Given below are three quantities named 1, 2 and 3. Based on the given information, you have to determine the relation among the three quantities. You should use the given data and knowledge of Mathematics to choose between the possible answer. The options represent the relations among three quantities. A. < B. > C. ≤ D. ≥ E. = Quantity 1: There is a square field of side 4 meter. A flower bed is to be prepared in its centre, leaving a gravelled path of uniform length around the flowerbed. The cost of preparing the flowerbed and gravelling the path is Rs. 100 and Rs. 200 per square meter respectively. If the total cost is Rs. 1756 then find the double of width of gravelled path. Quantity 2: The height of a cone is 45 cm. A small cone is cut from it parallel to the base. If the ratio of volume of small cone and the remaining part is 8 : 19 then what is the height from the base from where the cone was cut Quantity 3: A cylindrical container of radius 60 cm and height 150 cm is filled with ice cream. 10 cones of equal volume need to be prepared from the whole ice-cream. These cones has hemispherical top and the height of the conical portion is 4 times the radius of hemisphere. Find radius of the hemisphere. ?

Question

Given below are three quantities named 1, 2 and 3. Based on the given information, you have to determine the relation among the three quantities. You should use the given data and knowledge of Mathematics to choose between the possible answer. The options represent the relations among three quantities. A. < B. > C. ≤ D. ≥ E. = Quantity 1: There is a square field of side 4 meter. A flower bed is to be prepared in its centre, leaving a gravelled path of uniform length around the flowerbed. The cost of preparing the flowerbed and gravelling the path is Rs. 100 and Rs. 200 per square meter respectively. If the total cost is Rs. 1756 then find the double of width of gravelled path. Quantity 2: The height of a cone is 45 cm. A small cone is cut from it parallel to the base. If the ratio of volume of small cone and the remaining part is 8 : 19 then what is the height from the base from where the cone was cut Quantity 3: A cylindrical container of radius 60 cm and height 150 cm is filled with ice cream. 10 cones of equal volume need to be prepared from the whole ice-cream. These cones has hemispherical top and the height of the conical portion is 4 times the radius of hemisphere. Find radius of the hemisphere. ?

E, A
B, E
A, C
B, A
E, C

LohaHridoy

6 views

2025-01-04 04:42

2 Answers

AhmedNazir

Option 4 : B, A

Quantity 1:

Suppose X and Y are the areas of square flowerbed and gravelled path respectively,

⇒ Area of the square field = 4 × 4 = 16 m²

⇒ X + Y = 16 ---- (1)

Since the cost of preparing the flowerbed and gravelling the path is Rs. 100 and Rs. 200 per square meter respectively,

⇒ 100X + 200Y = 1756

From equation 1 and 2:

⇒ X = 14.44

⇒ Area of the square flowerbed = 14.44 m²

⇒ Side of square flowerbed = 3.8 m

⇒ Width of double of gravelled path = (4 – 3.8) = 0.2 m = 20 cm

Quantity 2:

Since the ratio of volume of small cone and the remaining part is 8 ∶ 19

⇒ Ratio of volume of small cone and full cone = 8 ∶ = 8 ∶ 27

Suppose the radius of small and large cone is ‘r’ and ‘R’ respectively and the heights are ‘h’ and ‘H’ respectively

⇒ ∶ = 8 ∶ 27

From the figure,

r/R = h/H

[ alt="F1 Vishnu 12.1.21 Pallavi D12" src="//storage.googleapis.com/tb-img/production/21/01/F1_Vishnu_12.1.21_Pallavi_D12.png" style="width: 201px; height: 197px;">

⇒ h³/H³ = 8/27

⇒ h/H = 2/3

Since H = 45 cm

⇒ h = 45 × 2/3 = 30 cm

⇒ Height from the base from where the cone was cut = 45 – 30 = 15 cm

Quantity 3:

Volume of the cylindrical container = 22/7 × 60 × 60 × 150 cm³

Volume of a cone with hemispherical top = πr²h/3 + 2πr³/3

Putting h = 4r,

Volume of a cone with hemispherical top = 4πr³/3 + 2πr³/3 = 2πr³

Volume of 10 such cones = 20πr³

⇒ 20πr³ = 22/7 × 60 × 60 × 150

⇒ r³ = 27000 cm³

⇒ r = 30 cm

∴ Quantity 1 > Quantity 2 < Quantity 3

6 views Answered 2022-09-04 04:29

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AhmedNazir · Accepted Answer · 2022-09-04T04:29:57+06:00

Option 4 : B, A

Quantity 1:

Suppose X and Y are the areas of square flowerbed and gravelled path respectively,

⇒ Area of the square field = 4 × 4 = 16 m²

⇒ X + Y = 16 ---- (1)

Since the cost of preparing the flowerbed and gravelling the path is Rs. 100 and Rs. 200 per square meter respectively,

⇒ 100X + 200Y = 1756

From equation 1 and 2:

⇒ X = 14.4

⇒ Area of the square flowerbed = 14.4 m²

⇒ Side of square flowerbed = 3.8 m

⇒ Width of double of gravelled path = (4 – 3.8) = 0.2 m = 20 cm

Quantity 2:

Since the ratio of volume of small cone and the remaining part is 8 ∶ 19

⇒ Ratio of volume of small cone and full cone = 8 ∶ = 8 ∶ 27

Suppose the radius of small and large cone is ‘r’ and ‘R’ respectively and the heights are ‘h’ and ‘H’ respectively

⇒ ∶ = 8 ∶ 27

From the figure,

r/R = h/H

[ alt="F1 Vaibhav.S 13-04-21 Savita D16" src="//storage.googleapis.com/tb-img/production/21/04/F1__Vaibhav.S_13-04-21_Savita_D16.png" style="width: 203px; height: 198px;">

⇒ h³/H³ = 8/27

⇒ h/H = 2/3

Since H = 45 cm

⇒ h = 45 × 2/3 = 30 cm

⇒ Height from the base from where the cone was cut = 45 – 30 = 15 cm

Quantity 3:

Volume of the cylindrical container = 22/7 × 60 × 60 × 150 cm³

Volume of a cone with hemispherical top = πr²h/3 + 2πr³/3

Putting h = 4r,

Volume of a cone with hemispherical top = 4πr³/3 + 2πr³/3 = 2πr³

Volume of 10 such cones = 20πr³

⇒ 20πr³ = 22/7 × 60 × 60 × 150

⇒ r³ = 27000 cm³

⇒ r = 30 cm

∴ Quantity 1 > Quantity 2 < Quantity 3