Milk in a container, which is in the form of a frustum of a cone of height n= 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 Rs 35 per litre and 880 litres of milk is needed daily for camp, find how many such containers of milk are needed for a camp?

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Solution

$V = \frac{π}{3} (R^{2} + R r + r^{2}) = \frac{22}{7} \times \frac{30}{3} (1600 + 800 + 400) = \frac{22}{7} \times 10 (2800) = 22 (10) (400) = 220 \times 400 = 88000 c m^{3} = 0.088 m^{3} = 88 l t$
volume of container =88lt
Total milk needed =880 lt
No.of containers $= \frac{880}{88} = 10$

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V=π3(R2+Rr+r2)=227×303(1600+800+400)=227×10(2800)=22(10)(400)=220×400=88000cm3=0.088m3=88lt volume of container =88lt Total milk needed =880 lt No.of containers =88088=10

$V = \frac{π}{3} (R^{2} + R r + r^{2}) = \frac{22}{7} \times \frac{30}{3} (1600 + 800 + 400) = \frac{22}{7} \times 10 (2800) = 22 (10) (400) = 220 \times 400 = 88000 c m^{3} = 0.088 m^{3} = 88 l t$
volume of container =88lt
Total milk needed =880 lt
No.of containers $= \frac{880}{88} = 10$