Volume And Area All Competence Aptitude

To find out how many cubes of 3 cm edge can be cut out of a cube of 18 cm edge, we need to calculate the volume of each cube and the volume of the larger cube, then divide the volume of the larger cube by the volume of each smaller cube.

The volume of a cube is calculated by the formula: \[ \text{Volume} = \text{edge}^3 \]

For the smaller cube with a 3 cm edge:\[ \text{Volume of smaller cube} = 3^3 = 27 \text{ cubic cm} \]

For the larger cube with an 18 cm edge:\[ \text{Volume of larger cube} = 18^3 = 5832 \text{ cubic cm} \]

Now, we can find out how many of the smaller cubes can fit into the larger cube:\[ \text{Number of smaller cubes} = \dfrac{\text{Volume of larger cube}}{\text{Volume of smaller cube}} \]\[ \text{Number of smaller cubes} = \dfrac{5832}{27} \]\[ \text{Number of smaller cubes} = 216 \]

So, 216 cubes of 3 cm edge can be cut out of a cube of 18 cm edge.

Therefore, the correct answer is 216.

To find the ratio of the total surface areas of the smaller cubes to the surface area of the large cube, we first need to calculate the surface areas of each cube.

For a cube with side \( s \), the surface area \( A \) is given by:\[ A = 6s^2 \]

For the smaller cubes:
1. Cube with side 3 cm: \( A_1 = 6(3^2) = 6(9) = 54 \) square cm
2. Cube with side 4 cm: \( A_2 = 6(4^2) = 6(16) = 96 \) square cm
3. Cube with side 5 cm: \( A_3 = 6(5^2) = 6(25) = 150 \) square cm

Now, let's find the total surface area of the smaller cubes:\[ \text{Total surface area of smaller cubes} = A_1 + A_2 + A_3 \]\[ \text{Total surface area of smaller cubes} = 54 + 96 + 150 = 300 \] square cm

Next, let's calculate the side of the large cube. The volume of the large cube should be equal to the sum of the volumes of the smaller cubes.

The volume of a cube with side \( s \) is given by:\[ V = s^3 \]

For the large cube:\[ V_{\text{large cube}} = (3^3) + (4^3) + (5^3) \]\[ V_{\text{large cube}} = 27 + 64 + 125 = 216 \] cubic cm

Now, to find the side length of the large cube, we take the cube root of its volume:\[ s_{\text{large cube}} = \sqrt[3]{216} = 6 \] cm

Now, we find the surface area of the large cube:\[ A_{\text{large cube}} = 6s^2 \]\[ A_{\text{large cube}} = 6(6^2) \]\[ A_{\text{large cube}} = 6(36) \]\[ A_{\text{large cube}} = 216 \] square cm

Now, we can find the ratio of the total surface areas:\[ \text{Ratio} = \dfrac{\text{Total surface area of smaller cubes}}{\text{Surface area of large cube}} \]\[ \text{Ratio} = \dfrac{300}{216} \]\[ \text{Ratio} = \dfrac{25}{18} \]

So, the ratio of the total surface areas of the smaller cubes to the surface area of the large cube is \( 25 : 18 \).

Therefore, the correct answer is \( 25 : 18 \).

When each edge of a cube is increased by 50%, the new length of each edge becomes \(1.5\) times its original length.

Let's denote the original length of each edge as \(s\). After increasing each edge by 50%, the new length becomes \(1.5s\).

The original surface area of the cube is given by:\[ A_{\text{original}} = 6s^2 \]

The new surface area of the cube with each edge increased by 50% is:\[ A_{\text{new}} = 6(1.5s)^2 = 6(2.25s^2) = 13.5s^2 \]

Now, let's calculate the percentage increase in surface area:

\[ \text{Percentage increase} = \left( \dfrac{\text{New surface area} - \text{Original surface area}}{\text{Original surface area}} \right) \times 100\% \]

\[ \text{Percentage increase} = \left( \dfrac{13.5s^2 - 6s^2}{6s^2} \right) \times 100\% \]

\[ \text{Percentage increase} = \left( \dfrac{7.5s^2}{6s^2} \right) \times 100\% \]

\[ \text{Percentage increase} = \left( \dfrac{7.5}{6} \right) \times 100\% \]

\[ \text{Percentage increase} = 125\% \]

So, the percentage increase in surface area is \(125\%\).

Therefore, the correct answer is \(125\%\).

To find the cost of plastering the walls and bottom of the tank, we first need to calculate the total surface area to be plastered.

The tank has six faces: four walls, a bottom, and a top (which won't be plastered). We'll calculate the surface area of the four walls and the bottom.

1. Surface area of each wall = length * depth = \(25 \times 6\) (since there are two walls of length 25m and two walls of length 12m)
2. Surface area of the bottom = length * width = \(25 \times 12\)

So, the total surface area to be plastered = 2 * (25 * 6) + 2 * (12 * 6) + 25 * 12

\[= 2 \times (150) + 2 \times (72) + 300\]\[= 300 + 144 + 300\]\[= 744 \text{ square meters}\]

Now, let's calculate the cost of plastering:

Cost per square meter = 75 paise = \(0.75\) rupees

Total cost = \(744 \times 0.75\)

\[= 558 \text{ rupees}\]

So, the cost of plastering its walls and bottom at 75 paise per square meter is Rs. 558.

Therefore, the correct answer is Rs. 558.

Let's denote the original radius of the cone as \( r \) and the original height as \( h \). 

When the radius and height are increased by 10%, the new radius becomes \( 1.1r \) and the new height becomes \( 1.1h \).

The formula for the volume of a cone is given by:

\[ V = \dfrac{1}{3} \pi r^2 h \]

For the original cone:\[ V_{\text{original}} = \dfrac{1}{3} \pi r^2 h \]

For the new cone:\[ V_{\text{new}} = \dfrac{1}{3} \pi (1.1r)^2 (1.1h) \]\[ V_{\text{new}} = \dfrac{1}{3} \pi (1.21r^2) (1.1h) \]\[ V_{\text{new}} = \dfrac{1}{3} \pi \times 1.21 \times r^2 \times h \times 1.1 \]

Now, let's calculate the percentage increase in volume:

\[ \text{Percentage increase} = \left( \dfrac{V_{\text{new}} - V_{\text{original}}}{V_{\text{original}}} \right) \times 100\% \]\[ \text{Percentage increase} = \left( \dfrac{\dfrac{1}{3} \pi \times 1.21 \times r^2 \times h \times 1.1 - \dfrac{1}{3} \pi r^2 h}{\dfrac{1}{3} \pi r^2 h} \right) \times 100\% \]\[ \text{Percentage increase} = \left( \dfrac{1.21 \times 1.1 - 1}{1} \right) \times 100\% \]\[ \text{Percentage increase} = (1.331 - 1) \times 100\% \]\[ \text{Percentage increase} = 0.331 \times 100\% \]\[ \text{Percentage increase} = 33.1\% \]

So, the percentage increase in volume is 33.1%.

Therefore, the correct answer is 33.1%.

Let's denote the width of the wall as \( w \) meters.

Given:
1. The height of the wall is 6 times its width.
2. The length of the wall is 7 times its height.

From the first statement, we can express the height \( h \) in terms of the width \( w \):\[ h = 6w \]

From the second statement, we can express the length \( l \) in terms of the height \( h \):\[ l = 7h \]

We know that the volume \( V \) of a cuboid is given by:\[ V = l \times w \times h \]

Substituting the expressions for \( l \) and \( h \) into the volume formula:\[ 16128 = (7h) \times w \times h \]\[ 16128 = 7wh^2 \]

Substituting the expression for \( h \):\[ 16128 = 7w(6w)^2 \]\[ 16128 = 7w(36w^2) \]\[ 16128 = 252w^3 \]

Divide both sides by 252:\[ w^3 = \dfrac{16128}{252} \]

\[ w^3 = 64 \]

Take the cube root of both sides:\[ w = \sqrt[3]{64} \]

\[ w = 4 \]

So, the width of the wall is 4 meters.

Therefore, the correct answer is 4m.

\[ h = 14 \text{ cm}, \ r = 7 \text{ cm} \]

So, \( \ell = \sqrt{7^2 + 14^2} = \sqrt{245} = 7\sqrt{5} \) cm.

Curved surface area \( = \pi rl + \pi r^2 \)

\[ = (227 \times 7 \times 7\sqrt{5} + 227 \times 7 \times 7) \text{ cm}^2 \]

\[ = 154(\sqrt{5} + 1) \text{ cm}^2 \]

\[ = 154 \times 3.236 \text{ cm}^2 \]

\[ = 498.35 \text{ cm}^2 \]

Therefore, the corrected total surface area of the cone is indeed \( 498.35 \text{ cm}^2 \).

Thank you for your patience, and I apologize for the earlier errors.

To find the length of the longest pole that can be placed in a room, we need to find the diagonal distance from one corner of the room to the opposite corner.

Using the formula for the diagonal of a rectangular prism:

\[ \text{Diagonal} = \sqrt{length^2 + width^2 + height^2} \]

Given:
Length (\( l \)) = 12 m
Width (\( w \)) = 8 m
Height (\( h \)) = 9 m

Plugging in the values:

\[ \text{Diagonal} = \sqrt{12^2 + 8^2 + 9^2} \]\[ \text{Diagonal} = \sqrt{144 + 64 + 81} \]\[ \text{Diagonal} = \sqrt{289} \]\[ \text{Diagonal} = 17 \text{ m} \]

Therefore, the length of the longest pole that can be placed in the room is 17 meters.

So, the correct answer is 17m.

To find the surface area of the new cube formed by melting the three solid cubes, we need to find the volume of all three cubes combined, and then find the side length of the new cube. Finally, we calculate the surface area of the new cube.

Given:
Side lengths of the cubes: \(1 \text{ cm}\), \(6 \text{ cm}\), and \(8 \text{ cm}\)

The total volume of the cubes combined is the sum of the volumes of each cube:\[ \text{Total Volume} = (1^3 + 6^3 + 8^3) \text{ cm}^3 \]\[ \text{Total Volume} = (1 + 216 + 512) \text{ cm}^3 \]\[ \text{Total Volume} = 729 \text{ cm}^3 \]

The side length of the new cube is the cube root of its total volume:\[ \text{Side Length} = \sqrt[3]{729} \text{ cm} \]\[ \text{Side Length} = 9 \text{ cm} \]

Now, let's find the surface area of the new cube:\[ \text{Surface Area} = 6 \times (\text{Side Length})^2 \]\[ \text{Surface Area} = 6 \times (9^2) \text{ cm}^2 \]\[ \text{Surface Area} = 6 \times 81 \text{ cm}^2 \]\[ \text{Surface Area} = 486 \text{ cm}^2 \]

Therefore, the surface area of the cube formed is \( 486 \text{ cm}^2 \).

So, the correct answer is 486.

To find the total area of the wet surface of the cistern, we first need to calculate the area of the four walls and the bottom of the cistern that are in contact with the water.

Given:
Length of the cistern (\(l\)) = 6 m
Width of the cistern (\(w\)) = 4 m
Depth of the water (\(h\)) = 1 m 25 cm = 1.25 m

The area of the four walls is the perimeter of the base multiplied by the depth of the water.

Perimeter of the base (\(P\)) = \(2(l + w)\)

\[ P = 2(6 + 4) = 2 \times 10 = 20 \text{ m} \]

The area of the four walls (\(A_{\text{walls}}\)) = \(P \times h\)

\[ A_{\text{walls}} = 20 \times 1.25 = 25 \text{ m}^2 \]

The area of the bottom of the cistern (\(A_{\text{bottom}}\)) = \(l \times w\)

\[ A_{\text{bottom}} = 6 \times 4 = 24 \text{ m}^2 \]

Now, let's calculate the total area of the wet surface by adding the areas of the four walls and the bottom:

Total area of wet surface = \(A_{\text{walls}} + A_{\text{bottom}}\)\[ = 25 \text{ m}^2 + 24 \text{ m}^2 = 49 \text{ m}^2 \]

Therefore, the total area of the wet surface of the cistern is 49 square meters.

So, the correct answer is 49.