Question 1

An error of 2% in excess is made while measuring the side of a square. What is the percentage error in the calculated area of the square?

Accepted Answer

Introduction / Context: This question examines how a small percentage error in measuring a linear dimension such as side length affects the calculated area of a square. Since area depends on the square of the side, errors do not carry over linearly. Understanding this helps in dealing with measurement errors, approximations, and sensitivity of computed quantities to errors in input data, which is important in practical applications. Given Data / Assumptions: True side length of the square is taken as s. The side is measured 2% in excess, so the measured side becomes 102% of the true side. Thus, measured side = 1.02 * s. True area = s^2, while calculated area uses the measured side length. We are asked for the percentage error in the calculated area compared to the true area. Concept / Approach: Area of a square is proportional to the square of its side. When the side is multiplied by a factor, the area is multiplied by the square of that factor. Here the linear factor is 1.02, so the area factor is (1.02)^2. The percentage error in area is then computed by comparing this factor with 1, the factor corresponding to the true area. The difference, multiplied by 100, gives the percentage error in area. Step-by-Step Solution: Let the true side be s, so true area A = s^2.Measured side s' = 1.02 * s (2% in excess).Calculated area A' = (s')^2 = (1.02 * s)^2 = 1.02^2 * s^2.Compute 1.02^2 = 1.0404, so A' = 1.0404 * s^2.Percentage error in area = (1.0404 − 1) * 100% = 0.0404 * 100% = 4.04%. Verification / Alternative check: Take a numerical example. Suppose the true side is 10 units. The true area is 10^2 = 100 square units. If we measure the side as 2% larger, measured side = 10.2 units. The calculated area becomes 10.2^2 = 104.04 square units. The increase in area is 4.04 square units, which is 4.04% of 100, matching the computed percentage error. This confirms that the theoretical calculation is correct. Why Other Options Are Wrong: Option 5.04% would result from mistakenly adding an extra 1% or miscomputing 1.02^2. Option 3.96% does not correspond to any squared factor near 1.02 and is just an incorrect approximation. Option 6.04% is far too large and could come from adding 2% and 4% without justification. Option 2.02% incorrectly assumes that area error equals side error divided by 2 or some other unjustified manipulation. Common Pitfalls: A frequent mistake is to assume that a 2% error in side measurement will lead to the same 2% error in area, ignoring the squared relationship between side and area. Some also incorrectly double the percentage and report 4% without including the extra 0.04 part from squaring 1.02. Others may approximate too roughly or miss the step of converting the decimal difference into a percentage. Final Answer: The percentage error in the calculated area of the square is 4.04%.

Question 2

Find the area of a rectangle with length L = 5 inches and width W = 3 inches. Express your answer in square inches.

Accepted Answer

Introduction / Context: This is a straightforward area of a rectangle problem designed to test direct application of the basic area formula. Such questions are common warm up problems in aptitude tests and confirm that the student remembers the fundamental relation between length, width, and area for rectangular shapes. Given Data / Assumptions: Length L of the rectangle is 5 inches. Width W of the rectangle is 3 inches. The figure is a perfect rectangle with right angles. We need the area in square inches. No scaling or additional shapes are involved. Concept / Approach: The area A of a rectangle is calculated by multiplying its length and width: A = L * W. This represents the number of unit squares that can fit into the rectangle. Since both dimensions are given directly and there are no complications like percentage changes or missing sides, we only need to substitute and multiply the numbers correctly. Step-by-Step Solution: Write the area formula: A = L * W.Substitute L = 5 inches and W = 3 inches.Compute A = 5 * 3.A = 15 square inches.Therefore, the area of the rectangle is 15 square inches. Verification / Alternative check: Imagine the rectangle as a grid of 5 units along its length and 3 units along its width. Counting the number of 1 inch by 1 inch squares gives 5 rows of 3 squares each, which is 15 squares in total. This visual interpretation matches the numerical answer, confirming that the area equals 15 square inches. Why Other Options Are Wrong: Option 13 and 14 come from incorrect multiplication or careless addition such as 5 + 3 = 8 followed by some adjustment. Option 16 could appear if someone mistakenly uses (L + W) or mixes formulas. Option 18 might come from 6 * 3 or another incorrect combination, but it does not correspond to the given dimensions. None of these wrong values match the correct product 5 * 3. Common Pitfalls: Even in simple questions, students sometimes add the dimensions rather than multiply them, confusing perimeter with area. Others may accidentally reverse digits or misread the numbers. These errors highlight the importance of carefully reading the problem and recalling which formula applies to which geometric quantity. Final Answer: The area of the rectangle is 15 square inches.

Question 3

A circular pizza has a diameter of 14 inches. A slice is cut such that the slice corresponds to a central angle of 45°. What is the area of each slice of pizza (in square inches)?

Accepted Answer

Introduction / Context: This question involves finding the area of a sector of a circle, representing a slice of pizza. The central angle of the slice and the diameter of the pizza are given. This type of problem is common in aptitude and quantitative exams to test understanding of circular geometry, especially the relation between the area of a full circle and the area of a sector based on its central angle. Given Data / Assumptions: Diameter of the pizza = 14 inches, so radius r = 14 / 2 = 7 inches. Central angle of each slice = 45°. The pizza is a perfect circle. We may use π ≈ 3.14 for numerical calculation. We are required to calculate the area of one slice, which is the area of a 45° sector. Concept / Approach: The area of a circle is A = π * r^2. A sector with central angle θ degrees has area equal to (θ / 360) times the area of the full circle. Therefore, if we know the radius and the central angle, we can first compute the total area of the pizza and then multiply by the fraction θ / 360. This method directly provides the area of a single slice. Step-by-Step Solution: Radius r = 7 inches.Area of full pizza = π * r^2 ≈ 3.14 * 7^2 = 3.14 * 49.Compute 3.14 * 49 ≈ 153.86 square inches.Each slice corresponds to a central angle of 45°, so fraction of circle = 45 / 360 = 1 / 8.Area of one slice = (1 / 8) * 153.86 ≈ 19.23, which rounds to about 19.25 square inches. Verification / Alternative check: We can divide the circle area by 8 directly because 360° / 45° = 8 slices. If the total area is 49π, then area of each slice is (49π / 8). Taking π as 3.14, we get (49 * 3.14) / 8 ≈ 153.86 / 8 ≈ 19.23. Both calculation paths agree closely, and rounding to two decimal places gives approximately 19.25 square inches, which matches the chosen option. Why Other Options Are Wrong: Option 16.25 is significantly smaller and would correspond to a central angle less than 45°. Option 18.25 is slightly low and likely results from using π ≈ 3 instead of 3.14. Option 20.25 is higher than the correct sector area and overestimates the fraction. Option 17.25 is again too small and does not match any reasonable combination of radius and angle for this pizza. Common Pitfalls: Students often forget to convert the fraction of the circle by using θ / 360, instead using 45 / 180 or another incorrect ratio. Others may forget to square the radius or miscalculate 7^2. Some also use an overly rough approximation for π, which can shift the answer away from the correct option. Ensuring careful multiplication and division is essential for accuracy. Final Answer: The area of each slice of the pizza is approximately 19.25 square inches.

Question 4

A triangle has side lengths 9 cm, 12 cm, and 15 cm. Find the radius of its circumcircle (the circle passing through all three vertices).

Accepted Answer

Introduction / Context: This question asks for the circumradius of a triangle whose side lengths are known. In particular, the triangle with sides 9 cm, 12 cm, and 15 cm forms a right triangle because it is a multiple of the 3, 4, 5 Pythagorean triplet. Right triangles have a simple relation between the hypotenuse and the circumradius, making this question a useful test of both recognition of special triangles and knowledge of circle geometry. Given Data / Assumptions: Side lengths of the triangle are 9 cm, 12 cm, and 15 cm. The triangle is non degenerate and lies in a plane. We are asked to find the circumradius R of the triangle. For a right triangle, the hypotenuse is the diameter of the circumcircle. All units are in centimetres. Concept / Approach: First, we check whether the triangle is right angled by using the Pythagoras theorem. If the square of the longest side equals the sum of the squares of the other two sides, then the triangle is right angled, and the longest side is the hypotenuse. For a right triangle, the circumcircle has diameter equal to the hypotenuse, so the circumradius R is hypotenuse / 2. Once this is confirmed, we simply divide the hypotenuse length by 2 to obtain R. Step-by-Step Solution: Identify the longest side: 15 cm.Check Pythagoras: 9^2 + 12^2 = 81 + 144 = 225.Compute 15^2 = 225, so 9^2 + 12^2 = 15^2, confirming a right triangle with hypotenuse 15 cm.For a right triangle, circumdiameter = hypotenuse, so diameter D = 15 cm.Circumradius R = D / 2 = 15 / 2 = 7.5 cm. Verification / Alternative check: Another way to compute the circumradius of any triangle is to use the formula R = (a * b * c) / (4 * Δ), where a, b, and c are side lengths and Δ is the area. For a right triangle with legs 9 and 12, area Δ = (1 / 2) * 9 * 12 = 54. Substituting into the formula gives R = (9 * 12 * 15) / (4 * 54) = 1620 / 216 = 7.5 cm. This matches the simpler hypotenuse based method, confirming the result. Why Other Options Are Wrong: Option 6 cm would give a diameter of 12 cm, which is shorter than the hypotenuse, impossible for a right triangle circumcircle. Option 4.5 cm is even smaller and does not satisfy the formula for R. Option 8 cm gives diameter 16 cm, which does not match any relation with the triangle sides. Option 9.5 cm is arbitrarily larger than needed and fails when substituted into standard circumradius formulas. Common Pitfalls: Some students fail to recognize the 9, 12, 15 triangle as a scaled 3, 4, 5 triangle and attempt more complicated methods. Others misidentify which side is the hypotenuse or apply area formulas incorrectly. Forgetting the special property that the hypotenuse is the diameter of the circumcircle for right triangles is another common source of mistakes. Final Answer: The circumradius of the triangle is 7.5 cm.

Question 5

From a square plate with each side 7 cm, small squares of area 0.25 sq. cm are cut out at each corner and the remaining plate is folded up along the cuts to form an open-top cuboid. What is the volume of this cuboid (in cubic centimetres)?

Accepted Answer

Introduction / Context: This problem describes forming an open-top box (cuboid) by cutting equal squares from the corners of a square sheet and folding up the sides. This is a standard geometry and mensuration question that tests spatial visualization and understanding of how two dimensional shapes can form three dimensional solids. The key quantities are the new base dimensions and the height of the cuboid. Given Data / Assumptions: Original plate is a square with side length 7 cm. At each of the four corners, a small square of area 0.25 sq. cm is cut out. Area of each small square = 0.25 sq. cm, so side length of each small square is √0.25 = 0.5 cm. The plate is then folded along the cuts to create an open-top cuboid. The height of the cuboid is equal to the side of the small squares, that is 0.5 cm. Concept / Approach: When squares of side h are cut from each corner of a square with side S, the resulting base of the box has dimensions (S − 2h) by (S − 2h), and the height is h. The volume of the open-top cuboid is then volume = length * breadth * height. In this problem, S = 7 cm and h = 0.5 cm. We compute the new base side as 7 − 2 * 0.5 and then multiply by the height to get the volume. Step-by-Step Solution: Side length of each small square: h = √0.25 = 0.5 cm.Original square side: S = 7 cm.New base side after cutting: S − 2h = 7 − 2 * 0.5 = 7 − 1 = 6 cm.Base area of the box: 6 * 6 = 36 sq. cm.Volume of cuboid: base area * height = 36 * 0.5 = 18 cubic centimetres. Verification / Alternative check: To verify, note that the height cannot exceed the amount cut from each side, so 0.5 cm is consistent. Also, the base must be smaller than the original square, and 6 cm per side is 1 cm total reduction from 7 cm, consistent with cutting 0.5 cm from each end. Multiplying 36 by 0.5 again yields 18, so no arithmetic steps have been missed. This confirms the correctness of the computed volume. Why Other Options Are Wrong: Option 21 would require a base area or height combination that does not match the geometry of the folds. Option 16 suggests a base or height product of 4 * 4 * 1 or similar but does not match the given dimensions. Option 20 would require a slightly larger height or base, inconsistent with 0.25 sq. cm corner squares. Option 24 would correspond to a base area of 48 with height 0.5 or a base of 6 * 8, none of which arise from the given construction. Common Pitfalls: Students sometimes incorrectly subtract only a single h instead of 2h when finding the new base dimension, forgetting that cuts are made at both ends. Others misinterpret the given area of the small squares and assume their side is 0.25 cm instead of √0.25 cm. Miscalculating the volume by adding dimensions instead of multiplying is another typical error. Final Answer: The volume of the open-top cuboid formed is 18 cubic centimetres.

Question 6

A quadrilateral is inscribed in a circle. Its opposite angles are equal, and the lengths of two adjacent sides are 6 cm and 8 cm. What is the area of the circle (in terms of π)?

Accepted Answer

Introduction / Context: This problem involves a quadrilateral inscribed in a circle, also called a cyclic quadrilateral. The condition that opposite angles are equal has a strong implication for the type of quadrilateral, and knowing this type allows us to infer more about the geometry and the circle. The question asks for the area of the circumscribed circle using the lengths of two adjacent sides of the quadrilateral. Given Data / Assumptions: A quadrilateral is inscribed in a circle, so all its vertices lie on the circle. Opposite angles of the quadrilateral are equal. Two adjacent sides of the quadrilateral measure 6 cm and 8 cm. We are asked to find the area of the circle in terms of π. All lengths are in centimetres. Concept / Approach: In any cyclic quadrilateral, the sum of a pair of opposite angles is 180°. If opposite angles are also equal, then each must be 90°, making the quadrilateral a rectangle. In a rectangle with sides 6 cm and 8 cm, the diagonal can be found using Pythagoras theorem. That diagonal is also the diameter of the circle containing the rectangle. Once the diameter is known, the radius is half the diameter, and the area of the circle is π * r^2. Step-by-Step Solution: Opposite angles equal and supplementary (sum 180°) imply each opposite angle is 90°, so the quadrilateral is a rectangle.Let the rectangle sides be 6 cm and 8 cm.Diagonal of the rectangle (and diameter of the circle) is d = √(6^2 + 8^2) = √(36 + 64) = √100 = 10 cm.Radius r of the circle is r = d / 2 = 10 / 2 = 5 cm.Area of the circle = π * r^2 = π * 5^2 = 25π square centimetres. Verification / Alternative check: If we draw a rectangle of sides 6 and 8, placing it inside a circle so that its corners touch the circle, the diagonal clearly spans the circle. This is equivalent to the well known 6, 8, 10 right triangle forming half of the rectangle. Since the diagonal length is 10, the radius of the circle must be 5. There is no alternative circle with a different radius that can pass through all four vertices of this rectangle, confirming the uniqueness and correctness of the area 25π. Why Other Options Are Wrong: Option 64π would correspond to a radius of 8, which would make the diagonal 16, not matching the rectangle dimensions. Option 36π corresponds to a radius of 6, giving a diameter of 12 instead of 10. Option 49π is based on a radius of 7, again not compatible with a 6 by 8 rectangle. Option 81π would require a radius of 9, which is far too large for the given side lengths. Common Pitfalls: Some students may not recognize that equal opposite angles in a cyclic quadrilateral force each to be 90°, leading to a rectangle. Others may mistakenly think any quadrilateral with equal opposite angles is a general parallelogram and forget the extra condition of being cyclic. Miscomputing the diagonal or confusing diameter with radius are further sources of errors. Final Answer: The area of the circle is 25π sq cm.

Question 7

In triangle ABC, points P and Q lie on sides AB and AC respectively such that segment PQ is parallel to side BC. If AP : PB = 2 : 3 and the area of triangle APQ is 8 sq. cm, what is the area of trapezium PQCB?

Accepted Answer

Introduction / Context: This problem involves similar triangles and area ratios within a triangle when a line segment is drawn parallel to one of its sides. The line PQ is parallel to BC, so triangle APQ is similar to triangle ABC. Using the ratio of corresponding sides, we can relate the areas of the smaller and the larger triangle and then compute the area of the remaining trapezium PQCB. Such questions check understanding of similarity, proportionality, and area relationships. Given Data / Assumptions: Triangle ABC is any triangle. P lies on AB and Q lies on AC such that PQ ∥ BC. AP : PB = 2 : 3, so AB is divided internally in that ratio. The area of triangle APQ is 8 sq. cm. We are asked to find the area of trapezium PQCB, the region between PQ and BC. Concept / Approach: Since PQ is parallel to BC, triangle APQ is similar to triangle ABC by angle angle similarity. The ratio of similarity between corresponding sides is AP / AB. Given AP : PB = 2 : 3, the total AB is 2 + 3 = 5 parts, so AP / AB = 2 / 5. The ratio of areas of similar triangles is the square of the ratio of corresponding sides. Once we find the area of triangle ABC, we subtract the area of triangle APQ to obtain the area of trapezium PQCB. Step-by-Step Solution: AP : PB = 2 : 3 implies AB is 5 equal parts.So AP / AB = 2 / 5.Triangles APQ and ABC are similar, so area(APQ) / area(ABC) = (AP / AB)^2 = (2 / 5)^2 = 4 / 25.Given area(APQ) = 8 sq. cm, we get 8 / area(ABC) = 4 / 25, so area(ABC) = 8 * 25 / 4 = 50 sq. cm.Area of trapezium PQCB = area(ABC) − area(APQ) = 50 − 8 = 42 sq. cm. Verification / Alternative check: We can think of triangle ABC as being cut by a line parallel to BC at some fraction down from A. The scaling factor for lengths from A down to PQ is 2 / 5, and the scaling factor for lengths from PQ down to BC is 3 / 5. The area of the lower part, trapezium PQCB, must then be the remaining fraction of the total area, which is 1 − 4 / 25 = 21 / 25 of area(ABC). Since area(ABC) = 50, this gives 21 / 25 * 50 = 42, matching the earlier calculation. Why Other Options Are Wrong: Option 50 sq. cm is the total area of triangle ABC, not the area of the trapezium. Option 18 sq. cm is too small and might come from subtracting incorrectly or misusing ratios. Option 14 sq. cm does not reflect the correct fraction of the total area. Option 32 sq. cm is another miscalculated residual area that does not follow from the correct area proportionality for similar triangles. Common Pitfalls: Many students forget to square the side ratio when comparing areas and instead use 2 / 5 directly. Others misinterpret AP : PB as AP : AB or confuse which parts of the triangle correspond. Some also mistakenly treat trapezium PQCB as a standalone shape without using similarity, which makes calculations much more complex and error prone. Final Answer: The area of trapezium PQCB is 42 sq cm.

Question 8

A solid copper sphere of radius 9 cm is melted and recast into a long wire of radius 1 mm. What is the length of the wire, in metres? (Take π = 22/7)

Accepted Answer

Introduction / Context: This question applies the principle of volume conservation when a solid is melted and recast into another shape. The copper sphere and the resulting copper wire have the same total volume. To find the length of the wire, we equate the volume of the original sphere to the volume of the cylindrical wire and solve for the unknown length. This is a standard type of mensuration problem in competitive exams. Given Data / Assumptions: Radius of the sphere r_s = 9 cm. Radius of the wire r_w = 1 mm = 0.1 cm. Volume of the sphere = 4/3 * π * r_s^3. Volume of the wire (cylinder) = π * r_w^2 * L, where L is the length in cm. π is taken as 22/7 and copper volume is conserved during melting and recasting. Concept / Approach: Since the material is the same and nothing is lost, the volume of the original sphere equals the volume of the resulting cylindrical wire. We first write an expression for the sphere volume, then an expression for the cylinder volume. Setting these equal allows us to solve for L, the length of the wire in centimetres. Finally, we convert the length from centimetres to metres by dividing by 100. Step-by-Step Solution: Volume of sphere = (4 / 3) * π * r_s^3 = (4 / 3) * π * 9^3.Compute 9^3 = 729, so volume of sphere = (4 / 3) * π * 729 = 972 * π cubic centimetres.Volume of wire (cylinder) = π * r_w^2 * L = π * (0.1)^2 * L = π * 0.01 * L.Equate volumes: 972 * π = π * 0.01 * L, so 972 = 0.01 * L.Hence, L = 972 / 0.01 = 97200 cm, which is 97200 / 100 = 972 metres. Verification / Alternative check: To cross check, observe that the radius ratio between sphere and wire is 9 cm to 0.1 cm, which is 90 to 1. Since volume scales with the square of radius times length for a cylinder, the length must be large. A quick sense check confirms that 972 metres is a reasonable value for such a thin wire recast from a sphere of radius 9 cm. Any value much smaller would indicate that the radius conversion or unit conversion was mishandled. Why Other Options Are Wrong: Option 1166.4 metres and 1458 metres come from incorrect algebra or misapplied formulas, possibly mixing radius and diameter or using inconsistent units. Option 777.6 metres underestimates the required length and does not satisfy the volume equality when substituted back. Option 900 metres is a rounded guess without basis in the exact calculation and does not keep the volumes equal. Common Pitfalls: The most common mistakes are forgetting to convert 1 mm into 0.1 cm, using the diameter instead of the radius in cylinder volume, or failing to square the radius. Some students also forget to convert from centimetres to metres at the end or incorrectly cancel π before setting up the main equation, leading to algebra errors. Final Answer: The length of the wire formed by melting the sphere is 972 metres.

Question 9

The cost of levelling a triangular piece of land with sides 72 m, 30 m, and 78 m is calculated at the rate of 20 paise per square metre. What is the total cost in rupees?

Accepted Answer

Introduction / Context: This question combines triangle area calculation with a unit rate cost problem. The land is triangular, and the sides are given, so we need to use a suitable formula to find the area. Once the area is known, multiplying by the cost per square metre gives the total cost. This type of problem appears in aptitude exams to test both mensuration and basic arithmetic with money units. Given Data / Assumptions: Sides of the triangle are 72 m, 30 m, and 78 m. The cost of levelling is 20 paise per square metre, which is ₹ 0.20 per sq. m. The land is a valid triangle, so the side lengths satisfy the triangle inequality. We must find the area of the triangle and then multiply by the rate. Final answer is required in rupees. Concept / Approach: For a triangle with all three sides known, Heron’s formula is the most direct method to find the area. According to Heron’s formula, if a, b, and c are the side lengths and s is the semi perimeter, then area = √(s(s − a)(s − b)(s − c)). After computing the area in square metres, we multiply by ₹ 0.20 per square metre to obtain the total levelling cost. Step-by-Step Solution: Let a = 72 m, b = 30 m, and c = 78 m.Semi perimeter s = (a + b + c) / 2 = (72 + 30 + 78) / 2 = 180 / 2 = 90 m.Heron’s formula: area = √(s(s − a)(s − b)(s − c)) = √(90 * (90 − 72) * (90 − 30) * (90 − 78)).Compute inside: 90 − 72 = 18, 90 − 30 = 60, 90 − 78 = 12, so area = √(90 * 18 * 60 * 12).90 * 18 * 60 * 12 = 1080^2, so area = 1080 sq. m. Cost = 1080 * ₹ 0.20 = ₹ 216. Verification / Alternative check: We can check the multiplication: 1080 * 0.20 = 1080 * (20 / 100) = (1080 * 20) / 100 = 21600 / 100 = 216. The square root step is validated because 1080^2 gives 90 * 18 * 60 * 12. Thus, there is no arithmetic slip, and the resulting cost of ₹ 216 is consistent and reasonable given the area and per metre rate. Why Other Options Are Wrong: Option ₹ 220 is slightly higher than the correct value and might come from rounding or misplacing a decimal point. Option ₹ 200 underestimates the total by ignoring part of the area or miscalculating the square root. Option ₹ 210 is close but still off, indicating a partial error in the area or cost multiplication. Option ₹ 240 overestimates the cost and would correspond to a larger area or higher rate, neither of which is correct. Common Pitfalls: Errors often arise from incorrect computation of the semi perimeter, forgetting to divide by two, or miscomputing the products inside the square root. Some students also misconvert 20 paise into rupees or accidentally use ₹ 20 per sq. m instead of ₹ 0.20 per sq. m. Careful step by step handling of both the area and the cost calculations is essential. Final Answer: The cost of levelling the triangular piece of land is ₹ 216.

Question 10

The angles of a triangle are in the ratio 1 : 1 : 4. If the perimeter of the triangle is k times its largest side, what is the value of k?

Accepted Answer

Introduction / Context: This problem links angle ratios in a triangle with the ratio of perimeter to the largest side. It tests knowledge of how triangle side lengths relate to angles via trigonometric functions, especially in non right triangles. Recognizing that side lengths are proportional to sines of opposite angles allows us to express perimeter in terms of the largest side and then determine the factor k. Given Data / Assumptions: Angles of the triangle are in ratio 1 : 1 : 4. Let the three angles be A, B, and C with C being the largest angle. Perimeter P is k times the largest side. We are asked to find k. All trigonometric values are taken in degrees, and the triangle is valid in Euclidean geometry. Concept / Approach: If the angles are in the ratio 1 : 1 : 4, we can set A = x, B = x, and C = 4x. Because A + B + C = 180°, we get 6x = 180°, so x = 30°. Thus the angles are 30°, 30°, and 120°, with 120° as the largest angle. In any triangle, side lengths are proportional to the sines of the opposite angles. Let the largest side, opposite the 120° angle, be denoted by a. Then the other two sides have lengths proportional to sin 30°. Using these proportions, we can express the perimeter in terms of a and simplify to get k. Step-by-Step Solution: Let angles be A = 30°, B = 30°, C = 120°.Let side a be opposite angle C = 120°, and sides b and c be opposite angles A and B respectively.By sine rule, a : b : c = sin 120° : sin 30° : sin 30°.Compute sin 120° = sin 60° = √3 / 2 and sin 30° = 1 / 2, so a : b : c = (√3 / 2) : (1 / 2) : (1 / 2) = √3 : 1 : 1.Let a = √3 units, then b = 1 and c = 1. Perimeter P = a + b + c = √3 + 1 + 1 = √3 + 2. Since P = k * a, we have k = (√3 + 2) / √3 = 1 + 2 / √3. Verification / Alternative check: To verify, compute k numerically. With √3 ≈ 1.732, we have P ≈ 1.732 + 2 = 3.732. The largest side a ≈ 1.732. Thus k ≈ 3.732 / 1.732 ≈ 2.155. Computing 1 + 2 / √3 gives 1 + 2 / 1.732 ≈ 1 + 1.155 ≈ 2.155, which matches. This confirms that the expression 1 + 2 / √3 correctly represents the ratio of perimeter to the largest side. Why Other Options Are Wrong: Option 1 − 2 / √3 would yield a number less than 1 and cannot represent the ratio of perimeter to the largest side, since perimeter is always larger than any individual side. Option 2 + 2 / √3 is too large and does not match the numeric check. Option 2 corresponds roughly to P ≈ 2a, which is not correct for this specific triangle. Option None of these is not needed because one of the given expressions matches the correct ratio. Common Pitfalls: Some students mistakenly treat the angle ratio as side ratio without using trigonometric relations. Others forget that the largest angle corresponds to the largest side and instead assign labels inconsistently. Errors also occur when computing sine values, especially for 120°, or when simplifying the algebraic expression for k. Careful use of the sine rule and correct substitution prevents such mistakes. Final Answer: The value of k is 1 + 2/√3.

Question 11

What is the area (in square centimetres) of an equilateral triangle with side length 6 cm?

Accepted Answer

Introduction / Context: This problem tests the standard formula for the area of an equilateral triangle. Since all sides and angles are equal, the area can be computed quickly using a known expression rather than decomposing it into right triangles every time. This is a fundamental result in geometry and frequently appears in aptitude questions. Given Data / Assumptions: The triangle is equilateral, so all sides are equal. Side length a = 6 cm. We need the area in square centimetres. No additional constructions or special angles are required beyond equilateral triangle properties. We may use the standard area formula for equilateral triangles. Concept / Approach: The area A of an equilateral triangle with side length a is given by A = (√3 / 4) * a^2. This formula can be derived by dropping an altitude and using Pythagoras theorem, but in exam situations it is usually taken as known. Substituting the given side length into this formula will yield the required area in simplest radical form. Step-by-Step Solution: Side length a = 6 cm.Area formula for an equilateral triangle: A = (√3 / 4) * a^2.Compute a^2 = 6^2 = 36.Substitute into the formula: A = (√3 / 4) * 36.Simplify: 36 / 4 = 9, so A = 9√3 square centimetres. Verification / Alternative check: Alternatively, drop an altitude from one vertex, which splits the triangle into two right triangles with hypotenuse 6 and base 3. Using Pythagoras theorem, altitude h satisfies 6^2 = 3^2 + h^2, so h^2 = 36 − 9 = 27 and h = 3√3. The area is then (1 / 2) * base * height = (1 / 2) * 6 * 3√3 = 9√3, matching the earlier calculation. Why Other Options Are Wrong: Option 36√3 is four times the correct area and corresponds to forgetting the factor 1 / 4 in the formula. Option 9 is missing the √3 factor and thus underestimates the area. Option 36 is equivalent to 6^2 and ignores the triangular shape entirely. Option 18√3 is twice the correct area and might result from using an incorrect base or height in the right triangle approach. Common Pitfalls: Some students mix up the formula for the area of a triangle with that of a square or rectangle. Others write (√3 / 2) * a^2 instead of (√3 / 4) * a^2. Another common mistake is to miscompute the altitude by taking it as half the side length rather than using Pythagoras theorem. Final Answer: The area of the equilateral triangle is 9√3 square centimetres.

Question 12

In a right triangle, the length of the side opposite the right angle (the hypotenuse) is 9√3 cm. What is the length of the side opposite the angle that measures 30°?

Accepted Answer

Introduction / Context: This question deals with a special right triangle that has angles 30°, 60°, and 90°. Such triangles have well known side ratios that greatly simplify calculations. Recognizing this pattern allows us to find the length of a specific side directly from the length of the hypotenuse, without using heavy trigonometry. Given Data / Assumptions: The triangle is a right triangle with one angle of 30° and another of 60°. The side opposite the right angle, the hypotenuse, is 9√3 cm. We are asked for the side opposite the 30° angle. The triangle follows standard Euclidean geometry. We may use the standard ratios for a 30°–60°–90° triangle. Concept / Approach: In a 30°–60°–90° triangle, the side opposite 30° is the shortest side and has length x, the side opposite 60° has length x√3, and the hypotenuse has length 2x. Here we are given the hypotenuse and need to find x. Once we find x, that value directly gives the side opposite 30°. The problem reduces to solving 2x equal to the known hypotenuse length. Step-by-Step Solution: Let the side opposite 30° be x.In a 30°–60°–90° triangle, hypotenuse = 2x.Given hypotenuse = 9√3 cm, so 2x = 9√3.Solve for x: x = (9√3) / 2.Therefore, the side opposite the 30° angle is (9√3) / 2 cm. Verification / Alternative check: We can check by computing the other leg. The side opposite 60° should be x√3 = ((9√3) / 2) * √3 = (9 * 3) / 2 = 27 / 2 = 13.5 cm. Now, apply Pythagoras theorem: (side opposite 30°)^2 + (side opposite 60°)^2 should equal hypotenuse^2. Compute ((9√3) / 2)^2 + (13.5)^2 and verify it equals (9√3)^2. This check confirms that the side lengths are consistent with a 30°–60°–90° triangle. Why Other Options Are Wrong: Option 9 cm would make the hypotenuse 18 cm, which is inconsistent with 9√3 cm. Option 3√3 cm is too small and does not satisfy the 2x relation with the hypotenuse. Option 6 cm would produce a hypotenuse of 12 cm, again inconsistent. Option (3√3)/2 cm is smaller still and does not fit the triangle side ratio system for the given hypotenuse. Common Pitfalls: Some students confuse which side corresponds to which angle and may assign the value to the wrong side. Others misremember the ratio as 1 : 2 : √3 instead of 1 : √3 : 2. A few may try to use sine or cosine without recognizing the standard pattern, leading to extra steps and potential arithmetic errors. Final Answer: The length of the side opposite the 30° angle is (9√3)/2 cm.

Question 13

The area of an equilateral triangle is 9√3 sq. cm. What is the height (altitude) of this triangle in centimetres?

Accepted Answer

Introduction / Context: This question requires reversing the standard area formula for an equilateral triangle in order to find its side length and then its height. Once the side length is known, the height follows from the basic relation in equilateral triangles. This type of reverse application helps reinforce understanding of formulas rather than only using them in forward direction. Given Data / Assumptions: The triangle is equilateral. Area A = 9√3 sq. cm. We need the height (altitude) of the triangle. All sides and angles are equal, with each angle equal to 60°. Standard formulas for equilateral triangles apply. Concept / Approach: The area of an equilateral triangle with side length a is A = (√3 / 4) * a^2. Given A, we can solve for a by rearranging this formula. Once a is found, we use the formula for the height h of an equilateral triangle, which is h = (√3 / 2) * a. Substituting the computed side length into this height formula gives the required altitude. Step-by-Step Solution: Given A = 9√3 sq. cm and A = (√3 / 4) * a^2.Set (√3 / 4) * a^2 = 9√3.Divide both sides by √3: (1 / 4) * a^2 = 9, so a^2 = 36.Thus side length a = 6 cm.Height h = (√3 / 2) * a = (√3 / 2) * 6 = 3√3 cm. Verification / Alternative check: We can verify by recomputing the area from the derived side and height. For a triangle with base a = 6 and height h = 3√3, area should be (1 / 2) * base * height = (1 / 2) * 6 * 3√3 = 9√3 sq. cm. This matches the given area, confirming that both the side and the height are correctly computed. Why Other Options Are Wrong: Option 6 cm is the side length, not the height, so it is larger than the correct altitude. Option 6√3 cm is too large and would give a much larger area than 9√3. Option 9 cm would correspond to a different area entirely. Option 4√3 cm would give a base and height combination inconsistent with the given area when substituted into area formulas. Common Pitfalls: Many students confuse the side length with the height and may stop after finding a = 6 cm. Others misapply the area formula or forget to multiply by 1 / 4 when isolating a^2. Some misremember the height formula and use a / 2 instead of (√3 / 2) * a, which leads to incorrect results. Final Answer: The height of the equilateral triangle is 3√3 cm.

Question 14

What is the area (in square centimetres) of a circle whose circumference is 26.4 cm?

Accepted Answer

Introduction / Context: This question connects the circumference and area of a circle. Given the circumference, we must first find the radius, and then use the area formula. It tests familiarity with both formulas and comfort in rearranging them to solve for the radius before substituting into the area expression. Given Data / Assumptions: Circumference C = 26.4 cm. Circumference formula: C = 2 * π * r. Area formula: A = π * r^2. We may use π ≈ 3.14 for numerical work. We seek the area in square centimetres. Concept / Approach: First, use the circumference formula to solve for the radius: r = C / (2 * π). Once r is known, substitute into the area formula A = π * r^2. Careful handling of division and squaring is important to avoid rounding errors. Using π consistently throughout the calculation provides a reliable approximate value for the area. Step-by-Step Solution: Circumference C = 26.4 cm.Radius r = C / (2 * π) ≈ 26.4 / (2 * 3.14) = 26.4 / 6.28.Compute r ≈ 4.21 cm (approximately).Area A = π * r^2 ≈ 3.14 * (4.21)^2.(4.21)^2 ≈ 17.7, so A ≈ 3.14 * 17.7 ≈ 55.5 sq. cm, close to 55.44 sq. cm, which matches the given option. Verification / Alternative check: We can match the exact pattern used in many exam solutions by noticing that if r is chosen so that C = 26.4, then solving with more precise intermediate steps yields A ≈ 55.44. Small rounding differences in r lead to minor changes in A, but the closest option by far is 55.44. The value fits within expected precision when π is approximated as 3.14. Why Other Options Are Wrong: Option 44.55 and 44.33 are much smaller than the value obtained from the formulas and reflect using too small a radius. Option 33.44 is far too small and could only arise from significant formula misuse. Option 60.84 is larger than the correctly computed area and does not result from reasonable approximations of π and r. Common Pitfalls: A common error is to plug the circumference directly into the area formula without first solving for the radius. Others may mistakenly use C^2 in place of r^2 or miscompute the denominator 2 * π. Some students also mix up π approximations, switching between 22/7 and 3.14 inconsistently across steps, which can introduce larger rounding errors. Final Answer: The area of the circle is approximately 55.44 square centimetres.

Question 15

What is the area (in square centimetres) of a circle whose circumference is 22 cm?

Accepted Answer

Introduction / Context: This is another question linking circumference and area of a circle, but with a circumference that simplifies nicely when using π = 22/7. The goal is to find the radius from the given circumference and then compute the area with minimal calculation. Recognizing when 22/7 is convenient saves time and reduces arithmetic complexity in exams. Given Data / Assumptions: Circumference C = 22 cm. Circumference formula: C = 2 * π * r. Area formula: A = π * r^2. Take π = 22/7, which is especially convenient here. We seek the area in square centimetres. Concept / Approach: We write 22 = 2 * π * r and substitute π = 22/7. This gives an equation for r that is easy to solve because the 22 terms cancel. Once r is known, we substitute into A = π * r^2 using the same π approximation. This yields a clean fractional value for the area, which can then be written as a decimal or left as a fraction depending on the options given. Step-by-Step Solution: Given C = 22 cm and C = 2 * π * r.Substitute π = 22/7: 22 = 2 * (22 / 7) * r.Simplify: 22 = (44 / 7) * r, so r = 22 * 7 / 44 = 154 / 44 = 3.5 cm.Now compute area A = π * r^2 = (22 / 7) * (3.5)^2.(3.5)^2 = 12.25, so A = (22 / 7) * 12.25 = 22 * 12.25 / 7 = 38.5 square centimetres. Verification / Alternative check: We can check the circumference from r = 3.5 cm to ensure consistency. Using C = 2 * π * r = 2 * (22 / 7) * 3.5 = (44 / 7) * 3.5. Since 3.5 = 7 / 2, the product is (44 / 7) * (7 / 2) = 44 / 2 = 22 cm, which matches the given circumference. This confirms that r = 3.5 cm and that the area 38.5 sq. cm is correct. Why Other Options Are Wrong: Option 77 sq. cm is double the correct area and would correspond to using radius 5 or misapplying the formula. Option 44 sq. cm is larger than the true area and does not fit with r = 3.5 cm. Option 88 sq. cm is far too large and more appropriate for a circle with much larger radius. Option 55 sq. cm is also too big and likely arises from rough approximation or ignoring the squared radius step. Common Pitfalls: Some students forget to substitute π as 22/7 and instead use 3.14, which complicates the arithmetic, though it can still give a close answer. Others solve incorrectly for r by not handling the fraction (44 / 7) properly. There is also confusion between circumference and area formulas, leading some to square the circumference directly or to omit the square on the radius in the area formula. Final Answer: The area of the circle is 38.5 square centimetres.

Question 16

Find the area of the triangle with vertices A(0, 8), O(0, 0), and B(5, 0). What is the area of triangle AOB (in square units)?

Accepted Answer

Introduction / Context: This question checks coordinate geometry and basic mensuration. When three points are given on the coordinate plane, the area of the triangle can often be found quickly by recognizing a right triangle (two perpendicular sides) or by using the coordinate area formula. Here, two vertices lie on the axes, which strongly suggests a right triangle whose legs are parallel to the x-axis and y-axis. Once we identify the base and height as perpendicular distances along the axes, the area is simply (1/2) * base * height. This is a common aptitude pattern because it avoids heavy computation if you observe the geometry correctly. Given Data / Assumptions: A = (0, 8) O = (0, 0) B = (5, 0) AO is vertical (x = 0), OB is horizontal (y = 0), so AO is perpendicular to OB Area of right triangle = (1/2) * base * height Concept / Approach: Use AO as height and OB as base because they lie on perpendicular axes. Compute their lengths from coordinate differences, then apply the triangle area formula. Step-by-Step Solution: Length OB = distance from (0, 0) to (5, 0) = 5 units Length AO = distance from (0, 0) to (0, 8) = 8 units Triangle AOB is right-angled at O (axes are perpendicular) Area = (1/2) * OB * AO = (1/2) * 5 * 8 Area = 20 square units Verification / Alternative check: Using the coordinate triangle area formula also works: Area = (1/2)*|x1(y2-y3) + x2(y3-y1) + x3(y1-y2)|. Substituting A(0,8), O(0,0), B(5,0) gives (1/2)*|0*(0-0) + 0*(0-8) + 5*(8-0)| = (1/2)*40 = 20. Same result confirms correctness. Why Other Options Are Wrong: 8 and 13 can come from confusing side lengths or using (5+8) instead of (1/2)*5*8. 40 is twice the correct value (missing the 1/2 factor). 10 can come from halving only one side incorrectly. Common Pitfalls: Forgetting the 1/2 in triangle area, misreading coordinates, or not noticing the right angle formed by the axes. Final Answer: The area of the triangle is 20 sq units.

Question 17

In triangle ABC, the three medians AD, BE, and CF intersect at G (the centroid). If the area of triangle ABC is 36 sq cm, what is the area of triangle CGE (in sq cm)?

Accepted Answer

Introduction / Context: This question tests a standard centroid (intersection of medians) area property in triangles. Medians divide a triangle into smaller regions of equal area. Specifically, the three medians together partition the original triangle into 6 small triangles of equal area. Because the centroid is the common intersection point, many sub-triangle areas become simple fractions of the total area, independent of the triangle’s exact shape. Once you know which of those 6 equal triangles corresponds to the requested region (triangle CGE), you can compute its area as a fraction of the total 36 sq cm. This is a classic geometry aptitude problem designed to reward recognizing the equal-area partition rather than doing coordinate geometry. Given Data / Assumptions: AD, BE, CF are medians of triangle ABC They intersect at centroid G Total area of triangle ABC = 36 sq cm Medians divide triangle ABC into 6 equal-area small triangles Concept / Approach: Use the known result: area(ABC) is split into 6 equal parts by the medians. Triangle CGE is one of these 6 equal triangles, so its area is (1/6) of the total area. Step-by-Step Solution: Medians partition triangle ABC into 6 equal-area triangles So each small triangle area = (Area of ABC) / 6 Each area = 36 / 6 = 6 sq cm Triangle CGE is one such small triangle Therefore, Area of triangle CGE = 6 sq cm Verification / Alternative check: You can confirm using a sample coordinate triangle: choose any triangle, compute centroid G, midpoint E, then compute the area of CGE and compare to total area. The ratio always comes out 1/6, showing the result is shape-independent and purely based on median properties. Why Other Options Are Wrong: 12 and 18 are too large and correspond to 1/3 or 1/2 of the total area, not 1/6. 9 corresponds to 1/4 of the area, which is not produced by median partitioning. 3 would be 1/12, which is smaller than any single small triangle formed by medians. Common Pitfalls: Assuming medians make 4 equal parts (they do not), mixing centroid with incenter, or guessing fractions without using the 6-equal-triangles fact. Final Answer: The area of triangle CGE is 6 sq cm.

Question 18

Two circles touch each other at point X. A common tangent touches them at two distinct points Y and Z. If another tangent through X meets line segment YZ at A and XA = 16 cm, what is the length of YZ (in cm)?

Accepted Answer

Introduction / Context: This problem tests a powerful tangent property from geometry: tangents drawn from the same external point to a circle have equal lengths. The setting involves two circles that touch at a point X, which means they share a common tangent line at X. Another common tangent touches the circles at points Y and Z. The tangent at X intersects the common tangent line (YZ) at point A. From point A, there are two tangents to the first circle (AX and AY), and similarly two tangents to the second circle (AX and AZ). Using the equal-tangents-from-an-external-point theorem, we can link AY, AZ, and AX directly. Then, because Y and Z lie on the same line with A between them in this configuration, YZ becomes the sum AY + AZ, which turns into 2*AX. This allows a clean numeric answer without needing radii or angles. Given Data / Assumptions: Two circles touch at X (common tangent at X exists) Common external tangent touches circles at Y and Z Tangent through X intersects line YZ at A XA = 16 cm Tangent theorem: from external point A, tangent lengths to a circle are equal Concept / Approach: Use equal tangents from A to each circle: For circle 1: AX = AY. For circle 2: AX = AZ. Then YZ = AY + AZ = AX + AX = 2*AX. Step-by-Step Solution: From point A, AY is tangent to the first circle at Y and AX is tangent at X So, AY = AX (tangents from same external point to a circle are equal) Similarly, AZ is tangent to the second circle at Z and AX is tangent at X So, AZ = AX Given AX = 16 cm, we get AY = 16 cm and AZ = 16 cm Point A lies on line YZ, so YZ = AY + AZ = 16 + 16 = 32 cm Verification / Alternative check: The result does not depend on the radii, only on the tangent equality property. Any valid diagram with the same XA will force AY and AZ to match XA, so the total YZ must be twice XA when A lies between Y and Z on the common tangent line. Why Other Options Are Wrong: 16 matches only one segment (AY or AZ), not the full YZ. 24 or 18 would imply unequal tangent lengths from A, violating the tangent theorem. 48 incorrectly treats YZ as 3*AX without geometric justification. Common Pitfalls: Forgetting that tangents from the same external point are equal, confusing chord lengths with tangent lengths, or adding the wrong segments along line YZ. Final Answer: The length of YZ is 32 cm.

Question 19

The length of a rectangular plot is 20 m more than its breadth. If 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 perimeter (fencing) with a linear relationship between length and breadth. Fencing a rectangular plot means fencing its boundary, which is the perimeter. The cost information gives the total perimeter because cost = (rate per metre) * (perimeter). Once the perimeter is known, we use the rectangle perimeter formula 2*(L + B). Along with the condition that length is 20 m more than breadth, we can solve for both dimensions. The final question asks only for the length, but we still compute breadth internally as part of solving the system. This is a classic two-step aptitude problem: first convert money to perimeter, then use algebra for rectangle dimensions. Given Data / Assumptions: Length L = B + 20 m Fencing rate = Rs 26.50 per m Total cost = Rs 5300 Perimeter P = 2*(L + B) Concept / Approach: Compute perimeter from cost: P = 5300 / 26.50. Then use 2*(L+B)=P with L=B+20 to find L. Step-by-Step Solution: Perimeter P = Total cost / Rate = 5300 / 26.50 5300 / 26.50 = 200 m So, 2*(L + B) = 200 => L + B = 100 Given L = B + 20 Substitute: (B + 20) + B = 100 => 2B + 20 = 100 2B = 80 => B = 40 L = B + 20 = 40 + 20 = 60 m Verification / Alternative check: Check perimeter with L=60 and B=40: 2*(60+40)=2*100=200 m. Cost would be 200*26.50=5300, matching the given cost exactly. Why Other Options Are Wrong: 40 m is the breadth, not the length. 20 m and 30 m are too small to satisfy both the perimeter and the +20 relation. 50 m would make breadth 30 m, giving L+B=80, not 100. Common Pitfalls: Using area instead of perimeter for fencing, forgetting to divide by the fencing rate, or forgetting the factor 2 in the perimeter formula. Final Answer: The length of the plot is 60 m.

Question 20

A sector of 120 degrees cut out from a circle has an area of 66/7 sq cm (the text may appear as 667 sq cm, meaning 66/7). Find the radius of the circle (in cm), assuming pi = 22/7.

Accepted Answer

Introduction / Context: This question tests the area-of-sector concept. A sector is a fraction of the full circle determined by its central angle. The sector area formula is: Sector area = (theta/360) * pi * r^2. Here theta is 120 degrees, so the sector is exactly one-third of the circle. That means the full circle’s area is 3 times the sector area. Once we convert sector area into full circle area, we can use the standard circle area formula A = pi*r^2 to solve for r. The problem is designed to work neatly with pi = 22/7 and an area written as 66/7 sq cm (which can sometimes be mistyped as 667). Given Data / Assumptions: Sector angle theta = 120 degrees Sector area = 66/7 sq cm pi = 22/7 Sector area formula: (theta/360)*pi*r^2 Concept / Approach: Since 120/360 = 1/3, we have (1/3)*pi*r^2 = 66/7. Multiply both sides by 3 to get pi*r^2, then solve for r using pi = 22/7. Step-by-Step Solution: (120/360) = 1/3 (1/3) * pi * r^2 = 66/7 pi * r^2 = 3 * (66/7) = 198/7 Substitute pi = 22/7: (22/7) * r^2 = 198/7 Cancel /7: 22*r^2 = 198 r^2 = 198/22 = 9 r = 3 cm Verification / Alternative check: Full circle area with r=3 is (22/7)*9 = 198/7. One-third of that (for 120 degrees) is (198/7)/3 = 66/7, matching the given sector area exactly. Why Other Options Are Wrong: 1 cm and 2 cm produce much smaller areas because area grows with r^2. 4 cm and 6 cm produce much larger sector areas than 66/7 for 120 degrees. Common Pitfalls: Forgetting the (theta/360) factor, using 120/180 instead of 120/360, or not squaring r properly when rearranging the equation. Final Answer: The radius of the circle is 3 cm.

Area Questions