Milk in a container, which is in the form of a frustum of a cone of height 30 cm and the radii of whose lower and upper circular ends are 20 cm and 40 cm respectively, is to be distributed in a camp for flood victims. If this milk is available at the rate of < 35 per litre and 880 litres of milk is needed daily for a camp, find how many such containers of milk are needed for a camp and what cost will it put on the donor agency for this. What value is indicated through this by the donor agency ?
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Given--Radius of lower end (R1) = 20cm
Radius of upper end (R2)= 40 cm
Height (H)= 30 cm
So the volume of container= 1/3 pi H ( R1² + R2² + R1R2)
= 1/3 * 22/7 * 30 ( (20)² + (40)² + 40* 20)
= 1/3 * 22/7 * 30 ( 400 + 1600 + 800)
= 1/3* 22/7 * 30 * 2800
= 88000 cm³ or 88 litres.
Amount of milk required = 880 litre
No. of containers needed will be= 880/88= 10 containers.
Cost of 1 litre milk= Rs. 30
So, Cost of 880 litre= 880 * 30 = Rs. 26400
Answered by
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Radius of the lower end of the container(r) = 20 cm
Radius of the bigger end of the container( R) = 40 cm
Height of the container(h) = 30 cm
Capacity of container = 1/3πh[R² + r² + Rr]
=1/3×22/7×30×[(40)² +(20)²+40×20]
=1/3×22/7×30×[1600 +400 +800]
=1/3×22/7×30×[2800]
=22×10×[400]
=88000 cm³
= 88000/1000
=88 litres
[1000 cm³ = 1 litres]
Amount of milk needed by the camp daily = 880 litres
Hence, number of containers needed = 880/88 = 10
cost of 1 litre milk = ₹35
Cost of 880 litres of milk = ₹ (35 × 880) =₹ 30800.
Radius of the bigger end of the container( R) = 40 cm
Height of the container(h) = 30 cm
Capacity of container = 1/3πh[R² + r² + Rr]
=1/3×22/7×30×[(40)² +(20)²+40×20]
=1/3×22/7×30×[1600 +400 +800]
=1/3×22/7×30×[2800]
=22×10×[400]
=88000 cm³
= 88000/1000
=88 litres
[1000 cm³ = 1 litres]
Amount of milk needed by the camp daily = 880 litres
Hence, number of containers needed = 880/88 = 10
cost of 1 litre milk = ₹35
Cost of 880 litres of milk = ₹ (35 × 880) =₹ 30800.
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