius and h is the height or length. Given h = 1 m and V_wₐₜₑᵣ = ⁸⁸⁄₇ cubic metres, we have ²²⁄₇ × r² × 1 = ⁸⁸⁄₇. Solving for r, we get 22r² = 88, which means r² = 4, so the inner radius r = 2 m.

The inner diameter dᵢ is 2r = 4 m. The problem states that the inner diameter is equal to the outer radius R, so R = 4 m. The volume of the material used to make the pipe is the difference between the outer and inner volumes, calculated using the formula Vₘₐₜₑᵣᵢₐₗ = π (R² - r²) h. Substituting the known values, we get Vₘₐₜₑᵣᵢₐₗ = ²²⁄₇ × (4² - 2²) × 1 = ²²⁄₇ × (16 - 4) = (22 × 12)/(7) = ²⁶⁴⁄₇ ≈ 37.71 cubic units.

Explanation

Let the speed of the boat in still water be u km/h and the speed of the stream be v km/h. The downstream speed is D = u + v and the upstream speed is U = u - v. According to the first condition, the boat covers 120 km upstream and 120 km downstream in 30 hours, so 120/U + 120/D = 30. Dividing by 30 gives the equation 4/U + 4/D = 1 or 1/U + 1/D = 1/4.

According to the second condition, the boat covers 25 km upstream and 40 km downstream in 7 hours, so 25/U + 40/D = 7. To solve this system, let x = 1/U and y = 1/D. The equations become x + y = 0.25 and 25x + 40y = 7. Multiplying the first equation by 40 yields 40x + 40y = 10. Subtracting the second equation from this result gives 15x = 3, which means x = 1/5, so U = 5 km/h. Substituting x = 0.2 into x + y = 0.25 gives y = 0.05, so D = 20 km/h.

The speed of the boat in still water is the average of the downstream and upstream speeds, u = (D + U) / 2 = (20 + 5) / 2 = 12.5 km/h. The distance covered by the boat in still water in 16 hours is calculated as Distance = Speed × Time = 12.5 × 16 = 200 km.

Explanation

Let the length and width of the rectangle be 3x and 2x respectively. The area of a rectangle is calculated using the formula Area = Length × Width. Substituting the given values into the formula, we get 3x × 2x = 150, which simplifies to 6x² = 150. Dividing both sides by 6 gives x² = 25, which means x = 5.

Using the value of x, the length of the rectangle is 3 × 5 = 15 cm and the width is 2 × 5 = 10 cm. The perimeter of a rectangle is determined by the formula Perimeter = 2(Length + Width). Substituting the dimensions results in Perimeter = 2(15 + 10) = 2 × 25 = 50 cm.

Explanation

From the given equation 2 cot θ = 3, we find cot θ = 3/2 and tan θ = 1/(cot θ) = 2/3. In a right-angled triangle where cot θ = (base)/(perpendicular), let the base be 3 and the perpendicular be 2. The hypotenuse is calculated as √(3² + 2²) = √13.

Using these dimensions, we determine sin θ = (perpendicular)/(hypotenuse) = 2/√13 and cos θ = (base)/(hypotenuse) = 3/√13. Substituting these values into the numerator of the given expression: √(13) sin θ - 3 tan θ = √(13) ( 2/√13 ) - 3 ( 2/3 ) = 2 - 2 = 0

Substituting into the denominator: 3 tan θ + √(13) cos θ = 3 ( 2/3 ) + √(13) ( 3/√13 ) = 2 + 3 = 5

The value of the fraction is 0/5 = 0.

Explanation

The expression incorporates the trigonometric identity for the cosine of the sum of two angles, defined as cos(A + B) = cos A cos B - sin A sin B.

In the given expression 2 + cos 49°cos 41° - sin 49°sin 41°, let A = 49° and B = 41°. By substituting these values into the identity, we get:

cos 49°cos 41° - sin 49°sin 41° = cos(49° + 41°) cos 49°cos 41° - sin 49°sin 41° = cos 90°

Since the value of cos 90° is 0, the entire expression becomes:

2 + 0 = 2

Explanation

The ratio of the number of junior school students to senior school students can be determined using the rule of alligation. Let n₁ be the number of junior students and n₂ be the number of senior students.

The junior charge is c₁ = 325 and the senior charge is c₂ = 400, with a weighted average A = 352. The ratio n₁ : n₂ is calculated as:

n₁ : n₂ = (c₂ - A) : (A - c₁) n₁ : n₂ = (400 - 352) : (352 - 325) n₁ : n₂ = 48 : 27

Dividing both sides by 3 simplifies the ratio to: n₁ : n₂ = 16 : 9

We are given that n₁ = 80. According to the ratio: 16 parts = 80 1 part = ⁸⁰⁄₁₆ = 5

The number of senior students corresponds to 9 parts: n₂ = 9 × 5 = 45

Explanation

The relative speed of the policeman with respect to the thief is the difference in their speeds because they are running in the same direction: Vᵣₑₗ = 20 km/h - 17 km/h = 3 km/h. To find the distance covered in meters over 8 minutes, convert the relative speed to meters per minute: Vᵣₑₗ = 3 × (1000 m)/(60 min) = 50 m/min. The distance by which the gap is reduced in 8 minutes is: Distance reduced = Relative speed × Time = 50 m/min × 8 min = 400 m. Initially, the distance between them was 500 m. The distance remaining after 8 minutes is: Distance remaining = Initial distance - Distance reduced = 500 m - 400 m = 100 m.

Explanation

In a circle, a perpendicular line drawn from the centre to a chord bisects the chord. Given the total length of the chord is 40 cm, the length of the half-chord is 40 / 2 = 20 cm.

The radius of the circle (r), the distance of the chord from the centre (d), and the half-chord length form a right-angled triangle where the radius is the hypotenuse. Using the Pythagoras theorem:

r² = 21² + 20² r² = 441 + 400 r² = 841 r = √(841) = 29 cm

This calculation corresponds to the Pythagorean triplet (20, 21, 29), where 20² + 21² = 29².

Explanation

Let the total work be the Least Common Multiple (LCM) of the time taken by P, Q, and R, which is LCM(10, 15, 20) = 60 units.

The efficiency of P is 60 / 10 = 6 units per day. The efficiency of Q is 60 / 15 = 4 units per day. The efficiency of R is 60 / 20 = 3 units per day.

When they work together, their combined efficiency is 6 + 4 + 3 = 13 units per day. The total time taken to finish the work is the total work divided by the combined efficiency, which is 60 / 13 days. Converting this improper fraction into a mixed fraction, we get 4 ⁸⁄₁₃ days.

Explanation

The total percentage difference between a 5% loss and a 5% gain is 5% - (-5%) = 10%. This 10% difference in the cost price (CP) is equivalent to the ₹300 increase in the selling price.

Let the cost price of the watch be x. 10% of x = 300 ¹⁰⁄₁₀₀ × x = 300 x = 3000

The question asks for the original selling price (SP) at which the watch was sold, which was at a 5% loss. SP = x × (100 - Loss%)/(100) SP = 3000 × ⁹⁵⁄₁₀₀ = 2850

Explanation

To simplify the given expression, apply the exponent rule xᵐ⁺ⁿ = xᵐ × xⁿ to factor out common terms. In the numerator, 4ᵃ⁺⁴ can be written as 4ᵃ⁺² × 4². Factoring out 4ᵃ⁺² gives: 4ᵃ⁺²(4² - 5) = 4ᵃ⁺²(16 - 5) = 11 × 4ᵃ⁺²

In the denominator, recognize that 2² = 4. This allows 4ᵃ to be factored out from both terms: 15 × 4ᵃ - 4 × 4ᵃ = 4ᵃ(15 - 4) = 11 × 4ᵃ

Substituting these simplified expressions back into the fraction results in: 11 × 4ᵃ⁺²{11 × 4ᵃ}

The constant 11 cancels out. Using the exponent division rule (xᵐ)/(xⁿ) = xᵐ⁻ⁿ, the expression further simplifies to: 4⁽ᵃ⁺²⁾⁻ᵃ = 4² = 16

Explanation

Using the BODMAS rule to solve the equation 720 ÷ 8 + 915 ÷ 15 - m + 32 × 5 = 1104 ÷ 16 × 111 ÷ 37, first perform the divisions and multiplications on both sides. On the left side, 720 ÷ 8 = 90, 915 ÷ 15 = 61, and 32 × 5 = 160, transforming the expression into 90 + 61 - m + 160. Summing the numerical values gives 90 + 61 + 160 = 311, so the left side is 311 - m. On the right side, the expression involves 1104 ÷ 16 = 69 and 111 ÷ 37 = 3, leading to 69 × 3 = 207. The simplified equation is 311 - m = 207. Rearranging the terms to isolate m gives m = 311 - 207, which yields m = 104.

Explanation

Let the time taken by P to complete the task be x days. Since P takes twice as long as Q and three times as long as R, the time taken by Q is x/2 days and the time taken by R is x/3 days.

The combined work rate of P, Q, and R is the sum of their individual rates. Given they complete the task together in 2 days, the equation is:

¹⁄ₓ + (1)/(x/2) + (1)/(x/3) = ½

¹⁄ₓ + ²⁄ₓ + ³⁄ₓ = ½

(1 + 2 + 3)/(x) = ½

⁶⁄ₓ = ½

x = 12

The time taken by Q to complete the task on his own is x/2 = 12/2 = 6 days. Alternatively, using the efficiency method, if P's efficiency is 1 unit/day, then Q's efficiency is 2 units/day and R's efficiency is 3 units/day. Their combined efficiency is 1 + 2 + 3 = 6 units/day. Total work is efficiency multiplied by time, which is 6 × 2 = 12 units. Q's time is 12 / 2 = 6 days.

Explanation

For an equilateral triangle where all sides are of length a, the circumradius R is the distance from the center of the triangle to any of its vertices. The formula to calculate the circumradius of an equilateral triangle is R = (a)/(√(3)). Given that each side of the triangle is 12 cm, we substitute a = 12 into the formula to find R = (12)/(√(3)). Rationalizing the denominator by multiplying both the numerator and the denominator by √3, we obtain R = 12 √3{3}. Simplifying this fraction results in R = 4√3 cm.

Explanation

Using the BODMAS rule, the expression is evaluated by solving brackets in the order of parentheses, then braces, then square brackets, followed by division, multiplication, addition, and subtraction.

First, evaluate the innermost parentheses: 10 - 4 + 3 = 6 + 3 = 9

Next, evaluate the curly braces using the result from the previous step: {6 - 9} = -3

Then, evaluate the square brackets: [9 - (-3)] = 9 + 3 = 12

The expression now simplifies to 22 - 12 ÷ 2 × 3. Following the order of operations, division and multiplication are performed from left to right: 12 ÷ 2 = 6 6 × 3 = 18

Finally, perform the subtraction: 22 - 18 = 4

Explanation

The thief gets a head start of 10 minutes, which is 10 / 60 = 1/6 hours. The distance covered by the thief during this head start period is Distance = Speed × Time = 6 km/h × (1/6) h = 1 km. To catch the thief, the policeman must close this 1 km gap. The relative speed of the policeman with respect to the thief is 8 km/h - 6 km/h = 2 km/h. The time required to close the gap is Time = (Gap)/(Relative Speed) = (1 km)/(2 km/h) = 0.5 hours. The distance from the starting point where the policeman catches the thief is the total distance the policeman travels in that time, which is 8 km/h × 0.5 h = 4 km. Alternatively, using the time difference formula where x is the distance, ˣ⁄₆ - ˣ⁄₈ = ¹⁰⁄₆₀ simplifies to ˣ⁄₂₄ = ⅙, resulting in x = 4 km.

Explanation

The algebraic expression a² + b² + c² - ab - bc - ca can be simplified using the standard identity: a² + b² + c² - ab - bc - ca = ½ [(a - b)² + (b - c)² + (c - a)²]

Substituting the given values a = 101, b = 102, and c = 103 into this identity: a - b = 101 - 102 = -1 b - c = 102 - 103 = -1 c - a = 103 - 101 = 2

Now, plug these differences back into the formula: ½ [(-1)² + (-1)² + (2)²] = ½ [1 + 1 + 4] = ½ × 6 = 3

Alternatively, when a, b, and c are in an arithmetic progression with a common difference d, the value of the expression a² + b² + c² - ab - bc - ca is always 3d². Here, the common difference d is 1, so the result is 3 × (1)² = 3.

Explanation

For a sum of money to become n times itself under simple interest, the interest earned must be (n - 1) times the principal P. Here, the sum becomes four times of itself, meaning the amount A = 4P. Therefore, the simple interest SI = A - P = 4P - P = 3P.

Using the standard simple interest formula SI = (P × R × T)/(100), where R is the rate of interest and T is the time in years, we substitute the given values: 3P = (P × R × 12)/(100)

By canceling P from both sides, the equation simplifies to: 3 = (12R)/(100) 12R = 300 R = ³⁰⁰⁄₁₂ R = 25%

Thus, the annual rate of interest is 25%.

Explanation

Let the monthly salary in January 2022 be S. After paying an EMI of 22%, the remaining salary is 100% - 22% = 78% of S. This remaining amount is divided between groceries and clothes in the ratio 7:5. The expenditure on clothes is given as ₹18,200. Using the ratio: Expenditure on clothes = ((5)/(7 + 5)) × 0.78S = 18200 (⁵⁄₁₂) × 0.78S = 18200 0.325S = 18200 S = (18200)/(0.325) = 56000

The salary in January 2022 was ₹56,000. In February 2022, the salary increased by 16%. The February salary is calculated as: February Salary = 56000 × (1 + ¹⁶⁄₁₀₀) February Salary = 56000 × 1.16 = 64960

Explanation

Substituting α = 15° into the identity tan 2α = (2tanα)/(1 - tan²α) gives: tan 30° = (2tan 15°)/(1 - tan² 15°) Let x = tan 15°. Given tan 30° = 1/√3, the equation becomes: 1/√3 = (2x)/(1 - x²) 1 - x² = 2√3x x² + 2√3x - 1 = 0 Solving this quadratic equation using the formula x = (-b ± √(b² - 4ac)) / (2a): x = (-2√3 ± √((2√3)² - 4(1)(-1))) / 2 = (-2√3 ± √16) / 2 = (-2√3 ± 4) / 2 = -√3 ± 2 Since 15° is in the first quadrant, tan 15° must be positive. Therefore, we select the root x = 2 - √3. Using the provided value √3 = 1.732: tan 15° = 2 - 1.732 = 0.268

Explanation

According to the Tangent-Secant Theorem, when a tangent segment and a secant segment are drawn to a circle from an exterior point, the square of the length of the tangent segment is equal to the product of the lengths of the whole secant segment and its external part. In this case, K is the external point, KH is the tangent, and KDE is the secant. The relationship is given by the formula KH² = KD · KE.

The external part of the secant KD is given as 9 cm and the chord DE is 7 cm. The length of the entire secant segment KE is the sum of these two parts, so KE = KD + DE = 9 + 7 = 16 cm. Substituting the known values into the formula yields KH² = 9 · 16. Multiplying these gives KH² = 144. To find the length of KH, take the square root of 144, which results in KH = 12 cm.

Explanation

Efficiency and time are inversely proportional when the total work is constant, which can be expressed by the formula E₁ × T₁ = E₂ × T₂. Let the efficiency of Virat be 100. Since Daniel is 60% more efficient than Virat, Daniel efficiency is 160.

Using the inverse proportion relationship: 100 × 30 = 160 × T_Dₐₙᵢₑₗ 3000 = 160 × T_Dₐₙᵢₑₗ T_Dₐₙᵢₑₗ = ³⁰⁰⁰⁄₁₆₀ T_Dₐₙᵢₑₗ = ³⁰⁰⁄₁₆

By simplifying the fraction by dividing both the numerator and the denominator by 4, we get: T_Dₐₙᵢₑₗ = ⁷⁵⁄₄

To convert the improper fraction ⁷⁵⁄₄ into a mixed fraction, we divide 75 by 4: 75 = 18 × 4 + 3 Therefore, T_Dₐₙᵢₑₗ = 18 ¾ days.

Explanation

The correct average can be calculated by adjusting the initial average with the net change caused by the misread value. The formula for the corrected average is:

Correct Average = Old Average + (Correct Value - Incorrect Value)/(n)

Given that the old average is 45, the number of observations n is 36, the correct value is 84, and the incorrect value is 48, we calculate the net difference in the sum:

Net Difference = 84 - 48 = 36

Now, determine the change in the average by dividing the net difference by the total number of observations:

Change in Average = ³⁶⁄₃₆ = 1

Applying this adjustment to the original average:

Correct Average = 45 + 1 = 46

Explanation

Given sin 23° = (a)/(b), we can determine cos 23° using the identity cos² θ = 1 - sin² θ. Thus, cos 23° = √(1 - ((a)/(b))²) = √(b² - a²){b}.

The expression to evaluate is sec 23° - sin 67°. Using the complementary angle identity sin(90° - θ) = cos θ, we have sin 67° = sin(90° - 23°) = cos 23°. Substituting this and sec 23° = (1)/(cos 23°) into the expression gives:

sec 23° - cos 23° = (1)/(cos 23°) - cos 23° = (1 - cos² 23°)/(cos 23°) = (sin² 23°)/(cos 23°)

Substituting the values of sin 23° and cos 23°:

((a)/(b))²{√(b² - a²){b}} = (a²)/(b²) · (b)/(√(b² - a²)) = (a²)/(b√(b² - a²))

Explanation

Let B, S, and A represent the number of votes received by Bhaskar, Sarai, and Arjun respectively. According to the problem, Sarai received 2% fewer votes than Bhaskar: S = B - 0.02B = 0.98B

The difference between the votes of Bhaskar and Sarai is given as 1296: B - S = 1296 B - 0.98B = 1296 0.02B = 1296 B = (1296)/(0.02) = 64800

Arjun received 50% more votes than Sarai: A = 1.5 × S = 1.5 × (0.98B) = 1.47B

The problem asks for half of the difference between the votes of Arjun and Bhaskar. We first find the difference in terms of B: |A - B| = |1.47B - B| = 0.47B

The required value is: ½ × 0.47B = 0.235B

Substituting the value of B calculated above: 0.235 × 64800 = 23.5 × 648 = 15228

Explanation

The number of boys in the class is 60 - 20 = 40. The average weight of the entire class is calculated using the weighted average formula:

Average weight = ((n₁ × w₁) + (n₂ × w₂))/(n₁ + n₂)

Substituting the given values, where n₁ = 40 boys with an average w₁ = 40 kg, and n₂ = 20 girls with an average w₂ = 35 kg:

Average weight = ((40 × 40) + (20 × 35))/(40 + 20)

Average weight = (1600 + 700)/(60)

Average weight = ²³⁰⁰⁄₆₀ = ²³⁰⁄₆ = 38.333...

Rounding to two decimal places, the average weight of the entire class is 38.33 kg.

Explanation

The trigonometric identity states that csc² θ - cot² θ = 1. This can be factored as (csc θ + cot θ)(csc θ - cot θ) = 1. Given that csc θ + cot θ = m, we substitute this to find csc θ - cot θ = ¹⁄ₘ.

By adding these two equations: 2 csc θ = m + ¹⁄ₘ = (m² + 1)/(m)

By subtracting the second equation from the first: 2 cot θ = m - ¹⁄ₘ = (m² - 1)/(m)

To find the value of (m² - 1)/(m² + 1), we divide the result of the subtraction by the result of the addition: (2 cot θ)/(2 csc θ) = ((m² - 1) / m)/((m² + 1) / m)

Simplifying both sides: (cos θ / sin θ)/(1 / sin θ) = (m² - 1)/(m² + 1) cos θ = (m² - 1)/(m² + 1)

Explanation

To simplify the ratio ⅘ : ⅚ : ⁹⁄₁₀, we multiply each term by the Least Common Multiple (LCM) of the denominators 5, 6, and 10, which is 30.

First term: ⅘ × 30 = 24 Second term: ⅚ × 30 = 25 Third term: ⁹⁄₁₀ × 30 = 27

The simplified ratio of the three numbers is 24 : 25 : 27. Let the numbers be 24x, 25x, and 27x. The smallest value is 24x and the greatest is 27x. The difference between them is given as 12:

27x - 24x = 12 3x = 12 x = 4

The number that is neither the smallest nor the greatest is the middle term, 25x. Substituting the value of x gives:

25 × 4 = 100

Explanation

In the triangle formed by the two radii (15 cm and 13 cm) and the distance between the centers (14 cm), the common chord is exactly twice the altitude drawn to the side representing the distance between the centers. This is because the line joining the centers of two intersecting circles is the perpendicular bisector of their common chord.

First, calculate the semi-perimeter s of the triangle with sides 15 cm, 13 cm, and 14 cm: s = (15 + 13 + 14)/(2) = 21 cm

Next, determine the area A of this triangle using Heron's formula: A = √(s(s-a)(s-b)(s-c)) = √(21(21-15)(21-13)(21-14)) A = √(21 × 6 × 8 × 7) = √(7056) = 84 cm²

The area of the triangle can also be expressed as ½ × base × height. Taking the distance between centers (14 cm) as the base and half the common chord (h) as the height: 84 = ½ × 14 × h 84 = 7h h = 12 cm

The total length of the common chord is 2h, which is 2 × 12 = 24 cm.

Explanation

The provided data is a cumulative frequency distribution of the less than type. In this format, the value associated with each class limit represents the total number of observations that fall below that specific value. The table explicitly states that the number of students who scored less than 20 marks is 5. This cumulative total includes all students scoring from zero up to, but not including, 20. Therefore, the count is taken directly from the second row of the marks distribution table.

Explanation

To find the value of the algebraic expression x² + y², we substitute the given numerical values of the variables into the formula. Substituting x = 1 and y = 2, the expression becomes 1² + 2². Evaluating the exponents first, we find that 1² = 1 × 1 = 1 and 2² = 2 × 2 = 4. Adding these individual results together yields 1 + 4 = 5. This straightforward substitution and arithmetic summation provide the result for the expression at the specified coordinates.

Explanation

The shopkeeper incurs a cost for only the weight he actually gives to the customer. Since the cost of 1000 g of millet is ₹16, the cost price (CP) of the 800 g delivered is:

CP = 16 × ⁸⁰⁰⁄₁₀₀₀ = ₹12.8

The shopkeeper sells this 800 g quantity for ₹20. The actual profit made on this transaction is the difference between the selling price and the cost of the quantity delivered:

Profit = 20 - 12.8 = ₹7.2

The actual profit percentage is calculated based on the cost of the goods sold (₹12.8):

Profit Percentage = (Profit)/(CP) × 100 Profit Percentage = (7.2)/(12.8) × 100 = ⁷²⁄₁₂₈ × 100

Simplifying the fraction ⁷²⁄₁₂₈ by dividing both the numerator and denominator by 8 gives ⁹⁄₁₆.

Profit Percentage = ⁹⁄₁₆ × 100 = 56.25%

Using the successive ratio method, the profit ratio due to price is ²⁰⁄₁₆ = ⁵⁄₄ and the profit ratio due to weight is ¹⁰⁰⁰⁄₈₀₀ = ⁵⁄₄. The total ratio is ⁵⁄₄ × ⁵⁄₄ = ²⁵⁄₁₆. The profit is 25 - 16 = 9 units over a base of 16 units, which results in 56.25%.

Explanation

Let the total work be the Least Common Multiple (LCM) of the given time periods 30 and 120, which is 120 units. The combined efficiency of A and B is E_A + E_B = ¹²⁰⁄₃₀ = 4 units per day. The efficiency of C alone is E_C = ¹²⁰⁄₁₂₀ = 1 unit per day.

The problem states that A takes 1.5 times as many days as B and C together. Since efficiency is inversely proportional to the time taken, the efficiency of A is (1)/(1.5) times the efficiency of B and C combined. This is expressed as E_A = (1)/(1.5)(E_B + E_C), which simplifies to 3E_A = 2(E_B + E_C).

Substituting E_A = 4 - E_B and E_C = 1 into the efficiency equation: 3(4 - E_B) = 2(E_B + 1) 12 - 3E_B = 2E_B + 2 10 = 5E_B E_B = 2 units per day.

The number of days B takes to complete the work alone is the total work divided by B efficiency: Days = ¹²⁰⁄₂ = 60 days.

Explanation

The rate of work is the ratio of total work completed to the time taken. Aarti's typing rate is calculated as R_A = (85 pages)/(10 hours) = 8.5 pages per hour. The combined typing rate of Aarti and Bina is R_A₊_B = (500 pages)/(40 hours) = 12.5 pages per hour.

To find Bina's individual typing rate, subtract Aarti's rate from the combined rate: R_B = R_A₊_B - R_A = 12.5 - 8.5 = 4 pages per hour. The time required for Bina to type 40 pages is then determined by dividing the total number of pages by her hourly rate: Time = (40 pages)/(4 pages/hour) = 10 hours.

Explanation

The expression is evaluated following the BODMAS rule, which dictates the order of operations as Brackets, Orders, Division, Multiplication, Addition, and Subtraction.

First, address the calculation within the inner parentheses: 6 + 6 = 12

Substitute this back into the main expression: [7 + 30 ÷ 6 - 12 + 7]

Next, perform the division operation: 30 ÷ 6 = 5

The expression now simplifies to: 7 + 5 - 12 + 7

Finally, perform the addition and subtraction from left to right: 7 + 5 = 12 12 - 12 = 0 0 + 7 = 7

Alternatively, by grouping all addition terms first: (7 + 5 + 7) - 12 = 19 - 12 = 7

Explanation

The simple interest SI is calculated using the formula SI = (P × R × T)/(100), where P is the principal, R is the rate of interest per annum, and T is the time in years. Since the simple interest for both cases is stated to be equal, we can equate the expressions for both scenarios:

(800 × 5.5 × 6)/(100) = (600 × 4 × T)/(100)

By eliminating the denominator and simplifying the numbers: 800 × 5.5 × 6 = 600 × 4 × T 4800 × 5.5 = 2400 × T

Dividing both sides by 2400: 2 × 5.5 = T T = 11

The value of T represents the required time period in years.

Explanation

To find the required ratio, first calculate the total number of students in class IX for schools Q and R. From the table, the number of students in class IX for school Q is 157 and for school R is 143. The sum is 157 + 143 = 300.

Next, calculate the total number of students in class VI for schools S and T. According to the table, school S has 142 students and school T has 144 students in class VI. The sum is 142 + 144 = 286.

The respective ratio is calculated as: Ratio = 300 : 286

Dividing both terms by their highest common factor, which is 2, the ratio simplifies to: 150 : 143

Explanation

In the second quadrant, the sine function is positive while the cosine, tangent, and cotangent functions are negative. Given tan x = -¹²⁄₅, we can consider the perpendicular P = 12 and the base B = -5. The hypotenuse H is calculated using the Pythagorean identity H = √(P² + B²) = √(12² + (-5)²) = √(144 + 25) = 13.

The trigonometric ratios for x in the second quadrant are: sin x = (P)/(H) = ¹²⁄₁₃ cot x = (1)/(tan x) = -⁵⁄₁₂

Substituting these values into the required expression sin x - cot x: ¹²⁄₁₃ - (-⁵⁄₁₂) = ¹²⁄₁₃ + ⁵⁄₁₂

To solve this, we find the common denominator, which is 13 × 12 = 156: (144 + 65)/(156) = ²⁰⁹⁄₁₅₆

Explanation

The least number divisible by a set of numbers is their Least Common Multiple (LCM). To find the LCM of 2, 3, 5, 6, 9, and 18, we perform prime factorization: 2 = 2¹ 3 = 3¹ 5 = 5¹ 6 = 2¹ × 3¹ 9 = 3² 18 = 2¹ × 3²

The LCM is the product of the highest powers of all prime factors involved: LCM = 2¹ × 3² × 5¹ = 2 × 9 × 5 = 90

For a number to be a perfect square, every prime factor in its prime factorization must have an even exponent. In the prime factorization of 90, which is 2¹ × 3² × 5¹, the factors 2 and 5 have odd exponents. To find the least perfect square, we multiply 90 by the missing prime factors to make their exponents even: Required number = 90 × 2¹ × 5¹ = 90 × 10 = 900

Thus, 900 is the smallest perfect square (30²) that is divisible by all the given numbers.