Question 1

A rectangular box measures internally 1.6 m in length, 1 m in breadth and 60 cm in depth. How many solid cubical boxes, each with edge 20 cm, can be packed completely inside the box?

Accepted Answer

Introduction / Context: Packing problems ask how many smaller solids fit inside a larger solid without overlapping. Here, a large rectangular box must be filled with smaller cubes. Because the edges align neatly, the number of cubes along each dimension is given by division of the corresponding lengths, and the total count is the product of these numbers. Given Data / Assumptions: Internal dimensions of big box: length = 1.6 m, breadth = 1 m, depth = 60 cm. Edge of each small cube = 20 cm. Cubes are packed without gaps, with edges parallel to box edges. Concept / Approach: We first convert all dimensions to a common unit, preferably centimetres, because the cube edge is given in centimetres. Then we find the number of cubes along each dimension by integer division (floor), and finally multiply these three counts to get the total number of cubes that fit inside. Step-by-Step Solution: Step 1: Convert dimensions to centimetres. Length = 1.6 m = 160 cm. Breadth = 1 m = 100 cm. Depth = 60 cm (already in centimetres). Edge of each cube = 20 cm. Step 2: Number of cubes along length = 160 / 20 = 8. Step 3: Number of cubes along breadth = 100 / 20 = 5. Step 4: Number of cubes along depth = 60 / 20 = 3. Step 5: Total cubes = 8 * 5 * 3 = 120. Verification / Alternative check: The volume method also confirms this result. Volume of big box = 160 * 100 * 60 = 960000 cm^3. Volume of each cube = 20 * 20 * 20 = 8000 cm^3. Dividing gives 960000 / 8000 = 120 cubes, which agrees with the count from dimensions. Why Other Options Are Wrong: 30 cubes and 60 cubes are too few and would fill only a fraction of the volume. 90 cubes fill three quarters of the full 120 cube capacity and would leave significant empty space. 150 cubes would require more volume than the box provides, so they cannot all fit without overlapping. Common Pitfalls: Some learners forget to convert all dimensions into the same units and may incorrectly divide metres by centimetres. Others may mistakenly add or average the dimensions instead of taking products. For packing problems with exact cube fitting, ensure that the box dimensions are multiples of the cube edge and always work in a consistent unit system. Final Answer: The maximum number of cubical boxes that can be packed is 120 cubes.

Question 2

The height of a solid rectangular wall is six times its width and the length of the wall is seven times its height. If the volume of the wall is 16128 cubic metres, find the width of the wall.

Accepted Answer

Introduction / Context: This question uses algebra with geometric dimensions. The wall is modelled as a rectangular solid (a cuboid) where the dimensions are linked by simple ratios. By expressing all dimensions in terms of a single variable, we can use the given volume to solve for that variable and hence determine the width of the wall. Given Data / Assumptions: Let width of wall be w metres. Height of wall h = 6w. Length of wall L = 7h = 7 * 6w = 42w. Volume of wall V = 16128 m^3. Volume formula for a cuboid: V = L * w * h. Concept / Approach: We substitute the expressions for L and h in terms of w into the volume formula. This gives us an equation with a single unknown w. Solving this equation yields w, and then we can check that the resulting height and length make sense with the given volume. Step-by-Step Solution: Step 1: Write volume in terms of w. V = L * w * h = (42w) * w * (6w) = 42 * 6 * w^3. Step 2: Compute the constant factor 42 * 6 = 252. So V = 252 * w^3. Step 3: Use the given volume. 252 * w^3 = 16128. Step 4: Solve for w^3. w^3 = 16128 / 252. Step 5: Perform the division. 252 * 64 = 16128, so w^3 = 64. Step 6: Take cube root: w = cube root of 64 = 4 metres. Verification / Alternative check: If w = 4 m, then h = 6 * 4 = 24 m and L = 7 * 24 = 168 m. The volume then is L * w * h = 168 * 4 * 24. Multiplying, 168 * 4 = 672, and 672 * 24 = 16128 m^3, which matches the given volume exactly. This confirms that the width of 4 m is correct. Why Other Options Are Wrong: If w = 5 m, then w^3 = 125 m^3, and V would be 252 * 125 = 31500 m^3, not 16128 m^3. For w = 6 m, V = 252 * 216 = 54432 m^3, which is much larger than given. For w = 7 m or 8 m, the computed volumes are even larger and do not match 16128 m^3. Common Pitfalls: A frequent error is misreading the relationships between dimensions, such as confusing length being 7 times the width rather than 7 times the height. Another pitfall is computational accuracy when dividing 16128 by 252 or when finding the cube root. It is helpful to factor numbers or to check small cube values like 4^3 = 64 to quickly identify the correct width. Final Answer: The width of the wall is 4 m.

Question 3

In measurement and geometry, which of the following units can correctly be used to represent volume?

Accepted Answer

Introduction / Context: Volume is a fundamental physical quantity that describes how much three dimensional space an object or substance occupies. Different measurement systems use different units for volume depending on the context. In school mathematics, science and aptitude tests, learners often encounter millilitres, cubic centimetres and cubic metres, and it is important to recognize that all of these can represent volume when used correctly. Given Data / Assumptions: Option (a): Millilitre (mL) is commonly used for liquids such as water, milk or medicine. Option (b): Cubic centimetre (cm^3) appears in geometry, physics and chemistry to measure smaller solid or liquid volumes. Option (c): Cubic metre (m^3) is used for larger volumes such as room space, water tanks or construction work. We must identify which of these units are valid volume units. Concept / Approach: Volume in the International System of Units (SI) is based on length. The derived SI unit of volume is the cubic metre (m^3), which is defined as the volume of a cube with side 1 metre. Smaller units are derived by scaling down the basic unit or by using related practical units for liquids: 1 m^3 = 10^6 cm^3. 1 litre = 1000 cm^3. 1 millilitre = 1 cm^3. Thus, millilitres, cubic centimetres and cubic metres all describe volume; they simply operate at different scales. Step-by-Step Solution: Step 1: Consider cubic metre (m^3). It measures large volumes, for example the interior of a room or a tank. It is a standard SI unit of volume. Step 2: Consider cubic centimetre (cm^3). It measures smaller volumes and is suitable for laboratory measurements or small solid objects. It is directly derived from the SI base unit of length, the centimetre. Step 3: Consider millilitre (mL). It is a commonly used unit for liquid volume. By definition, 1 mL equals 1 cm^3, so it is also a volume unit even though it is often used in daily life rather than pure geometry. Step 4: Because each of the three units is a way of measuring volume, the correct choice is to select all of them. Verification / Alternative check: We can verify the relationships numerically. If we pour 1 mL of water into a small rectangular container with internal dimensions 1 cm by 1 cm by 1 cm, it fills the container exactly. That shows that 1 mL equals 1 cm^3. If we pour 1000 litres (1,000,000 mL) of water into a cubic tank of side 1 m, it fills the 1 m^3 tank entirely. These real world interpretations confirm that millilitres, cubic centimetres and cubic metres are all legitimate volume units. Why Other Options Are Wrong: Choosing only millilitre or only cubic centimetre or only cubic metre ignores the fact that volume can be expressed in many consistent units connected by conversion factors. The option that says none of these is clearly incorrect, because all three are well known and widely used volume units in science and everyday life. Common Pitfalls: Learners sometimes confuse area and volume units, for example mixing up cm^2 with cm^3 or treating millilitres as if they were units of mass. Another common confusion is between litre and millilitre or between cubic metre and square metre. Remember that volume units always involve three dimensions and therefore carry a power of 3, and that liquid measurement units like litre and millilitre have exact relationships with cubic centimetres and cubic metres. Final Answer: All three listed units represent volume, so the correct choice is All of the above.

Question 4

The radius of the base and the height of a right circular cone are both increased by 10 percent. By what percentage does the volume of the cone increase?

Accepted Answer

Introduction / Context: This problem illustrates how percentage changes in dimensions affect the volume of a three dimensional solid. Because volume often involves a length dimension raised to a power, a small percentage change in each dimension can produce a larger percentage change in volume. The cone is a good example, because its volume depends on the square of the radius and directly on the height. Given Data / Assumptions: Original radius of base = r. Original height = h. Both radius and height are increased by 10 percent. Volume formula for a cone: V = (1 / 3) * π * r^2 * h. We are asked for the percentage increase in volume. Concept / Approach: If a quantity x is increased by 10 percent, it is multiplied by 1.10. Here both radius and height are multiplied by 1.10. Because volume involves r^2 and h, the new volume factor relative to old volume is: New volume factor = (new r / old r)^2 * (new h / old h). So we have: New volume factor = (1.10)^2 * 1.10 = (1.10)^3. Once we compute this factor, we subtract 1 and convert to a percentage to obtain the percentage increase in volume. Step-by-Step Solution: Step 1: Express new dimensions. New radius r1 = 1.10 * r. New height h1 = 1.10 * h. Step 2: Write original and new volumes. Original volume V = (1 / 3) * π * r^2 * h. New volume V1 = (1 / 3) * π * r1^2 * h1. Step 3: Substitute r1 and h1. V1 = (1 / 3) * π * (1.10 * r)^2 * (1.10 * h). V1 = (1 / 3) * π * (1.21 * r^2) * (1.10 * h) = (1.21 * 1.10) * (1 / 3) * π * r^2 * h. Step 4: Compute the factor 1.21 * 1.10 = 1.331. So V1 = 1.331 * V. Step 5: Percentage increase = (1.331 - 1) * 100% = 0.331 * 100% = 33.1%. Verification / Alternative check: We can test with concrete values. Let r = 10 units and h = 10 units. Then V = (1 / 3) * π * 10^2 * 10 = (1 / 3) * π * 1000. New radius = 11, new height = 11. New volume V1 = (1 / 3) * π * 11^2 * 11 = (1 / 3) * π * 1331. The ratio V1 / V = 1331 / 1000 = 1.331, which again indicates a 33.1 percent increase. Why Other Options Are Wrong: 33.5% and 32.1% are close but do not match the exact factor of 1.331 and represent rounding or calculation errors. 53.1% is much too high and would correspond to a much larger change in one of the dimensions. 21% suggests adding the two 10 percent increases instead of considering the cubic effect on volume. Common Pitfalls: A common mistake is to add the percentage increases directly (10% for radius plus 10% for height) and conclude a 20 percent increase in volume, which ignores the squared dependence on radius. Another error is to compute only (1.10)^2 and forget that height also changes. Recognizing that volume scales as r^2 * h and applying the percentage factor to each dimension separately is essential to obtain the right answer. Final Answer: The volume of the cone increases by 33.1%.

Question 5

A single circular pipe of diameter x centimetres is to be replaced by six smaller circular pipes, each of diameter 12 cm, to convey a liquid in a laboratory. If the flow speed of the liquid is kept the same in all pipes, what is the value of x?

Accepted Answer

Introduction / Context: This question explores how multiple smaller pipes can replace a single larger pipe while maintaining the same flow rate. When the liquid speed is the same in each pipe, the volume flow rate is proportional to the total cross sectional area of the pipes. Therefore, the area of the single large pipe must equal the combined area of the six smaller pipes. Given Data / Assumptions: Single pipe has diameter x cm and radius x / 2 cm. Each of the six smaller pipes has diameter 12 cm and radius 6 cm. Speed of the liquid is the same in all pipes. Flow rate is directly proportional to cross sectional area. Concept / Approach: The cross sectional area A of a circular pipe is: A = π * r^2. For equivalent flow, the area of the single large pipe must equal the sum of the areas of the six smaller pipes: Area of large pipe = 6 * area of one small pipe. Expressing both sides in terms of π and radii, we can solve for x, the diameter of the large pipe. Step-by-Step Solution: Step 1: Area of one small pipe. Radius of small pipe r_s = 12 / 2 = 6 cm. Area_s = π * 6^2 = 36π cm^2. Step 2: Total area of six small pipes. Total area_small = 6 * 36π = 216π cm^2. Step 3: Area of large pipe. Radius of large pipe r_L = x / 2. Area_L = π * (x / 2)^2 = π * x^2 / 4. Step 4: Equate total areas. π * x^2 / 4 = 216π. Step 5: Cancel π from both sides and solve for x^2. x^2 / 4 = 216 ⇒ x^2 = 864. Step 6: Take square root. x = sqrt(864) = sqrt(144 * 6) = 12 * sqrt(6). Using sqrt(6) ≈ 2.449, x ≈ 12 * 2.449 ≈ 29.39 cm. Verification / Alternative check: We can verify by substituting x ≈ 29.39 cm back. Radius r_L ≈ 14.695 cm. Area_L ≈ π * 14.695^2 ≈ π * 216, which matches the total area of six small pipes, 216π cm^2. Therefore the diameter of about 29.39 cm is consistent with the requirement for equal flow. Why Other Options Are Wrong: 14.69 cm is approximately the radius, not the diameter, and would give an area only about one quarter of the required total area. 18.65 cm and 22.21 cm are too small and would yield total flow capacity less than that of six smaller pipes combined. 25.00 cm is closer but still short; its area is π * (12.5)^2, which is significantly less than 216π cm^2. Common Pitfalls: A common mistake is to directly equate diameters or to sum diameters instead of equating areas. Another pitfall is forgetting that area depends on the square of the radius, so combining six pipes does not mean just multiplying the diameter by six. Careful use of the area formula and attention to radius versus diameter are essential here. Final Answer: The required diameter of the single equivalent pipe is approximately 29.39 cm.

Question 6

For a cube, find the ratio of the surface area of the sphere inscribed in the cube to the surface area of the sphere circumscribed about the cube.

Accepted Answer

Introduction / Context: This problem compares two spheres related to the same cube. One sphere is inscribed in the cube, touching each face, and the other is circumscribed about the cube, passing through all eight vertices. The question asks for the ratio of their surface areas. Understanding the relationships between these radii and the cube side is a useful exercise in three dimensional geometry. Given Data / Assumptions: Consider a cube with side length a. Inscribed sphere touches all six faces of the cube. Circumscribed sphere passes through all eight vertices of the cube. Surface area of a sphere with radius r is 4 * π * r^2. Concept / Approach: For the inscribed sphere, the diameter equals the side length of the cube because it just fits between opposite faces. For the circumscribed sphere, the diameter equals the space diagonal of the cube. By expressing both radii in terms of a, we can compute the ratio of their surface areas. Constants like 4 and π will cancel in the ratio, leaving a simple relationship. Step-by-Step Solution: Step 1: Radius of inscribed sphere. Diameter = side of cube = a. So radius r_in = a / 2. Step 2: Radius of circumscribed sphere. Diameter = space diagonal of cube = a * sqrt(3). So radius r_out = (a * sqrt(3)) / 2. Step 3: Surface area formulas. Surface area of inscribed sphere S_in = 4 * π * (a / 2)^2 = 4 * π * a^2 / 4 = π * a^2. Surface area of circumscribed sphere S_out = 4 * π * (a * sqrt(3) / 2)^2. Compute radius squared: (a * sqrt(3) / 2)^2 = 3 * a^2 / 4. So S_out = 4 * π * (3 * a^2 / 4) = 3 * π * a^2. Step 4: Ratio of surface areas. S_in : S_out = π * a^2 : 3 * π * a^2 = 1 : 3. Verification / Alternative check: We can check numerically by taking a convenient side length, say a = 2 units. Then r_in = 1, r_out = sqrt(3). Surface areas become S_in = 4 * π * 1^2 = 4π, and S_out = 4 * π * (sqrt(3))^2 = 4 * π * 3 = 12π. The ratio is 4π : 12π = 1 : 3, confirming our algebraic result. Why Other Options Are Wrong: 1 : 2, 1 : 4 and 1 : 5 correspond to incorrect relationships between the radii and ignore the exact geometric connection through the space diagonal. 2 : 3 suggests the circumscribed sphere is only 1.5 times the surface area of the inscribed sphere, which contradicts the explicit calculation showing a factor of 3. Common Pitfalls: Students sometimes confuse face diagonal with space diagonal, or misapply Pythagoras theorem. The face diagonal is a * sqrt(2), whereas the space diagonal is a * sqrt(3). Another frequent error is to compare radii directly instead of areas; radius ratio r_in : r_out is 1 : sqrt(3), but for areas the ratio involves the square, leading to 1 : 3. Remember that surface area depends on the square of the radius, not the radius itself. Final Answer: The required ratio of surface areas is 1 : 3.

Question 7

The volume of a rectangular solid (rectangular prism) is to be increased by 50 percent without changing the area of its base. By what percentage must the height of the solid be increased?

Accepted Answer

Introduction / Context: This question tests understanding of how the dimensions of a prism relate to its volume. A rectangular solid has a base area and a height, and the volume is the product of these two quantities. If the base area is kept constant, any change in volume must come entirely from changing the height. The problem asks how much the height must increase to increase volume by 50 percent under this condition. Given Data / Assumptions: Let base area be B (constant). Let original height be h. Original volume V = B * h. New volume is 50 percent greater than original, so V_new = 1.5 * V. Base area does not change. Concept / Approach: Since V = B * h and B stays the same, any percentage change in volume must be equal to the percentage change in height. If V_new = B * h_new, then: B * h_new = 1.5 * B * h. The base area B cancels out, leaving a simple proportion between h_new and h. From this, we can directly deduce the required percentage increase in height. Step-by-Step Solution: Step 1: Express original and new volumes. V = B * h. V_new = 1.5 * V = 1.5 * B * h. Step 2: Write new volume in terms of new height. V_new = B * h_new. Step 3: Equate the two expressions for V_new. B * h_new = 1.5 * B * h. Step 4: Cancel B on both sides. h_new = 1.5 * h. Step 5: The new height is 1.5 times the original height. Step 6: Percentage increase in height = (1.5 - 1) * 100% = 0.5 * 100% = 50%. Verification / Alternative check: Take a simple example: base area B = 10 cm^2 and original height h = 10 cm. Then original volume V = 10 * 10 = 100 cm^3. A 50 percent increase in volume means V_new = 150 cm^3. If we keep base area at 10 cm^2, the new height must be h_new = 150 / 10 = 15 cm. This is 5 cm more than the original height, which is a 50 percent increase (5 is half of 10). Why Other Options Are Wrong: 40%, 30% and 20% increases in height would produce smaller percentage increases in volume because the base area is constant. For example, a 20 percent increase in height would yield only a 20 percent increase in volume. 10% is clearly too small; it would yield a volume of 110 percent of the original, not 150 percent. Common Pitfalls: Some learners mistakenly think that the volume increase will be more complicated because volume is three dimensional. However, when the base area is fixed, volume is directly proportional to height. Another common misunderstanding is to think that part of the volume change might be absorbed by the base, but the problem explicitly states that the base is unchanged. Focusing on the direct proportionality between volume and height with constant base area avoids these errors. Final Answer: The height of the solid must be increased by 50%.

Question 8

A store owner packs identical rectangular books into a larger carton that measures 25 inches by 42 inches by 60 inches internally. Each book measures 7 inches by 6 inches by 5 inches. What is the maximum number of books that can be placed in the car…

Accepted Answer

Introduction / Context: Packing optimization problems are common in aptitude tests and practical logistics. Here, we have a large rectangular carton and smaller rectangular books. Because the books fit exactly in integer counts along each dimension, the maximum number of books can be found by seeing how many fit along the length, breadth and height, and then multiplying these counts. Given Data / Assumptions: Carton dimensions: 25 in by 42 in by 60 in. Book dimensions: 7 in by 6 in by 5 in. Books are identical and rigid. Books are oriented with edges parallel to the carton edges, but we can choose the orientation that gives the maximum count. Books cannot overlap or be bent. Concept / Approach: We must consider all possible orientations of the book in the carton, since placing different edges along the carton dimensions can change how many books fit. For each orientation, we compute: Number along length = floor(L_carton / L_book). Number along breadth = floor(B_carton / B_book). Number along height = floor(H_carton / H_book). We then multiply these three counts to get the total books for that orientation and choose the maximum among all orientations. Step-by-Step Solution: Let carton dimensions be 25, 42, 60. Book dimensions are 7, 6, 5. Consider several orientations: Orientation 1: 7 along 25, 6 along 42, 5 along 60. Counts: floor(25 / 7) = 3, floor(42 / 6) = 7, floor(60 / 5) = 12. Total = 3 * 7 * 12 = 252. Orientation 2: 5 along 25, 7 along 42, 6 along 60. Counts: floor(25 / 5) = 5, floor(42 / 7) = 6, floor(60 / 6) = 10. Total = 5 * 6 * 10 = 300. Other orientations give totals such as 240, 256, 280 or 288, all less than 300. Therefore, the maximum possible number of books is 300. Verification / Alternative check: Consider Orientation 2 in more detail. Along the 25 in side, 5 books of thickness 5 in fit exactly. Along the 42 in side, 6 books of width 7 in fit exactly. Along the 60 in side, 10 books of height 6 in fit exactly. No further books can be added in any direction. This packing is efficient and uses the space well, giving confidence that 300 is truly the maximum. Why Other Options Are Wrong: 175, 225 and 265 books are all less than the achievable 300 and correspond to suboptimal orientations or incomplete use of space. 345 books would require more volume than the carton has and would not fit given the integer constraints on counts along each dimension. Common Pitfalls: A common mistake is to calculate only one orientation and assume it is optimal without checking others. Another error is to divide by dimensions incorrectly or to forget to take the floor (integer part) of each division, accidentally assuming fractional books are possible. Systematically checking different orientations and always using whole number counts ensures an accurate answer. Final Answer: The maximum number of books that can be packed into the carton is 300 books.

Question 9

How many small squares with sides 1/2 inch long are needed to completely cover a square region that measures 4 ft by 4 ft?

Accepted Answer

Introduction / Context: This question connects unit conversion with area concepts. We have a large square region measured in feet, and we want to know how many smaller squares measured in inches are needed to cover it without overlap or gaps. This is a classic tiling problem that reinforces the idea that the number of tiles equals the ratio of total area to the area of one tile when there is perfect coverage. Given Data / Assumptions: Large square side length = 4 ft. Small square side length = 1/2 inch. Squares cover the region exactly with no overlap and no uncovered space. Concept / Approach: We first convert the large square side from feet to inches because the tile side is given in inches. Then we can either compute the areas and divide, or more simply count how many small squares fit along each side and then square that count, since the region is also a square. The key conversion is: 1 ft = 12 inches. Step-by-Step Solution: Step 1: Convert the side of the large square. Side length = 4 ft = 4 * 12 = 48 inches. Step 2: Side of each small square = 1/2 inch. Step 3: Number of small squares along one side of the large square. Count along a side = 48 / (1/2) = 48 * 2 = 96. Step 4: Since the region is square, there are also 96 small squares along the perpendicular side. Total small squares = 96 * 96 = 9216. Verification / Alternative check: Using areas: area of large square = 4 ft * 4 ft = 16 ft^2. In square inches, 1 ft^2 = 144 in^2, so area = 16 * 144 = 2304 in^2. Area of one small square = (1/2 in)^2 = 1/4 in^2. The number of tiles is 2304 / (1/4) = 2304 * 4 = 9216. This matches the earlier computation, confirming the result. Why Other Options Are Wrong: 10246, 12345 and 7527 are arbitrary large numbers that do not correspond to any natural square of an integer number of tiles along each side. 4096 equals 64^2 and would correspond to only 64 tiles along each side, which would happen if the side were 32 inches instead of 48 inches, so it is inconsistent with the given dimensions. Common Pitfalls: Students sometimes convert 4 ft incorrectly or forget that 1/2 inch tiles give twice as many along each side compared to 1 inch tiles. Others mistakenly divide area in feet squared by area in inches squared without proper conversion, leading to incorrect results. Working consistently in one unit system and deciding whether to use side lengths or areas at the start helps avoid confusion. Final Answer: A total of 9216 squares with side 1/2 inch are needed to cover the 4 ft by 4 ft square region.

Question 10

A volume of 88 cubic centimetres of silver is drawn into a uniform wire 1 mm in diameter. What will be the length of the wire in metres?

Accepted Answer

Introduction / Context: This problem converts a given volume of metal into the length of a cylindrical wire of specified diameter. Such questions are common in physics and aptitude tests and rely on the volume formula for a cylinder. Because the metal is simply reshaped, its volume is conserved, so the volume of the wire equals the original volume of silver. Given Data / Assumptions: Volume of silver V = 88 cm^3. Wire is uniform and cylindrical. Diameter of wire = 1 mm. We want wire length in metres. No loss of material is assumed when drawing the wire. Concept / Approach: The volume of a cylinder is given by: V = π * r^2 * h, where r is radius and h is height or, in this context, wire length. We must convert the diameter from millimetres to centimetres so that all length units in the formula are consistent. Once we have r in centimetres and V in cubic centimetres, we can solve for the length h in centimetres and then convert to metres. Step-by-Step Solution: Step 1: Convert diameter to radius in centimetres. Diameter = 1 mm. Since 10 mm = 1 cm, 1 mm = 0.1 cm. Radius r = 0.1 / 2 = 0.05 cm. Step 2: Use cylinder volume formula V = π * r^2 * h. 88 = π * (0.05)^2 * h. Step 3: Compute r^2 = (0.05)^2 = 0.0025. So 88 = π * 0.0025 * h. Step 4: Solve for h. h = 88 / (π * 0.0025). Take π ≈ 22 / 7, so denominator ≈ (22 / 7) * 0.0025. It is easier to rewrite in fraction form: 0.0025 = 25 / 10000 = 1 / 4000. So π * 0.0025 ≈ (22 / 7) * (1 / 4000) = 22 / 28000. Then h ≈ 88 / (22 / 28000) = 88 * (28000 / 22) = 4 * 28000 = 112000 cm. Step 5: Convert centimetres to metres. 1 m = 100 cm, so h = 112000 / 100 = 1120 m? But note we must recheck: a simpler exact calculation gives h = 35200 / π cm, and with π ≈ 22 / 7, this becomes 35200 * 7 / 22 = 11200 cm. So the correct h = 11200 cm = 112 m. Verification / Alternative check: Using the simplified exact form directly: V = π * r^2 * h with r = 0.05 cm and V = 88 cm^3 gives h = 88 / (π * 0.0025) = 35200 / π cm. With π ≈ 3.14, h ≈ 35200 / 3.14 ≈ 11210 cm, very close to 11200 cm, confirming that 112 m is accurate when standard approximations are used. Why Other Options Are Wrong: 84 m, 96 m, 108 m and 128 m correspond to different assumed lengths that, if substituted back into the volume formula, would not yield 88 cm^3 of silver. They therefore do not conserve the original volume and cannot be correct. Common Pitfalls: The main sources of error are incorrect unit conversions between millimetres, centimetres and metres, and miscalculations involving small decimal squares like 0.05^2. Some learners also forget that the radius is half the diameter. Working carefully with fractions instead of decimals often reduces arithmetic mistakes in such questions. Final Answer: The length of the wire is 112 metres.

Question 11

A rectangular block has dimensions 5 cm by 6 cm by 7 cm and is gently lowered into a cylindrical vessel of radius 6 cm containing a liquid. As a result, the level of the liquid in the cylinder rises by 4 cm. Assuming no overflow, what is the volume …

Accepted Answer

Introduction / Context: This problem connects volume displacement with a rise in liquid level inside a cylindrical vessel. When a solid block is lowered into the liquid, it displaces some liquid, causing the liquid level to rise. The extra liquid level, combined with the cross sectional area of the cylinder, directly gives the volume of liquid displaced, regardless of the detailed shape of the block. Given Data / Assumptions: Cylindrical vessel radius R = 6 cm. Initial liquid level is not needed explicitly. Rise in liquid level after block is inserted = 4 cm. No overflow of liquid occurs. We are asked for the volume of liquid displaced in cubic centimetres. Concept / Approach: In a right circular cylinder, the volume corresponding to a small change in height is equal to the base area multiplied by that change in height. The base area of a cylinder is: Base area = π * R^2. If the height increases by h, then the additional volume is: Volume displaced = π * R^2 * h. Here, R and h are given, so we can compute the displaced volume directly. Step-by-Step Solution: Step 1: Compute the base area of the cylindrical vessel. Base area = π * R^2 = π * 6^2 = π * 36. Step 2: Use the rise in height of the liquid. Rise in level h = 4 cm. Step 3: Compute the volume displaced. Volume displaced = base area * rise = π * 36 * 4. Volume displaced = 144π cubic centimetres. Verification / Alternative check: If we approximate π as 22 / 7, the volume becomes 144 * 22 / 7 = 3168 / 7 ≈ 452.6 cm^3. This is a reasonable volume for a modest rise of 4 cm in a vessel with radius 6 cm. The presence of the rectangular block in the problem statement simply provides the physical cause of the displacement but does not change the computation based on the cylinder geometry and the observed rise in liquid level. Why Other Options Are Wrong: 72π corresponds to only half the correct rise in liquid level and would match a 2 cm increase instead of 4 cm. 108π corresponds to three quarters of the correct displaced volume and would result from using an incorrect height or radius. 216π and 288π are larger volumes that would require a greater rise in liquid level or a larger radius than given in the problem. Common Pitfalls: A common mistake is to attempt to use the dimensions of the rectangular block directly, rather than focusing on the observed rise in the cylinder. While Archimedes principle states that the volume of displaced liquid equals the volume of the submerged part of the block, the easiest route here is to work with the cylinder geometry. Another pitfall is forgetting to square the radius when computing the base area or miscalculating 6^2 as 12 rather than 36. Final Answer: The volume of liquid displaced in the cylindrical vessel is 144π cubic centimetres.

Question 12

A volume of 66 cubic centimetres of silver is drawn into a uniform wire 1 mm in diameter. What will be the length of the wire in metres?

Accepted Answer

Introduction / Context: This problem is similar to other wire drawing questions, where a known volume of metal is reshaped into a cylindrical wire. The key idea is that the volume of metal remains constant during the reshaping, so the volume of the wire equals the given volume. Knowing the diameter, we can use the cylinder volume formula to find the length of the wire. Given Data / Assumptions: Volume of silver V = 66 cm^3. Wire is cylindrical and uniform. Diameter of wire = 1 mm. We want the wire length in metres. No loss of material occurs during drawing. Concept / Approach: The volume of a cylinder is: V = π * r^2 * h, where r is radius in centimetres and h is length in centimetres when we want volume in cubic centimetres. The diameter is given in millimetres, so we must convert to centimetres before using the formula. After finding h in centimetres, we convert it to metres. Step-by-Step Solution: Step 1: Convert diameter to radius in centimetres. Diameter = 1 mm = 0.1 cm. Radius r = 0.1 / 2 = 0.05 cm. Step 2: Use V = π * r^2 * h with V = 66 cm^3. 66 = π * (0.05)^2 * h. Step 3: Compute r^2 = (0.05)^2 = 0.0025. So 66 = π * 0.0025 * h. Step 4: Solve for h. h = 66 / (π * 0.0025). Rewrite 0.0025 as 1 / 4000, so π * 0.0025 = π / 4000. Then h = 66 / (π / 4000) = 66 * 4000 / π = 264000 / π cm. Step 5: Use π ≈ 22 / 7. h ≈ 264000 * 7 / 22 = (264000 / 22) * 7. 264000 / 22 = 12000, so h ≈ 12000 * 7 = 84000 cm. Step 6: Convert centimetres to metres. h = 84000 cm = 84000 / 100 = 840 m? This intermediate suggests rechecking. A simpler exact approach gives h = 26400 / π cm, and with π ≈ 22 / 7, h = 26400 * 7 / 22 = 8400 cm. Thus correct h = 8400 cm = 84 m. Verification / Alternative check: Using the form h = 26400 / π cm and taking π ≈ 3.14, we get h ≈ 26400 / 3.14 ≈ 8417 cm, which is close to 8400 cm. The small difference arises from rounding π. Therefore, the length of about 84 m is consistent with the given volume and diameter. Why Other Options Are Wrong: 90 m is slightly larger than the correct value and would give a volume larger than 66 cm^3 when substituted back into the formula. 168 m and 336 m are much too large and would correspond to volumes several times the given 66 cm^3. 42 m is half of 84 m and would yield a volume only half of what is required. Common Pitfalls: As with similar problems, unit conversion between millimetres, centimetres and metres is a common source of error. Squaring the radius and handling the small decimal 0.05 requires care. Some learners also forget to convert the final length into metres or misplace decimal points when doing so. Working patiently with fractions and checking results using an approximate value of π helps ensure accuracy. Final Answer: The length of the wire formed is 84 metres.

Question 13

In mensuration, what is the total surface area of a right circular cone with height 14 cm and base radius 7 cm?

Accepted Answer

Introduction / Context: This aptitude question tests the formula for the total surface area of a right circular cone, which often appears in competitive exams under the topic of mensuration and geometry. Learners are expected to know how to combine the curved surface area and the base area of a cone, and how to compute the slant height using the Pythagoras theorem. Understanding this concept helps in solving many real life problems that involve conical shapes, such as ice cream cones, funnels, and lampshades. Given Data / Assumptions: • Height h of the cone = 14 cm• Radius r of the base = 7 cm• The cone is right circular, so the axis is perpendicular to the base• Use the standard formula for total surface area of a cone Concept / Approach: The total surface area of a right circular cone is the sum of its curved surface area and the area of its circular base. The curved surface area is given by π * r * l, where r is the radius and l is the slant height. The base area is π * r^2. The slant height l is found using the Pythagoras theorem in the right triangle formed by the height, radius, and slant height: l^2 = h^2 + r^2. Step-by-Step Solution: Step 1: Compute the slant height l.l^2 = h^2 + r^2 = 14^2 + 7^2 = 196 + 49 = 245.l = sqrt(245) cm.Step 2: Write the formula for total surface area (TSA) of the cone.TSA = π * r * l + π * r^2.Step 3: Substitute r = 7 cm and l = sqrt(245).TSA = π * 7 * sqrt(245) + π * 7^2.Step 4: Simplify numerically using π approximately equal to 3.14.TSA ≈ 3.14 * 7 * 15.65 + 3.14 * 49.Step 5: This gives a total surface area close to 498.35 square centimetres. Verification / Alternative check: As a quick check, the curved surface area is roughly π * 7 * 15.65 which is a bit more than 340 square centimetres, and the base area π * 49 is about 154 square centimetres. Adding these two approximate values gives around 494 to 500 square centimetres, which confirms that an answer close to 498 square centimetres is reasonable. This supports the selection of 498.35 sq. cm as the most accurate option among those given. Why Other Options Are Wrong: Option A: 344.35 sq. cm is too small because it is close only to the curved surface area but ignores the base area contribution.Option B: 462 sq. cm underestimates the total area and does not match the sum of curved and base areas for the given dimensions.Option D: None of these is incorrect because one of the numerical options does match the calculated result quite closely. Common Pitfalls: Candidates sometimes forget to include the base area and use only π * r * l, which gives only the curved surface area. Another frequent mistake is to treat the height as the slant height or to apply the Pythagoras theorem incorrectly. Rounding errors can also cause confusion, so it is important to keep enough decimal places while doing intermediate calculations. Final Answer: The total surface area of the right circular cone is approximately 498.35 sq. cm.

Question 14

How many rectangular meshes are there in 1 square metre of wire gauge if each mesh opening is 8 mm long and 5 mm wide?

Accepted Answer

Introduction / Context: This aptitude problem checks understanding of area, unit conversion, and counting identical small regions inside a larger region. Such questions are common in topics like mensuration and practical mathematics, where students must convert between square metres and square millimetres and then use the idea of repeated patterns, here called meshes, to compute how many fit into a given area. Given Data / Assumptions: • Total area of the wire gauge sheet = 1 square metre.• Each mesh is a small rectangle of dimensions 8 mm by 5 mm.• All meshes are identical and arranged without gaps or overlaps.• We need the number of such meshes that exactly fit into 1 square metre. Concept / Approach: The number of small identical rectangles that can fit into a large rectangle is equal to the total area of the large rectangle divided by the area of one small rectangle, provided everything is neatly arranged. Here, the large area is given in square metres, while the mesh dimensions are in millimetres. Therefore, the key steps are to convert the large area into square millimetres, compute the area of one mesh in square millimetres, and then perform a simple division. Step-by-Step Solution: Step 1: Convert 1 square metre to square millimetres.1 metre = 1000 millimetres, so 1 square metre = 1000 mm * 1000 mm = 1000000 mm^2.Step 2: Compute the area of one mesh.Area of one mesh = length * width = 8 mm * 5 mm = 40 mm^2.Step 3: Compute the number of meshes.Number of meshes = total area / area of one mesh = 1000000 / 40.Step 4: Perform the division.1000000 / 40 = 25000 meshes. Verification / Alternative check: We can check the order of magnitude. If each mesh were roughly 10 mm by 10 mm (100 mm^2), we would expect about 1000000 / 100 = 10000 meshes in one square metre. Our mesh area is 40 mm^2, which is smaller, so more meshes should fit. 25000 is indeed larger than 10000, which is consistent with this rough estimate, confirming that the answer is reasonable. Why Other Options Are Wrong: Option A: 2500 is too small and would correspond to a much larger mesh area than 40 mm^2.Option C: 12500 represents half of the correct count and would result from using an incorrect area or conversion factor.Option D: 15400 is not supported by the exact area calculations and does not match any logical intermediate step. Common Pitfalls: Common mistakes include converting only line units but forgetting that area scales with the square of the conversion factor, for example treating 1 metre as 100 centimetres but then not squaring 100. Another error is to reverse the division and divide the small area by the large area, which gives a number less than one rather than a count of meshes. Careful handling of units and correct order of operations help avoid these errors. Final Answer: The number of meshes in 1 square metre of wire gauge is 25000.

Question 15

In a triangle with side lengths a = 38 units, b = 36 units and c = 18 units, what is the perimeter of the triangle?

Accepted Answer

Introduction / Context: This problem checks a very basic but important concept in geometry, namely the perimeter of a triangle. In aptitude and school level mathematics, learners must be comfortable adding side lengths and interpreting what a perimeter represents. The question provides all three sides explicitly and asks for the sum, which makes this an example of a straightforward recall and application question. Given Data / Assumptions: • Side a of the triangle = 38 units.• Side b of the triangle = 36 units.• Side c of the triangle = 18 units.• The triangle is valid and follows the triangle inequality, so a perimeter can be meaningfully computed. Concept / Approach: The perimeter of a triangle is defined as the total distance around the triangle. Mathematically, if the three side lengths are a, b, and c, then perimeter P is given by P = a + b + c. There are no square roots or advanced formulas required here; the problem mainly tests accuracy in simple addition and reading the values correctly from the question. Step-by-Step Solution: Step 1: Write the formula for the perimeter of a triangle.P = a + b + c.Step 2: Substitute the given side lengths.P = 38 + 36 + 18.Step 3: Add the first two numbers.38 + 36 = 74.Step 4: Add the remaining side length.74 + 18 = 92.Step 5: Conclude that the perimeter is 92 units. Verification / Alternative check: We can verify the arithmetic in a different grouping: add 36 and 18 first, which gives 54, and then add 38. So P = 38 + 54 = 92. Since both methods give the same total, the addition is consistent. Additionally, the sides satisfy the triangle inequality because each side is less than the sum of the other two, which reassures us that the side lengths are reasonable. Why Other Options Are Wrong: Option A: 74 units is only the sum of 38 and 36 and ignores the third side of 18 units.Option C: 56 units appears unrelated to any direct combination of the three sides and does not represent the full perimeter.Option D: 54 units is only the sum of 36 and 18 and again omits one side. Common Pitfalls: A frequent mistake is to forget one of the sides when performing the addition, especially in timed tests. Another error can occur if a learner misreads one of the numbers, for example reading 38 as 36 or vice versa. Writing the formula and substituting carefully helps avoid these issues. Learners should also remember that perimeter is always a simple linear sum and not an area or product based calculation. Final Answer: The perimeter of the triangle is 92 units.

Question 16

The diagonal of the floor of a rectangular closet is 7.5 feet and the shorter side of the closet is 4.5 feet. What is the area of the closet in square feet?

Accepted Answer

Introduction / Context: This question involves a rectangular floor, its diagonal, and a side length. Problems like this are common in coordinate geometry and mensuration for aptitude tests. The key idea is that the diagonal of a rectangle forms the hypotenuse of a right angled triangle with the length and breadth as the other two sides. Once both side lengths are known, the area of the rectangle can be calculated easily as length multiplied by breadth. Given Data / Assumptions: • The floor is rectangular.• The diagonal of the floor = 7.5 feet.• The shorter side (breadth) = 4.5 feet.• We assume standard right angle rectangle properties so that Pythagoras theorem applies. Concept / Approach: In a rectangle, if one side is a, the other side is b, and the diagonal is d, then Pythagoras theorem gives d^2 = a^2 + b^2. Here, we know d and one side, so we can solve for the unknown side. After we find both side lengths, the area A of the rectangle is A = length * breadth. The main skills tested are the use of Pythagoras theorem and careful handling of squares and square roots. Step-by-Step Solution: Step 1: Let the longer side be L and the shorter side be B.B = 4.5 ft, d = 7.5 ft.Step 2: Apply Pythagoras theorem.d^2 = L^2 + B^2.7.5^2 = L^2 + 4.5^2.Step 3: Compute the squares.7.5^2 = 56.25 and 4.5^2 = 20.25.Step 4: Rearrange to find L^2.L^2 = 56.25 − 20.25 = 36.Step 5: Take the square root.L = sqrt(36) = 6 ft.Step 6: Compute the area of the rectangular floor.Area = L * B = 6 * 4.5 = 27 square feet. Verification / Alternative check: We can verify that 6 ft, 4.5 ft, and 7.5 ft indeed form a right triangle. Check using Pythagoras theorem: 6^2 + 4.5^2 = 36 + 20.25 = 56.25 and 7.5^2 is also 56.25. Since both sides of the equation match, the dimensions are consistent. The area 27 square feet also makes sense because both side lengths are moderate values and their product falls within a reasonable range for such a closet. Why Other Options Are Wrong: Option A: 5.5 is far too small and cannot be a product of two lengths that are both greater than 4 feet.Option B: 13.5 would correspond to length 3 ft and breadth 4.5 ft, which would not give the correct diagonal of 7.5 ft.Option D: 37 square feet is too large for the given side and diagonal and does not match the computed product of the side lengths. Common Pitfalls: Learners sometimes confuse which value is the diagonal and which ones are the sides, or they may incorrectly add rather than subtract when applying Pythagoras theorem. Another issue is rounding too early or miscalculating simple squares like 7.5^2. Writing each step clearly and verifying each arithmetic operation helps avoid these mistakes in exams. Final Answer: The area of the closet floor is 27 square feet.

Question 17

Calculate the number of bricks, each measuring 25 cm × 15 cm × 8 cm, required to construct a brick wall of dimensions 10 m × 4 m × 0.5 m when 10% of its volume is occupied by concrete.

Accepted Answer

Introduction / Context: This question belongs to practical mensuration and volume calculations, often used in civil engineering and construction related aptitude tests. It asks how many standard rectangular bricks are needed to build a brick wall when part of the wall volume is occupied by concrete or mortar. The problem combines unit conversion between metres and centimetres, computation of volumes, and an adjustment for the percentage of concrete used instead of bricks. Given Data / Assumptions: • Dimensions of each brick = 25 cm × 15 cm × 8 cm.• Wall dimensions = 10 m × 4 m × 0.5 m (length × height × thickness).• 10% of the wall volume is occupied by concrete or mortar.• Remaining 90% of the wall volume is filled by bricks. Concept / Approach: The number of bricks is given by the effective volume of the wall that must be filled by bricks divided by the volume of a single brick. Since the wall dimensions are in metres and the brick dimensions are in centimetres, we first convert the wall dimensions into centimetres. After computing the total wall volume in cubic centimetres, we take 90% of it to account for the space taken up by concrete. Finally, we divide this effective brick volume by the volume of a single brick to get the required number of bricks. Step-by-Step Solution: Step 1: Convert wall dimensions into centimetres.10 m = 1000 cm, 4 m = 400 cm, 0.5 m = 50 cm.Step 2: Compute total volume of the wall.Wall volume = 1000 * 400 * 50 cubic centimetres.Wall volume = 1000000 * 20 = 20000000 cm^3.Step 3: Compute effective volume to be filled by bricks.Only 90% is brick volume, so effective volume = 0.9 * 20000000 = 18000000 cm^3.Step 4: Compute volume of one brick.Brick volume = 25 * 15 * 8 = 3000 cm^3.Step 5: Compute the number of bricks.Number of bricks = 18000000 / 3000 = 6000 bricks. Verification / Alternative check: We can simplify the ratio before division: 18000000 / 3000 is equal to 18000 / 3 when we divide numerator and denominator by 1000. This equals 6000, which confirms the earlier calculation. The number also looks reasonable for a wall of the given size, since a typical construction uses thousands of bricks for even moderate walls. Why Other Options Are Wrong: Option B: 5400 would correspond to a different percentage of concrete or a different thickness and does not match the given 10% concrete assumption.Option C: 3800 is much too low for the volume in question and would not fill 90% of the wall volume.Option D: 4700 is also inconsistent with the computed ratio of volumes and cannot be justified by correct calculations. Common Pitfalls: Many learners forget to convert the wall dimensions from metres to centimetres before calculating volume, which leads to a large numerical error. Another frequent mistake is to apply the 10% factor incorrectly, for example subtracting 10 in absolute terms instead of taking 10% of the volume. It is also easy to miscalculate the volume of a brick if the multiplication is rushed. Careful unit handling and stepwise calculation prevent these mistakes. Final Answer: The number of bricks required to construct the wall is 6000.

Question 18

If the areas of three adjacent faces of a cuboidal box are 120 cm², 72 cm² and 60 cm² respectively, what is the volume of the box in cubic centimetres?

Accepted Answer

Introduction / Context: This is a classic cuboid problem where areas of three mutually perpendicular faces are given, and the task is to determine the volume. Such questions appear often in aptitude exams and test understanding of three dimensional geometry. Instead of giving the edge lengths directly, the problem provides the products of pairs of edges, which requires some algebraic manipulation to extract the volume. Given Data / Assumptions: • Let the sides of the cuboid be a, b, and c (in centimetres).• Area of face with sides a and b = ab = 120 cm^2.• Area of face with sides b and c = bc = 72 cm^2.• Area of face with sides c and a = ca = 60 cm^2.• We need the volume V = a * b * c. Concept / Approach: The key idea is to multiply the three face areas ab, bc, and ca. This product equals (ab) * (bc) * (ca) = a^2 * b^2 * c^2 = (abc)^2. Taking the square root of this product gives abc, which is the volume. Thus, instead of solving separately for a, b, and c, we can directly calculate the volume from the given face areas. This method is efficient and widely used in similar cuboid problems. Step-by-Step Solution: Step 1: Write the face areas in terms of a, b, and c.ab = 120, bc = 72, ca = 60.Step 2: Multiply the three equations.(ab) * (bc) * (ca) = 120 * 72 * 60.Step 3: Recognise that the left side is (abc)^2.(abc)^2 = 120 * 72 * 60.Step 4: Compute the numerical product.120 * 72 * 60 = 518400.Step 5: Take the square root to obtain the volume.abc = sqrt(518400) = 720.Step 6: Therefore, the volume V = 720 cubic centimetres. Verification / Alternative check: We can try to factor 518400 to ensure that the square root is correct. Notice that 518400 = 5184 * 100 and 5184 is 72^2. So sqrt(518400) = sqrt(5184) * sqrt(100) = 72 * 10 = 720. This confirms that there was no arithmetic error and validates the computed volume of the cuboid. Why Other Options Are Wrong: Option A: 640 cub. cm does not equal the square root of 518400 and would not satisfy the given face areas.Option C: 860 cub. cm is larger than the correct volume and cannot be obtained from the current set of face areas.Option D: 945 cub. cm is unrelated to the calculated value and reflects incorrect handling of multiplication or square rooting. Common Pitfalls: Learners sometimes attempt to solve for a, b, and c individually, which is more time consuming and error prone. Another mistake is to add the face areas instead of multiplying them, or to forget that taking the square root is necessary after obtaining (abc)^2. Careful algebraic reasoning and checking the square of the final answer against the intermediate product help avoid these issues. Final Answer: The volume of the cuboidal box is 720 cub. cm.

Question 19

A metallic sheet is in the shape of a rectangle with dimensions 48 m by 36 m. From each corner a square of side 8 m is cut off to make an open rectangular box. What is the volume of the box in cubic metres?

Accepted Answer

Introduction / Context: This problem belongs to the topic of solids formed by folding metal sheets. A flat rectangular sheet is converted into an open box by cutting congruent squares from each corner and folding up the flaps. Such questions test spatial understanding and the ability to translate two dimensional dimensions into a three dimensional volume using basic geometry and mensuration formulas. Given Data / Assumptions: • Original sheet dimensions = 48 m by 36 m.• Squares of side 8 m are cut from each of the four corners.• An open box is formed by folding up the remaining parts.• We are asked to find the volume of this box in cubic metres. Concept / Approach: When a square of side 8 m is cut from each corner, the length and breadth of the base of the box are reduced by 2 times 8 m, because cuts are made at both ends along each dimension. The height of the box equals the side of the cut square, that is 8 m. Once we know the new length, breadth, and height of the box, we apply the volume formula for a cuboid, V = length * breadth * height. Step-by-Step Solution: Step 1: Compute the base dimensions of the box.Original length = 48 m, original breadth = 36 m.After cutting squares of side 8 m from both ends, new length = 48 − 2 * 8 = 48 − 16 = 32 m.New breadth = 36 − 2 * 8 = 36 − 16 = 20 m.Step 2: Determine the height of the box.Height = side of the cut square = 8 m.Step 3: Use the volume formula for a cuboid.V = length * breadth * height.V = 32 * 20 * 8.Step 4: Multiply step by step.32 * 20 = 640.V = 640 * 8 = 5120 cubic metres. Verification / Alternative check: We can check the reasonableness of the answer by noting that the original sheet area is 48 * 36 = 1728 square metres. The base of the box has area 32 * 20 = 640 square metres, which is less than the original area due to the removed corner squares. A height of 8 m gives a volume of 5120 cubic metres, which is consistent with the base area and height. There is no obvious arithmetic error in the calculations. Why Other Options Are Wrong: Option A: 4830 is not a product of simple integer dimensions and suggests incorrect reduction of length or breadth.Option C: 6420 is too large and would require at least one dimension to be miscomputed.Option D: 8960 assumes a larger volume than is possible with the given sheet and cut size. Common Pitfalls: Learners may forget that the cut happens at both ends of each side, so they subtract 8 m only once instead of twice, leading to incorrect base dimensions. Others may confuse the original sheet area with the volume, or mistakenly add instead of multiplying dimensions. Carefully drawing a sketch and systematically reducing the length and breadth by twice the cut size helps to avoid these errors. Final Answer: The volume of the open rectangular box is 5120 cubic metres.

Question 20

Suppose that the medians BD, CE and AF of triangle ABC meet at the centroid G. What is the ratio AG : GF?

Accepted Answer

Introduction / Context: This conceptual geometry question is about medians and the centroid of a triangle. It tests a standard property known in coordinate geometry and Euclidean geometry: the centroid divides each median of a triangle in a fixed ratio. Understanding this property is very useful for solving many triangle related problems in competitive examinations. Given Data / Assumptions: • ABC is any triangle.• BD, CE, and AF are medians of triangle ABC.• All three medians meet at point G, which is called the centroid.• We need the ratio AG : GF along the median from vertex A to side BC. Concept / Approach: A median of a triangle joins a vertex to the midpoint of the opposite side. In any triangle, the three medians are concurrent; they meet at a single point called the centroid. A fundamental result in geometry states that the centroid divides each median in the ratio 2:1, with the longer segment always between the vertex and the centroid. Therefore, on median AF from vertex A to midpoint F of BC, the centroid G lies such that AG:GF = 2:1. Step-by-Step Solution: Step 1: Recall the definition of a median.A median connects a vertex to the midpoint of the opposite side.Step 2: Recall the property of the centroid.The centroid is the common point of intersection of the three medians.Step 3: Use the known ratio property.For any median, the centroid divides the median such that the vertex to centroid segment is twice the centroid to midpoint segment.Step 4: Apply this to median AF.Hence, AG:GF = 2:1. Verification / Alternative check: In coordinate geometry, if the vertices of a triangle are (x1, y1), (x2, y2), and (x3, y3), then the centroid has coordinates ((x1 + x2 + x3) / 3, (y1 + y2 + y3) / 3). If we compute the midpoint of BC and use the formula for the centroid, we see that the vector from A to G is two thirds of the vector from A to F, and the vector from G to F is one third. This gives the same ratio of 2:1, confirming the property. Why Other Options Are Wrong: Option A: 1:2 reverses the ratio and would imply the centroid is closer to the midpoint, which contradicts the known theorem.Option C: 1:3 does not correspond to any standard median division property.Option D: 2:3 is not supported by geometric theory for medians and centroids. Common Pitfalls: Some learners misremember the ratio as 1:2 or confuse it with other properties such as side ratios in special triangles. Others may incorrectly think the centroid divides the median into equal halves. To avoid this, it is helpful to remember that the centroid is always two thirds of the way from a vertex to the midpoint of the opposite side, giving a fixed 2:1 division along every median. Final Answer: The ratio AG : GF is 2:1.

Volume and Surface Area Questions