Question 1

The length of a rectangle is increased by 25%. By what percentage must the breadth be decreased so that the area of the rectangle remains unchanged?

Accepted Answer

Introduction / Context: This problem is a standard percentage and area question about rectangles. When one dimension of a rectangle changes, the other dimension must adjust in the opposite direction if the area is to remain the same. Here the length increases, so the breadth must decrease. The question asks for the exact percentage decrease required in the breadth to offset a 25 percent increase in length. Given Data / Assumptions: Original length = L. Original breadth = B. Original area A = L * B. New length = L + 25 percent of L = 1.25L. Let new breadth be B'. New area must equal old area: 1.25L * B' = L * B. Concept / Approach: When dealing with relative changes and constant products, it is often easiest to work with multiplicative factors. An increase of 25 percent corresponds to multiplying by 1.25. To keep the area constant, the product of the new length factor and the new breadth factor must equal 1. Therefore, if the length is multiplied by 1.25, the breadth must be multiplied by 1 / 1.25. Once this factor is found, it can be converted into a percentage decrease from the original breadth. Step-by-Step Solution: Old area A = L * B. New area A' = 1.25L * B'. Set A' = A to keep area unchanged: 1.25L * B' = L * B. Divide both sides by L: 1.25 * B' = B. So B' = B / 1.25. 1.25 = 5/4, so B' = B * (4/5) = 0.8B. This means the new breadth is 80 percent of the original breadth, a 20 percent reduction. Verification / Alternative check: Assume some simple values to confirm. Let L = 100 units and B = 100 units, so area = 10000 units. Increase length by 25 percent: new length = 125 units. If breadth is reduced by 20 percent, new breadth = 80 units. New area = 125 * 80 = 10000 units, which is exactly the original area, verifying that a 20 percent decrease is correct. Why Other Options Are Wrong: 25% and 30%: These changes would make the area smaller than the original. 40%: This is a very large reduction and would reduce the area significantly below the original. 15%: This reduction is too small and would result in an area larger than the original. Common Pitfalls: Adding or subtracting percentage changes instead of using multiplicative factors. Assuming that a 25 percent increase must be cancelled by a 25 percent decrease, which is not correct. Forgetting that the product of the two change factors must equal 1 to keep the area constant. Final Answer: The breadth must be decreased by 20%

Question 2

The total area of the four walls of a cuboid room is 400 sq cm. Its length is 15 cm and its height is 8 cm. What is the breadth of the cuboid in centimetres?

Accepted Answer

Introduction / Context: This question uses the lateral surface area of a cuboid, sometimes described as the area of the four walls of a room. Given this lateral area and two of the dimensions, you must find the third dimension, which is the breadth. It tests your ability to apply and rearrange the lateral surface area formula for a rectangular prism. Given Data / Assumptions: Lateral surface area (area of 4 walls) = 400 sq cm. Length l = 15 cm. Height h = 8 cm. Breadth b is unknown. Formula for area of 4 walls of a cuboid: 2h(l + b). Concept / Approach: The area of the four walls of a cuboid excludes the top and bottom faces and equals the perimeter of the base multiplied by the height. The perimeter of the base is 2(l + b), and when multiplied by h, it gives 2h(l + b). We are given that this equals 400 sq cm. By substituting the known values of l and h into this formula, we can solve a simple linear equation for b, the breadth. Step-by-Step Solution: Use lateral surface area formula: 2h(l + b) = 400. Substitute h = 8 cm and l = 15 cm. 2 * 8 * (15 + b) = 400. 16 * (15 + b) = 400. Divide both sides by 16: 15 + b = 400 / 16 = 25. So b = 25 - 15 = 10 cm. Verification / Alternative check: Verify by recomputing the lateral surface area with length 15 cm, breadth 10 cm and height 8 cm. Perimeter of base = 2 * (15 + 10) = 2 * 25 = 50 cm. Lateral surface area = perimeter of base * height = 50 * 8 = 400 sq cm, which matches the given area, confirming that b = 10 cm is correct. Why Other Options Are Wrong: 12, 20, 24 and 8: Substituting any of these values for b into 2 * 8 * (15 + b) does not give 400 sq cm. They produce larger or smaller lateral surface areas. Common Pitfalls: Using the total surface area formula instead of the lateral surface area formula. Forgetting to multiply by 2 when forming 2h(l + b). Arithmetic mistakes when dividing 400 by 16 or subtracting to find b. Final Answer: The breadth of the cuboid is 10 cm

Question 3

Triangles ΔABC and ΔPQR are similar. The ratio of the perimeter of ΔABC to the perimeter of ΔPQR is 5 : 9. If side PQ = 45 cm, what is the length of side AB in centimetres?

Accepted Answer

Introduction / Context: This question is about similar triangles and how ratios of their perimeters relate to ratios of corresponding sides. When two triangles are similar, every linear dimension of one triangle is in the same ratio to the corresponding dimension of the other triangle. The question gives you the ratio of perimeters and one side of the second triangle, and asks for the corresponding side of the first triangle. Given Data / Assumptions: ΔABC is similar to ΔPQR. Perimeter ratio P(ABC) : P(PQR) = 5 : 9. Corresponding sides are AB ↔ PQ, BC ↔ QR, AC ↔ PR. PQ = 45 cm. We need to find AB. Concept / Approach: For similar triangles, the ratio of perimeters equals the ratio of corresponding side lengths. That means AB : PQ = 5 : 9. With PQ known, AB can be found by scaling 45 cm by the ratio 5/9. This avoids the need to know the actual perimeters and uses the similarity property directly and efficiently. Step-by-Step Solution: Given P(ABC) : P(PQR) = 5 : 9. For similar triangles, AB : PQ = 5 : 9. So AB / PQ = 5 / 9. PQ = 45 cm, so AB / 45 = 5 / 9. AB = 45 * (5 / 9) = 45 * 5 / 9. AB = 5 * 5 = 25 cm. Verification / Alternative check: The scale factor from ΔPQR to ΔABC for sides is 5/9. If PQ = 45 cm, the full perimeter of ΔPQR would be some multiple of 45, and the corresponding perimeter of ΔABC would be that multiple times 5/9. Although we do not need the full perimeters to answer the question, this reasoning is fully consistent with the side ratio used above, confirming that AB = 25 cm is correct. Why Other Options Are Wrong: 15 cm and 20 cm: These correspond to smaller side ratios than 5/9 and would give perimeter ratios different from 5 : 9. 16 cm: Not a simple fraction of 45 in the ratio 5/9. 30 cm: This corresponds to a side ratio of 2/3 and would imply a perimeter ratio 2 : 3 rather than 5 : 9. Common Pitfalls: Mixing up the direction of the ratio and using 9/5 instead of 5/9. Thinking that perimeter ratio cannot be used directly for side ratios in similar triangles. Arithmetic mistakes in simplifying 45 * 5 / 9. Final Answer: The length of AB is 25 cm

Question 4

In triangle ΔABC, points D and E lie on sides AB and AC respectively. Segment DE is parallel to BC. If AD : DB = 2 : 5 and the area of ΔABC is 98 sq cm, what is the area (in sq cm) of quadrilateral BDEC?

Accepted Answer

Introduction / Context: This geometry question involves similar triangles formed by drawing a line parallel to the base of a triangle. You are given a ratio on one side and the total area of the main triangle, and you must find the area of a quadrilateral formed between the parallel line and the base. It tests your mastery of area ratios in similar triangles and simple subtraction of areas. Given Data / Assumptions: ΔABC is a triangle. D lies on AB and E lies on AC. DE is parallel to BC. AD : DB = 2 : 5. Area of ΔABC = 98 sq cm. We need the area of quadrilateral BDEC. Concept / Approach: When a line segment is drawn parallel to the base of a triangle and cuts the other two sides, the smaller triangle at the top is similar to the original triangle. The ratio of similarity equals the ratio of corresponding sides. Because area scales as the square of the linear scale factor, we can find the area of the smaller top triangle and subtract it from the total area to get the area of the remaining quadrilateral BDEC. Step-by-Step Solution: Given AD : DB = 2 : 5. Then AB = AD + DB corresponds to 2 + 5 = 7 parts, so AD : AB = 2 : 7. Since DE is parallel to BC, triangle ADE is similar to triangle ABC. The linear scale factor (ADE : ABC) = AD / AB = 2 / 7. Area scale factor for similar triangles is the square of the side ratio, so Area(ADE) / Area(ABC) = (2 / 7)^2 = 4 / 49. Area of ΔADE = (4 / 49) * 98 = 4 * 2 = 8 sq cm. Area of quadrilateral BDEC = Area(ABC) - Area(ADE) = 98 - 8 = 90 sq cm. Verification / Alternative check: It is useful to cross check the fractions. If the top triangle occupies 4/49 of the full area, then the remaining lower region, which includes quadrilateral BDEC, should occupy 45/49 of the full area. 45/49 of 98 = 45 * 2 = 90 sq cm, which matches the earlier computation exactly, confirming the answer. Why Other Options Are Wrong: 98: This is the area of the entire triangle, not the quadrilateral. 94 and 86: These correspond to subtracting incorrect fractions of the total area, indicating a wrong area ratio. 88: Another incorrect subtraction, not supported by the similarity and area ratio rules. Common Pitfalls: Using AD : DB directly as the area ratio instead of converting to AD : AB. Forgetting to square the side ratio when moving to area ratios. Subtracting the wrong part of the area or confusing triangle ADE with the quadrilateral. Final Answer: The area of quadrilateral BDEC is 90 sq cm

Question 5

In a trapezium, the lengths of the two parallel sides are 8 cm and 4 cm. M and N are the midpoints of the two diagonals of the trapezium. What is the length of segment MN in centimetres?

Accepted Answer

Introduction / Context: This is a property based question about a trapezium, a quadrilateral with exactly one pair of parallel sides. It involves the midpoints of the diagonals and a known formula for the segment joining these midpoints. Instead of computing coordinates or angles, you can use a neat geometric result that links the parallel sides directly to the length of the segment between the midpoints of the diagonals. Given Data / Assumptions: The quadrilateral is a trapezium. Parallel sides have lengths 8 cm and 4 cm. M and N are the midpoints of the two diagonals. We need the length of segment MN. Standard geometric property: in any trapezium, the segment joining the midpoints of the diagonals has length equal to half the difference of the parallel sides. Concept / Approach: For a trapezium with parallel sides a and b (with a greater than b), there is a known result that the line segment connecting the midpoints of its diagonals has length (1/2) * |a - b|. This comes from vector geometry or coordinate geometry analysis of diagonals and midpoints. Using this direct formula allows you to avoid complicated calculations and quickly compute MN from the given side lengths. Step-by-Step Solution: Let the lengths of the parallel sides be a = 8 cm and b = 4 cm. Length of segment MN = (1/2) * |a - b|. Compute difference: |a - b| = |8 - 4| = 4 cm. Hence MN = (1/2) * 4 = 2 cm. Verification / Alternative check: Though more involved, this result can be derived using coordinate geometry. Place the trapezium with parallel sides on a horizontal axis, compute the coordinates of diagonal midpoints using average formulas, then use distance formula to find MN. Doing so leads back to the same expression involving half the difference of the parallel sides, which confirms the formula and the numeric answer 2 cm for the present values. Why Other Options Are Wrong: 6 cm and 12 cm: These are based on adding or taking half the sum of sides instead of half the difference. 1 cm and 4 cm: These correspond to misapplied differences or forgetting the half factor. Common Pitfalls: Confusing this formula with the one for the mid segment parallel to the bases of a trapezium, which is half the sum of the parallel sides. Using the sum of the sides instead of the difference. Forgetting to take absolute value, although here the longer side is obvious. Final Answer: The length of MN is 2 cm

Question 6

An equilateral triangle of side 6 cm is inscribed in a circle. What is the radius of the circle (the circumradius) in centimetres?

Accepted Answer

Introduction / Context: This question deals with the circumradius of an equilateral triangle. When an equilateral triangle is inscribed in a circle, the circle is its circumcircle and the radius is called the circumradius. There is a well known relationship between the side length of an equilateral triangle and its circumradius, which lets you solve such problems very quickly. Given Data / Assumptions: The triangle is equilateral. Side length a = 6 cm. The triangle is inscribed in a circle, so the circle is the circumcircle of the triangle. We need to find the radius R of this circle. Standard formula for an equilateral triangle: R = a / sqrt(3). Concept / Approach: The circumradius of a general triangle can be expressed in terms of sides and angles, but for an equilateral triangle the formula simplifies greatly. Using geometry or trigonometry, one can show that R = a / sqrt(3). Here, we substitute a = 6 and then rationalise or simplify the result. Knowing this special case formula is very helpful in many geometry problems involving equilateral triangles and circles. Step-by-Step Solution: Side of the equilateral triangle a = 6 cm. For an equilateral triangle, circumradius R = a / sqrt(3). So R = 6 / sqrt(3). Multiply numerator and denominator by sqrt(3) to simplify: R = (6 * sqrt(3)) / 3. Simplify 6 / 3 to get R = 2 * sqrt(3) cm. Verification / Alternative check: You can verify the relationship using the fact that the centroid, circumcentre and incentre of an equilateral triangle coincide. Drawing radii from the centre to each vertex divides the triangle into three congruent smaller triangles. Working with right triangles formed by dropping an altitude and using basic trigonometric ratios for 30 and 60 degrees also leads to the same circumradius formula, confirming that R = 2 * sqrt(3) cm when a = 6 cm. Why Other Options Are Wrong: 3√2 cm and 4√3 cm: These are larger than the correct radius and would not fit a side length of only 6 cm. √3 cm: This is half the correct value and corresponds to a radius that is too small. 3 cm: This would correspond to R = a / 2, which is not the circumradius for an equilateral triangle. Common Pitfalls: Mixing up circumradius and inradius formulas. The inradius r for an equilateral triangle is a * sqrt(3) / 6, which is not the same as R. Using a generic triangle formula without exploiting the equilateral symmetry. Not rationalising the denominator correctly when dividing by sqrt(3). Final Answer: The radius of the circle is 2√3 cm

Question 7

Triangle ΔABC is an isosceles right angled triangle with ∠C = 90°. Point D is any point on the hypotenuse AB. What is the value of AD^2 + BD^2 in terms of CD^2?

Accepted Answer

Introduction / Context: This is a geometry identity question involving a special triangle, namely an isosceles right angled triangle, and a point on its hypotenuse. It explores a lesser known but elegant relationship between the squares of distances from that point to the endpoints of the hypotenuse and the distance from the point to the right angle vertex. Such identities often arise in coordinate geometry or vector geometry and are useful in advanced problem solving. Given Data / Assumptions: ΔABC is an isosceles right angled triangle. ∠C = 90°, so C is the right angle. CA = CB, so the legs adjacent to the right angle are equal. AB is the hypotenuse. D is any point on AB. We are asked to express AD^2 + BD^2 in terms of CD^2. Concept / Approach: The cleanest way to analyse this is to place the triangle in a coordinate system. Because the triangle is isosceles and right angled at C, we can put C at the origin, A on the x axis, and B on the y axis with equal leg lengths. Then any point D on the hypotenuse has parametric coordinates depending on a single parameter. By computing squared distances AD^2, BD^2 and CD^2 in terms of this parameter, we can look for a pattern that links AD^2 + BD^2 to CD^2. For an isosceles right triangle, a very neat simplification occurs. Step-by-Step Solution: Place C at (0, 0), A at (a, 0) and B at (0, a) so CA = CB = a. Then AB is the segment from (a, 0) to (0, a). Let D be any point on AB, for example D = (t, a - t) for some real t between 0 and a. Compute AD^2 = (t - a)^2 + (a - t)^2 = 2(t - a)^2. Compute BD^2 = t^2 + (a - t - a)^2 = t^2 + (-t)^2 = 2t^2. So AD^2 + BD^2 = 2(t - a)^2 + 2t^2 = 4t^2 - 4at + 2a^2. Now compute CD^2 = t^2 + (a - t)^2 = 2t^2 - 2at + a^2. Notice that AD^2 + BD^2 = 2 * CD^2. Verification / Alternative check: You can plug in specific values for a and t to check the relationship numerically. For instance, let a = 1 and t = 0.2. Compute CD^2 and AD^2 + BD^2 numerically and you will find that AD^2 + BD^2 is very close to twice CD^2, accounting for rounding. Since the derivation above is algebraic and general, it holds for any valid position of D on AB, confirming that the identity AD^2 + BD^2 = 2CD^2 is true for this triangle. Why Other Options Are Wrong: CD^2: This would mean AD^2 + BD^2 equals CD^2, which is not supported by the coordinate derivation. 3CD^2 and 4CD^2: These factors overstate the combined squared distances and do not match any consistent algebraic identity for the situation described. AD^2: This does not involve CD at all and is unrelated to the requested expression. Common Pitfalls: Assuming D is a special point such as the midpoint of AB rather than any point on AB. Trying to use Pythagoras directly without setting up a coordinate or vector framework. Not recognising that symmetry in an isosceles right triangle often leads to simple numeric factors like 2. Final Answer: For any point D on AB, we have AD^2 + BD^2 = 2CD^2

Question 8

ABCDEF is a regular hexagon with each side equal to 12 cm. What is the area, in square centimetres, of triangle ECD?

Accepted Answer

Introduction / Context: This problem involves a regular hexagon and asks for the area of a triangle formed by three of its vertices. Regular polygons have strong symmetry, and in the case of a regular hexagon, many internal triangles are equilateral or easily related to equilateral triangles. By using these symmetries and basic area formulas, we can find the area of triangle ECD without needing heavy coordinate calculations. Given Data / Assumptions: ABCDEF is a regular hexagon. Each side length a = 12 cm. Vertices are in order around the hexagon. We need the area of triangle ECD. In a regular hexagon, all internal angles are 120 degrees and all sides are equal. Concept / Approach: It helps to visualise the regular hexagon with centre O and vertices A, B, C, D, E, F equally spaced on a circle. Consecutive vertices form equilateral triangles with the centre. For triangle ECD, vertices C and D are adjacent, while E is adjacent to D on the other side. Thus triangle CDE has two sides that are hexagon sides and an included interior angle of 120 degrees at D. We can use the formula for the area of a triangle with two sides and the included angle: (1/2) * a * b * sin(angle). Step-by-Step Solution: In regular hexagon ABCDEF, side length a = 12 cm. Consider triangle CDE with vertices C, D and E on the hexagon. Sides CD and DE are both sides of the hexagon, so CD = DE = 12 cm. The interior angle at D in the hexagon is 120 degrees, and this is the angle between CD and DE in triangle CDE. Area of a triangle given two sides and included angle: Area = (1/2) * a * b * sin(theta). So Area(CDE) = (1/2) * 12 * 12 * sin(120 degrees). Compute sin(120 degrees) = sin(60 degrees) = square root of 3 / 2. Area = (1/2) * 144 * (square root of 3 / 2) = 72 * (square root of 3 / 2) = 36 * square root of 3. Verification / Alternative check: Another way is to recognise that a regular hexagon can be divided into six congruent equilateral triangles of side 12 cm by drawing all radii from the centre. The area of each equilateral triangle is (square root of 3 / 4) * a^2 = (square root of 3 / 4) * 144 = 36 * square root of 3. In a regular hexagon, some triangles formed by non adjacent vertices coincide with these fundamental triangles. Triangle CDE is such a fundamental equilateral based triangle, so its area equals 36 * square root of 3 sq cm, which matches the previous method. Why Other Options Are Wrong: 18√3 and 24√3: These are smaller than the actual area and would arise from halving the result or using a shorter side length. 42√3 and 30√3: These do not correspond to any simple multiple of the equilateral triangle area with side 12 cm and are inconsistent with regular hexagon geometry. Common Pitfalls: Misidentifying which triangles in the hexagon are equilateral and which are not. Using the wrong interior angle at D, such as 60 degrees instead of 120 degrees. Forgetting that sin(120 degrees) equals sin(60 degrees) and not cos(60 degrees). Final Answer: The area of triangle ECD is 36√3 square centimetres

Question 9

In which of the following quadrilaterals are all four sides congruent (that is, equal in length)?

Accepted Answer

Introduction / Context: This is a conceptual geometry question about the basic properties of common quadrilaterals. It focuses on which standard quadrilateral has all four sides congruent, meaning all sides have exactly the same length. Recognising these defining properties is essential for quickly classifying quadrilaterals and understanding their symmetry and angle relationships in more advanced problems. Given Data / Assumptions: Options include Rectangle, Parallelogram, Rhombus, Kite and Trapezium. We are asked to identify the quadrilateral whose defining property is that all four sides are equal in length. We consider standard Euclidean geometry definitions for these shapes. Concept / Approach: Each type of quadrilateral has characteristic properties. A rectangle has four right angles and opposite sides equal, but not necessarily all four sides equal. A general parallelogram has opposite sides parallel and equal, but adjacent sides can be different. A rhombus is a special parallelogram with all four sides equal. A kite has two pairs of adjacent equal sides, not all four necessarily equal. A trapezium in general has at least one pair of parallel sides and no requirement on equal side lengths. Therefore, only the rhombus matches the property described in the question. Step-by-Step Solution: Consider a rectangle: opposite sides are equal, but adjacent sides are usually different, so not all four sides are equal. Consider a general parallelogram: opposite sides are equal, but there is no requirement that all four sides are the same length. Consider a rhombus: by definition, it is a quadrilateral with all four sides equal and opposite sides parallel. Angles need not be right angles. Consider a kite: it has two pairs of adjacent equal sides. The two pairs can be of different lengths, so not all four sides are equal. Consider a trapezium: it has at least one pair of parallel sides but no requirement of side equality. Thus, among the listed options, only the rhombus always has all four sides congruent. Verification / Alternative check: As a useful mental check, remember that a square is a special type of rhombus and also a special type of rectangle. A square has all four sides equal and all angles equal to 90 degrees. If you drop the condition of right angles but keep all sides equal, you get a general rhombus. This confirms that rhombus is the most general quadrilateral type whose key property is four equal side lengths. Why Other Options Are Wrong: Rectangle: Requires all angles 90 degrees and opposite sides equal but does not require adjacent sides to be equal unless it is specifically a square. Parallelogram: Only opposite sides are equal and parallel; adjacent sides can differ in length. Kite: Has equal pairs of adjacent sides, not all four sides equal. Trapezium: Defined by having at least one pair of parallel sides, with no condition on side equality. Common Pitfalls: Confusing a rectangle with a square and assuming every rectangle has four equal sides. Thinking that all parallelograms must have equal adjacent sides, which is not true in general. Not distinguishing clearly between the definitions of kite, rhombus and square. Final Answer: The quadrilateral with all four sides congruent is the rhombus

Question 10

The area of a circular cricket ground is 24.64 hectares. Find the cost of making a rope boundary all around the ground at the rate of ₹5.40 per metre.

Accepted Answer

Introduction / Context: This problem links area and circumference of a circle with a real life application involving cost. You are given the area of a circular cricket ground in hectares and asked to compute the cost of placing a rope boundary around it. The steps involve unit conversion, finding the radius from the area, calculating the circumference and then multiplying by the given rate per metre. Given Data / Assumptions: Area of the circular ground A = 24.64 hectares. 1 hectare = 10,000 square metres. Rate of rope per metre = ₹5.40. We assume π ≈ 3.14 for circular calculations. Area of a circle A = πr^2, circumference C = 2πr. Concept / Approach: First convert the area from hectares to square metres. Then use the area formula A = πr^2 to find the radius r. After that, find the circumference C = 2πr, which represents the length of the rope required to surround the ground. Finally, calculate the total cost by multiplying C by the given rate per metre. Approximations using π = 3.14 are standard for such multiple choice questions. Step-by-Step Solution: Convert area to square metres: A = 24.64 hectares = 24.64 * 10000 = 246400 sq m. Use A = πr^2 with π ≈ 3.14: 246400 ≈ 3.14 * r^2. So r^2 ≈ 246400 / 3.14. Compute r^2 ≈ 78471.3 (approximate value). Take square root: r ≈ 280 m (approximately 280.1 m). Circumference C ≈ 2 * 3.14 * 280 ≈ 1759 m. Cost = C * rate ≈ 1759 * 5.40 ≈ ₹9504 (approximate). Verification / Alternative check: The key is to be consistent with the approximations. Using r ≈ 280.1 m rather than exactly 280 m produces nearly the same circumference and thus cost. The values cluster very close around ₹9500, and among the options, ₹9,504 is the closest and clearly intended answer. Small rounding differences are normal in such questions and are accounted for in the options provided. Why Other Options Are Wrong: ₹9,600: Slightly larger and would correspond to a slightly bigger radius than justified by the given area. ₹9,802 and ₹9,876: These values are further away from the consistent rounding with π ≈ 3.14 and the given area. ₹9,300: Too low compared to the carefully calculated approximate cost. Common Pitfalls: Forgetting to convert hectares to square metres before applying the area formula. Using an incorrect value of π or inconsistent rounding that leads to large errors. Mistakes in taking the square root of r^2 or in the final multiplication with the rate per metre. Final Answer: The cost of making the rope boundary is approximately ₹9,504

Question 11

The diameter of a circle is 28 cm. Using π = 22/7, find the area of the circle in square centimetres.

Accepted Answer

Introduction / Context: This is a straightforward circle mensuration problem. You are given the diameter of a circle and asked to find its area using the common approximation π = 22/7. It tests your ability to convert diameter to radius and correctly apply the area formula for a circle with the specified value of π. Given Data / Assumptions: Diameter of circle d = 28 cm. Radius r = d / 2. Use π = 22/7. Area of a circle A = πr^2. Concept / Approach: The first step is to find the radius, which is half of the diameter. The area formula requires the radius squared multiplied by π. Using the value π = 22/7 keeps the arithmetic with denominators manageable because the radius will be a multiple of 7, leading to easy cancellation and an exact integer result for the area in this case. Step-by-Step Solution: Diameter d = 28 cm. Radius r = d / 2 = 28 / 2 = 14 cm. Area A = πr^2 = (22/7) * 14^2. Compute 14^2 = 196. So A = (22/7) * 196. Simplify 196 / 7 = 28. Therefore A = 22 * 28 = 616 sq cm. Verification / Alternative check: As a quick check, note that 28 is 4 times 7, so r = 14 is 2 times 7. Squaring r gives 196, and multiplying 196 by 22/7 is equivalent to dividing 196 by 7 to get 28 and then multiplying by 22. There are no fractions in the final step, which reduces the chance of arithmetic errors. This confirms the exact result of 616 square centimetres. Why Other Options Are Wrong: 308 sq cm and 154 sq cm: These correspond to halving or quartering the correct area, possibly from misusing diameter instead of radius or missing a factor of 2. 77 sq cm: Much too small and likely comes from confusing circumference with area or misusing the formula. 704 sq cm: Larger than the correct area and does not result from any correct application of A = πr^2 with the given numbers. Common Pitfalls: Using the diameter directly in the area formula instead of the radius. Forgetting to square the radius before multiplying by π. Miscalculating 14^2 or failing to cancel 196 with 7 correctly. Final Answer: The area of the circle is 616 sq cm

Question 12

The measures of the four interior angles of a quadrilateral are in the ratio 3 : 4 : 5 : 6. What is the measure, in degrees, of the largest angle of the quadrilateral?

Accepted Answer

Introduction / Context: This question tests your understanding of angle sums in polygons and how to use given ratios to find individual angle measures. A quadrilateral has four interior angles whose sum is fixed. When the angles are in a given ratio, you can treat them as multiples of a common factor, solve for that factor, and then determine each angle, including the largest one. Given Data / Assumptions: The quadrilateral has four interior angles. The ratio of the angles is 3 : 4 : 5 : 6. Sum of interior angles of a quadrilateral is 360 degrees. Let the common factor be k, so angles are 3k, 4k, 5k and 6k. Concept / Approach: For any quadrilateral in Euclidean geometry, the sum of the four interior angles equals 360 degrees. When the angles are given in a ratio, you express each angle as a multiple of a common unknown k. The sum of these expressions must equal 360 degrees, giving a simple linear equation in k. Once k is known, each angle can be computed. The largest angle corresponds to the largest ratio term, which in this case is 6k. Step-by-Step Solution: Let the angles be 3k, 4k, 5k and 6k degrees. Sum of interior angles of a quadrilateral is 360 degrees. So 3k + 4k + 5k + 6k = 360. Add the coefficients: (3 + 4 + 5 + 6)k = 18k. Thus 18k = 360. k = 360 / 18 = 20 degrees. Angles are therefore 3k = 60°, 4k = 80°, 5k = 100° and 6k = 120°. The largest angle is 6k = 120°. Verification / Alternative check: Check that the calculated angles sum correctly: 60° + 80° + 100° + 120° = 360°, which matches the known total for a quadrilateral. The ratio 3 : 4 : 5 : 6 is preserved since 60 : 80 : 100 : 120 simplifies to 3 : 4 : 5 : 6 when divided by the common factor 20. This confirms both the ratio and the total angle sum, and thus the largest angle of 120 degrees is consistent. Why Other Options Are Wrong: 100°: This corresponds to the third largest angle (5k), not the largest one. 80° and 60°: These are the smaller angles 4k and 3k respectively. 140°: This angle does not appear in the set and would break the given ratio and total sum of 360 degrees. Common Pitfalls: Forgetting that the angle sum of a quadrilateral is 360 degrees, not 180 degrees as in a triangle. Adding the ratio numbers incorrectly or miscomputing 360 / 18. Choosing 100 degrees because it looks large without checking that 120 degrees is actually larger and valid. Final Answer: The measure of the largest angle is 120°

Question 13

A rectangular tank has internal dimensions 25 m in length, 12 m in width and 6 m in depth. If only the four side walls and the bottom of the tank are to be plastered, calculate the total cost of plastering at the rate of 75 paise (Rs. 0.75) per squa…

Accepted Answer

Introduction / Context: This aptitude question is based on mensuration of three dimensional shapes, specifically a rectangular tank. The task is to find the cost of plastering only the internal walls and the bottom surface, given the internal dimensions and the plastering rate per square metre. Understanding how to compute total surface area of selected faces is the key concept here. Given Data / Assumptions: Length of tank (L) = 25 m. Width of tank (W) = 12 m. Depth or height (H) = 6 m. Plastering is to be done on the four side walls and the bottom only, not on the top opening. Cost of plastering = Rs. 0.75 per sq m (75 paise per square metre). Concept / Approach: A rectangular tank is like a cuboid. The area to be plastered is the sum of the area of the bottom and the areas of the four vertical walls. The side walls form two pairs of equal rectangles. Once the total area is obtained, we multiply this area by the cost per square metre to obtain the total plastering cost. Careful separation of bottom and wall areas avoids overcounting. Step-by-Step Solution: Area of bottom = L * W = 25 * 12 = 300 sq m.Two longer walls each have area = L * H = 25 * 6 = 150 sq m, so together = 2 * 150 = 300 sq m.Two shorter walls each have area = W * H = 12 * 6 = 72 sq m, so together = 2 * 72 = 144 sq m.Total area to be plastered = bottom + four walls = 300 + 300 + 144 = 744 sq m.Cost per sq m = Rs. 0.75, so total cost = 744 * 0.75.Compute: 744 * 0.75 = 744 * 3 / 4 = 558.Therefore, total cost of plastering = Rs. 558. Verification / Alternative check: As a quick check, note that the total wall area is 444 sq m (300 + 144), and adding the bottom area 300 sq m gives 744 sq m. Multiplying by 0.75 can be checked by first finding three quarters of 700 (which is 525) and three quarters of 44 (which is 33), giving 525 + 33 = 558. This independently confirms the computed cost. Why Other Options Are Wrong: Rs. 458 and Rs. 608 come from using an incorrect total area such as missing one pair of walls or miscalculating the product 25 * 12. Rs. 658 and Rs. 758 are larger values that would imply an area greater than 744 sq m or an incorrect rate. None of these are consistent with the correct area and rate multiplication. Common Pitfalls: A common error is to include the top face of the tank, which is open and not to be plastered. Another mistake is to calculate the surface area of the whole cuboid and then forget to subtract the top. Students also sometimes convert 75 paise incorrectly; remembering that 75 paise equals Rs. 0.75 simplifies the multiplication. Always identify clearly which faces are to be covered before computing the area. Final Answer: Rs. 558

Question 14

If the radius of a circle is decreased by 50%, find the percentage decrease in the area of the circle.

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Introduction / Context: This question tests understanding of how the area of a circle depends on its radius. When the radius changes, the area does not change linearly but according to a square relationship. Here the radius is reduced by 50 percent, and we must determine the resulting percentage decrease in area. This is a standard conceptual problem in percentage change and geometry. Given Data / Assumptions: Initial radius of the circle = r (some positive value). Radius is decreased by 50 percent, so new radius = r / 2. Area formula for a circle = pi * r^2. We need the percentage decrease in area, not the new area itself alone. Concept / Approach: The area of a circle is proportional to the square of the radius. When the radius becomes half, the new area becomes (1/2)^2 of the original area, that is one quarter. After computing the new area as a fraction of the original, we can translate that fraction into a percentage decrease. This method avoids needing the actual numerical value of the radius. Step-by-Step Solution: Let original radius be r.Original area A1 = pi * r^2.New radius r2 = r / 2.New area A2 = pi * (r2)^2 = pi * (r / 2)^2 = pi * r^2 / 4.So A2 = (1/4) * A1.This means the new area is 25 percent of the original area.Percentage decrease in area = 100% - 25% = 75%. Verification / Alternative check: We can use simple numbers to verify. Suppose original radius r = 2 units. Then original area A1 = pi * 2^2 = 4pi. New radius r2 = 1 unit, giving area A2 = pi * 1^2 = pi. The decrease in area = 4pi - pi = 3pi. Percentage decrease = (3pi / 4pi) * 100% = 75%. This numerical example confirms the earlier algebraic reasoning. Why Other Options Are Wrong: 65% and 35% do not correspond to any standard squared ratio from halving. A 50% decrease would occur only if the quantity were directly proportional to the radius, not to the square of the radius. A 25% value would be the fraction of area remaining, not the decrease. Therefore only 75% correctly represents the reduction in area when the radius is halved. Common Pitfalls: Many students mistakenly assume that if radius is reduced by 50 percent, area also reduces by 50 percent. This ignores the fact that area depends on r^2, not on r. Another common mistake is to compute r - r/2 = r/2 and treat that as the area difference. Remember always to square the new radius when computing the new area, then compare areas to find the percentage change. Final Answer: 75%

Question 15

The length of a rectangle is 12 cm. If the area of the rectangle is 72 square centimetres, what is the width (breadth) of the rectangle?

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Introduction / Context: This is a basic mensuration problem involving a rectangle. For a rectangle, the area is the product of its length and width. Here, the length and area are given, and the width must be found. It tests the ability to rearrange the area formula and perform simple division, which is very common in quantitative aptitude exams. Given Data / Assumptions: Length L = 12 cm. Area A = 72 square centimetres. Formula for area of a rectangle: A = L * W, where W is width (breadth). We assume the rectangle has right angles and standard geometric properties. Concept / Approach: The approach is to use the area formula A = L * W. With A and L known, we can solve for W by dividing the area by the length. In symbols, W = A / L. This straightforward algebraic rearrangement allows us to find the unknown dimension directly. Once W is computed, we compare it with the given options to select the correct one. Step-by-Step Solution: Area A = L * W.Given A = 72 sq cm and L = 12 cm.So 72 = 12 * W.To find W, divide both sides by 12: W = 72 / 12.Compute 72 / 12 = 6.Therefore, width of the rectangle W = 6 cm. Verification / Alternative check: We can verify by recomputing the area using the found width. Take L = 12 cm and W = 6 cm. Then A = 12 * 6 = 72 sq cm, which matches the given area. This confirms that the width 6 cm is correct. Any other width would produce a different area and would not fit the given data. Why Other Options Are Wrong: A width of 5 cm would give area 12 * 5 = 60 sq cm, which is too small. A width of 3 cm gives area 36 sq cm, even smaller. A width of 4 cm results in area 48 sq cm, still not equal to 72. A width of 8 cm would give area 96 sq cm, which is larger than required. Therefore, only 6 cm produces the correct area of 72 sq cm. Common Pitfalls: Some students mix up the roles of length and width and mistakenly divide the length by the area. Others misinterpret the units, forgetting that area is measured in sq cm while length and width are in cm. Keeping the formula clear and performing the correct division ensures the right answer. Always check by multiplying the found width back with the length to see if the original area is obtained. Final Answer: 6 cm

Question 16

Find the perimeter of a rectangle whose length is 5 feet and width is 2 feet.

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Introduction / Context: This question checks understanding of the perimeter formula for a rectangle. Perimeter represents the total distance around the boundary of the shape. Given the length and width in feet, we must compute the perimeter in feet. Such problems are common in basic geometry and aptitude tests, helping build confidence in fundamental formulas. Given Data / Assumptions: Length L = 5 ft. Width W = 2 ft. Rectangle has four right angles and opposite sides are equal. Perimeter formula: P = 2(L + W). Concept / Approach: The perimeter of a rectangle is twice the sum of its length and width because there are two sides of each type. We first add L and W, then multiply the result by 2. The calculation is straightforward and involves only basic arithmetic operations. Keeping units consistent is important when expressing the final answer. Step-by-Step Solution: Length L = 5 ft, width W = 2 ft.Perimeter P = 2(L + W).Compute inside the brackets: L + W = 5 + 2 = 7.Now P = 2 * 7 = 14 ft.Thus, the perimeter of the rectangle is 14 feet. Verification / Alternative check: Another way is to add all sides individually. The sides are 5 ft, 2 ft, 5 ft, and 2 ft. Adding gives 5 + 2 + 5 + 2 = 14 ft. This matches the result from the formula approach. Using two consistent methods helps confirm that there is no arithmetic mistake. Why Other Options Are Wrong: 11 ft, 13 ft, and 15 ft can appear if one wrongly adds length and width only once or miscalculates the sum. For example, 5 + 2 + 2 = 9 or 5 + 5 + 2 = 12 could lead to nearby incorrect choices. 10 ft would arise if someone incorrectly used 2 * 5 instead of adding width. None of these match the correct total boundary length of the rectangle. Common Pitfalls: Common errors include forgetting to double the sum of length and width, or multiplying length and width instead of adding them. Some learners also confuse perimeter with area. Remember that perimeter is a one dimensional measure around the figure, while area measures the space inside. Visualising the rectangle and walking around its sides can help keep the concept clear. Final Answer: 14 ft

Question 17

Find the perimeter of a square whose side length s is 3 cm.

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Introduction / Context: This problem focuses on the perimeter of a square. A square is a special rectangle where all four sides are equal. Knowing the side length allows us to find the perimeter easily. This is a fundamental concept in geometry and a common question type in school level aptitude tests. Given Data / Assumptions: Side length of square s = 3 cm. All four sides of a square are equal in length. Perimeter of a square is the total length around its four sides. Concept / Approach: For a square with side length s, the perimeter P is 4 times the side, because there are four equal sides. Therefore P = 4 * s. The strategy is simply to multiply the given side length by 4 and express the result in centimetres. No further geometric reasoning is required beyond recognising the properties of a square. Step-by-Step Solution: Given side s = 3 cm.Perimeter P of a square = 4 * s.So P = 4 * 3 = 12 cm.Therefore the perimeter of the square is 12 centimetres. Verification / Alternative check: We can imagine the four sides explicitly: each side is 3 cm, and there are four of them. Adding them directly gives 3 + 3 + 3 + 3 = 12 cm. This confirms the formula based calculation and ensures there is no arithmetic mistake. Both methods yield the same perimeter, which increases confidence in the answer. Why Other Options Are Wrong: 13 cm, 14 cm, and 15 cm represent values close to the correct answer and could result from adding numbers incorrectly, such as 3 + 4 + 5 or 3 + 3 + 3 + 4. 9 cm might come from multiplying 3 by 3 instead of by 4, confusing area with perimeter. None of these values equal four times the side length, so they do not represent the true perimeter of the square. Common Pitfalls: Learners sometimes confuse the formulas for area and perimeter. Area of a square is s^2, whereas perimeter is 4 * s. Mixing them up leads to answers like 9 cm or 9 sq cm in this problem. Remember that perimeter is a linear measure around the boundary, while area covers the interior space. Keeping units in mind also helps: perimeter is in cm, area in sq cm. Final Answer: 12 cm

Question 18

The ratio of the area of a rectangle to its perimeter is 60:11. The length and breadth of the rectangle are in the ratio 6:5. Find the length of the rectangle in units.

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Introduction / Context: This question links the concepts of area, perimeter, and ratios for a rectangle. Instead of giving direct dimensions, it provides two different ratios: the ratio of area to perimeter, and the ratio of length to breadth. By expressing length and breadth in terms of a single variable, we can use these ratios to solve for the actual dimensions of the rectangle, specifically its length. Given Data / Assumptions: Area : Perimeter = 60 : 11. Length : Breadth = 6 : 5. Let length = 6x and breadth = 5x for some positive x. Area A = length * breadth = 6x * 5x. Perimeter P = 2(length + breadth) = 2(6x + 5x). Concept / Approach: Area and perimeter are both expressed in terms of x. Their ratio A:P must equal 60:11. By setting A / P = 60 / 11 and substituting A and P in terms of x, we obtain an equation that can be solved for x. Once x is found, the actual length 6x can be computed and compared with the options. This approach uses ratio algebra and basic mensuration formulas. Step-by-Step Solution: Let length L = 6x and breadth B = 5x.Area A = L * B = 6x * 5x = 30x^2.Perimeter P = 2(L + B) = 2(6x + 5x) = 2 * 11x = 22x.Given area to perimeter ratio A : P = 60 : 11, so A / P = 60 / 11.A / P in terms of x is (30x^2) / (22x) = (30x) / 22 = 15x / 11.Set 15x / 11 = 60 / 11.Therefore 15x = 60, so x = 60 / 15 = 4.Length L = 6x = 6 * 4 = 24 units. Verification / Alternative check: With x = 4, breadth B = 5 * 4 = 20 units. Area A = 24 * 20 = 480 square units. Perimeter P = 2(24 + 20) = 2 * 44 = 88 units. The ratio A : P = 480 : 88 simplifies by dividing both by 8, giving 60 : 11, exactly as given. This confirms that L = 24 units is consistent with all the information in the question. Why Other Options Are Wrong: If L = 40 units, breadth would be 40 * (5 / 6) ≈ 33.33 units, giving an area and perimeter ratio that does not simplify to 60:11. Lengths of 30 or 13 units also fail when the ratio of area to perimeter is computed. A length of 20 units would force breadth to be less than required by the 6:5 ratio. Only 24 units satisfies both given ratios simultaneously. Common Pitfalls: Some students misread the ratio 60:11 and set A = 60 and P = 11 directly, ignoring the scaling. Others mistake which ratio applies to which pair of quantities or forget to reduce A / P in terms of x properly. Carefully writing out A and P in terms of x and then equating the ratios ensures a systematic solution. Always verify the final dimensions by recomputing both area and perimeter ratios. Final Answer: 24 units

Question 19

The base of a parallelogram is twice its height. If the area of the parallelogram is 338 square centimetres, find the height of the parallelogram in centimetres.

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Introduction / Context: This question involves the area formula for a parallelogram and a simple algebraic relationship between base and height. The base is given as twice the height, and the area is known. By expressing the base in terms of height, we can form an equation and solve for the height. Such problems reinforce the connection between geometry and basic algebra. Given Data / Assumptions: Let height of the parallelogram = h cm. Base of the parallelogram = 2h cm (twice the height). Area of the parallelogram A = 338 sq cm. Area formula: A = base * height. Concept / Approach: The area of a parallelogram is given by the product of its base and height. Here, both base and height are expressed in terms of h. Substituting into the area formula, we get a simple quadratic equation in h. Solving this equation by taking a square root gives the height. Since height is a length, we take the positive root only. Step-by-Step Solution: Let height = h cm. Then base = 2h cm.Area A = base * height = 2h * h = 2h^2.Given A = 338 sq cm, so 2h^2 = 338.Divide both sides by 2: h^2 = 338 / 2 = 169.Therefore h = sqrt(169) = 13 cm (taking the positive value).Thus, the height of the parallelogram is 13 cm. Verification / Alternative check: With height h = 13 cm, base = 2h = 26 cm. Area = base * height = 26 * 13. Compute 26 * 13 = (20 * 13) + (6 * 13) = 260 + 78 = 338 sq cm, which matches the given area. This confirms that the height of 13 cm is consistent with all the data in the question. Why Other Options Are Wrong: If h = 14 cm, base would be 28 cm and area = 28 * 14 = 392 sq cm, too large. If h = 16 cm, area would be 32 * 16 = 512 sq cm. A height of 12 cm would give area 24 * 12 = 288 sq cm, which is too small. A height of 18 cm gives 36 * 18 = 648 sq cm. None of these areas equal 338 sq cm, so these options are incorrect. Common Pitfalls: A common mistake is to add base and height instead of multiplying them, or to misinterpret the phrase base is twice the height. Some learners also forget to divide by 2 when solving 2h^2 = 338, resulting in an incorrect value for h^2. Always write down the area formula clearly, substitute, simplify, and then take the square root carefully to avoid these errors. Final Answer: 13 cm

Question 20

What is the measure of each interior angle of a regular decagon (a regular polygon with 10 equal sides)?

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Introduction / Context: This question checks knowledge of the relationship between the number of sides of a regular polygon and its interior angles. A regular decagon is a polygon with 10 equal sides and 10 equal interior angles. Using the standard formula for the interior angle of a regular n sided polygon, we can find the required measure in degrees. Given Data / Assumptions: The polygon is a regular decagon, so n = 10 sides. All interior angles of a regular polygon are equal. Sum of interior angles of an n sided polygon = (n - 2) * 180 degrees. Each interior angle of a regular polygon = sum of interior angles / n. Concept / Approach: First we compute the total sum of interior angles using the formula (n - 2) * 180. Then we divide this sum by n to get each interior angle, since in a regular polygon all interior angles are equal. Substituting n = 10 for a decagon leads to the required value. The key is to apply the formula carefully and simplify the fraction correctly. Step-by-Step Solution: Number of sides n = 10.Sum of interior angles S = (n - 2) * 180 = (10 - 2) * 180 = 8 * 180.Compute 8 * 180 = 1440 degrees.Each interior angle of a regular decagon = S / n = 1440 / 10.1440 / 10 = 144 degrees.Therefore each interior angle is 144°. Verification / Alternative check: Another way is to recall the formula for each interior angle directly: interior angle = ((n - 2) * 180) / n. Substituting n = 10 gives interior angle = (8 * 180) / 10 = 1440 / 10 = 144 degrees. Both the stepwise method and the direct formula produce the same result, confirming that 144° is correct. Why Other Options Are Wrong: 120° is the interior angle of a regular hexagon (n = 6), not a decagon. 140° and 150° do not correspond to any standard regular polygon with 10 sides. 160° would give a sum of 160 * 10 = 1600 degrees, which would imply (n - 2) * 180 = 1600 and does not yield an integer n. Only 144° matches the decagon formula exactly. Common Pitfalls: Students often confuse interior and exterior angles. For a regular polygon, each exterior angle is 360 / n, and sometimes learners mistakenly compute this instead. Others forget to divide the total sum by the number of sides. It is helpful to remember that interior angle plus exterior angle at each vertex equals 180 degrees, which for n = 10 gives 144 + 36 = 180, a useful consistency check. Final Answer: 144°

Area Questions