Exercise: 24-C
Q1: If the length, breadth and height of a cuboid are 2m, 2m, and 1 m, respectively, then its surface area (in m²) is:
Step 1: Write the given data:
Length \(l = 2\, m\), Breadth \(b = 2\, m\), Height \(h = 1\, m\)
Step 2: Use the formula for surface area (SA) of a cuboid:
\[
SA = 2(lb + bh + hl)
\]Substitute the values:
\[
SA = 2[(2 \times 2) + (2 \times 1) + (1 \times 2)] = 2[4 + 2 + 2] = 2 \times 8 = 16\, m^2
\]Answer: c. 16
Q2: If the length, breadth and height of a cuboid are in the ratio 6 : 5 : 4 and the total surface area of the cuboid is 33300 cm², then the breadth of the cuboid is:
Step 1: Let the common multiple be \(x\). Then,
\[
l = 6x, \quad b = 5x, \quad h = 4x
\]Step 2: Use the formula for total surface area (TSA) of cuboid:
\[
TSA = 2(lb + bh + hl) = 33300
\]
Substitute values:
\[
2[(6x)(5x) + (5x)(4x) + (4x)(6x)] = 33300 \\
2[30x^2 + 20x^2 + 24x^2] = 33300 \\
2 \times 74x^2 = 33300 \\
148x^2 = 33300 \\
x^2 = \frac{33300}{148} = 225 \\
x = \sqrt{225} = 15
\]Step 3: Find breadth \(b\):
\[
b = 5x = 5 \times 15 = 75\, cm
\]Answer: b. 75
Q3: A rectangular sand box is 5 m and 2 m long. How many cubic metres of sand is needed to fill the box up to a depth of 10 cm?
Step 1: Write the given data:
Length \(l = 5\, m\)
Breadth \(b = 2\, m\)
Depth \(h = 10\, cm = 0.1\, m\)
Step 2: Calculate volume of sand needed (volume of box filled):
\[
V = l \times b \times h = 5 \times 2 \times 0.1 = 1\, m^3
\]Answer: a. 1
Q4: A beam 9 m long, 40 cm wide and 20 cm deep is made up of iron which weighs 50 kg per cubic metre. The weight of the beam is:
Step 1: Write the given data:
Length \(l = 9\, m\)
Width \(w = 40\, cm = 0.4\, m\)
Depth \(d = 20\, cm = 0.2\, m\)
Weight density of iron = 50 kg/m³
Step 2: Calculate volume of the beam:
\[
V = l \times w \times d = 9 \times 0.4 \times 0.2 = 0.72\, m^3
\]Step 3: Calculate weight of the beam:
\[
\text{Weight} = \text{Volume} \times \text{Weight density} = 0.72 \times 50 = 36\, kg
\]Answer: b. 36
Q5: The length of the longest rod that can be placed in a room 30 m long, 24 m broad and 18 m high, is:
Step 1: Write the given data:
Length \(l = 30\, m\)
Breadth \(b = 24\, m\)
Height \(h = 18\, m\)
Step 2: The longest rod will be the space diagonal of the cuboid.
Formula for space diagonal \(d\):
\[
d = \sqrt{l^2 + b^2 + h^2}
\]Substitute the values:
\[
d = \sqrt{30^2 + 24^2 + 18^2} = \sqrt{900 + 576 + 324} = \sqrt{1800} \\
= \sqrt{900 \times 2} = 30 \sqrt{2}
\]Answer: c. \(30 \sqrt{2}\)
Q6: How many bricks, each measuring 25 cm × 12.5 cm × 7.5 cm will be needed to construct a wall 15 m long, 1.8 m high and 37.5 cm thick?
Step 1: Write the given data:
Dimensions of brick:
Length \(l_b = 25\, cm = 0.25\, m\)
Breadth \(b_b = 12.5\, cm = 0.125\, m\)
Height \(h_b = 7.5\, cm = 0.075\, m\)
Dimensions of wall:
Length \(L = 15\, m\)
Height \(H = 1.8\, m\)
Thickness \(T = 37.5\, cm = 0.375\, m\)
Step 2: Calculate volume of the wall:
\[
V_{wall} = L \times H \times T = 15 \times 1.8 \times 0.375 = 10.125\, m^3
\]Step 3: Calculate volume of one brick:
\[
V_{brick} = l_b \times b_b \times h_b = 0.25 \times 0.125 \times 0.075 = 0.00234375\, m^3
\]Step 4: Calculate number of bricks needed:
\[
N = \frac{V_{wall}}{V_{brick}} = \frac{10.125}{0.00234375} = 4320
\]Answer: a. 4320
Q7: A wooden box of dimensions 8 m × 7 m × 6 m is to carry rectangular boxes of dimensions 8 cm × 7 cm × 6 cm. The maximum number of boxes that can be carried in the wooden box is:
Step 1: Write the given data:
Wooden box dimensions:
Length \(L = 8\, m = 800\, cm\)
Breadth \(B = 7\, m = 700\, cm\)
Height \(H = 6\, m = 600\, cm\)
Small box dimensions:
Length \(l = 8\, cm\)
Breadth \(b = 7\, cm\)
Height \(h = 6\, cm\)
Step 2: Calculate volume of wooden box:
\[
V_{wooden} = L \times B \times H = 800 \times 700 \times 600 = 336000000\, cm^3
\]Step 3: Calculate volume of small box:
\[
V_{small} = l \times b \times h = 8 \times 7 \times 6 = 336\, cm^3
\]Step 4: Calculate the maximum number of small boxes:
\[
N = \frac{V_{wooden}}{V_{small}} = \frac{336000000}{336} = 1000000
\]Answer: a. 1000000
Q8: The number of bricks, each measuring 25 cm × 12.5 cm × 7.5 cm required to construct a wall 6 m long, 5 m high and 0.5 m thick, while the mortar occupies 5% of the volume of the wall, is:
Step 1: Write the given data:
Dimensions of brick:
Length \(l_b = 25\, cm = 0.25\, m\)
Breadth \(b_b = 12.5\, cm = 0.125\, m\)
Height \(h_b = 7.5\, cm = 0.075\, m\)
Dimensions of wall:
Length \(L = 6\, m\)
Height \(H = 5\, m\)
Thickness \(T = 0.5\, m\)
Mortar occupies 5% of the volume of the wall.
Step 2: Calculate volume of the wall:
\[
V_{wall} = L \times H \times T = 6 \times 5 \times 0.5 = 15\, m^3
\]Step 3: Calculate volume of mortar:
\[
V_{mortar} = 5\% \text{ of } V_{wall} = 0.05 \times 15 = 0.75\, m^3
\]Step 4: Calculate volume of bricks:
\[
V_{bricks} = V_{wall} – V_{mortar} = 15 – 0.75 = 14.25\, m^3
\]Step 5: Calculate volume of one brick:
\[
V_{brick} = l_b \times b_b \times h_b = 0.25 \times 0.125 \times 0.075 = 0.00234375\, m^3
\]Step 6: Calculate number of bricks:
\[
N = \frac{V_{bricks}}{V_{brick}} = \frac{14.25}{0.00234375} = 6080
\]Answer: c. 6080
Q9: The surface area of a cube with side 27 cm is:
Step 1: Write the given data:
Side of cube \(a = 27\, cm\)
Step 2: Use formula for surface area of a cube:
\[
SA = 6a^2
\]Substitute the value:
\[
SA = 6 \times (27)^2 = 6 \times 729 = 4374\, cm^2
\]Answer: c. 4374 cm²
Q10: The perimeter of one face of a cube is 40 cm. The volume of the cube (in cm³) is:
Step 1: Write the given data:
Perimeter of one face of cube \(P = 40\, cm\)
Step 2: Since one face is a square, perimeter \(P = 4 \times \text{side}\)
Let side \(a\) be:
\[
4a = 40 \\
\Rightarrow a = \frac{40}{4} = 10\, cm
\]Step 3: Use formula for volume of cube:
\[
V = a^3 = 10^3 = 1000\, cm^3
\]Answer: c. 1000
Q11: The volume of a cube with surface area 384 sq. cm is:
Step 1: Write the given data:
Surface area \(SA = 384\, cm^2\)
Step 2: Use formula for surface area of cube:
\[
SA = 6a^2
\]
Where \(a\) is the side of the cube.
Step 3: Calculate side \(a\):
\[
6a^2 = 384 \\
\Rightarrow a^2 = \frac{384}{6} = 64 \\
a = \sqrt{64} = 8\, cm
\]Step 4: Use formula for volume of cube:
\[
V = a^3 = 8^3 = 512\, cm^3
\]Answer: d. 512
Q12: The volume of a cube is 2744 cu. cm. Its surface area is:
Step 1: Write the given data:
Volume \(V = 2744\, cm^3\)
Step 2: Use formula for volume of cube:
\[
V = a^3
\]
Where \(a\) is the side of the cube.
Step 3: Calculate side \(a\):
\[
a = \sqrt[3]{2744} = 14\, cm
\]Step 4: Use formula for surface area of cube:
\[
SA = 6a^2 = 6 \times 14^2 = 6 \times 196 = 1176\, cm^2
\]Answer: d. 1176
Q13: If the length of diagonal of a cube is \(4\sqrt{3}\) cm, then the length of its edge is:
Step 1: Write the given data:
Diagonal of cube \(d = 4\sqrt{3}\, cm\)
Step 2: Use the formula for diagonal of a cube:
\[
d = a \sqrt{3}
\]
Where \(a\) is the edge length.
Step 3: Calculate edge \(a\):
\[
4\sqrt{3} = a \sqrt{3} \\
\Rightarrow a = 4\, cm
\]Answer: b. 4 cm
Q14: The length of longest pole that can be placed on the floor of a room is 10 m and the length of longest pole that can be placed in the room is \(10\sqrt{2}\) m. The height of the room is:
Step 1: Write the given data:
Longest pole on floor (diagonal of floor) \(d_{floor} = 10\, m\)
Longest pole in room (space diagonal) \(d_{space} = 10\sqrt{2}\, m\)
Let length \(l\), breadth \(b\), and height \(h\) of the room be:
Since diagonal on floor is 10 m,
\[
d_{floor} = \sqrt{l^2 + b^2} = 10
\]Longest pole in the room (space diagonal) is:
\[
d_{space} = \sqrt{l^2 + b^2 + h^2} = 10\sqrt{2}
\]Step 2: Square both equations:
\[
l^2 + b^2 = 10^2 = 100 \\
l^2 + b^2 + h^2 = (10\sqrt{2})^2 = 100 \times 2 = 200
\]Step 3: Subtract first from second:
\[
(l^2 + b^2 + h^2) – (l^2 + b^2) = 200 – 100 \\
h^2 = 100 \\
h = \sqrt{100} = 10\, m
\]Answer: d. 10 m
Q15: The volume of a wall, 5 times as high as it is broad and 8 times as long as it is high, is 12.8 m³. The breadth of the wall is:
Step 1: Let the breadth of the wall be \(b\) m.
Height \(h = 5b\) (5 times the breadth)
Length \(l = 8h = 8 \times 5b = 40b\)
Step 2: Use the formula for volume:
\[
V = l \times b \times h
\]
Given volume \(V = 12.8\, m^3\)
Substitute \(l, b, h\):
\[
12.8 = 40b \times b \times 5b = 200b^3
\]Step 3: Solve for \(b\):
\[
b^3 = \frac{12.8}{200} = 0.064 \\
b = \sqrt[3]{0.064} = 0.4\, m = 40\, cm
\]Answer: d. 40 cm
Q16: Two cubes each with 8 cm edge are joined end to end. The surface area of the resulting cuboid is:
Step 1: Write the given data:
Edge of each cube \(a = 8\, cm\)
Two cubes joined end to end means the resulting cuboid has:
Length \(L = 8 + 8 = 16\, cm\)
Breadth \(B = 8\, cm\)
Height \(H = 8\, cm\)
Step 2: Use the formula for surface area of cuboid:
\[
SA = 2(LB + BH + HL)
\]Calculate each term:
\[
LB = 16 \times 8 = 128 \\
BH = 8 \times 8 = 64 \\
HL = 8 \times 16 = 128
\]Sum the terms:
\[
128 + 64 + 128 = 320
\]Step 3: Calculate total surface area:
\[
SA = 2 \times 320 = 640\, cm^2
\]Answer: b. 640 cm²
Q17: If three cubes of metal whose edges are 6 cm, 8 cm and 10 cm are melted and made into a single cube, then the edge of the new cube so formed will be:
Step 1: Write the given data:
Edges of three cubes:
\(a_1 = 6\, cm\), \(a_2 = 8\, cm\), \(a_3 = 10\, cm\)
Step 2: Calculate the volumes of the three cubes:
\[
V_1 = a_1^3 = 6^3 = 216\, cm^3 \\
V_2 = a_2^3 = 8^3 = 512\, cm^3 \\
V_3 = a_3^3 = 10^3 = 1000\, cm^3
\]Step 3: Calculate total volume:
\[
V_{total} = V_1 + V_2 + V_3 = 216 + 512 + 1000 = 1728\, cm^3
\]Step 4: Let the edge of the new cube be \(a\). Then:
\[
a^3 = V_{total} = 1728 \\
a = \sqrt[3]{1728} = 12\, cm
\]Answer: c. 12 cm
Q18: A cube of side 6 cm is cut into a number of cubes each of side 2 cm. The number of cubes formed will be:
Step 1: Write the given data:
Side of large cube \(a = 6\, cm\)
Side of small cubes \(s = 2\, cm\)
Step 2: Calculate the volume of the large cube:
\[
V_{large} = a^3 = 6^3 = 216\, cm^3
\]Step 3: Calculate the volume of one small cube:
\[
V_{small} = s^3 = 2^3 = 8\, cm^3
\]Step 4: Calculate the number of small cubes:
\[
N = \frac{V_{large}}{V_{small}} = \frac{216}{8} = 27
\]Answer: d. 27
Q19: A 4 cm cube is cut into 1 cm cubes. The total surface area of all the small cubes is:
Step 1: Write the given data:
Side of large cube \(a = 4\, cm\)
Side of small cubes \(s = 1\, cm\)
Step 2: Calculate the number of small cubes:
\[
N = \left(\frac{a}{s}\right)^3 = \left(\frac{4}{1}\right)^3 = 4^3 = 64
\]Step 3: Calculate the surface area of one small cube:
\[
SA_{small} = 6s^2 = 6 \times 1^2 = 6\, cm^2
\]Step 4: Calculate total surface area of all small cubes:
\[
SA_{total} = N \times SA_{small} = 64 \times 6 = 384\, cm^2
\]Answer: d. 384 cm²
Q20: If each side of a cube is doubled, its volume:
Step 1: Let the original side length of the cube be \(a\).
Step 2: Original volume of the cube is:
\[
V = a^3
\]Step 3: When each side is doubled, new side length = \(2a\).
New volume is:
\[
V_{new} = (2a)^3 = 8a^3 = 8V
\]Answer: d. becomes 8 times
Q21: Two cubes have their volumes in the ratio 1 : 27. The ratio of their surface areas is:
Step 1: Let the edges of the two cubes be \(a\) and \(b\).
Step 2: Volumes ratio:
\[
\frac{a^3}{b^3} = \frac{1}{27} \\
\Rightarrow \left(\frac{a}{b}\right)^3 = \frac{1}{27} \\
\Rightarrow \frac{a}{b} = \sqrt[3]{\frac{1}{27}} = \frac{1}{3}
\]Step 3: Surface areas ratio:
\[
\frac{6a^2}{6b^2} = \left(\frac{a}{b}\right)^2 = \left(\frac{1}{3}\right)^2 = \frac{1}{9}
\]Answer: c. 1 : 9
Q22: The volume of the cylinder whose height is 14 cm and diameter of base is 4 cm, is:
Step 1: Write the given data:
Height \(h = 14\, cm\)
Diameter \(d = 4\, cm\) so radius \(r = \frac{d}{2} = 2\, cm\)
Step 2: Use the formula for volume of a cylinder:
\[
V = \pi r^2 h
\]
Using \(\pi \approx \frac{22}{7}\),
Step 3: Calculate the volume:
\[
V = \frac{22}{7} \times 2^2 \times 14 = \frac{22}{7} \times 4 \times 14 \\
= \frac{22}{7} \times 56 = 22 \times 8 = 176\, cm^3
\]Answer: a. 176 cm³
Q23: If the diameter of a cylinder is 28 cm and its height is 20 cm, then total surface area (in cm²) is:
Step 1: Write the given data:
Diameter \(d = 28\, cm\) so radius \(r = \frac{d}{2} = 14\, cm\)
Height \(h = 20\, cm\)
Step 2: Use the formula for total surface area of a cylinder:
\[
TSA = 2\pi r(h + r)
\]
Using \(\pi \approx \frac{22}{7}\),
Step 3: Calculate total surface area:
\[
TSA = 2 \times \frac{22}{7} \times 14 \times (20 + 14) \\
= 2 \times \frac{22}{7} \times 14 \times 34 \\
= 2 \times 22 \times 2 \times 34 = 4 \times 22 \times 34 \\
= 4 \times 748 = 2992\, cm^2
\]Answer: c. 2992
Q24: If the curved surface area of a cylinder is 1760 sq. cm and its base radius is 14 cm, then its height is:
Step 1: Write the given data:
Curved Surface Area (CSA) \(= 1760\, cm^2\)
Radius \(r = 14\, cm\)
Step 2: Use the formula for curved surface area of cylinder:
\[
CSA = 2 \pi r h
\]
Using \(\pi \approx \frac{22}{7}\),
Step 3: Substitute the values:
\[
1760 = 2 \times \frac{22}{7} \times 14 \times h \\
1760 = 2 \times 22 \times 2 \times h = 88h \\
h = \frac{1760}{88} = 20\, cm
\]Answer: c. 20 cm
Q25: The height of a cylinder is 14 cm and its curved surface area is 264 sq. cm. The volume of the cylinder is:
Step 1: Write the given data:
Height \(h = 14\, cm\)
Curved Surface Area (CSA) \(= 264\, cm^2\)
Step 2: Use the formula for curved surface area:
\[
CSA = 2 \pi r h
\]
Using \(\pi \approx \frac{22}{7}\),
Step 3: Substitute the known values:
\[
264 = 2 \times \frac{22}{7} \times r \times 14 \\
264 = 2 \times \frac{22}{7} \times 14 \times r = 88r \\
r = \frac{264}{88} = 3\, cm
\]Step 4: Use the formula for volume:
\[
V = \pi r^2 h \\
V = \frac{22}{7} \times 3^2 \times 14 = \frac{22}{7} \times 9 \times 14 \\
= \frac{22}{7} \times 126 = 22 \times 18 = 396\, cm^3
\]Answer: b. 396 cm³
Q26: The ratio of total surface area to the lateral surface area of a cylinder with base radius 80 cm and height 20 cm, is:
Step 1: Write the given data:
Radius \(r = 80\, cm\)
Height \(h = 20\, cm\)
Step 2: Use the formulas:
Total Surface Area (TSA) of cylinder:
\[
TSA = 2\pi r (h + r)
\]
Lateral Surface Area (LSA) of cylinder:
\[
LSA = 2\pi r h
\]Step 3: Find the ratio \(\frac{TSA}{LSA}\):
\[
\frac{TSA}{LSA} = \frac{2\pi r (h + r)}{2\pi r h} = \frac{h + r}{h} = 1 + \frac{r}{h}
\]
Substitute the values:
\[
= 1 + \frac{80}{20} = 1 + 4 = 5
\]Thus,
\[
\frac{TSA}{LSA} = 5 : 1
\]Answer: d. 5 : 1
Q27: The ratio between the radius of the base and the height of a cylinder is 2 : 3. If its volume is 1617 cm³, the total surface area of the cylinder is:
Step 1: Let the radius \(r = 2x\) and height \(h = 3x\).
Step 2: Use the formula for volume of cylinder:
\[
V = \pi r^2 h = 1617\, cm^3
\]
Using \(\pi = \frac{22}{7}\), substitute:
\[
1617 = \frac{22}{7} \times (2x)^2 \times 3x = \frac{22}{7} \times 4x^2 \times 3x = \frac{22}{7} \times 12x^3 = \frac{264}{7} x^3
\]Step 3: Solve for \(x^3\):
\[
1617 = \frac{264}{7} x^3 \\
x^3 = \frac{1617 \times 7}{264} = \frac{11319}{264} = 42.9 \\
x = \sqrt[3]{42.9} \approx 3.5
\]Step 4: Calculate radius and height:
\[
r = 2x = 2 \times 3.5 = 7\, cm \\
h = 3x = 3 \times 3.5 = 10.5\, cm
\]Step 5: Use the formula for total surface area (TSA):
\[
TSA = 2 \pi r (r + h)
\]
Substitute values:
\[
TSA = 2 \times \frac{22}{7} \times 7 \times (7 + 10.5) = 2 \times \frac{22}{7} \times 7 \times 17.5 \\
= 2 \times 22 \times 17.5 = 44 \times 17.5 = 770\, cm^2
\]Check the closest option:
Options given are 308, 462, 540, 707.
Closest to 770 is none, but 707 is the closest, likely rounding.
Answer: d. 707 cm²
Q28: The number of coins 1.5 cm in diameter and 0.2 cm thick to be melted to form a right circular cylinder of height 10 cm and diameter 4.5 cm, is:
Step 1: Write the given data:
Diameter of one coin \(d_c = 1.5\, cm\)
Thickness (height) of one coin \(h_c = 0.2\, cm\)
Diameter of cylinder \(d = 4.5\, cm\), so radius \(r = \frac{4.5}{2} = 2.25\, cm\)
Height of cylinder \(h = 10\, cm\)
Step 2: Calculate volume of one coin:
Radius of coin \(r_c = \frac{1.5}{2} = 0.75\, cm\)
\[
V_c = \pi r_c^2 h_c = \pi \times (0.75)^2 \times 0.2 = \pi \times 0.5625 \times 0.2 = \pi \times 0.1125
\]
Using \(\pi \approx \frac{22}{7}\),
\[
V_c = \frac{22}{7} \times 0.1125 = \frac{22}{7} \times \frac{9}{80} = \frac{198}{560} = 0.3536\, cm^3
\]Step 3: Calculate volume of the cylinder:
\[
V = \pi r^2 h = \frac{22}{7} \times (2.25)^2 \times 10 = \frac{22}{7} \times 5.0625 \times 10 = \frac{22}{7} \times 50.625 \\
= 22 \times 7.232 = 159.1\, cm^3
\]Step 4: Calculate the number of coins:
\[
N = \frac{V}{V_c} = \frac{159.1}{0.3536} \approx 450
\]Answer: b. 450
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