Question 1

A towel, when bleached, is found to have lost 20% of its length and 10% of its breadth. What is the percentage decrease in its area?

Accepted Answer

Introduction / Context:This question is a good example of percentage change applied to two dimensions of a rectangle. The towel is treated as a rectangle with length and breadth. When each dimension changes by a certain percentage, the area changes by a combined effect, which is not simply the arithmetic sum of the two percentages. Such problems appear in aptitude exams to test understanding of compound percentage and area relationships. Given Data / Assumptions:• Original length of the towel is L.• Original breadth of the towel is B.• Length decreases by 20%, so new length is 80% of L.• Breadth decreases by 10%, so new breadth is 90% of B.• We seek percentage decrease in area. Concept / Approach:The original area A1 of the towel is L * B. After bleaching, the new length is 0.8L and the new breadth is 0.9B. The new area A2 is therefore 0.8L * 0.9B = 0.72L * B. This means the new area is 72% of the original area. The percentage decrease is then 100% − 72% = 28%. This demonstrates how combined percentage effects on dimensions multiply rather than simply add. Step-by-Step Solution:Step 1: Let original area A1 = L * B.Step 2: Compute new length.Length decreases by 20%, so new length = 0.8L.Step 3: Compute new breadth.Breadth decreases by 10%, so new breadth = 0.9B.Step 4: Compute new area A2.A2 = 0.8L * 0.9B = 0.72L * B.Step 5: Express A2 as a percentage of A1.A2 / A1 = (0.72L * B) / (L * B) = 0.72.So new area is 72% of the original area.Step 6: Compute percentage decrease.Decrease = 100% − 72% = 28%. Verification / Alternative check:We can also think in terms of a percentage formula: the approximate net percentage change when two changes x% and y% affect area is x + y + (x * y) / 100, where decreases are taken as negative. Here, x = −20 and y = −10. So net change ≈ −20 + (−10) + (−20 * −10) / 100 = −30 + 2 = −28%. This confirms that the area decreases by 28% overall, matching the previous calculation. Why Other Options Are Wrong:Option A: 10% treats only one dimension or incorrectly averages the effect.Option B: 10.08% comes from misapplying percentage formulas or calculation errors.Option C: 20% simply adds one of the decreases and ignores the other dimension. Common Pitfalls:A common mistake is to add 20% and 10% and conclude that the area decreases by 30%, which is incorrect because changes in area are multiplicative. Another error is to apply the percentage decrease directly to area without considering that both dimensions are changed. Using either the product of factors (0.8 and 0.9) or the net percentage change formula helps avoid these misunderstandings. Final Answer:The percentage decrease in the area of the towel is 28%.

Question 2

What is the least number of square tiles required to pave completely the floor of a room that is 15 m 17 cm long and 9 m 2 cm broad?

Accepted Answer

Introduction / Context:This problem explores the idea of tiling a rectangular floor with identical square tiles so that there is no cutting or wastage. It is common in aptitude exams under the heading of least common multiple and greatest common divisor, applied in a geometric setting. The goal is to find the largest possible tile size that exactly divides both dimensions, and then compute how many such tiles are required. Given Data / Assumptions:• Length of the room = 15 m 17 cm.• Breadth of the room = 9 m 2 cm.• Tiles are square and all have the same side length.• Tiles must be placed without overlapping and without leaving gaps. Concept / Approach:First, convert both dimensions into centimetres to work with whole numbers. Then, find the greatest common divisor of the two lengths in centimetres. This greatest common divisor is the side of the largest possible square tile that can exactly fit along both length and breadth. The number of tiles will then be equal to the floor area divided by the area of one tile. This approach ensures the least number of tiles, since a larger tile size means fewer tiles. Step-by-Step Solution:Step 1: Convert dimensions into centimetres.Length = 15 m 17 cm = 1500 cm + 17 cm = 1517 cm.Breadth = 9 m 2 cm = 900 cm + 2 cm = 902 cm.Step 2: Find the greatest common divisor of 1517 and 902.Using the Euclidean algorithm, gcd(1517, 902) = 41 cm.Step 3: Use 41 cm as the side of the largest possible square tile.Step 4: Compute the number of tiles along the length.1517 / 41 = 37 tiles.Step 5: Compute the number of tiles along the breadth.902 / 41 = 22 tiles.Step 6: Compute the total number of tiles.Total tiles = 37 * 22 = 814. Verification / Alternative check:We can confirm that 41 divides both dimensions exactly since 41 * 37 = 1517 and 41 * 22 = 902. No larger square side can divide both 1517 and 902 because 41 is the greatest common divisor. If we check a larger number like 43 or 50, it does not divide at least one of the dimensions exactly. Hence, 41 is the correct tile side and 814 tiles is the minimal count. Why Other Options Are Wrong:Option B: 820 would correspond to a different non maximal tile size or to rounding errors.Option C: 840 is too large and assumes smaller tiles, leading to more tiles than necessary.Option D: 844 also indicates a smaller tile side and does not reflect the greatest common divisor approach. Common Pitfalls:Students sometimes convert only one dimension into centimetres or make mistakes in converting metres and centimetres. Others look for the least common multiple instead of the greatest common divisor, which is incorrect here because we want the largest tile that fits exactly. Using a systematic method to find the greatest common divisor and carefully converting units helps avoid these typical mistakes. Final Answer:The least number of square tiles required is 814.

Question 3

A rectangular field is to be fenced on three sides, leaving one side of length 20 feet uncovered. If the area of the field is 680 square feet, how many feet of fencing are required?

Accepted Answer

Introduction / Context:This question is a mix of area and perimeter concepts for rectangles. It is common in aptitude tests where candidates must relate the area of a field to its sides and then determine the length of fencing required for part of its boundary. Here, only three sides of the rectangular field are fenced, and one side is left open, which adds a small twist to the usual perimeter calculation. Given Data / Assumptions:• The field is rectangular.• One side of the rectangle, left unfenced, has length 20 ft.• Area of the field = 680 square feet.• We need to find the total length of fencing for the remaining three sides. Concept / Approach:Let the length of the field be L and the breadth be B. The area is L * B. One side of 20 ft is given as the side left uncovered. We assume that this side corresponds to length L = 20 ft. Then the breadth B can be found from the area relation. The total fencing required is the sum of the other three sides: one more length and two breadths, so fencing = L + 2B. We simply compute B using the area and then compute this expression. Step-by-Step Solution:Step 1: Assign the known side.Let L = 20 ft be the side left unfenced.Step 2: Use the area formula.Area = L * B = 680 sq. ft.20 * B = 680.Step 3: Solve for B.B = 680 / 20 = 34 ft.Step 4: Compute the length of fencing required.Fencing length = L + 2B = 20 + 2 * 34.2 * 34 = 68.Fencing length = 20 + 68 = 88 ft. Verification / Alternative check:We can verify the area using the calculated breadth. With L = 20 ft and B = 34 ft, the area is 20 * 34 = 680 square feet, which matches the given data. The three fenced sides are 34 ft, 34 ft, and 20 ft, whose sum is 88 ft. This confirms that the fencing length is computed correctly and is internally consistent with the problem statement. Why Other Options Are Wrong:Option A: 34 ft corresponds only to one breadth and ignores the other sides.Option B: 40 ft does not match any reasonable combination of sides for a field of area 680 sq. ft.Option C: 68 ft accounts only for the two breadths and neglects the remaining length that must be fenced. Common Pitfalls:Some learners mistakenly compute the full perimeter 2(L + B) instead of just three sides, which would give 2(20 + 34) = 108 ft and does not match any option. Others may mix up which side is left unfenced. Carefully interpreting the phrase "fenced on three sides" and explicitly writing the expression L + 2B avoids these misunderstandings. Final Answer:The total length of fencing required is 88 feet.

Question 4

Find the area of a square in square metres if one of its diagonals is 3.8 m long.

Accepted Answer

Introduction / Context:This question uses a well known property of a square relating its diagonal to its side. In many quantitative aptitude tests, candidates are required to compute either side or area when the diagonal is given. The square is a special case of a rectangle where all sides are equal and all angles are right angles, which simplifies the relationship between side length and diagonal. Given Data / Assumptions:• The figure is a square.• The length of the diagonal is 3.8 m.• We need to find the area of the square in square metres. Concept / Approach:For a square with side s and diagonal d, by Pythagoras theorem we have d^2 = s^2 + s^2 = 2s^2. Thus s^2 = d^2 / 2. The area A of the square is s^2, so A = d^2 / 2. Once the diagonal is known, we can square it, divide by 2 and obtain the area directly. This formula avoids separate computation of the side and is very convenient in exams. Step-by-Step Solution:Step 1: Write the relation between diagonal and side.d^2 = 2s^2, so s^2 = d^2 / 2.Step 2: Compute d^2.d = 3.8 m, so d^2 = 3.8 * 3.8 = 14.44.Step 3: Compute the area using A = d^2 / 2.A = 14.44 / 2.Step 4: Perform the division.A = 7.22 square metres. Verification / Alternative check:We can also approximate 3.8 m as slightly less than 4 m. A square with diagonal 4 m has area 4^2 / 2 = 16 / 2 = 8 sq. m. Since our diagonal is 3.8 m, which is slightly smaller, the area should be slightly less than 8 sq. m. The value 7.22 sq. m is consistent with this expectation and therefore seems reasonable. Why Other Options Are Wrong:Option A: 4.22 sq. m is too small and would correspond to a much shorter diagonal.Option B: 5.22 sq. m also underestimates the area for a diagonal of 3.8 m.Option C: 6.22 sq. m is still smaller than the correct value and does not satisfy A = d^2 / 2. Common Pitfalls:Some candidates mistakenly use the formula A = d^2 instead of d^2 / 2 or they treat the diagonal as if it were a side. Others miscalculate 3.8^2 due to errors in decimal multiplication. Writing the relation d^2 = 2s^2 clearly and carrying out the arithmetic slowly helps to avoid errors in this type of problem. Final Answer:The area of the square is 7.22 sq. m.

Question 5

One side of a triangle has length 8 units and a second side has length 5 units. Which of the following values could be the area of the triangle?

Accepted Answer

Introduction / Context:This conceptual question checks a candidate understanding of how the area of a triangle depends on the included angle between two sides. Given two fixed side lengths, the maximum possible area occurs when the angle between them is 90 degrees. This type of problem appears in aptitude tests to test reasoning about possible and impossible values, not just direct calculations. Given Data / Assumptions:• Two sides of a triangle have lengths 8 units and 5 units.• The third side and the included angle are not specified.• We must determine which listed area values could occur for some configuration of the triangle. Concept / Approach:For a triangle with sides a and b and included angle θ, the area A is given by A = (1/2) * a * b * sin(θ). For fixed a and b, the sine function ranges from 0 to 1, so the area ranges from 0 to a maximum of (1/2) * a * b when sin(θ) = 1. Here, a = 8 and b = 5, so maximum area is (1/2) * 8 * 5 = 20 square units. Therefore, any possible area must be less than or equal to 20 and greater than 0. Among the given options, only 20 satisfies this requirement. Step-by-Step Solution:Step 1: Use the area formula with two sides and included angle.A = (1/2) * a * b * sin(θ).Step 2: Substitute a = 8 and b = 5.A = (1/2) * 8 * 5 * sin(θ).Step 3: Simplify the constant part.(1/2) * 8 * 5 = 20.So A = 20 * sin(θ).Step 4: Note the range of sin(θ).0 < sin(θ) ≤ 1 for a non degenerate triangle.Step 5: Conclude the possible area range.0 < A ≤ 20.Step 6: From the options 24, 20, 26, 34, only 20 falls in this range. Verification / Alternative check:We can verify that the area 20 actually occurs when θ = 90 degrees. In that case the triangle is right angled with legs 8 and 5, and the area formula for a right triangle is A = (1/2) * base * height = (1/2) * 8 * 5 = 20. Thus 20 is not just theoretically possible but exactly realised by a right triangle with these sides as legs. Why Other Options Are Wrong:Option A: 24 is greater than the maximum area 20 and so cannot occur.Option C: 26 is even larger and is impossible for any angle between sides 8 and 5.Option D: 34 greatly exceeds the maximum area and is clearly impossible. Common Pitfalls:Some learners incorrectly assume that any positive area is possible as long as the triangle inequality holds, without considering the maximum area for given sides. Others may attempt to use Heron formula without knowing the third side, which is not practical here. Remembering that the largest area for fixed sides is obtained with a right angle, and computing that maximum value, provides a quick and reliable way to answer these questions. Final Answer:The area that could be the area of the triangle is 20 square units.

Question 6

Two cylindrical buckets have their diameters in the ratio 3 : 1 and their heights in the ratio 1 : 3. What is the ratio of their volumes?

Accepted Answer

Introduction / Context:This question examines proportional reasoning and the volume formula for cylinders. Instead of giving actual dimensions, it provides ratios of diameters and heights of two cylindrical buckets. Candidates must know that the volume of a cylinder depends on the square of the radius and the height, and then apply ratio rules correctly. Such questions are frequent in aptitude tests dealing with similar solids. Given Data / Assumptions:• Ratio of diameters of the two buckets = 3 : 1.• Therefore, ratio of radii is also 3 : 1.• Ratio of heights of the two buckets = 1 : 3.• We are asked to find the ratio of their volumes. Concept / Approach:The volume of a cylinder is V = π * r^2 * h, where r is radius and h is height. When comparing two cylinders, we can use ratios rather than actual numbers. If r1 : r2 = 3 : 1, then r1^2 : r2^2 = 9 : 1. If h1 : h2 = 1 : 3, we combine these to get the volume ratio: V1 : V2 = (r1^2 * h1) : (r2^2 * h2). Constants like π cancel out in the ratio, making calculations simpler. Step-by-Step Solution:Step 1: Let the radii be r1 and r2 with r1 : r2 = 3 : 1.Then r1^2 : r2^2 = 9 : 1.Step 2: Let the heights be h1 and h2 with h1 : h2 = 1 : 3.Step 3: Write the volume ratio.V1 : V2 = (π * r1^2 * h1) : (π * r2^2 * h2).Step 4: Cancel π from the ratio.V1 : V2 = (r1^2 * h1) : (r2^2 * h2).Step 5: Substitute the ratios.V1 : V2 = 9 * 1 : 1 * 3 = 9 : 3.Step 6: Simplify the ratio.9 : 3 = 3 : 1. Verification / Alternative check:We can assign actual values that respect the ratio. For example, let bucket 1 have radius 3 units and height 1 unit, and bucket 2 have radius 1 unit and height 3 units. Then V1 = π * 3^2 * 1 = 9π and V2 = π * 1^2 * 3 = 3π. The ratio 9π : 3π simplifies to 3 : 1, confirming the result. Why Other Options Are Wrong:Option A: 2:1 does not reflect the combined effect of squaring the radius and the height ratio.Option C: 4:1 would require different diameter or height ratios.Option D: 5:1 is not supported by any correct combination of the given ratios. Common Pitfalls:Some learners forget to square the radius and treat volume as proportional to r * h instead of r^2 * h. Others might mistakenly multiply the diameter ratio by the height ratio directly, getting 3:1 * 1:3 = 1:1, which is wrong. Remembering that volume for a cylinder involves the square of the radius is crucial to avoid these errors. Final Answer:The ratio of the volumes of the two buckets is 3:1.

Question 7

The cost of paint is Rs. 36.50 per kg. If 1 kg of paint covers 16 square feet, how much will it cost to paint the outside surface of a solid cube having each edge 8 feet long?

Accepted Answer

Introduction / Context:This is a practical mensuration problem involving the surface area of a cube and a unit rate for painting. It appears often in aptitude tests that combine geometry with cost calculations. The key steps are to find the total area to be painted, determine how many kilograms of paint are needed, and then multiply by the given cost per kilogram. Given Data / Assumptions:• The solid is a cube.• Edge length of the cube = 8 feet.• Cost of paint = Rs. 36.50 per kg.• Coverage of paint = 1 kg covers 16 square feet.• Entire external surface area of the cube is to be painted once. Concept / Approach:The surface area of a cube with side a is given by 6a^2, because it has six identical square faces. Once we compute the total area, we divide by the coverage per kilogram to find the quantity of paint required. Finally, we multiply the number of kilograms by the cost per kilogram. If the result is a whole number of kilograms, the cost will be a simple multiplication; otherwise, we might need to consider fractions, but in this case the numbers work out neatly. Step-by-Step Solution:Step 1: Compute total surface area of the cube.Side a = 8 ft.Surface area = 6 * a^2 = 6 * 8^2.8^2 = 64, so area = 6 * 64 = 384 square feet.Step 2: Compute how many kilograms of paint are required.1 kg covers 16 sq. ft., so paint required = total area / coverage per kg.Paint required = 384 / 16 = 24 kg.Step 3: Compute the cost.Cost per kg = Rs. 36.50.Total cost = 24 * 36.50.24 * 36.50 = 876.So total cost = Rs. 876. Verification / Alternative check:We can check the multiplication by breaking 36.50 into 36 and 0.50. Then 24 * 36 = 864 and 24 * 0.50 = 12. Adding them gives 864 + 12 = 876, which matches the earlier product. The paint requirement of 24 kg also makes sense because 384 sq. ft. divided by 16 sq. ft. per kg is exactly 24, with no remainder, confirming the coverage calculation. Why Other Options Are Wrong:Option A: Rs.962 comes from an incorrect multiplication or a different assumed coverage.Option B: Rs.672 is too low and ignores part of the surface area or uses a wrong cost per kilogram.Option C: Rs.546 also does not match the correct product of 24 and 36.50. Common Pitfalls:Students sometimes forget that a cube has six faces and mistakenly use 4a^2 or 2a^2 instead of 6a^2. Others may divide the total area by the cost per kilogram rather than the coverage, or mix up units. Double checking which numbers represent area, coverage, and cost helps prevent these mistakes, as does rewriting the steps clearly before calculating. Final Answer:The cost of painting the outside of the cube is Rs.876.

Question 8

A cuboid with dimensions 24 cm × 9 cm × 8 cm is melted and re-cast into smaller solid cubes each of side 3 cm. How many such cubes can be formed?

Accepted Answer

Introduction / Context:This question deals with conversion of one solid shape into another by melting and re casting, a common topic in mensuration and volume based aptitude problems. The total volume is conserved in the process. Here, a cuboid is melted and turned into smaller cubes, and the number of cubes is determined by dividing the volume of the original cuboid by the volume of one cube. Given Data / Assumptions:• Original solid is a cuboid of dimensions 24 cm × 9 cm × 8 cm.• New solids are cubes each with side 3 cm.• There is no loss of material during melting and re casting.• We need the number of cubes formed. Concept / Approach:The key idea is that volume is conserved. Let V1 be the volume of the original cuboid and V2 be the volume of one cube. The number of cubes N is N = V1 / V2. For the cuboid, V1 = length * breadth * height. For each cube, V2 = side^3. By substituting the given dimensions, we can calculate the value of N directly. Step-by-Step Solution:Step 1: Compute volume of the cuboid.V1 = 24 * 9 * 8 cubic centimetres.24 * 9 = 216.V1 = 216 * 8 = 1728 cm^3.Step 2: Compute volume of one small cube.Side of cube = 3 cm.V2 = 3^3 = 3 * 3 * 3 = 27 cm^3.Step 3: Compute the number of cubes.N = V1 / V2 = 1728 / 27.Step 4: Perform the division.1728 / 27 = 64. Verification / Alternative check:We can factor 1728 as 64 * 27, so the division 1728 / 27 equals 64 exactly. This confirms that the number of cubes is an integer and matches our calculation. Since there is no leftover volume, the assumption of volume conservation is satisfied and the result is logically consistent. Why Other Options Are Wrong:Option B: 56 implies an incorrect division or a mistaken volume for either the cuboid or the cube.Option C: 48 undercounts the cubes and would require a larger cube side or a smaller cuboid volume.Option D: 40 is much smaller than the correct answer and does not arise from any correct arithmetic based on the given dimensions. Common Pitfalls:Learners sometimes use the sum of dimensions instead of their product when calculating volume, or they forget to cube the side for the cube volume. Another mistake is to reverse the division, computing 27 / 1728 instead of 1728 / 27. Writing the formulas clearly and performing the arithmetic carefully helps avoid such errors. Final Answer:The number of cubes that can be formed is 64.

Question 9

The base of a pyramid is a rectangle whose length and breadth are 16 cm and 12 cm respectively. All lateral edges from the vertex to the base vertices are 26 cm long, and the pyramid is a right rectangular pyramid. What is the volume of the pyramid …

Accepted Answer

Introduction / Context:This question involves a three dimensional solid called a right rectangular pyramid. The base is a rectangle, and all lateral edges from the vertex to the base vertices are equal, which simplifies some geometry. The task is to find the volume of this pyramid. Problems like this combine knowledge of the Pythagoras theorem in space with the volume formula for pyramids and are common in more advanced aptitude and engineering exams. Given Data / Assumptions:• Base is a rectangle with length 16 cm and breadth 12 cm.• All lateral edges (from vertex to each base vertex) are 26 cm long.• The pyramid is a right rectangular pyramid, so the vertex lies vertically above the centre of the base.• We need the volume of the pyramid. Concept / Approach:The volume V of a pyramid is V = (1/3) * base area * height. The base area is easy to compute as length times breadth. The challenge is to find the vertical height of the pyramid. Because it is a right rectangular pyramid, the foot of the perpendicular from the vertex to the base is at the centre of the rectangle. The line from the centre of the base to any vertex of the base, together with the vertical height and the lateral edge, forms a right triangle. Using Pythagoras theorem in this triangle allows us to find the height. Step-by-Step Solution:Step 1: Compute the base area.Base area = length * breadth = 16 * 12 = 192 cm^2.Step 2: Find the distance from the centre of the rectangle to a base vertex.The rectangle diagonal = sqrt(16^2 + 12^2) = sqrt(256 + 144) = sqrt(400) = 20 cm.The centre is at the midpoint of the diagonal, so the distance from the centre to a vertex is half of 20, which is 10 cm.Step 3: Use Pythagoras theorem in the triangle formed by the height, this 10 cm segment, and the lateral edge.Let h be the vertical height and the lateral edge be 26 cm.Then 26^2 = h^2 + 10^2.676 = h^2 + 100.h^2 = 676 − 100 = 576.h = sqrt(576) = 24 cm.Step 4: Compute the volume of the pyramid.V = (1/3) * base area * height = (1/3) * 192 * 24.192 * 24 = 4608.V = 4608 / 3 = 1536 cubic centimetres. Verification / Alternative check:We can double check the arithmetic: 192 * 24 can be computed as 200 * 24 − 8 * 24 = 4800 − 192 = 4608. Dividing 4608 by 3 gives 1536, since 3 * 1500 = 4500 and 3 * 36 = 108, which sums to 4608. The height 24 cm is also reasonable because it is slightly less than the lateral edge 26 cm but significantly larger than the base radius 10 cm, consistent with a right pyramid geometry. Why Other Options Are Wrong:Option B: 1024 is too small and would arise only if either the height or base area were miscalculated.Option C: 718 does not match the required one third factor and suggests multiple arithmetic errors.Option D: 2072 is too large compared to the correct volume and contradicts the relation V = (1/3) * base area * height. Common Pitfalls:Students often mistake the lateral edge for the vertical height and directly use 26 cm as height, which gives an incorrect volume. Others miscalculate the distance from the centre of the base to a vertex or fail to divide by 3 in the volume formula for a pyramid. Drawing a clear diagram of the pyramid and marking the right triangles involved helps in avoiding these errors. Final Answer:The volume of the pyramid is 1536 cubic centimetres.

Question 10

If triangle ABC is an equilateral triangle with each side 16 cm long, what is the length of its altitude?

Accepted Answer

Introduction / Context:This problem is about an equilateral triangle and its altitude. In an equilateral triangle, all sides are equal and all angles are 60 degrees. The altitude has a special relationship with the side length because it also acts as a median and an angle bisector. Knowing the formula for the altitude of an equilateral triangle is very useful for geometry and trigonometry questions in aptitude tests. Given Data / Assumptions:• Triangle ABC is equilateral.• Each side of triangle ABC = 16 cm.• We are required to find the length of an altitude of this triangle. Concept / Approach:In an equilateral triangle with side a, the altitude h can be found using Pythagoras theorem. When the altitude is drawn from a vertex to the opposite side, it splits the triangle into two congruent right angled triangles. Each has hypotenuse a and base a/2. Therefore, h^2 + (a/2)^2 = a^2. Solving gives h = (sqrt(3) / 2) * a. We can directly apply h = (sqrt(3) / 2) * a to compute the altitude length for side 16 cm. Step-by-Step Solution:Step 1: Use the altitude formula for an equilateral triangle.h = (sqrt(3) / 2) * a.Step 2: Substitute the given side length a = 16 cm.h = (sqrt(3) / 2) * 16.Step 3: Simplify the expression.16 / 2 = 8.So h = 8 * sqrt(3) cm. Verification / Alternative check:We can verify by using Pythagoras theorem directly. After drawing the altitude, we have two right triangles with hypotenuse 16 cm and one leg 8 cm (half of the base). Then h^2 = 16^2 − 8^2 = 256 − 64 = 192. So h = sqrt(192). Factoring 192 as 64 * 3, we get h = sqrt(64 * 3) = 8 * sqrt(3), which agrees with our formula based calculation. Why Other Options Are Wrong:Option A: 2√3 cm would correspond to a much smaller triangle and does not match side 16 cm.Option B: 4√3 cm is exactly half the correct altitude and would match a triangle with side 8 cm.Option D: 5√3 cm does not result from any correct computation based on the given side length. Common Pitfalls:Candidates may confuse the formula for altitude with that of the median in other triangles or may misremember the coefficient, writing a/2 instead of (sqrt(3) / 2) * a. Errors also occur when simplifying the square root of 192. Remembering that in an equilateral triangle the altitude is greater than half the side and using h = (sqrt(3) / 2) * a helps prevent such mistakes. Final Answer:The length of the altitude of the equilateral triangle is 8√3 cm.

Question 11

In triangle PQR, PQ = PR = 18 cm. Point A is the mid-point of QR. Segments AB and AC are drawn parallel to PR and PQ respectively. What is the perimeter, in centimetres, of quadrilateral ABPC?

Accepted Answer

Introduction / Context:This geometry question involves an isosceles triangle and a quadrilateral formed by drawing lines parallel to the sides. It tests understanding of similar triangles and symmetry. The triangle PQR has two equal sides, and a mid point on the base is used together with parallels to create a smaller quadrilateral ABPC inside. The task is to find the perimeter of this quadrilateral. Given Data / Assumptions:• Triangle PQR is isosceles with PQ = PR = 18 cm.• A is the mid point of QR.• AB is drawn through A parallel to PR.• AC is drawn through A parallel to PQ.• Points B and C lie on PQ and PR respectively.• Quadrilateral ABPC is formed, and we need its perimeter. Concept / Approach:Because PQR is isosceles, the configuration is symmetric about the altitude from P to QR. Point A is at the centre of the base, and lines AB and AC are drawn parallel to the equal sides. Using coordinate geometry or symmetry, we find that AB, BC, CP, and PA are all equal in length. In particular, each side of quadrilateral ABPC turns out to be 9 cm, which is half of the equal sides in the original triangle. Therefore, the perimeter is four times this common side length. Step-by-Step Solution:Step 1: Represent triangle PQR in a coordinate system for clarity.Let QR be horizontal with Q at (−a, 0) and R at (a, 0). Let P be at (0, h). The equal sides PQ and PR each have length 18 cm.Step 2: Since A is the mid point of QR, A is at (0, 0).Step 3: The line through A parallel to PR will intersect PQ at B, and the line through A parallel to PQ will intersect PR at C.Step 4: Using similarity of triangles or direct coordinate calculations, one can show that the segments AB, BP, PC, and CA are all equal and each equals 9 cm.Step 5: Hence quadrilateral ABPC has four equal sides each of 9 cm.Step 6: Perimeter of ABPC = 4 * 9 = 36 cm. Verification / Alternative check:A more algebraic approach confirms the same result. Taking concrete values for the base and height that produce side length 18 cm and solving using the section formula and parallel line properties, we again find each side of the quadrilateral is 9 cm. In either method, the symmetry of the isosceles triangle and the use of parallels from the base midpoint ensure the quadrilateral behaves like a smaller rhombus inside the triangle. Why Other Options Are Wrong:Option A: 18 would correspond to all four sides being 4.5 cm each, which does not match the geometry.Option B: 28 gives an average side length of 7 cm, inconsistent with the similarity ratio derived from the original triangle.Option C: 32 implies each side is 8 cm, which is still not half of 18 cm and does not arise from the correct construction. Common Pitfalls:Students may attempt to compute side lengths directly from incomplete geometric reasoning or may incorrectly assume that some sides of the quadrilateral match either the base or the original sides. Others might overlook the symmetry and not recognise that all four sides are equal. Drawing a clear labelled diagram and using the properties of parallel lines and mid points in an isosceles triangle helps reach the correct conclusion. Final Answer:The perimeter of quadrilateral ABPC is 36 centimetres.

Question 12

A spherical balloon has its volume increased by 700%. By what percentage does its surface area increase?

Accepted Answer

Introduction / Context: This aptitude question tests how the volume and surface area of a sphere change when the radius is scaled. It is a classic example of geometric similarity, where solid figures with proportional dimensions have related ratios for volume and surface area. Understanding this relationship is useful in many real life situations, such as inflating balloons, designing spherical tanks, or scaling models. The key idea is that volume of a sphere depends on the cube of the radius, while surface area depends on the square of the radius. So, a percentage change in volume does not translate directly into the same percentage change in surface area, and we must carefully work with ratios and powers of the scale factor. Given Data / Assumptions: • Initial volume of the spherical balloon is V₁ (unknown but not needed numerically). • Final volume V₂ is obtained by increasing the volume by 700%. • A 700% increase means the new volume is 8 times the old volume. • Let the initial radius be r₁ and the final radius be r₂. • Surface area is proportional to the square of the radius for a sphere. Concept / Approach: For a sphere, volume is proportional to r³ and surface area is proportional to r². If we know how many times the volume has increased, we can find the scale factor of the radius using the cube root. Once the new radius factor is known, we square that factor to obtain the factor by which surface area changes. Finally, we convert this factor into a percentage increase relative to the original surface area. The percentage increase is given by (new area - old area) / old area multiplied by 100. Step-by-Step Solution: Step 1: A 700% increase in volume means the new volume is V₂ = V₁ + 7V₁ = 8V₁. Step 2: Since volume of a sphere is proportional to r³, we have V ∝ r³, so V₂ / V₁ = (r₂³) / (r₁³). Step 3: Substitute the volume ratio: 8 = (r₂³) / (r₁³), so r₂³ = 8 r₁³. Step 4: Take the cube root of both sides: r₂ = 2 r₁, so the radius doubles. Step 5: Surface area of a sphere is proportional to r², so the area ratio is A₂ / A₁ = (r₂²) / (r₁²) = (2²) / 1² = 4. Step 6: A factor of 4 means the surface area becomes four times the original, which is a 300% increase, because (4 - 1) * 100 = 300. Verification / Alternative check: We can verify with simple numbers. Assume the initial radius r₁ is 1 unit. Then the initial volume is proportional to 1³ = 1. If the volume becomes 8, the new radius r₂ must be 2, because 2³ = 8. For surface area, initially it is proportional to 1² = 1, and after the change it is proportional to 2² = 4. So the surface area has become four times the original. Converting this factor to a percentage increase gives (4 - 1) / 1 * 100 = 300%. This confirms the earlier reasoning and shows that the exact initial numerical values do not matter, only the ratios and the power relationships are important. Why Other Options Are Wrong: 100% is wrong because it would mean the surface area only doubled, which would correspond to a radius multiplied by sqrt(2), not a factor of 2 in radius. 200% is incorrect because it suggests the area becomes three times the original, while the radius doubling makes area four times the original. 400% and 500% are too large, implying area factors of 5 and 6. These would require radius scale factors greater than 2, which contradicts the volume relationship that gave r₂ = 2 r₁. Therefore, only 300% matches the correct geometric scaling. Common Pitfalls: A common mistake is to assume that if volume increases by 700%, surface area must also increase by 700%, which is not true because they depend on different powers of the radius. Another error is to treat percentage increase as a simple ratio without converting 700% into a factor of 8. Some learners also confuse area and volume formulas or forget that volume scales with the cube of the linear dimension and area with the square. Careful attention to the exponents and the meaning of percentage increase avoids these issues. Final Answer: The percentage increase in the surface area of the spherical balloon is 300%.

Question 13

In a square of side length a, two circular arcs are drawn with centres at opposite vertices, joining the other two vertices and forming two overlapping sectors inside the square. What is the area common to both sectors?

Accepted Answer

Introduction / Context: This geometry question deals with the intersection area formed by two circular sectors inside a square. It uses ideas from circles, sectors, and symmetry. The configuration creates a lens shaped region that is common to two quarter circles drawn with centres at opposite corners of the square. Problems like this are important for building spatial reasoning and for understanding how standard formulas for areas of sectors and triangles can be combined to find areas of complex shapes. Given Data / Assumptions: • The figure is a square of side length a. • Opposite vertices of the square are used as centres of two equal circles of radius a. • Each circular arc is drawn inside the square and passes through the remaining two vertices. • We are asked to find the common area of the two circular sectors inside the square. • The distance between opposite vertices of the square is a√2, which is the distance between the two circle centres. Concept / Approach: Each circle has radius a and the distance between their centres is a√2. The lens shaped common region of the two circles is symmetric with respect to the line joining their centres. For two equal circles of radius r whose centres are distance d apart, the common area can be expressed in terms of circular sectors minus the area of an isosceles triangle. In this special case, the geometry simplifies because the angle between the radii at the centre is a right angle. We can therefore work with one circle and then double the area of a single sector based component to obtain the total intersection area. Step-by-Step Solution: Step 1: Let the square be ABCD with side a. Take A and C as the centres of two circles of radius a. The other vertices B and D lie on both circles. Step 2: In each circle, the angle subtended at the centre by chord BD is 90 degrees, since angle BAD and angle BCD in a square are right angles. Step 3: Therefore, in each circle, the relevant region corresponding to the intersection is based on a sector of angle 90 degrees, that is π/2 radians. Step 4: For one circle, the area of the sector with angle π/2 is (π/2) * a^2. Step 5: The triangle formed by the two radii and the chord BD inside that sector is a right isosceles triangle with legs of length a, so its area is (1/2) * a * a = a^2 / 2. Step 6: The area of the circular segment in one circle is area of sector minus area of triangle = (π/2) * a^2 - a^2 / 2. Step 7: The common lens shaped area is made of two identical segments, one from each circle, so the total common area is 2 * [(π/2) * a^2 - a^2 / 2] = a^2 * (π/2 - 1). Verification / Alternative check: We can verify the formula using the general expression for the intersection area of two equal circles of radius r whose centres are distance d apart. In general, the common area is 2 r^2 cos⁻¹(d / (2r)) - (d / 2) * √(4r^2 - d^2). Here r = a and d = a√2. We get d / (2r) = (a√2) / (2a) = √2 / 2, so cos⁻¹(d / (2r)) is 45 degrees or π/4 radians. Substituting, the area becomes 2 a^2 * (π/4) - (a√2 / 2) * √(4a^2 - 2a^2). This simplifies to (a^2 π) / 2 - a^2, which is exactly a^2 * (π/2 - 1). This matches the earlier reasoning and confirms that the chosen formula is correct. Why Other Options Are Wrong: a^2 * (π/2 + 1) is too large because it uses a plus sign where a subtraction is required, effectively adding the triangular area instead of subtracting it. a^2 is incorrect because it ignores the circular geometry and treats the region as equivalent to a simple square region of area a^2. a^2 * (π - 1) overestimates the area, roughly corresponding to a sector of 180 degrees rather than 90 degrees in each circle. a^2 * (π/4) underestimates the area by taking only half of the needed sector contribution and ignoring the triangular part. None of these match the exact lens area expression. Common Pitfalls: Learners often confuse the geometry and compute the area of a single sector instead of the overlapping region. Another frequent mistake is forgetting to subtract the triangular area from the sector to obtain the segment, which leads to overestimation. Some also forget that there are two identical segments contributing to the total common area. Finally, mixing up degrees and radians or misusing the diagonal length of the square can also lead to errors. Keeping the picture clear and working systematically with one circle at a time helps avoid these issues. Final Answer: The common area of the two sectors inside the square is a^2 * (π/2 - 1).

Question 14

The area of a semicircle is 1925 square centimetres. What is the perimeter of the semicircle (in cm)? (Use π = 22/7.)

Accepted Answer

Introduction / Context: This problem checks understanding of the relationship between the area and perimeter of a semicircle. It requires working backwards from the given area to find the radius, and then using that radius to compute the total boundary length, which includes both the curved part and the diameter. Such questions are common in aptitude exams and help reinforce how circle formulas are connected and how to correctly use π approximations such as 22/7 for calculation. Given Data / Assumptions: • Area of the semicircle is 1925 square centimetres. • π is to be taken as 22/7. • Let the radius of the semicircle be r centimetres. • Perimeter of a semicircle consists of the curved arc plus the diameter, so it is πr + 2r. Concept / Approach: The area of a full circle is πr², so the area of a semicircle is (1/2) * πr². We are given this area and must solve for r. Once the radius is known, the perimeter of the semicircle is the sum of the curved length (half the circumference, which is πr) and the straight diameter (2r). The steps therefore involve rearranging the area formula, substituting π = 22/7, solving for r² and r, and then computing the perimeter expression. Step-by-Step Solution: Step 1: Use the semicircle area formula: (1/2) * π * r² = 1925. Step 2: Substitute π = 22/7 to get (1/2) * (22/7) * r² = 1925. Step 3: Simplify the constant factor: (1/2) * (22/7) = 11/7, so we have (11/7) * r² = 1925. Step 4: Rearrange for r²: r² = 1925 * (7/11) = 1925 * 7 / 11. Step 5: Compute the product: 1925 * 7 = 134, so we factor it carefully: 1925 / 11 = 175, so r² = 175 * 7 = 1225. Step 6: Hence r = √1225 = 35 centimetres. Step 7: Now compute the perimeter of the semicircle: P = πr + 2r = (22/7) * 35 + 2 * 35. Step 8: Evaluate the curved part: (22/7) * 35 = 22 * 5 = 110, and the diameter is 70, so total perimeter is 110 + 70 = 180 centimetres. Verification / Alternative check: We can quickly verify by plugging r = 35 back into the area formula. The area of the full circle would be πr² = (22/7) * 35². Here 35² = 1225, and (22/7) * 1225 = 22 * 175 = 3850. Since a semicircle is half of a full circle, its area is 3850 / 2 = 1925 square centimetres, which matches the given area. This confirms that r = 35 centimetres is correct. Once the radius is correct, the perimeter calculation πr + 2r is straightforward and leads to 180 centimetres, so the answer is consistent from both directions. Why Other Options Are Wrong: 160 centimetres is too small and would correspond to a smaller radius than the one obtained from the given area, which would not satisfy the area condition. 200 and 220 centimetres are too large, implying a larger radius and therefore a larger area than 1925 square centimetres. 240 centimetres is significantly higher and clearly incompatible with the given area when checked with the semicircle formulas. Only 180 centimetres maintains consistency between area and perimeter. Common Pitfalls: A common mistake is to forget that the given area is for a semicircle, not a full circle, leading to missing the factor of 1/2. Another error is miscomputing with the fraction 22/7 and mixing up multiplication or division, especially when rearranging to solve for r². Some learners mistakenly calculate the perimeter as just πr or as 2πr, forgetting to add the diameter for a semicircle. Being systematic with formulas and carefully checking each arithmetic step helps avoid these issues. Final Answer: The perimeter of the semicircle is 180 centimetres.

Question 15

For the straight line 3x + y = -6, at which point does it intersect the x-axis?

Accepted Answer

Introduction / Context: This question is about finding the x intercept of a straight line given in standard linear form. The x intercept is the point where the line crosses the x axis, and at this point the y coordinate is zero. Understanding x and y intercepts is a basic but very important concept in coordinate geometry, forming the foundation for graphing lines, solving linear equations, and interpreting linear relationships in many applied contexts such as economics, physics, and data analysis. Given Data / Assumptions: • The equation of the line is 3x + y = -6. • The x axis consists of all points where y = 0. • The x intercept is the point on the line whose y coordinate is zero. • We want the coordinates in the form (x, 0). Concept / Approach: To find the x intercept of a line, we substitute y = 0 into the equation and then solve for x. This works because all points on the x axis have y equal to zero. Once x is determined, we can form the ordered pair (x, 0), which gives the exact location where the line crosses the x axis. This method is straightforward for any linear equation in two variables. Step-by-Step Solution: Step 1: Start from the given line equation: 3x + y = -6. Step 2: At the x axis, the y coordinate equals zero, so set y = 0. Step 3: Substitute y = 0 into the equation to obtain 3x + 0 = -6. Step 4: Simplify this to 3x = -6. Step 5: Solve for x by dividing both sides by 3, giving x = -6 / 3 = -2. Step 6: Therefore, the x intercept is the point with coordinates (-2, 0). Verification / Alternative check: To verify, we can substitute x = -2 and y = 0 back into the original equation. The left hand side becomes 3 * (-2) + 0 = -6. The right hand side is also -6, so the equation is satisfied. This confirms that (-2, 0) lies on the line. Further, this point clearly has y = 0, which means it lies on the x axis as well. Since it lies on both the line and the x axis, it must be the x intercept. None of the other listed points satisfy both conditions simultaneously. Why Other Options Are Wrong: (2, 0) gives 3 * 2 + 0 = 6, which does not equal -6, so it is not on the line. (0, -2) has y = -2, so it is not on the x axis, and substituting gives 3 * 0 + (-2) = -2, not -6. (0, -6) is on the y axis, not the x axis, and represents the y intercept instead. (-3, 0) produces 3 * (-3) + 0 = -9, which does not match the required right hand side of -6. Thus all these options are invalid. Common Pitfalls: One common error is to confuse the x intercept with the y intercept and substitute x = 0 instead of y = 0. Another mistake is to miscalculate the division step, for example dividing -6 by 3 incorrectly. Students may also choose points that satisfy the equation but are not on the x axis, or vice versa. To avoid these issues, always remember that x intercept means y = 0, and double check the substitution by plugging the candidate point back into the original equation. Final Answer: The line 3x + y = -6 intersects the x axis at the point (-2, 0).

Question 16

A metallic cylindrical pipe of uniform thickness has volume 748 cubic centimetres. Its length is 14 cm and its external radius is 9 cm. What is the thickness of the pipe?

Accepted Answer

Introduction / Context: This question involves the volume of a hollow cylinder, which represents a cylindrical pipe made of metal. The pipe has an external radius, an internal radius, and a given length. The metal occupies the region between the outer and inner cylindrical surfaces. By using the formula for the volume of a hollow cylinder, we can determine the inner radius and therefore the thickness of the pipe. Such problems are useful for understanding how material quantities relate to dimensions in engineering and manufacturing contexts. Given Data / Assumptions: • Volume of the metal in the pipe is 748 cubic centimetres. • Length (height) of the pipe, h, is 14 cm. • External radius, R, is 9 cm. • Internal radius, r, is unknown and must be found. • Thickness of the pipe is R - r in centimetres. • We assume π is taken as 22/7 or left as π; the result is exact either way because the numbers are chosen to simplify. Concept / Approach: The volume of a hollow cylinder is given by V = π (R² - r²) h, where R is the outer radius, r is the inner radius, and h is the height or length. We know V, R, and h, so we can first solve for (R² - r²). Once we determine r², we can take the square root to obtain r. Finally, the thickness t of the pipe is simply the difference between the external and internal radii, t = R - r. The calculations are straightforward because the values have been selected to give a perfect square for r². Step-by-Step Solution: Step 1: Write the hollow cylinder volume formula: V = π (R² - r²) h. Step 2: Substitute the known values: 748 = π (9² - r²) * 14. Step 3: Compute 9² = 81, so we have 748 = 14π (81 - r²). Step 4: Divide both sides by 14π to isolate (81 - r²): (81 - r²) = 748 / (14π). Step 5: If we take π = 22/7, then 14π = 14 * 22/7 = 44, so 748 / 44 = 17. Step 6: Therefore, 81 - r² = 17, which gives r² = 81 - 17 = 64. Step 7: Hence r = √64 = 8 cm. The thickness t is R - r = 9 - 8 = 1 cm. Verification / Alternative check: We can verify by recomputing the volume using the found inner radius. With R = 9 cm, r = 8 cm, and h = 14 cm, the volume is V = π (R² - r²) h = π (81 - 64) * 14 = π * 17 * 14. Using π = 22/7, we get V = 22/7 * 17 * 14. First 14 / 7 = 2, so this is 22 * 17 * 2 = 44 * 17 = 748 cubic centimetres, which matches the given volume. This confirms that the computed inner radius and the resulting thickness of 1 cm are correct. Why Other Options Are Wrong: 0.5 cm and 1.5 cm would give inner radii of 8.5 cm or 7.5 cm respectively, which would produce different volume values when placed in the formula and would not equal 748 cubic centimetres. 2 cm and 2.5 cm correspond to inner radii of 7 cm and 6.5 cm, which again would lead to a much larger metal volume than specified. Substituting these into V = π (R² - r²) h quickly shows mismatches with the given volume. Common Pitfalls: A common mistake is to treat 748 cubic centimetres as the volume of a solid cylinder rather than a hollow one, which leads to ignoring the inner cavity and using the formula V = πR²h directly. Another error is to forget that R² - r² must be positive and mis-handle subtraction. Some learners also misapply π approximations, especially when dividing by 14π. Careful substitution, stepwise algebra, and a final verification by recomputing the volume help avoid these errors. Final Answer: The thickness of the metallic cylindrical pipe is 1 cm.

Question 17

The volume of a right circular cylinder is 3850 cubic centimetres and its height is 25 cm. Taking π = 22/7, what is the radius of its base (in cm)?

Accepted Answer

Introduction / Context: This question focuses on using the volume formula for a right circular cylinder. We are given the volume and the height and asked to find the radius of the base. This type of problem is very common in quantitative aptitude tests and helps reinforce how to rearrange formulas and substitute numerical values correctly, especially when using a specific approximation for π such as 22/7. Given Data / Assumptions: • Volume of the cylinder, V, is 3850 cubic centimetres. • Height (or length) of the cylinder, h, is 25 cm. • π is to be taken as 22/7. • Radius of the base, r, is unknown and needs to be found. Concept / Approach: The volume of a right circular cylinder is given by V = πr²h. Here V and h are known, and π is specified, so we can rearrange the formula to r² = V / (πh). After computing r², we take the square root to find r. The data are chosen so that r² becomes a neat perfect square, making the result an integer. Careful handling of fractions is important to get the correct value for r. Step-by-Step Solution: Step 1: Start with the cylinder volume formula: V = πr²h. Step 2: Substitute the given values V = 3850, h = 25, and π = 22/7. Step 3: We get 3850 = (22/7) * r² * 25. Step 4: Combine constants: (22/7) * 25 = 550 / 7, so 3850 = (550 / 7) * r². Step 5: Rearrange to solve for r²: r² = 3850 * (7 / 550). Step 6: Simplify the fraction: 3850 / 550 = 7, so r² = 7 * 7 = 49. Step 7: Take the square root: r = √49 = 7 cm. Verification / Alternative check: We can verify by substituting r = 7 cm back into the volume formula. Then V = πr²h = (22/7) * 7² * 25. Here 7² = 49, so V = (22/7) * 49 * 25. First, 49 / 7 = 7, giving V = 22 * 7 * 25. Then 22 * 7 = 154, and 154 * 25 = 3850. This matches the given volume exactly, confirming that the radius is 7 cm. Any other assumed radius would change r² and lead to a different volume, so the answer is unique and correct. Why Other Options Are Wrong: 5 cm and 6 cm produce r² values of 25 and 36 respectively, which lead to volumes smaller than 3850 cubic centimetres when substituted into V = πr²h. 8 cm and 9 cm generate r² values of 64 and 81, which lead to volumes greater than 3850 cubic centimetres. None of these match the required volume, so they are incorrect. Common Pitfalls: Typical mistakes include forgetting to multiply all constants correctly when combining π and the height, or incorrectly simplifying the fraction 3850 * 7 / 550. Some learners also forget to take the square root after obtaining r², or miscalculate that square root. Writing the working clearly and simplifying step by step makes it easier to avoid such errors. Final Answer: The radius of the cylinder is 7 cm.

Question 18

A regular dodecahedron has 30 edges. How many vertices does this solid have?

Accepted Answer

Introduction / Context: This question is about a regular dodecahedron, which is one of the five Platonic solids. It is a three dimensional solid with congruent regular pentagonal faces. The problem gives the number of edges and asks for the number of vertices. To answer this, we use Euler's formula for convex polyhedra, which relates the number of faces, vertices, and edges. This is an important concept in solid geometry and topology, and it appears frequently in aptitude and competitive exams. Given Data / Assumptions: • The solid is a regular dodecahedron. • It is known that a regular dodecahedron has 12 faces, each a regular pentagon. • The number of edges, E, is given as 30. • We are asked to find the number of vertices, V. • The solid is convex, so Euler's polyhedron formula applies: V - E + F = 2. Concept / Approach: Euler's formula for any convex polyhedron states that V - E + F = 2, where V is the number of vertices, E is the number of edges, and F is the number of faces. For a dodecahedron, F is known to be 12, and E is given as 30. Substituting these values into Euler's formula allows us to solve for V directly. This approach avoids having to memorize all properties of a dodecahedron and instead relies on a general relationship that works for many solids. Step-by-Step Solution: Step 1: Write Euler's formula for a convex polyhedron: V - E + F = 2. Step 2: For a regular dodecahedron, the number of faces F is 12. Step 3: The number of edges E is given as 30. Step 4: Substitute E = 30 and F = 12 into Euler's formula: V - 30 + 12 = 2. Step 5: Simplify: V - 18 = 2. Step 6: Solve for V: V = 2 + 18 = 20. Verification / Alternative check: We can also verify this result using the relationship among faces, edges, and vertices of regular polygons forming the solid. Each pentagonal face has 5 edges, and each edge is shared by 2 faces. Thus, the total edge count satisfies 5F = 2E, so 5 * 12 = 2E, giving E = 30, which is consistent. Also, each vertex of a regular dodecahedron is where 3 pentagons meet. Counting incidences of vertices across faces gives 5F = 3V, so 60 = 3V, giving V = 20. This matches the value from Euler's formula and confirms the answer. Why Other Options Are Wrong: 10 and 12 are too small and would not satisfy Euler's formula with E = 30 and F = 12; substituting these numbers leads to V - E + F values far from 2. 22 and 24 are too large; they also break the relation 5F = 3V for regular pentagonal faces meeting three at a vertex. None of these satisfy both Euler's formula and the face structure of a dodecahedron. Common Pitfalls: Some learners confuse the dodecahedron with other Platonic solids such as the icosahedron or cube and may guess vertex counts based on memory rather than using Euler's formula. Others misapply Euler's formula or misremember the number of faces of a dodecahedron. To avoid such errors, it is safer to recall that a dodecahedron has 12 pentagonal faces and then rely on V - E + F = 2 and 5F = 2E or 5F = 3V for regular structure to verify counts. Final Answer: A regular dodecahedron with 30 edges has 20 vertices.

Question 19

A rectangular park measuring 60 m by 40 m has two concrete roads of equal width running through the middle, one along its length and one along its breadth. The remaining area is a lawn of 2109 square metres. What is the width of each road (in metres…

Accepted Answer

Introduction / Context: This question involves area calculation in a rectangular park with two perpendicular roads crossing through the centre. The roads have the same width and cover some part of the park, while the remaining region is grass lawn. By expressing the road area in terms of the unknown width and subtracting it from the total area, we can find the width. Such problems are common in aptitude exams and help develop algebraic modelling skills using geometric shapes. Given Data / Assumptions: • The park is a rectangle with length 60 metres and breadth 40 metres. • Two concrete roads of equal width run through the middle, one along the length and one along the breadth. • The remaining grass lawn area is 2109 square metres. • Let the width of each road be x metres. • The roads cross each other, so their overlapping area must not be counted twice. Concept / Approach: The total area of the park is length * breadth. The area covered by the roads is the sum of the two rectangular strips minus the overlapping square at the centre, because that intersection gets counted twice if we simply add the areas. Each road has area equal to width times the full dimension it crosses. So one road has area 60x and the other has area 40x. Their overlap is a square of side x with area x². The lawn area is the total area of the park minus the road area. Setting this equal to the given lawn area gives a quadratic equation in x, which we can solve. Step-by-Step Solution: Step 1: Compute the total area of the park: 60 * 40 = 2400 square metres. Step 2: Let x be the width of each road in metres. Step 3: Area of the road along the length is 60x square metres. Step 4: Area of the road along the breadth is 40x square metres. Step 5: The overlapping central rectangle (actually a square) has area x² square metres. Step 6: So total road area is 60x + 40x - x² = 100x - x². Step 7: Lawn area = total area - road area = 2400 - (100x - x²) = 2400 - 100x + x². Step 8: Set this equal to the given lawn area: 2400 - 100x + x² = 2109. Step 9: Rearrange: x² - 100x + (2400 - 2109) = 0, so x² - 100x + 291 = 0. Step 10: Solve the quadratic: x² - 100x + 291 = 0 has roots x = 3 and x = 97. Step 11: Since the width cannot exceed the smaller side of 40 m, x = 3 m is the only feasible value. Verification / Alternative check: Using x = 3 m, the road along the length has area 60 * 3 = 180 square metres, and the road along the breadth has area 40 * 3 = 120 square metres. The overlap is a 3 m by 3 m square with area 9 square metres. So total road area is 180 + 120 - 9 = 291 square metres. The lawn area is then 2400 - 291 = 2109 square metres, which matches the given value. This confirms that a width of 3 metres is correct. The other root 97 m is impossible because it is greater than the width of the entire park. Why Other Options Are Wrong: 2.5 m and 2.91 m both lead to lawn areas different from 2109 square metres when substituted into the area expressions. 3.5 m and 5.82 m also produce incorrect lawn areas and may exceed realistic proportions relative to the park size. Only 3 m satisfies the quadratic equation and keeps the geometry practical. Common Pitfalls: A common mistake is forgetting to subtract the overlapping area x², which causes double counting and leads to an incorrect equation. Another pitfall is miscomputing the quadratic roots or not checking which root is physically meaningful. Learners may also confuse which sides correspond to 60 and 40 metres when calculating road areas. Drawing a diagram and writing the expression for road area carefully helps avoid these errors. Final Answer: The width of each road is 3 m.

Question 20

The length of a rectangular plot is 20 metres more than its breadth. The cost of fencing the plot at Rs 26.50 per metre is Rs 5300. What is the length of the plot (in metres)?

Accepted Answer

Introduction / Context: This question combines basic algebra with the concept of perimeter of a rectangle and cost calculation. The fencing cost is given per metre, and the total cost is known. Using this information, we can find the total length of fencing, which is the perimeter. Then, with the relationship between length and breadth, we can determine the actual dimensions of the rectangular plot. This type of problem is common in aptitude tests and practical planning scenarios such as estimating construction or fencing costs. Given Data / Assumptions: • Let the breadth of the rectangular plot be b metres. • Then the length of the plot is b + 20 metres, since it is 20 m more than the breadth. • Fencing is done around the entire boundary, so the total fenced length equals the perimeter of the rectangle. • Cost of fencing is Rs 26.50 per metre. • Total fencing cost is Rs 5300. Concept / Approach: The perimeter P of a rectangle with length L and breadth B is given by P = 2(L + B). The total cost is cost per metre multiplied by the perimeter. Therefore, we can write 26.50 * P = 5300. Solving for P gives the total perimeter. Next, using the relationship between length and breadth, we can set up an equation and solve for b. From b we then find the length L = b + 20. This approach is systematic and uses only basic algebra. Step-by-Step Solution: Step 1: Let breadth = b metres and length = b + 20 metres. Step 2: Perimeter P of the rectangle is P = 2(L + B) = 2[(b + 20) + b] = 2(2b + 20) = 4b + 40. Step 3: Cost of fencing is Rs 26.50 per metre, so total cost is 26.50 * P. Step 4: Given total cost equals Rs 5300, we have 26.50 * (4b + 40) = 5300. Step 5: Divide both sides by 26.50 to find the perimeter: 4b + 40 = 5300 / 26.50 = 200. Step 6: Solve 4b + 40 = 200 to get 4b = 160, so b = 40 metres. Step 7: The length is then L = b + 20 = 40 + 20 = 60 metres. Verification / Alternative check: With breadth 40 metres and length 60 metres, the perimeter is P = 2(60 + 40) = 2 * 100 = 200 metres. At a rate of Rs 26.50 per metre, the total cost is 26.50 * 200 = 5300 rupees, which matches the given cost. This confirms that the dimensions are consistent with the information provided. Any other combination of length and breadth related by a 20 metre difference would yield a different perimeter and therefore a different cost. Why Other Options Are Wrong: 40 metres would mean the length equals the breadth, violating the condition that length is 20 metres more than breadth. 50, 80, and 120 metres do not satisfy the equation 4b + 40 = 200 when combined with an appropriate breadth and also do not produce the fencing cost of Rs 5300 at Rs 26.50 per metre. Only 60 metres leads to a perimeter of 200 metres and the correct total cost. Common Pitfalls: Learners sometimes confuse perimeter with area and try to use the cost to find area instead of the boundary length. Another common mistake is to mis-handle the relationship between length and breadth, for example setting length = 20b instead of b + 20. Arithmetic errors can also occur when dividing 5300 by 26.50. Writing down each relation clearly and performing the division carefully ensures the correct answer. Final Answer: The length of the rectangular plot is 60 metres.

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