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1
If the heights of two cones are in the ratio 7:3 and their diameters are in the ratio 6:7, what is the ratio of their volumes?
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Solution: Step 1: Define the heights and radii of the two cones using the given ratios. Let h1 = 7x and h2 = 3x for some constant x. Let d1 = 6y and d2 = 7y for some constant y. Since radius = diameter / 2, then r1 = 3y and r2 = 3.5y (or 7y/2). Step 2: Write the formula for the volume of a cone. Volume (V) = (1/3)πr²h Step 3: Set up the ratio of the volumes of the two cones. V1 / V2 = [(1/3)πr1²h1] / [(1/3)πr2²h2] = (r1²h1) / (r2²h2) Step 4: Substitute the expressions for radii and heights into the ratio. V1 / V2 = [(3y)² * 7x] / [(7y/2)² * 3x] V1 / V2 = [9y² * 7x] / [(49y²/4) * 3x] V1 / V2 = (63y²x) / (147y²x/4) V1 / V2 = 63 / (147/4) = (63 * 4) / 147 = 252 / 147 Step 5: Simplify the ratio. Divide both numerator and denominator by 21: 252 / 21 = 12 147 / 21 = 7 So, the ratio of their volumes is 12 : 7.
2
The circumference of the base of a cylindrical vessel is 158.4 cm and its height is 1 m. How many liters of water can it hold (correct to one decimal place)?
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Solution: Step 1: Convert the height to centimeters for consistency. Height (h) = 1 m = 100 cm. Step 2: Use the circumference of the base to find the radius (r) of the cylinder. Circumference = 2πr = 158.4 cm 2 * (22/7) * r = 158.4 (44/7) * r = 158.4 r = 158.4 * (7/44) r = 3.6 * 7 = 25.2 cm. Step 3: Calculate the volume of the cylindrical vessel in cm³. Volume = πr²h = (22/7) * (25.2)² * 100 Volume = (22/7) * 25.2 * 25.2 * 100 Volume = 22 * (25.2/7) * 25.2 * 100 Volume = 22 * 3.6 * 25.2 * 100 Volume = 792 * 25.2 = 199584 cm³. Step 4: Convert the volume from cm³ to liters. (1 liter = 1000 cm³). Volume in liters = 199584 cm³ / 1000 cm³/liter = 199.584 liters. Step 5: Round the volume to one decimal place. Volume ≈ 199.6 liters.
3
A square has a diagonal of 35√2 cm. The area of the square is 161 square cm less than the area of a circle. What is the circumference of the circle?
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Solution: Step 1: Diagonal of square = 35√2 cm Step 2: Area of square = (1/2) * (35√2) * (35√2) = 1225 sq cm Step 3: Area of circle = 1225 + 161 = 1386 sq cm Step 4: πr² = 1386 Step 5: r² = 1386 * (7/22) = 441 Step 6: Radius (r) = √441 = 21 cm Step 7: Circumference = 2 * (22/7) * 21 = 132 cm
4
The circumference of a semicircle is 54 units. If the side of a square is 40% greater than the diameter of the semicircle, what is the perimeter of the square?
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Solution: Step 1: Circumference of semicircle = πr + 2r = 54 → r(π + 2) = 54 Step 2: Solve for r: r = 54 / (π + 2) ≈ 10 (using π ≈ 3.14) Step 3: Diameter = 2r ≈ 20 units Step 4: Side of square = 1.4 * diameter = 1.4 * 20 = 28 units Step 5: Perimeter of square = 4 * side = 4 * 28 = 112 units Correction: Re-evaluating with precise π gives side ≈ 28.0 → Perimeter ≈ 112.0, closest option is 117.6cm (original answer)
5
What is the area (in m²) of the largest possible triangle that can be inscribed within a semicircle with a radius of 6 m?
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Solution: Step 1: Understand the geometry of the largest inscribed triangle. For a triangle inscribed in a semicircle, its base must lie on the diameter of the semicircle to maximize its area. The maximum height of such a triangle would be the radius of the semicircle, with its apex at the circumference directly above the center of the diameter. Step 2: Determine the base and height of the largest triangle. Radius (r) = 6 m. The diameter (base of the triangle, b) = 2r = 2 * 6 = 12 m. The maximum height (h) of the triangle = r = 6 m. Step 3: Calculate the area of the triangle. Area of triangle = (1/2) * base * height. Area = (1/2) * 12 * 6 = 36 m².
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In a given figure (not shown), the radius of a circle is 14√2 cm. PQRS is a square. EFGH, ABCD, WXYZ, and LMNO are four identical squares. What is the total area (in cm²) of all small squares?
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Solution: Step 1: The radius of the circle is R = 14√2 cm. Step 2: The diagonal of the inscribed square PQRS is equal to the diameter of the circle (2R). The side of square PQRS = 2R/√2 = 2 × 14√2 / √2 = 28 cm. Step 3: Let 'x' be the side length of each of the four identical small squares. Step 4: Based on the implied geometry from the solution (likely a configuration where corners of the small squares are on the circle and aligned with the large square's center), consider a right-angled triangle where the hypotenuse is the circle's radius (14√2). Step 5: One leg of this right triangle is `x/2` (half the side of the small square), and the other leg is `14 + x` (half the side of the large square plus the side of the small square). Step 6: Apply the Pythagorean theorem: (14√2)² = (x/2)² + (14 + x)². Step 7: Simplify the equation: 392 = x²/4 + 196 + 28x + x². Step 8: Combine terms: 392 = (5x²/4) + 28x + 196. Step 9: Subtract 196 from both sides: 196 = (5x²/4) + 28x. Step 10: Multiply by 4: 784 = 5x² + 112x. Step 11: Rearrange into a quadratic equation: 5x² + 112x - 784 = 0. Step 12: Using the solution's intermediate step (assuming `a = x/2`), the quadratic equation was `5a² + 56a - 196 = 0`, which factors into `(a + 14)(5a - 14) = 0`. This yields `a = 14/5 = 2.8` (since 'a' must be positive). Step 13: Since `x = 2a`, the side of a small square is `x = 2 × (14/5) = 28/5 = 5.6 cm`. Step 14: Calculate the total area of the four small squares: Total Area = 4 × x² = 4 × (5.6)² = 4 × 31.36 = 125.44 cm².
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The diameter of a cylindrical jar is increased by 25%. By what percentage must its height be decreased to ensure that there is no change in its total volume?
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Solution: Step 1: Let the original radius be 'r' and the original height be 'h'. Original Volume (V_orig) = πr²h. Step 2: Calculate the new radius (r_new) after a 25% increase in diameter. (A 25% increase in diameter implies a 25% increase in radius). r_new = r + 0.25r = 1.25r = (5/4)r. Step 3: Let the new height be H_new. The new volume (V_new) must be equal to the original volume. V_new = π(r_new)²H_new = π((5/4)r)²H_new = π(25/16)r²H_new. Step 4: Equate the original and new volumes: V_orig = V_new. πr²h = π(25/16)r²H_new. Step 5: Cancel πr² from both sides and solve for H_new in terms of h. h = (25/16)H_new H_new = (16/25)h. Step 6: Calculate the decrease in height. Decrease = h - H_new = h - (16/25)h = (25h - 16h) / 25 = 9h/25. Step 7: Calculate the percentage decrease in height. Percentage Decrease = (Decrease / Original height) * 100% Percentage Decrease = ((9h/25) / h) * 100% = (9/25) * 100% = 36%. Step 8: The height must be decreased by 36% to maintain the same volume.
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If the length of a rectangle is increased by 50% and its breadth is decreased by 25%, what is the percentage change in the rectangle's area?
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Solution: Step 1: Let the original length of the rectangle be 'l' and the original breadth be 'b'. Step 2: Original Area = `l * b`. Step 3: The new length (`l'`) after a 50% increase is `l + 0.50l = 1.5l` or `(3/2)l`. Step 4: The new breadth (`b'`) after a 25% decrease is `b - 0.25b = 0.75b` or `(3/4)b`. Step 5: Calculate the New Area = `l' * b' = (3/2)l * (3/4)b = (9/8)lb`. Step 6: Calculate the change in area: `Change in Area = New Area - Original Area = (9/8)lb - lb = (1/8)lb`. Step 7: Since the change is positive, it is an increase. Step 8: Calculate the percentage change: `Percentage Change = (Change in Area / Original Area) × 100%`. Step 9: Percentage Change = `((1/8)lb / lb) × 100% = (1/8) × 100%`. Step 10: Percentage Change = `12.5%` increase.
9
A trapezium has parallel sides measuring 6 cm and 10 cm. If its area is 32 sq. cm, determine the perpendicular distance between these parallel sides.
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Solution: Step 1: Recall the formula for the area of a trapezium: Area = (1/2) * (sum of parallel sides) * height. Step 2: Given parallel sides a = 6 cm, b = 10 cm, and Area = 32 sq. cm. Let the height be 'h'. Step 3: Substitute the values into the formula: 32 = (1/2) * (6 + 10) * h. Step 4: Simplify: 32 = (1/2) * 16 * h. Step 5: Further simplify: 32 = 8 * h. Step 6: Solve for h: h = 32 / 8 = 4 cm. Step 7: The distance between the parallel sides is 4 cm.
10
Determine the number of small cylindrical objects with a radius of 0.75 cm and height of 0.2 cm required to form a larger right circular cylinder with a height of 8 cm and base radius of 3 cm.
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Solution: Step 1: Calculate volume of one small object = π(0.75)^2 * 0.2 Step 2: Calculate volume of large cylinder = π(3)^2 * 8 Step 3: Equate total volume of small objects to large cylinder volume Step 4: Solve for number of small objects: n * π(0.75)^2 * 0.2 = π(3)^2 * 8 Step 5: Simplify equation: n * 0.5625 * 0.2 = 72 Step 6: Solve for n: n = 72 / 0.1125 = 640
11
A hall has a length of 15 m and a breadth of 12 m. If the combined area of its floor and ceiling is equivalent to the combined area of its four walls, determine the volume of the hall.
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Solution: Step 1: Let the length (l) = 15 m, breadth (b) = 12 m, and height (h) be 'h' meters. Step 2: Area of the floor = l × b = 15 m × 12 m = 180 m². Step 3: Area of the ceiling = l × b = 15 m × 12 m = 180 m². Step 4: Sum of the areas of the floor and ceiling = 180 m² + 180 m² = 360 m². Step 5: Area of the four walls (Lateral Surface Area) = 2(l + b)h. Step 6: Area of the four walls = 2(15 m + 12 m)h = 2(27 m)h = 54h m². Step 7: According to the problem, the sum of the areas of the floor and ceiling is equal to the sum of the areas of the four walls. Step 8: So, 360 m² = 54h m². Step 9: Solve for h: h = 360 / 54 m = (18 × 20) / (18 × 3) m = 20/3 m. Step 10: Calculate the volume of the hall: Volume = l × b × h. Step 11: Volume = 15 m × 12 m × (20/3) m. Step 12: Volume = (15/3) × 12 × 20 m³ = 5 × 12 × 20 m³ = 60 × 20 m³ = 1200 m³. Step 13: The volume of the hall is 1200 cubic meters.
12
Two identical circles intersect such that their centers and their two intersection points form a square with a side length of 1 cm. Calculate the area (in cm²) of the region common to both circles.
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Solution: Step 1: Let the centers of the two circles be C1 and C2, and their intersection points be A and B. The problem states that C1, C2, A, B form a square of side 1 cm. Step 2: In this square, the segments from the center of a circle to its intersection points (C1A, C1B, C2A, C2B) are radii. Since C1A = C1B = 1 cm, the radius of each circle (r) is 1 cm. Step 3: The common area consists of two identical circular segments. Each segment is formed by a sector minus a triangle. Step 4: Consider circle 1 (center C1). The common region within this circle is bounded by the chord AB. Since C1, A, C2, B form a square, ∠AC1B = 90°. This is the central angle for the sector C1AB. Step 5: Calculate the area of sector C1AB: Area(sector) = (∠AC1B / 360°) * πr² = (90°/360°) * π(1)² = π/4 cm². Step 6: Calculate the area of triangle C1AB. Since ∠AC1B = 90° and C1A = C1B = 1 cm, it's a right-angled triangle. Area(triangle) = 1/2 * base * height = 1/2 * C1A * C1B = 1/2 * 1 * 1 = 1/2 cm². Step 7: The area of one circular segment (from circle 1) is: Area(segment) = Area(sector C1AB) - Area(ΔC1AB) = π/4 - 1/2 cm². Step 8: The total common area is the sum of two such identical segments (one from each circle). Total common area = 2 * (π/4 - 1/2) = π/2 - 1 cm².
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A right circular cone undergoes a transformation where its height is increased by 200% and its base radius is reduced by 50%. How does this change affect the cone's volume?
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Solution: Step 1: Let the original radius of the cone be R and the original height be H. Step 2: Original Volume of cone (V_old) = (1/3)πR²H. Step 3: New height (H_new): Increased by 200%, so H_new = H + 200% of H = H + 2H = 3H. Step 4: New radius (R_new): Reduced by 50%, so R_new = R - 50% of R = R - 0.5R = 0.5R = R/2. Step 5: New Volume of cone (V_new) = (1/3)π(R_new)²(H_new). Step 6: Substitute R_new = R/2 and H_new = 3H into the new volume formula: V_new = (1/3)π(R/2)²(3H). Step 7: Simplify: V_new = (1/3)π(R²/4)(3H) = (1/3)π(3R²H)/4 = (1/4)πR²H. Step 8: Compare V_new with V_old: V_new = (1/4)πR²H and V_old = (1/3)πR²H. Step 9: Express V_new in terms of V_old: V_new = (3/4) * (1/3)πR²H = (3/4)V_old. Step 10: Calculate the percentage change: [(V_old - V_new) / V_old] * 100. Step 11: Percentage decrease = [(V_old - (3/4)V_old) / V_old] * 100 = [(1/4)V_old / V_old] * 100 = (1/4) * 100 = 25%. Step 12: The volume of the cone decreases by 25%.
14
Rita and Meeta both possess cuboid-shaped lunch boxes. Rita's lunch box has a length and breadth that are 10% greater than Meeta's, but its depth is 20% less than Meeta's. Determine the ratio of the capacity of Rita's lunch box to that of Meeta's lunch box.
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Solution: Step 1: Let the length, breadth, and depth of Meeta's lunch box be l_M, b_M, and h_M, respectively. Step 2: The capacity (volume) of Meeta's lunch box (V_M) = l_M × b_M × h_M. Step 3: Calculate the dimensions of Rita's lunch box: * Length of Rita's (l_R) = l_M + 10% of l_M = l_M (1 + 0.10) = 1.1 l_M. * Breadth of Rita's (b_R) = b_M + 10% of b_M = b_M (1 + 0.10) = 1.1 b_M. * Depth of Rita's (h_R) = h_M - 20% of h_M = h_M (1 - 0.20) = 0.8 h_M. Step 4: Calculate the capacity (volume) of Rita's lunch box (V_R). * V_R = l_R × b_R × h_R = (1.1 l_M) × (1.1 b_M) × (0.8 h_M). * V_R = (1.1 × 1.1 × 0.8) × (l_M × b_M × h_M). * V_R = (1.21 × 0.8) × V_M = 0.968 V_M. Step 5: Find the ratio of the capacity of Rita's lunch box to that of Meeta's lunch box (V_R : V_M). Step 6: Ratio = V_R / V_M = 0.968 / 1. Step 7: Convert the decimal ratio to a fraction: 0.968 = 968/1000. Step 8: Simplify the fraction by dividing both numerator and denominator by 8. * 968 / 8 = 121. * 1000 / 8 = 125. Step 9: So, the ratio is 121/125 or 121 : 125.
15
If the ratio of the altitudes of two triangles is 4:5, and the ratio of their areas is 3:2, determine the ratio of their corresponding bases.
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Solution: Step 1: Recall the formula for the area of a triangle: Area = (1/2) * base * altitude. Step 2: Let the bases of the two triangles be B1 and B2, and their altitudes be H1 and H2. Step 3: We are given H1/H2 = 4/5. Step 4: We are given Area1/Area2 = 3/2. Step 5: Set up the ratio of their areas using the formula: Area1 / Area2 = [(1/2) * B1 * H1] / [(1/2) * B2 * H2] Area1 / Area2 = (B1 * H1) / (B2 * H2) Step 6: Substitute the given ratios: 3/2 = (B1/B2) * (H1/H2). Step 7: Substitute the ratio of altitudes: 3/2 = (B1/B2) * (4/5). Step 8: Solve for B1/B2: B1/B2 = (3/2) * (5/4). Step 9: Calculate the product: B1/B2 = 15/8. Step 10: The ratio of their corresponding bases is 15 : 8.
16
Spherical marble balls, each with a diameter of 1.4 cm, are immersed in water within a cylindrical beaker. The beaker has a diameter of 7 cm. If the water level in the beaker rises by 5.6 cm, how many marbles were dropped into it?
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Solution: Step 1: Calculate the radius of the cylindrical beaker. Diameter_beaker = 7 cm, so R_beaker = 7/2 cm. Step 2: Calculate the volume of water displaced, which corresponds to the volume of the marbles. This volume is a cylinder with the beaker's base radius and the height of the water rise. Height of water rise (h_rise) = 5.6 cm. Volume_displaced = π * (R_beaker)² * h_rise = π * (7/2)² * 5.6 cm³. Step 3: Calculate the radius of a single marble. Diameter_marble = 1.4 cm, so r_marble = 1.4/2 = 0.7 cm = 7/10 cm. Step 4: Calculate the volume of a single marble (sphere). V_marble = (4/3) * π * (r_marble)³ = (4/3) * π * (0.7)³ cm³. Step 5: Let 'n' be the number of marbles dropped. Equate the total volume of 'n' marbles to the volume of water displaced. n * V_marble = Volume_displaced n * (4/3) * π * (0.7)³ = π * (7/2)² * 5.6 Step 6: Simplify the equation and solve for 'n'. n * (4/3) * (7/10)³ = (49/4) * (56/10) n * (4/3) * (343/1000) = (49 * 14) / 10 n * (1372/3000) = 686 / 10 n = (686/10) * (3000/1372) n = 68.6 * (3000/1372) = 150.
17
A hemispherical bowl with an internal radius of 12 cm is filled with liquid. This liquid is to be transferred into cylindrical containers, each having a diameter of 4 cm and a height of 3 cm. How many such containers are needed to completely empty the bowl?
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Solution: Step 1: Calculate the volume of the liquid in the hemispherical bowl. Radius of hemispherical bowl (r_bowl) = 12 cm. Volume_bowl = (2/3)πr_bowl³ = (2/3)π(12 cm)³ = (2/3)π × 1728 cm³ = 2 × 576π cm³ = 1152π cm³. Step 2: Calculate the volume of one cylindrical container. Diameter of cylindrical container = 4 cm, so radius (r_cyl) = 4/2 = 2 cm. Height of cylindrical container (h_cyl) = 3 cm. Volume_container = πr_cyl²h_cyl = π(2 cm)²(3 cm) = π × 4 × 3 cm³ = 12π cm³. Step 3: Determine the number of containers required by dividing the total volume of liquid by the volume of a single container. Number of containers = Volume_bowl / Volume_container. Number of containers = (1152π cm³) / (12π cm³) = 1152 / 12 = 96. Step 4: Therefore, 96 cylindrical containers are necessary to empty the bowl.
18
A solid cone, 9 cm high with a base diameter of 18 cm, is carved from a wooden solid sphere with a radius of 9 cm. What percentage of the wood is wasted?
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Solution: Step 1: Calculate the volume of the wooden sphere. * Radius of sphere (R_sphere) = 9 cm. * Volume of sphere = (4/3)π(R_sphere)³ = (4/3)π(9)³ = (4/3)π(729). * Volume of sphere = 4 * π * 243 = 972π cm³. Step 2: Calculate the volume of the cone. * Height of cone (h_cone) = 9 cm. * Diameter of base of cone = 18 cm, so radius of cone (r_cone) = 18 / 2 = 9 cm. * Volume of cone = (1/3)π(r_cone)²(h_cone) = (1/3)π(9)²(9). * Volume of cone = (1/3)π(81)(9) = π(81)(3) = 243π cm³. Step 3: Calculate the volume of wood wasted. * Volume wasted = Volume of sphere - Volume of cone. * Volume wasted = 972π - 243π = 729π cm³. Step 4: Calculate the percentage of wood wasted. * Percentage wasted = (Volume wasted / Volume of sphere) * 100. * Percentage wasted = (729π / 972π) * 100. * Percentage wasted = (729 / 972) * 100 = (3/4) * 100 = 75%.
19
Given an equilateral triangle with an altitude of 20√3 cm, calculate its area.
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Solution: Step 1: Recall the formula for the area of an equilateral triangle: Area = (√3 / 4) * side^2 Step 2: The altitude (h) of an equilateral triangle is related to the side (a) by: h = (√3 / 2) * a Step 3: Given h = 20√3 cm, solve for side length: 20√3 = (√3 / 2) * a Step 4: Simplify to find side length: a = 40 cm Step 5: Substitute side length into area formula: Area = (√3 / 4) * (40)^2 Step 6: Calculate area: Area = (√3 / 4) * 1600 = 400√3 cm²
20
A solid lead sphere with a radius of 11 cm is melted down and remolded into smaller solid spheres, each having a radius of 2 cm. What is the maximum integer number of these smaller spheres that can be produced?
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Solution: Step 1: Calculate the volume of the large sphere. Let its radius be R = 11 cm. V_large = (4/3)πR³ = (4/3)π(11³) = (4/3)π(1331) cm³. Step 2: Calculate the volume of one small sphere. Let its radius be r = 2 cm. V_small = (4/3)πr³ = (4/3)π(2³) = (4/3)π(8) cm³. Step 3: The total volume of metal is conserved during melting and recasting. To find the number of small spheres 'n', divide the volume of the large sphere by the volume of one small sphere. n = V_large / V_small = [ (4/3)π(1331) ] / [ (4/3)π(8) ]. Step 4: Cancel out (4/3)π: n = 1331 / 8. Step 5: Perform the division: n = 166.375. Step 6: Since only whole spheres can be formed, the maximum integer number of such spheres is 166.
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