Arithmatical - Volume and Surface Areas
DIRECTIONS : Problems based on Volume&Surface Areas.
| 1. |
50 men took a dip in water tank 40 m long and 20 m broad on a religious day. If the average displacements of water by a man is 4 m3, then the rise in the water level in the tank will be |
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A. 20 cm |
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B. 25 cm |
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C. 35 cm |
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D. 50 cm |
| Solution |
| Total number of water displaced | = (4 ×50)m3 |
| ‹=› 200m3 |
| Rise in water level | = (200 / 40×20)m |
| = 0.25 m |
| ‹=› 25 cm |
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| 2. |
A cube of edge 5 cm is cut into cubes of each edge 1 cm. The ratio of the total surface area of one of the small cubes to that of the large cube is equal to |
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A. 1 : 5 |
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B. 1 : 25 |
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C. 1 : 125 |
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D. 1 : 625 |
| Solution |
| Required ratio | =(6x1x1 / 6x5x5) |
| ‹=› 1/ 25 |
| ‹=› 1 : 125. |
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| 3. |
The cost of painting the whole surface area of a cube at the rate of 13 paise per Sq.cm is Rs. 343.98. Then the volume of the cube is |
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A. 8500 cm3 |
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B. 9000 cm3 |
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C. 9250 cm3 |
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D. 9261 cm3 |
| Solution |
| Surface area | = (34398 / 13) |
| ‹=›2646cm3 |
| ‹=›6a2= 2646 |
| ‹=›a2= 441 |
| ‹=›a = 21. |
| So,volume =(21x21x21)cm3= 9261cm3. |
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