surface area and volume problems with solutions for SSC and bank exams

Most of the students are facing difficulties in surface area and volume problems in competitive exams. But it happens, while solving questions. Don't panic, if you face any difficulty in surface area and volume problems, you can learn here, how to solve surface area and volume problems with the solution for SSC and bank exams.

With the help of these surface area and volume problems with solutions, you can increase your confidence or performance level through the practice of differential equations. So, keep practice with these solutions for the better rank in your competitive exams. 

You can practice surface area and volume formulas with examples also. 

Surface Area & Volume Problems with Solutions for Competitive Exams

Q.1.The dimensions of an open box are 50 cm, 40 cm and 23 cm. Its thickness is 3 cm. If 1 cubic cm of metal used in the box weight 0.5 gms, find the weight of the box.

Solution 

Volume of the metal used in the box = External Volume – Internal Volume

                                                                        = [(50×40×23) – (44×34×20)] cm³

                                                                        = 16080 cm³

$$∴ Weight \ of \ the \ metal=\left({16080×0.5\over1000} \right)\ kg = 8.04 \ kg. $$

Q.2. The diagonal of a cube is 6√3 cm. Find the its volume and surface area.

Solution 

Let the edge of the cube be a.

∴ √3 a = 6√3 → a = 6.

So, Volume = a³ = (6×6×6) cm³ = 216 cm³

Surface area = 6a² = (6×6×6) cm² = 216 cm²

Q.3. The surface area of a cube is 1734 sq. cm. Find its volume.

Solution 

Let the edge of the cube be a. Then,

6a² = 1734 → a² = 289 → a = 17 cm

∴ Volume = a³ = (17)³ = 4913 cm³

Q.4. A rectangular block 6 cm by 12 cm by 15 cm is cut up into an exact number of equal cubes. Find the least possible number of cubes.

Solution 

Volume of the block = (6×12×15) cm³ = 1080 cm³

Side of the largest cube = H.C.F of 6 cm, 12 cm, 15 cm = 3 cm.

Volume of this cube = (3×3×3) cm³ = 27 cm³

$$ Number \ of \ cubes = \left({1080\over27} \right) = 40 $$

Q.5. A cube of edge 15 cm is immersed completely in a rectangular vessel containing water. If the dimensions of the base of vessel are 20cm × 15cm, find the rise in water level.

Solution 

Increase in volume = Volume of the cube = (15×15×15) cm³.

$$ ∴ Rise \ in \ water \ level = \left({Volume\over Area} \right) =\left({15×15×15\over 20×15} \right) cm = 11.25 \ cm. $$

Q.6. Three solid cubes of sides 1 cm, 6 cm, and 8 cm are melted to form a new cube. Find the surface area of the cube so formed.

Solution 

Volume of new cube = (1³+6³+8³) cm³ = 729 cm³.

Edge of new cube = $$ = \ ^{3}\sqrt {729 } \ cm = 9 \ cm$$

∴ Surface area of the new cube = (6×9×9) cm². = 486 cm².

Q.7. If each edge of a cube is increased by 50% find the percentage increases in its surface area.

Solution 

Let, original length of each edge = a.

Then, original surface area = 6a²

$$ New \ edge \ = (150 \% \ of \ a) = \left({150\over100}a \right)={3a\over2}$$

  $$ New \ surface \ ara = 6×\left({3a\over2} \right)^2={27\over2}a^2$$

Increase percent in surface area   $$ =\left({15\over2}a^2×{1\over6a^2}×100\right)\% = 125 \%$$

Q.8. Two cubes have their volumes in the 1 : 27. Find the ratio of their surface areas.

Solution 

Let their edges be a and b. Then, 

$${a^3\over b^3}={1\over27} or \left({a\over b} \right)^3=\left({1\over3} \right)^3or{a\over b}={1\over 3} $$

∴ Ratio of their surface areas 

$$={6a^2\over 6 b^2}={a^2\over b^2}=\left({a\over b} \right)^2 = {1\over9},i.e.,1:9. $$

Q.9. If the capacity of a cylindrical tank is 1848 m³ and the diameter of its base is 14 m, then find the depth of the tank.

Solution 

Let the depth of the tank be h metres. Then,

$$ \pi×(7)^2×h=1848⟺h=\left(1848×{7\over22}×{1\over 7×7} \right)=12 \ m  $$

Q.10. 2.2 cubic dm of lead is to be drawn into a cylindrical wire 0.50 cm in diameter. Find the length of the wire in metres.

Solution 

Let the length of the wire be h metres. Then,

$$ \pi×\left({0.50\over2×100} \right)^2×h={2.2\over1000}⟺h=\left({2.2\over1000}×{100×100\over0.25×0.25}×{7\over22} \right)= 112 \ m $$

So, these are the surface area and volume problems with solutions which are helpful for your competitive exams. These solutions help you in your preparation also. You can practice from surface area and volume questions with answers also. 

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