Surface areas and volumes

Correct option: 2 — 16

Correct option: 2 — 71π cm³

Correct option: 2 — 210

Correct option: 4 — The volume of the solids is the same but the surface area of the solids is different.

Correct option: 2 — 1098π cm³

Correct option: 1 — ⅚

Correct option: 3 — 10 cm

Correct option: 2 — 160π cm³

• Takes height as h and radius as r. Writes h = r. Writes 2πrh + 2πr² = 144π cm². Writes 4πr² = 144π cm². [0.5]

• Solves 4πr² = 144π cm² to get r = 6 cm. [0.5]

• Equates the volumes of the cone and the sphere as: ⅓πr²h = (4/3)πr³ [0.5]

• Simplifies the equation in the above step to find the required ratio as r/h = ¼. [0.5]

• Writes s³ = 6s², where s is the side length of the cube. [0.5]

• Solves the above equation to get s = 6 units. [0.5]

• Writes that 14 × Volume of 1 cylindrical cup = Volume of cubical container. => 14 × (π × 2² × h) = 10³ where, h is the height of the cups. [0.5]

• Solves the above equation to get h as 5.68 cm. [0.5]

• Finds the slant length of conical part using pythagoras theorem as 61 cm. [0.5]

• Finds the CSA of the conical part as: (π × 11 × 61) = 671π cm². [0.5]

• Finds exposed area of the cuboidal part as: [2 × (22 × 22 + 22 × 8 + 22 × 8)] − (π × 11²) = (1672 − 121π) cm². [0.5]

• Finds total surface area to be painted red as: (671π + 1672 − 121π) = (1672 + 550π) cm² = 3399 cm². [0.5]

• Finds the volume of the hemispherical scoop on the top as (⅔ × π × 2³) = 16π/3 cm³. [0.5]

• Finds the volume of the cone as (⅓ × π × 2² × 10) = 40π/3 cm³. [0.5]

• Finds the total volume of the ice cream as Volume of the hemispherical scoop + Volume of the cone = 56π/3 cm³. Finds that an equal share of the ice cream is 28π/3 cm³. [0.5]

• Concludes that Swapnil got an equal share of the ice cream. [0.5]

• Finds the volume of the conical part as: (⅓ × π × r² × r) = πr³/3 [0.5]

• Finds the volume of the cylindrical part as: (π × r² × 2r) = 2πr³ [0.5]

• Finds the total volume of the solid as: (πr³/3 + 2πr³) = 7πr³/3 [0.5]

• Writes 7πr³/3 = 198. Solves the equation to get r = 3 cm. [0.5]

• Finds the surface area of the cube as (6 × 4²) = 96 cm². [0.5]

• Finds curved surface area of the hemisphere as (2 × π × 2²) = 8π cm². [0.5]

• Finds the surface area of the circular base of the hemisphere to be subtracted from the surface area of the cube as (π × 2²) = 4π cm². [0.5]

• Finds the total surface area of the paper weight to be polished as (96 + 8π − 4π) = (96 + 4π) cm² = 108.56 cm². [0.5]

• Assumes the radius of the cone as r units, and hence, writes the length of edge of the cube as 2r units and the height of the cone as 2r units. [0.5]

• Finds the volume of the cube as (2r)³ = 8r³. [1]

• Finds the volume of the cone as (⅓ × π × r² × 2r) = ⅔πr³. [1]

• Uses the above step to find the ratio as 42 : 11. [0.5]

• Finds the volume of the regular size container as (⅓ × π × 8² × 30) = 640π cm³. [1]

• Finds the volume of 5 regular size portions as (5 × 640π) = 3200π cm³. [0.5]

• Finds the volume of the family size container as (π × 10² × 30) = 3000π cm³. [1]

• Mentions that Yamir and his friends should get 5 regular size portions. [0.5]

• Finds the volume of the cone as: (⅓ × π × 7² × 5) = 245π/3 m³. [0.5]

• Finds volume of the cylinder as: (π × 7² × 15) = 735π m³. [0.5]

• Finds the volume of the silo as: (2 × 245π/3 + 735π) = 2695π/3 m³. [0.5]

• Finds the volume of grains in the silo as: Volume of cone + Volume of cylinder => (245π/3 + 735π) = 2450π/3 m³. (Award full marks for this step if volume of grains is directly calculated.) [1]

• Finds the fraction of silo that is filled with grains as (Volume of grains/Volume of silo) = 10/11. [0.5]

• i) Finds the CSA of the cone as (π × 3 × 14) = 42π cm². [0.5]

• Finds the surface area of the circular base of the cone as (π × 3²) = 9π cm². [0.5]

• Finds the surface area of the circular base of the cylinder as (π × 2²) = 4π cm². [0.5]

• Finds the area to be painted green as (42π + 9π − 4π) = 47π = 147.71 cm². [0.5]

• ii) Finds the CSA of the cylinder as (2 × π × 2 × 7) = 28π cm². [0.5]

• Finds the area to be painted brown as (28π + 4π) = 32π = 100.57 cm². [0.5]

• Finds the total surface area of the cuboidal table top as: 2 × (90 × 4 + 90 × 50 + 4 × 50) = 10120 cm² [1]

• Finds the lateral surface area of 4 wooden legs as: 4 × (5 × 60 + 5 × 60) = 2400 cm² [1]

• Finds the area of the base of the wooden legs where they connect with the table top as: 4 × (5 × 5) = 100 cm² [0.5]

• Finds the total surface area of the table to be laminated as: (10120 − 100 + 2400) = 12420 cm² [0.5]

• Finds the volume of the wooden cube to be s³ cm³, where s is the length of the side of the cube. [0.5]

• Finds the volume of cylinder as: {π × (s/2)² × s} = (πs³/4) cm³ [0.5]

• Finds the radius of of the hemisphere as ½ × ⅘ × s = ⅖s. [0.5]

• Finds the volume of the hemisphere carved out of the cylinder as: {⅔ × π × (2s/5)³} = (16s³π/375) cm³ [1]

• Finds total volume of the bird feeder as: (s³π/4 − 16s³π/375) = (311s³π/1500) cm³ [1]

• Mentions percentage of wooden cube left as: (Volume of bird feeder/Volume of wooden cube) × 100 [0.5]

• Simplifies the fraction in the above step to get percentage of wooden cube left as 65%. [1]

• i) Finds the surface area of the cube as (6 × 30²) = 5400 cm². [0.5]

• Finds the CSA of the 21 hemispherical dents as (21 × 2 × π × 4²) = 2112 cm². [0.5]

• Finds the surface area of the bases of the hemispherical dents to subtract from the surface area of the cube as (21 × π × 4²) = 1056 cm². [0.5]

• Finds total surface area of the plush toy, i.e., cloth required to make the plush toy as (5400 + 2112 − 1056) = 6456 cm². [1]

• ii) Finds the volume of the cube as 30³ = 27000 cm³. [0.5]

• Finds the volume of the 21 hemispherical dents as (21 × ⅔ × π × 4³) = 2816 cm³. [0.5]

• Finds the total volume of the plush toy, i.e., the volume of cotton required to stuff the plush toy as 27000 − 2816 = 24184 cm³. [1]

• iii) Finds the cost of the cloth and cotton required to make the plush toy as (0.01 × 6456 + 0.02 × 24184) = Rs 548.24. [0.5]

• Calculates the volume of the cylindrical container as πr²h = (22/7)(12.5)²(28) = 13750 cm³. [1]

• Finds the height till which lemonade is poured in the cup as 75% of 10 cm = 7.5 cm. Finds the amount of lemonade in the container as 10 × 1000 = 10000 cm³. [0.5]

• Calculates the volume of lemonade poured in the cylindrical paper cup as: Volume of lemonade poured in the cup = πr²h = 22/7 × (2.8)² × 7.5 = 184.8 cm³. [0.5]

• Writes the maximum amount of cups sold by her as {Total volume of lemonade/Volume of 1 cup of lemonade)} = 10000/184.8. [0.5]

• Solves the equation in the previous step to get 54.11 and rounds it to 54 cups. Calculates the maximum amount made by Kinjal as 54 x 10 = Rs. 540. [0.5]

• Writes new height of cup = 75% of 12 cm = 9 cm. Hence, the height of cylindrical part is 6 cm and the height of conic part is 3 cm. [0.5]

• Calculates the volume of lemonade filled in the cup as: Volume of Cylinder + Volume of cone = πr²H + ⅓πr²h = 22/7 × (2.8)² × 6 + ⅓ × 22/7 × (2.8)² × 3 = 172.48 cm³. [1.5]