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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, - Mathematics

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Question

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 Rs 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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Solution

Height of container, h = 30 cm

Lower radius,\[r_1 = 20 cm\]

Upper radius, \[r_2 = 40 cm\]

Volume of container =\[\frac{1}{3}\pi\left( {r_1}^2 + {r_2}^2 + r_1 r_2 \right)h\]

\[= \frac{1}{3} \times \frac{22}{7}\left( {20}^2 + {40}^2 + 20 \times 40 \right)30\]

\[ = \frac{1}{3} \times \frac{22}{7}\left( 400 + 1600 + 800 \right)30\]

\[ = \frac{1}{3} \times \frac{22}{7} \times 2800 \times 30\]

\[ = 88000 {cm}^3 = 88 L\]

Amount of milk needed daily for the camp = 880 L

\[= \frac{1}{3} \times \frac{22}{7}\left( {20}^2 + {40}^2 + 20 \times 40 \right)30\]
\[ = \frac{1}{3} \times \frac{22}{7}\left( 400 + 1600 + 800 \right)30\]
\[ = \frac{1}{3} \times \frac{22}{7} \times 2800 \times 30\]
\[ = 88000 {cm}^3 = 88 L\]

Number of containers required = \[\frac{880 L}{88 L} = 10\]

Thus, 10 such containers of milk are needed for a camp.
Also, milk is available at the rate of Rs 35 per litre.
∴ Cost of milk = \[Rs 35 \times 880 = Rs 30, 800\]
Here, the donor agency indicated the following values:
a) Charity
b) Sympathy
c) Humanity
 
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2014-2015 (March) Foreign Set 1
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