Question 1

In triangle PQR, angle Q is 90 degrees (right angled at Q) and angle R is 60 degrees. If side RQ has length 4√3 centimetres, what is the length of the hypotenuse PR in centimetres?

Accepted Answer

Introduction / Context: This problem uses the 30 60 90 special right triangle once again. When a right angled triangle has one 60 degree angle and one 30 degree angle, its side lengths follow a fixed ratio. Being comfortable with this ratio allows you to quickly find missing sides without using full trigonometric calculations every time. Given Data / Assumptions: - Triangle PQR is right angled at Q, so angle Q is 90 degrees. - Angle R is 60 degrees, so angle P is 30 degrees since the sum of angles in a triangle is 180 degrees. - Side RQ has length 4√3 cm. - We need to find PR, which is the hypotenuse of the triangle. Concept / Approach: In a 30 60 90 triangle, if we denote the side opposite 30 degrees as k, then the side opposite 60 degrees is k√3, and the hypotenuse opposite 90 degrees is 2k. In triangle PQR, side QR is opposite angle P, which is 30 degrees. Therefore QR corresponds to k. Once we identify k from the given length, we can directly compute the hypotenuse PR as 2k. Step-by-Step Solution: Step 1: Recognise that angle P is 30 degrees because 30 + 60 + 90 = 180 degrees. Step 2: Identify side QR as the side opposite angle P, which is 30 degrees, so QR corresponds to k. Step 3: Given QR = 4√3 cm, we set k = 4√3. Step 4: The hypotenuse PR in a 30 60 90 triangle is 2k. Step 5: Substitute k = 4√3 to get PR = 2 * 4√3 = 8√3 cm. Verification / Alternative check: We can cross check using a trigonometric ratio. For angle R = 60 degrees, cos 60 degrees = adjacent / hypotenuse = RQ / PR. Since cos 60 degrees is 1 / 2, we get 1 / 2 = (4√3) / PR. Therefore PR = 8√3 cm, consistent with the special triangle ratio approach. Why Other Options Are Wrong: 8 cm is too small, because if PR were 8 cm, then cos 60 degrees would be (4√3) / 8 = √3 / 2, not 1 / 2. 4 cm and 8 / √3 cm are both smaller than the given leg 4√3 cm, which is impossible for a hypotenuse. 12 cm does not satisfy the 30 60 90 side ratio when QR is 4√3 cm. Common Pitfalls: Learners sometimes incorrectly assign the role of k or k√3 to the wrong side by mixing up which side is opposite each angle. Another mistake is to try to use Pythagoras without identifying all sides, which can lead to extra steps and errors. Always mark the angles clearly and then map each side to the correct ratio in the 30 60 90 triangle pattern. Final Answer: The hypotenuse PR has length 8√3 cm.

Question 2

Triangle ABC is isosceles with AB = AC and angle A = 40 degrees. Side BC is produced at both ends to points D and E, forming exterior angles ∠ABD and ∠ACE. The bisectors of these exterior angles meet at point O. What is the measure of angle BOC?

Accepted Answer

Introduction / Context: This problem involves an isosceles triangle, exterior angles and angle bisectors intersecting at a point outside the triangle. It is more challenging than direct angle sum questions because it combines several geometry ideas. Solving it requires careful analysis of interior and exterior angle relationships and an understanding of symmetry in isosceles triangles. Given Data / Assumptions: - Triangle ABC is isosceles with AB = AC. - Angle A is 40 degrees. - Side BC is extended beyond B to D and beyond C to E. - Exterior angles at B and C are ∠ABD and ∠ACE respectively. - The bisectors of these exterior angles meet at point O. - We need to find the measure of angle BOC at the intersection of these bisectors. Concept / Approach: In an isosceles triangle with AB = AC, the base angles at B and C are equal. The interior angle sum gives the base angles when angle A is known. Each exterior angle is supplementary to its corresponding interior angle. The bisectors of the exterior angles at B and C form equal angles with each side involved. Due to symmetry of the isosceles triangle, point O lies on the perpendicular bisector of BC, which simplifies the angle at O. We can use these facts and some geometric reasoning to find angle BOC. Step-by-Step Solution: Step 1: In triangle ABC, sum of angles is 180 degrees, so A + B + C = 180 degrees. Step 2: Given AB = AC, triangle ABC is isosceles with equal base angles B and C. Step 3: Angle A is 40 degrees, so B + C = 180 - 40 = 140 degrees. Therefore B = C = 70 degrees. Step 4: Exterior angle ∠ABD at B is supplementary to interior angle B, so ∠ABD = 180 - 70 = 110 degrees. Similarly, exterior angle ∠ACE at C is also 110 degrees. Step 5: The bisectors of these 110 degree exterior angles divide each into two angles of 55 degrees. So each bisector makes a 55 degree angle with AB or AC and a 55 degree angle with BC extended. Step 6: Because the triangle is symmetric about the axis through A and the midpoint of BC, the two bisectors are symmetric and intersect on this axis. When you compute the angle at their intersection BOC using geometry or a coordinate approach, you obtain ∠BOC = 70 degrees. Verification / Alternative check: An alternative way is to use a coordinate geometry construction. Place B and C symmetrically on the x axis and A above the axis so that AB = AC and angle A is 40 degrees. Compute the equations of the exterior angle bisectors at B and C and find their intersection point O. Then find angle BOC using vector dot product and inverse cosine. This method also yields 70 degrees, confirming the exact value found by geometric reasoning. Why Other Options Are Wrong: Angles 80, 100, 110 and 55 degrees do not satisfy the symmetric configuration and the way the exterior bisectors intersect. For example, 110 degrees is the size of the exterior angles themselves before bisection, not the angle at O. 55 degrees appears inside the construction but is not the final central angle BOC. Common Pitfalls: Learners often confuse interior and exterior angles or forget that the bisector divides the exterior angle, not the interior one. Another mistake is to assume that the angle at the intersection of bisectors is directly equal to one of the base angles without proper reasoning. Drawing a clear symmetric diagram and systematically tracking each angle helps avoid these errors. Final Answer: The measure of angle BOC is 70 degrees.

Question 3

A right triangular prism has an equilateral triangle as its base. The side of the equilateral triangular base is 15 centimetres and the height of the prism is 20√3 centimetres. What is the volume of the prism in cubic centimetres?

Accepted Answer

Introduction / Context: This question deals with the volume of a right triangular prism whose base is an equilateral triangle. The volume of any prism is the area of its base multiplied by its height. When the base is an equilateral triangle, we use a specific formula for the area of an equilateral triangle in terms of its side length. Such problems are common in solid geometry and engineering applications. Given Data / Assumptions: - The base of the prism is an equilateral triangle. - Side length of the equilateral triangle is 15 cm. - Height of the prism is 20√3 cm. - The prism is right, meaning the lateral edges are perpendicular to the base. - We need to find the volume in cubic centimetres. Concept / Approach: Volume of a prism is given by: Volume = area of base * height For an equilateral triangle with side a, the area is: Area = (√3 / 4) * a^2 After computing the area of the triangular base using a = 15 cm, we multiply this area by the prism height 20√3 cm to obtain the volume. Step-by-Step Solution: Step 1: Compute the area of the equilateral triangular base using a = 15 cm. Step 2: Area of base = (√3 / 4) * a^2 = (√3 / 4) * 15^2 = (√3 / 4) * 225. Step 3: Simplify this to get area of base = (225√3) / 4 square centimetres. Step 4: Volume of the prism is area of base multiplied by height: Volume = (225√3 / 4) * 20√3. Step 5: Multiply √3 * √3 = 3, so Volume = (225 * 20 * 3) / 4 = (225 * 60) / 4 = 225 * 15 = 3375 cubic centimetres. Verification / Alternative check: We can also group the constants differently: (225 / 4) * 20 = 225 * 5 = 1125. Then Volume = 1125 * (√3 * √3) = 1125 * 3 = 3375 cm3. This provides a quick mental cross check and shows that there was no arithmetic slip in the original calculation. Why Other Options Are Wrong: 1125 cm3 is simply the intermediate product before multiplying by the additional factor of 3 from √3 * √3. 4500 cm3 and 6750 cm3 result from miscalculating the multiplication or missing one of the factors in the formula. 2250 cm3 arises if someone incorrectly uses half the height or misapplies the area formula of the equilateral triangle. Common Pitfalls: A common mistake is to forget the √3 factor in the area of an equilateral triangle or to use the formula for the area of a right triangle instead. Another pitfall is to treat the prism as a cylinder or some other solid. Always remember that volume of any prism is the base area times the height and then compute the base area correctly based on the given shape. Final Answer: The volume of the right triangular prism is 3375 cm3.

Question 4

The radius and height of a solid right circular cylinder are both increased by 2 percent. Approximately what is the percentage increase in the volume of the cylinder?

Accepted Answer

Introduction / Context: This question tests your understanding of percentage change in a compound situation where more than one dimension of a solid is changed. The volume of a cylinder depends on both its radius and its height. When both are changed by the same percentage, the effect on volume is not simply the sum of the percentage changes. Instead, we must use the idea of successive percentage changes or apply the formula directly with the new dimensions. Given Data / Assumptions: - Original radius of the cylinder is r. - Original height of the cylinder is h. - Radius is increased by 2 percent, so new radius is 1.02r. - Height is also increased by 2 percent, so new height is 1.02h. - Volume of a cylinder is V = π * r^2 * h. - We need the approximate percentage increase in volume. Concept / Approach: The original volume is V = π * r^2 * h. After the changes, the new volume V new is: V new = π * (1.02r)^2 * (1.02h) This simplifies to V new = π * r^2 * h * (1.02^2 * 1.02) = V * (1.02^3). So the factor by which the volume increases is 1.02^3. We can either compute 1.02^3 directly or use the successive percentage change concept to estimate the overall percentage increase. Step-by-Step Solution: Step 1: Write original volume V = π * r^2 * h. Step 2: After a 2 percent increase, the new radius is r new = 1.02r and new height is h new = 1.02h. Step 3: New volume V new = π * (1.02r)^2 * (1.02h) = π * r^2 * h * 1.02^3. Step 4: Compute 1.02^3 = 1.02 * 1.02 * 1.02 ≈ 1.0404 * 1.02 ≈ 1.061208. Step 5: The factor 1.061208 means the volume has increased by about 6.1208 percent, which rounds to approximately 6.12 percent. Verification / Alternative check: You can also use the idea of successive percentage change in three steps. First, increasing r by 2 percent increases r^2 by approximately 4.04 percent (since area of a circle depends on r^2). Then increasing the height by 2 percent multiplies the volume by 1.02. So total factor is approximately 1.0404 * 1.02 ≈ 1.061208, giving about 6.12 percent increase, which matches the earlier calculation. Why Other Options Are Wrong: 5.88 and 6.76 are close but result from rounding at intermediate steps or incorrect compounding. 3.34 and 4.04 arise if someone simply adds percentages incorrectly or only accounts for the change in radius or in radius squared without including the height change. Common Pitfalls: A very common mistake is to just add the percentage changes, for example 2 percent for radius and 2 percent for height, to get 4 percent, which ignores the compounding effect. Others might increase the radius squared incorrectly or forget that volume depends on r^2 and h. Always convert percentage increases into multipliers, apply them to the formula and then convert the final factor back into a percentage. Final Answer: The approximate percentage increase in the volume of the cylinder is 6.12 percent.

Question 5

A solid sphere of radius 21 centimetres is cut by three mutually perpendicular planes through its centre, one along each coordinate axis, so that it is divided into 8 identical parts. What is the total surface area of each part, in square centimetre…

Accepted Answer

Introduction / Context: This question involves a sphere that is divided into eight equal parts by three mutually perpendicular planes through its centre. Each part is often called an octant of the sphere. The surface area of each part includes a portion of the original curved surface and several flat circular pieces created by the cuts. Understanding how these areas are distributed is key to finding the total surface area of each part. Given Data / Assumptions: - Radius of the sphere r = 21 cm. - The sphere is cut by three planes through its centre, each perpendicular to the others, similar to the coordinate planes x = 0, y = 0 and z = 0. - These cuts divide the sphere into 8 congruent parts or octants. - We need the total surface area of one such part, including both curved and planar surfaces. Concept / Approach: The total surface area of the full sphere is: S sphere = 4 * π * r^2 When the sphere is divided into 8 octants by three perpendicular planes through the centre, each octant has: - One eighth of the original curved surface area of the sphere. - Three flat faces, each coming from one of the circular cross sections created by the planes. Each plane cuts the sphere along a great circle of radius r, whose area is π * r^2. For a single octant, each of these three disks contributes one quarter of its area, because each disk is divided into four equal sectors by the other two planes. So for each octant, total area = (curved area for that octant) + (area of three quarter disk sectors). Step-by-Step Solution: Step 1: Compute total surface area of the sphere: S sphere = 4 * π * r^2. Step 2: With r = 21 cm and π taken as 22 / 7, S sphere = 4 * 22 / 7 * 21^2 = 4 * 22 / 7 * 441. Step 3: Simplify 441 / 7 = 63, so S sphere = 4 * 22 * 63 = 88 * 63 = 5544 square centimetres. Step 4: The curved surface area for one octant is one eighth of this: 5544 / 8 = 693 square centimetres. Step 5: Each great circle formed by a cutting plane has area π * r^2 = 22 / 7 * 21^2 = 22 / 7 * 441 = 22 * 63 = 1386 square centimetres. Each such disk is divided into four equal sectors by the other two planes, so one octant includes one quarter of each disk. Step 6: The area from one disk in a single octant is 1386 / 4 = 346.5 square centimetres. Since each octant touches three such disks, total planar area per octant is 3 * 346.5 = 1039.5 square centimetres. Step 7: Add the curved and planar areas: total area per part = 693 + 1039.5 = 1732.5 square centimetres. Verification / Alternative check: An alternative perspective is to note that the total new area created by the cuts across all octants must equal the combined areas of the three great circle disks, each of area 1386. Each disk is cut into four sectors and each octant contains three of these sectors. Accounting for all eight octants, you recover exactly the three full disks. Combining this with the fact that the curved area is evenly distributed among the eight octants again leads to the same total area per octant. Why Other Options Are Wrong: 844.5 and 1039.5 represent partial contributions from either the curved portion or the planar faces alone, not both together. 1115.6 is not consistent with any clean fraction of the known areas. 1386 is the area of one complete great circle and does not represent the total area of an octant. Common Pitfalls: Learners often forget to include the flat faces introduced by the cutting planes and only count the curved surface, or they miscount how many sectors from each disk belong to one octant. Another error is to assume that only one disk contributes to each part, instead of three. Carefully visualising how the planes partition the sphere helps avoid these mistakes. Final Answer: The total surface area of each part is 1732.5 square centimetres.

Question 6

The ratio between the length and the breadth of a rectangular park is 3 : 2. A man cycles along the boundary of the park at a speed of 12 kilometres per hour and completes one round in 8 minutes. What is the area of the park in square metres?

Accepted Answer

Introduction / Context: This question combines speed, time and distance concepts with basic geometry of a rectangle. The man cycling around the boundary covers the perimeter of the park. Once we know the perimeter and the ratio of the sides, we can find the actual length and breadth and then compute the area. Such mixed concept questions are very common in aptitude exams. Given Data / Assumptions: - Ratio of length to breadth of the rectangular park is 3 : 2. - Cycling speed of the man is 12 km per hour. - Time to complete one round along the boundary is 8 minutes. - He travels exactly one full perimeter in that time. - We need to find the area of the park in square metres. Concept / Approach: The distance travelled when moving at constant speed is: Distance = speed * time Here, the distance for one round equals the perimeter of the rectangle. Once we find the perimeter, say 2(L + B), we use the given ratio L : B = 3 : 2 to express L and B in terms of a common factor k. Then we solve for k and compute the area L * B in square metres. Step-by-Step Solution: Step 1: Convert cycling speed into metres per minute. Speed is 12 km per hour = 12 * 1000 metres per 60 minutes = 12000 / 60 = 200 metres per minute. Step 2: Time for one round is 8 minutes, so distance travelled in one round is 200 * 8 = 1600 metres. Step 3: This distance is the perimeter of the park, so 2(L + B) = 1600, which gives L + B = 800 metres. Step 4: Using the ratio L : B = 3 : 2, let L = 3k and B = 2k. Then L + B = 3k + 2k = 5k = 800, so k = 160. Step 5: Therefore length L = 3 * 160 = 480 metres and breadth B = 2 * 160 = 320 metres. Area of the park is L * B = 480 * 320 = 153600 square metres. Verification / Alternative check: We can quickly verify the perimeter: 2(L + B) = 2(480 + 320) = 2 * 800 = 1600 metres, which matches the distance cycled in 8 minutes at 200 metres per minute. This confirms that the side lengths and thus the area are consistent with the speed and time data. Why Other Options Are Wrong: 15360 and 30720 are too small by factors of 10 or 5, arising from errors in unit conversion or forgetting that both length and breadth are large values. 307200 is double the correct area, often obtained when someone uses the perimeter directly in the area formula incorrectly. 76800 is exactly half the correct area and may appear when one of the side lengths is mistakenly taken as k rather than 3k or 2k. Common Pitfalls: A typical mistake is to neglect converting time into hours or minutes consistently, leading to wrong perimeter. Another common error is misusing the 3 : 2 ratio and assigning the wrong multiples to length and breadth. Ensure proper unit conversion and carefully apply the ratio to avoid these issues. Final Answer: The area of the park is 153600 square metres.

Question 7

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

Accepted Answer

Introduction / Context: This problem examines how a small percentage error in measuring a linear dimension affects the percentage error in an area calculation. The area of a square depends on the square of its side length, so a small error in side measurement can produce a larger error in the area. Understanding this relationship is important not only in exams but also in practical measurement tasks and engineering calculations. Given Data / Assumptions: - Let the true side length of the square be s units. - There is an error of 2 percent in excess while measuring the side, so measured side is 1.02s. - True area of the square is A true = s^2. - Measured area of the square is A measured = (1.02s)^2. - We need the percentage error in area, defined as (A measured - A true) / A true * 100 percent. Concept / Approach: When a measurement is changed by a small percentage, any quantity that depends on the square of that measurement changes approximately by twice that percentage, plus a small quadratic term. In this case, since area = side^2, we expect the area to increase by roughly 4 percent when the side increases by 2 percent. To be precise, we must square the factor 1.02 and then subtract 1 to find the exact relative increase in area and convert it into a percentage. Step-by-Step Solution: Step 1: Express the measured side as s measured = 1.02s. Step 2: True area is A true = s^2. Step 3: Measured area is A measured = (1.02s)^2 = 1.02^2 * s^2. Step 4: Compute 1.02^2 = 1.0404. Step 5: Therefore A measured = 1.0404 * s^2, meaning the area has increased by a factor of 1.0404. Step 6: The percentage error in area is (1.0404 - 1) * 100 percent = 0.0404 * 100 percent = 4.04 percent. Verification / Alternative check: As a rough mental check, recall that for small errors, the approximate error in area is about twice the error in side, so 2 percent in side suggests around 4 percent in area. The exact computation adds a small term equal to (2 percent)^2 = 0.04 percent, giving 4.04 percent, which matches our detailed calculation. This shows that the approximation rule is close but not exact, and the precise result is slightly higher than 4 percent. Why Other Options Are Wrong: 2 percent is just the error in the side, not the area, and therefore underestimates the effect on area. 2.02 percent and 3.96 percent are incorrect approximations that do not correspond to squaring the factor 1.02. 4 percent ignores the additional small term from squaring 1.02 and is only an approximation, not the exact error asked in the question. Common Pitfalls: Many students assume the percentage error in area is simply double the percentage error in side and answer 4 percent without checking the exact calculation. Others square 2 instead of 1.02 and get completely wrong values. Always represent the new measurement as a multiplicative factor, compute the new derived quantity and then convert the relative change into a percentage. Final Answer: The percentage error in the calculated area of the square is 4.04 percent.

Question 8

In a rectangle, the ratio of its perimeter to its breadth is 5 : 1. If the area of the rectangle is 216 sq. cm, what is the length of the rectangle?

Accepted Answer

Introduction / Context: This problem is a standard mensuration question on rectangles. It tests how well you can connect perimeter, area, length and breadth, and form simple algebraic equations from the given ratio. Such questions are common in aptitude exams because they require both conceptual understanding and careful manipulation of formulas. Given Data / Assumptions: - The figure is a rectangle with length and breadth as its sides. - Ratio of perimeter to breadth is 5 : 1. - Area of the rectangle is 216 sq. cm. - We are required to find the length of the rectangle in centimetres. - All measurements are in centimetres and the rectangle is a normal plane figure with right angles. Concept / Approach: The key formulas for a rectangle are: - Perimeter = 2 * (length + breadth). - Area = length * breadth. We use the ratio between perimeter and breadth to create one equation and the given area to create a second equation. Solving these simultaneously gives the actual numerical values of length and breadth. Step-by-Step Solution: Step 1: Let breadth = b cm and length = l cm.Step 2: Perimeter P = 2 * (l + b). The ratio P : b = 5 : 1, so 2 * (l + b) / b = 5.Step 3: This gives 2 * (l + b) = 5b, so 2l + 2b = 5b, hence 2l = 3b and l = 3b / 2.Step 4: Area A = l * b = 216 sq. cm. Substitute l = 3b / 2 to get (3b / 2) * b = 216.Step 5: This simplifies to 3b^2 / 2 = 216, so 3b^2 = 432 and b^2 = 144.Step 6: Therefore b = 12 cm (we take the positive value for a length).Step 7: Now l = 3b / 2 = 3 * 12 / 2 = 18 cm. Verification / Alternative check: Check perimeter: 2 * (18 + 12) = 2 * 30 = 60 cm. Ratio P : b = 60 : 12 = 5 : 1, so the ratio condition is satisfied. Area check: 18 * 12 = 216 sq. cm, which matches the given area. Both conditions hold, so the solution is consistent. Why Other Options Are Wrong: 24 cm would give area 24 * 9 (if breadth were adjusted to keep ratio) which does not equal 216 sq. cm. 12 cm cannot be the length because it is actually the breadth obtained from the equations. 20 cm does not satisfy both the ratio and area conditions simultaneously. Data inadequate is incorrect because the given information is sufficient to determine unique values for length and breadth. Common Pitfalls: A common mistake is to treat the ratio 5 : 1 as length : breadth or perimeter : length instead of perimeter : breadth, which completely changes the equations. Another frequent error is mishandling the algebra when solving 2l + 2b = 5b. Some students also forget to verify both the area and the ratio, leading to incorrect answers that partially fit the data. Final Answer: Therefore, the length of the rectangle is 18 cm.

Question 9

The difference between the length and breadth of a rectangle is 23 m. If its perimeter is 206 m, then what is the area of the rectangle in square metres?

Accepted Answer

Introduction / Context: This question involves a rectangle where you know the difference between length and breadth and the total perimeter. From these two relationships you must find the actual dimensions and then compute the area. It tests your ability to translate verbal information into equations and use basic mensuration formulas, which is very important for aptitude and competitive exams. Given Data / Assumptions: - The figure is a rectangle with length l and breadth b measured in metres. - Length minus breadth is 23 m. - Perimeter of the rectangle is 206 m. - We are required to find the area in square metres. - All sides meet at right angles as in a standard rectangle. Concept / Approach: For a rectangle, perimeter P and area A are given by: - P = 2 * (l + b). - A = l * b. The difference l - b = 23 provides one equation, and the perimeter gives the second. Solving these two linear equations provides l and b. Once the dimensions are known, area is found by direct multiplication. Step-by-Step Solution: Step 1: Let length = l m and breadth = b m.Step 2: Given l - b = 23.Step 3: Perimeter P = 2 * (l + b) = 206, so l + b = 206 / 2 = 103.Step 4: We now have two equations: l - b = 23 and l + b = 103.Step 5: Add the two equations to eliminate b: (l - b) + (l + b) = 23 + 103 which gives 2l = 126.Step 6: So l = 126 / 2 = 63 m. Substitute into l + b = 103 to get 63 + b = 103, hence b = 40 m.Step 7: Area A = l * b = 63 * 40 = 2520 sq. m. Verification / Alternative check: Check the perimeter with the found dimensions: 2 * (63 + 40) = 2 * 103 = 206 m, which matches the given perimeter. Check the difference: 63 - 40 = 23 m, which matches the given difference. Thus, the values are correct and the area 2520 sq. m is consistent. Why Other Options Are Wrong: 1520 sq. m, 2420 sq. m, 2480 sq. m and 2620 sq. m do not arise from any pair of integer dimensions that simultaneously satisfy both the perimeter and the length minus breadth conditions. They are merely close distractors meant to trap approximate or incorrect calculations. Common Pitfalls: One common mistake is to divide the perimeter by 4 to get a single side, which only works for a square. Another is to mis-handle the pair of equations and compute l + b or l - b incorrectly. Forgetting to verify both conditions can also lead to accepting an incorrect pair of dimensions. Always check both the difference and the perimeter before finalising the area. Final Answer: Therefore, the area of the rectangle is 2520 sq. m.

Question 10

The length of a rectangle is halved while its breadth is tripled. By what percentage does the area of the rectangle change?

Accepted Answer

Introduction / Context: This problem checks your understanding of how proportional changes in dimensions affect the area of a rectangle. Instead of calculating with actual numbers, you work algebraically with factors. These types of percentage change questions are very common in quantitative aptitude tests because they require clear conceptual thinking rather than just formula memorisation. Given Data / Assumptions: - Initial length of the rectangle is L units. - Initial breadth of the rectangle is B units. - Length is reduced to half of its original value. - Breadth is increased to three times its original value. - We must find the percentage change in area, indicating increase or decrease. Concept / Approach: The area of a rectangle is given by Area = length * breadth. When dimensions change by certain factors, the new area becomes the product of the new length and new breadth. The ratio new area / old area gives the overall factor by which area changes. Converting this factor into a percentage change tells us whether area increases or decreases and by how much. Step-by-Step Solution: Step 1: Original area A_original = L * B.Step 2: New length = L / 2 (because length is halved).Step 3: New breadth = 3B (because breadth is tripled).Step 4: New area A_new = (L / 2) * (3B) = (3L * B) / 2.Step 5: Express new area as a multiple of old area: A_new / A_original = (3L * B / 2) / (L * B) = 3 / 2.Step 6: The factor 3 / 2 equals 1.5, which means the new area is 1.5 times the original area.Step 7: An increase from 1 to 1.5 is an increase of 0.5, that is 50 percent. Verification / Alternative check: You can also pick simple numbers. Suppose initially L = 2 units and B = 2 units, then the original area is 4 square units. After the change, new length is 1 unit and new breadth is 6 units, so new area is 6 square units. From 4 to 6 is an increase of 2 on 4, which is 2 / 4 * 100 percent = 50 percent increase. This numerical check confirms the algebraic reasoning. Why Other Options Are Wrong: 25 percent increase or 25 percent decrease correspond to factors like 1.25 or 0.75, which are not obtained here. A 50 percent decrease would mean the area becomes half of the original, which is also not correct. No change in area would require the product of the dimension changes to be 1, but here the product factor is 1.5, not 1. Common Pitfalls: Students often incorrectly add or subtract percentage changes of individual dimensions instead of multiplying the factors. For example, they may think half and triple cancel out in some way. The correct method is always to treat each dimension change as a multiplicative factor and then multiply these factors to get the final effect on area. Confusing increases and decreases or forgetting to convert fractions into percentages are also common errors. Final Answer: Therefore, the area of the rectangle undergoes a 50% increase.

Question 11

A rectangular tank is 25 m long, 12 m wide and 6 m deep. If its walls and bottom are to be plastered at the rate of 75 paise per sq. m, what will be the total cost of plastering?

Accepted Answer

Introduction / Context: This mensuration question deals with the surface area of a rectangular tank and the cost of plastering that surface. It checks whether you can correctly identify which surfaces need to be plastered and then compute the total cost using a given rate per unit area. Such real life cost calculation problems are frequently used in exams to test practical application of geometry. Given Data / Assumptions: - Length of the tank = 25 m. - Width of the tank = 12 m. - Depth (or height) of the tank = 6 m. - Plastering is to be done on the four vertical walls and the horizontal bottom only. - Rate of plastering = 75 paise per sq. m, that is Rs. 0.75 per sq. m. Concept / Approach: The total plastered area is the sum of the areas of the four walls and the bottom. For a rectangular tank, opposite walls have equal areas. Once total area in square metres is found, multiply it by the cost per square metre to obtain the total expenditure. No plastering is done on the open top surface, so it is not included in the calculation. Step-by-Step Solution: Step 1: Area of the bottom = length * width = 25 * 12 = 300 sq. m.Step 2: There are two walls of size 25 m by 6 m each. Combined area of these two walls = 2 * (25 * 6) = 2 * 150 = 300 sq. m.Step 3: The other two walls are 12 m by 6 m each. Combined area of these walls = 2 * (12 * 6) = 2 * 72 = 144 sq. m.Step 4: Total plastered area = bottom area + area of all four walls = 300 + 300 + 144 = 744 sq. m.Step 5: Rate of plastering = Rs. 0.75 per sq. m.Step 6: Total cost = 744 * 0.75.Step 7: 744 * 0.75 = 744 * 3 / 4 = 558, so the total cost is Rs. 558. Verification / Alternative check: You can quickly verify by checking individual wall areas again: two large walls contribute 300 sq. m, two smaller walls add 144 sq. m and the base adds another 300 sq. m, confirming 744 sq. m. Multiplying by 0.75 can also be checked by first finding 75 percent of 744, which is indeed 558, because 50 percent is 372 and 25 percent is 186, and 372 + 186 = 558. Why Other Options Are Wrong: Rs. 456 and Rs. 502 are obtained if parts of the surface are mistakenly omitted or if the rate is applied incorrectly. Rs. 600 and Rs. 744 result from using the wrong rate or incorrectly taking one full rupee instead of 75 paise per sq. m or confusing area with perimeter. None of these match the correct combined effect of both area and rate. Common Pitfalls: Students often forget that the top surface is open and should not be plastered, while the bottom must be included. Another mistake is to add only one wall of each type instead of both equal walls, which halves their contribution. Finally, misinterpreting 75 paise as Rs. 75 or Rs. 1 can cause large numerical errors. Carefully reading the units and surfaces involved prevents such mistakes. Final Answer: Hence, the total cost of plastering the tank is Rs. 558.

Question 12

A cistern 6 m long and 4 m wide contains water up to a depth of 1 m 25 cm. What is the total area of the wet surface inside the cistern in square metres?

Accepted Answer

Introduction / Context: This question is about the concept of wet surface area in a container partially filled with water. It tests whether you can distinguish between the surfaces that are in contact with water and those that are dry, and then compute the total area of those wetted surfaces. This is a practical application of surface area of cuboids in mensuration. Given Data / Assumptions: - The cistern is rectangular with length 6 m and width 4 m. - Water depth is 1 m 25 cm, which is 1.25 m. - The bottom of the cistern is fully covered by water. - The water wets the side walls up to a height of 1.25 m from the bottom. - We need the total area of all surfaces in contact with water in square metres. Concept / Approach: The wet surface includes the base area and the area of the vertical sides up to the water level. The top horizontal surface at the water level is not part of the wall or bottom, so it is not counted as surface area; it is simply the water surface. For a rectangular cistern, opposite walls are equal, and we can group them for easier calculation. Step-by-Step Solution: Step 1: Convert the depth to metres: 1 m 25 cm = 1.25 m.Step 2: Bottom wet area = length * width = 6 * 4 = 24 sq. m.Step 3: Two longer side walls each have dimensions 6 m by 1.25 m. Combined area of these walls = 2 * (6 * 1.25) = 2 * 7.5 = 15 sq. m.Step 4: The other two side walls each have dimensions 4 m by 1.25 m. Combined area of these walls = 2 * (4 * 1.25) = 2 * 5 = 10 sq. m.Step 5: Total wet surface area = bottom area + area of all four wetted walls = 24 + 15 + 10.Step 6: 24 + 15 + 10 = 49 sq. m. Verification / Alternative check: As a quick check, note that the total side area up to the water line should equal perimeter of the base multiplied by depth. Perimeter of base = 2 * (6 + 4) = 20 m. Side area up to water level = 20 * 1.25 = 25 sq. m. This equals 15 + 10 from the individual wall calculation, confirming the side area is correct. Adding the bottom area 24 gives 24 + 25 = 49 sq. m again. Why Other Options Are Wrong: 50, 51, 52 and 55 sq. m all result from small miscalculations, often caused by rounding or by mistakenly adding the water surface at the top. They may also occur if a student includes the total tank height instead of the actual water depth or omits one pair of walls when computing the wetted area. Common Pitfalls: A frequent error is to calculate the total surface area of the cistern, including dry portions, instead of only the wetted surfaces. Another mistake is failing to convert 1 m 25 cm correctly into 1.25 m, or forgetting that there are two walls of each type. Careful identification of which surfaces actually touch water and accurate unit conversion are essential to avoid these mistakes. Final Answer: The total area of the wet surface inside the cistern is 49 sq. m.

Question 13

A boat having a length of 3 m and a breadth of 2 m is floating on a lake. When a man steps into the boat, it sinks by 1 cm. What is the mass of the man, assuming the density of water is 1000 kg per cubic metre?

Accepted Answer

Introduction / Context: This question is a practical application of the principle of flotation. When a man steps into the boat, the boat sinks a little more to displace extra water whose weight is equal to the weight of the man. By computing the extra volume of water displaced, we can find the mass of the man. This type of problem links basic physics with simple volume calculations. Given Data / Assumptions: - Length of the boat = 3 m. - Breadth of the boat = 2 m. - Boat sinks by 1 cm when the man steps in. - 1 cm = 0.01 m. - Density of water is taken as 1000 kg per cubic metre. - The boat base is assumed to be approximately rectangular and in full contact with the water surface area considered for displacement. Concept / Approach: According to the principle of flotation, the weight of water displaced equals the total weight of the floating body and any additional load. Here, the extra volume of water displaced when the man enters the boat multiplied by the density of water gives the mass of the man. Volume displaced is calculated as base area of the boat multiplied by the additional sinking depth. Step-by-Step Solution: Step 1: Base area of the boat A = length * breadth = 3 * 2 = 6 sq. m.Step 2: Additional sinking depth h = 1 cm = 0.01 m.Step 3: Extra volume of water displaced V = A * h = 6 * 0.01 = 0.06 cubic metres.Step 4: Mass of water displaced m_water = density * volume = 1000 * 0.06.Step 5: m_water = 60 kg.Step 6: By flotation, the mass of the man must equal the mass of this extra displaced water.Step 7: Therefore, mass of the man = 60 kg. Verification / Alternative check: As a sense check, note that a volume of 0.06 cubic metres is 60 litres of water, because 1 cubic metre equals 1000 litres. The mass of 60 litres of water is approximately 60 kg, which is a very reasonable value for a human adult. This supports the correctness of the computed mass. Why Other Options Are Wrong: 12 kg would correspond to a volume displaced of only 0.012 cubic metres, which does not match the given sinking depth. Values like 48 kg, 72 kg and 90 kg would require different sinking depths or different boat dimensions. They are distractors that result from miscalculating area, forgetting to convert centimetres to metres or using the wrong density. Common Pitfalls: Common mistakes include using 1 cm as 0.1 m instead of 0.01 m, or forgetting that density is 1000 kg per cubic metre, not 1 kg per cubic metre. Some learners also mistakenly multiply area and depth in centimetres without converting to metres, which leads to a volume that is off by a factor of 10000. Always keep units consistent and remember that displacement volume times density gives mass. Final Answer: Thus, the mass of the man is 60 kg.

Question 14

A pyramid is constructed on the base of a cube such that its base coincides with the base of the cube and its vertex is at the centre of the cube. What fraction of the volume of the cube is occupied by this pyramid?

Accepted Answer

Introduction / Context: This problem involves three dimensional geometry and compares the volume of a pyramid to the volume of a cube. The pyramid has the same square base as the cube, and its vertex is at the centre of the cube. This configuration appears frequently in solid geometry, and the question tests whether you can correctly use the formula for the volume of a pyramid and relate it to the volume of the cube. Given Data / Assumptions: - The solid is a cube with side length a units. - A pyramid is built with base equal to the base face of the cube. - The vertex of the pyramid is located at the centre of the cube. - The height of the pyramid is the distance from the base of the cube to its centre, which is half the side length of the cube. - We are required to find the fraction of the cube volume that the pyramid occupies. Concept / Approach: The volume of a cube with side a is V_cube = a^3. The volume of a pyramid is given by V_pyramid = (1 / 3) * base area * height. In this configuration, the base area of the pyramid is the area of the square base of the cube, and its height is half the side of the cube. By writing both volumes in terms of a and forming a ratio, we can find the fraction of the cube volume that belongs to the pyramid. Step-by-Step Solution: Step 1: Let the side length of the cube be a units.Step 2: Volume of the cube V_cube = a^3.Step 3: Area of the square base of the cube = a^2.Step 4: The centre of the cube lies midway between its top and bottom faces, so the height of the pyramid h = a / 2.Step 5: Volume of the pyramid V_pyramid = (1 / 3) * base area * height = (1 / 3) * a^2 * (a / 2).Step 6: Simplify V_pyramid = (1 / 3) * (a^3 / 2) = a^3 / 6.Step 7: Fraction of the cube volume occupied by the pyramid = V_pyramid / V_cube = (a^3 / 6) / a^3 = 1 / 6. Verification / Alternative check: You can assign a simple numerical value to the cube side, for example a = 2 units. Then V_cube = 8 cubic units. Height of the pyramid is 1 unit, and its base area is 4 square units, so its volume is (1 / 3) * 4 * 1 = 4 / 3 cubic units. The ratio 4 / 3 divided by 8 equals (4 / 3) * (1 / 8) = 4 / 24 = 1 / 6, confirming the symbolic calculation. Why Other Options Are Wrong: Fractions like 1 / 3, 1 / 4, 1 / 5 and 1 / 8 would correspond to different combinations of base area and height. They might arise if someone mistakenly takes the height as the full side length of the cube, uses the wrong base area or forgets the factor 1 / 3 in the pyramid volume formula. None of these match the correct ratio derived from the proper formula and geometry of the situation. Common Pitfalls: Many students forget the factor 1 / 3 in the volume of a pyramid, incorrectly using base area * height instead of (1 / 3) * base area * height. Others may assume the height of the pyramid equals the full side of the cube rather than half, which happens because they misinterpret where the vertex lies. Carefully identifying base and height and recalling the correct formula avoids these mistakes. Final Answer: Thus, the pyramid occupies 1/6 of the volume of the cube.

Question 15

The surface areas of three faces of a cuboid that meet at a common vertex are 20 sq. m, 32 sq. m and 40 sq. m respectively. What is the volume of the cuboid in cubic metres?

Accepted Answer

Introduction / Context: This problem deals with a cuboid whose three pairwise perpendicular faces have known areas. The faces share a common vertex, so their areas correspond to products of the three edge lengths. The question tests your ability to use these face areas to deduce the edge lengths and thereby find the volume of the cuboid. It is a classic aptitude question on three dimensional geometry. Given Data / Assumptions: - Let the three edges meeting at a vertex be of lengths a, b and c metres. - Areas of the three faces at that vertex are: ab = 20 sq. m, bc = 32 sq. m and ca = 40 sq. m. - All angles between the edges are right angles as in a standard cuboid. - We need to find the volume of the cuboid in cubic metres, which is a * b * c. Concept / Approach: Given ab, bc and ca, we want abc. Notice that (ab) * (bc) * (ca) = a^2 * b^2 * c^2 = (abc)^2. So, by multiplying the three given face areas, we obtain (abc)^2, and taking the positive square root gives abc, the volume. This approach avoids finding each individual edge separately and is both elegant and efficient. Step-by-Step Solution: Step 1: Let ab = 20, bc = 32 and ca = 40 (all in sq. m).Step 2: Multiply the three expressions: (ab) * (bc) * (ca) = 20 * 32 * 40.Step 3: Left side becomes a^2 * b^2 * c^2 = (abc)^2.Step 4: Compute the product on the right: 20 * 40 = 800, and 800 * 32 = 25600.Step 5: Thus (abc)^2 = 25600.Step 6: Take the positive square root: abc = square root of 25600.Step 7: 25600 = 160^2, so abc = 160 cubic metres. Verification / Alternative check: We can individually recover one set of edge lengths. From ab = 20 and ca = 40, divide to get (ca) / (ab) = c / b = 40 / 20 = 2. Thus c = 2b. From bc = 32, substitute to get b * 2b = 32 so 2b^2 = 32 and b^2 = 16, giving b = 4 m. Then c = 8 m and a = 20 / b = 5 m. Volume = a * b * c = 5 * 4 * 8 = 160 cubic metres, confirming the earlier method. Why Other Options Are Wrong: 120, 140, 184 and 200 cubic metres do not satisfy the relationships implied by the three face areas. If you pick any of those volumes and attempt to factor them into edges that reproduce ab = 20, bc = 32 and ca = 40, you will fail. These values arise mainly from arithmetic mistakes in the multiplication or in extracting the square root. Common Pitfalls: Learners sometimes try to guess side lengths directly without using the elegant (abc)^2 relationship, which can lead to confusion. Another frequent mistake is miscalculating the product 20 * 32 * 40 or taking an incorrect square root of 25600. Ensuring careful multiplication and recognising 160^2 prevents errors. Remember that in physical geometry we only take the positive square root because edge lengths and volumes are positive quantities. Final Answer: Therefore, the volume of the cuboid is 160 cubic metres.

Question 16

In a right-angled triangle, the two sides forming the right angle are 3 cm and 4 cm long. What is the area of its circumcircle in square centimetres?

Accepted Answer

Introduction / Context: This geometry question combines properties of right angled triangles with the concept of a circumcircle. You are given the two perpendicular sides of the triangle and are asked to find the area of its circumcircle. It tests knowledge of the relationship between the hypotenuse and the circumradius in a right triangle, as well as correct use of the area formula for a circle. Given Data / Assumptions: - The triangle is right angled. - The legs forming the right angle measure 3 cm and 4 cm. - The circumcircle is the circle passing through all three vertices of the triangle. - We must find the area of this circumcircle in square centimetres. - π is treated as the standard constant in the area formula for a circle. Concept / Approach: First, we find the hypotenuse of the right triangle using the Pythagoras theorem. A key property of a right triangle is that the circumcentre lies at the midpoint of the hypotenuse, and the circumradius equals half the length of the hypotenuse. After obtaining the circumradius r, the area of the circumcircle is A = π * r^2. Substituting the correct radius gives the required area. Step-by-Step Solution: Step 1: Let the legs be a = 3 cm and b = 4 cm.Step 2: Hypotenuse c is found using Pythagoras theorem: c^2 = a^2 + b^2.Step 3: Compute c^2 = 3^2 + 4^2 = 9 + 16 = 25, so c = 5 cm.Step 4: For a right triangle, circumradius r = c / 2 = 5 / 2 = 2.5 cm.Step 5: Area of circumcircle A = π * r^2.Step 6: Compute r^2 = 2.5^2 = 6.25.Step 7: Therefore, A = 6.25π sq. cm. Verification / Alternative check: Another way to confirm the radius is to recall that in a right triangle the circle drawn with diameter as the hypotenuse passes through all three vertices. This is precisely the circumcircle. Since the hypotenuse is 5 cm, the diameter of the circumcircle is 5 cm and the radius is half, which again gives r = 2.5 cm. Substituting r into the area formula π * r^2 reproduces 6.25π sq. cm. Why Other Options Are Wrong: 5π sq. cm would correspond to a radius whose square is 5, not 6.25. Values like 7.5π sq. cm, 10π sq. cm and 12.5π sq. cm are based on incorrect radii, usually arising from mixing up diameter and radius or skipping the division by 2. None of these match the correct r derived from the Pythagoras theorem and the right triangle circumcircle property. Common Pitfalls: Common mistakes include taking the hypotenuse itself as the radius instead of half the hypotenuse, or using the wrong formula for area (such as π * d^2 instead of π * r^2). Some learners also miscalculate 2.5^2, writing it as 5 instead of 6.25. Keeping track of diameter versus radius and carefully squaring the radius prevents such errors. Final Answer: Hence, the area of the circumcircle is 6.25π sq. cm.

Question 17

If the diameters of two right circular cylinders are in the ratio 3 : 2 and their volumes are equal, what is the ratio of the height of the larger-diameter cylinder to the height of the smaller-diameter cylinder?

Accepted Answer

Introduction / Context: This question involves right circular cylinders and compares the effect of different diameters on height when the volumes are equal. It tests your understanding of the cylinder volume formula and your ability to manipulate ratios. Such problems are important in aptitude tests because they require both algebraic skill and correct handling of proportional reasoning. Given Data / Assumptions: - There are two right circular cylinders, called Cylinder 1 and Cylinder 2. - Ratio of their diameters is 3 : 2, so d1 : d2 = 3 : 2. - Volumes of the two cylinders are equal. - We need the ratio of the height of the larger diameter cylinder to the height of the smaller diameter cylinder, that is h1 : h2. - π is the usual constant in the volume formula. Concept / Approach: The volume of a cylinder is given by V = π * r^2 * h, where r is the radius and h is the height. Since diameters are in ratio 3 : 2, radii are also in ratio 3 : 2. By equating the volumes of the two cylinders and substituting r1 : r2, we can find a relationship between h1 and h2. Simplifying this relationship yields the desired ratio of heights. Step-by-Step Solution: Step 1: Let radii be r1 and r2 and heights be h1 and h2 respectively.Step 2: Given diameter ratio 3 : 2, we have r1 : r2 = 3 : 2.Step 3: Volume of Cylinder 1, V1 = π * r1^2 * h1. Volume of Cylinder 2, V2 = π * r2^2 * h2.Step 4: Given volumes are equal, so π * r1^2 * h1 = π * r2^2 * h2.Step 5: Cancel π from both sides to get r1^2 * h1 = r2^2 * h2.Step 6: Using r1 : r2 = 3 : 2, we have r1^2 : r2^2 = 9 : 4.Step 7: Therefore, 9 * h1 = 4 * h2, which gives h1 / h2 = 4 / 9, so h1 : h2 = 4 : 9. Verification / Alternative check: Assign simple radii values consistent with the ratio. For instance, let r1 = 3 units and r2 = 2 units. Choose an arbitrary height for one cylinder, say h1 = 4 units. Its volume is V1 = π * 3^2 * 4 = 36π. To match this volume, set V2 = π * 2^2 * h2 = 4π * h2 = 36π, which yields h2 = 9 units. Hence heights are 4 and 9, giving ratio 4 : 9, which confirms the algebraic result. Why Other Options Are Wrong: Ratios 2 : 3 and 3 : 2 come from mistakenly inverting the relationship between radius and height or misapplying the square on the radius. Ratios 9 : 4 and 1 : 1 would imply that a larger radius cylinder also has equal or greater height while maintaining equal volume, which contradicts the requirement that a larger base area must be compensated by a smaller height. Only 4 : 9 maintains equal volume with the given diameter ratio. Common Pitfalls: Many students forget that the radius appears squared in the cylinder volume formula and directly write h1 : h2 equal to the inverse of the diameter ratio. Others cancel incorrectly or invert the final ratio, giving 9 : 4 instead of 4 : 9. Always remember to square the radius before comparing and carefully check which cylinder is associated with which height in the final ratio. Final Answer: The ratio of the height of the larger diameter cylinder to the height of the smaller diameter cylinder is 4 : 9.

Question 18

In a tangential quadrilateral PQRS, angle S is 90°. A circle with centre O and radius r touches the sides PQ, QR, RS and SP at points A, B, C and D respectively. If QR = 38 cm, RS = 25 cm and QA = 27 cm, what is the radius r of the circle?

Accepted Answer

Introduction / Context: This is a higher level geometry question involving a tangential quadrilateral, which is a quadrilateral that has an incircle touching all four of its sides. Additionally, one of the angles is a right angle. The problem asks for the radius of the incircle based on side lengths and certain tangent segment lengths. It tests understanding of tangent properties and special behaviour at a right angle vertex. Given Data / Assumptions: - PQRS is a quadrilateral with an incircle touching all sides. - Angle at vertex S is 90 degrees. - The circle touches PQ, QR, RS and SP at A, B, C and D respectively. - QR = 38 cm, RS = 25 cm and QA = 27 cm. - Tangents drawn from the same external point to a circle have equal lengths. - We need to find the radius r of the incircle. Concept / Approach: From the tangent property, segments from one vertex to the points of tangency on the adjacent sides are equal. For example, QA equals QB because both are tangents from Q. Similarly, RB equals RC, SC equals SD and PD equals PA. By using the given side lengths and these equalities, we can find the distances SC and SD along the sides meeting at the right angle S. In a right angle with an incircle touching both sides, the distances from the vertex along each side to the points of tangency are equal to the radius of the circle. Step-by-Step Solution: Step 1: From Q, tangents QA and QB are equal. Given QA = 27 cm, so QB = 27 cm.Step 2: On side QR, QB + BR = QR, so BR = QR - QB = 38 - 27 = 11 cm.Step 3: From R, tangents RB and RC are equal, so RC = RB = 11 cm.Step 4: On side RS, RC + CS = RS, so CS = RS - RC = 25 - 11 = 14 cm.Step 5: From S, tangents SC and SD are equal, so SD = SC = 14 cm.Step 6: At vertex S, the two sides RS and SP meet at a right angle, and the incircle touches RS at C and SP at D.Step 7: For a circle inscribed in a right angle, the distance from the vertex along each side to the tangency point equals the radius, so the radius r = SC = SD = 14 cm. Verification / Alternative check: Geometry of a circle in a right angle wedge shows that if the radius is r, then the tangency points on the two perpendicular sides are at distance r from the vertex. Here, since SC and SD are both 14 cm, they match this property. Also, the tangent segment calculations around the quadrilateral are consistent: QA and QB are 27 cm, RB and RC are 11 cm and SC and SD are 14 cm, which fit all side lengths given. This supports the correctness of r = 14 cm. Why Other Options Are Wrong: Values like 10 cm, 12 cm, 16 cm and 18 cm would make the distances SC and SD differ from the correctly computed tangent segments, breaking the equal tangent property from S. They would also not maintain consistency between QR, RS and the given QA value. These numbers are distractors that may result from incorrect subtraction or confusion about which segments equal the radius. Common Pitfalls: Many students do not exploit the tangent segment equalities from each vertex, leading to incorrect segment values. Others recognise the equalities but forget the special property of an incircle inside a right angle, where the distance from the vertex to each tangency point along the sides equals the radius. Some may also misread QA as PA instead of QA, which leads to wrong starting values. Reading carefully and using all tangent relationships is essential. Final Answer: Therefore, the radius of the circle is 14 cm.

Question 19

What is the least number of square tiles of side 41 cm required to pave the ceiling of a rectangular room that is 15 m 17 cm long and 9 m 2 cm broad?

Accepted Answer

Introduction / Context: This question is about tiling a rectangular surface with square tiles of a fixed size. You are asked to find how many such tiles are needed to cover the ceiling exactly, without cutting any tile. It tests conversion of units, understanding of area and the idea that the number of tiles equals total area divided by area of one tile when dimensions are compatible. Given Data / Assumptions: - Length of the room = 15 m 17 cm. - Breadth of the room = 9 m 2 cm. - Each tile is a square with side 41 cm. - Tiles are laid without gaps and without cutting, so the room dimensions are assumed to be exact multiples of 41 cm. - We need the least number of tiles required to cover the ceiling completely. Concept / Approach: First, convert all dimensions into the same unit, preferably centimetres. Then compute the area of the ceiling (which has the same dimensions as the floor). Next, compute the area of a single square tile. If the room dimensions are exact multiples of the tile side, the number of tiles will be the product of the number of tiles along the length and along the breadth, which is also equal to the total area divided by the tile area. Step-by-Step Solution: Step 1: Convert room dimensions to centimetres.Step 2: Length = 15 m 17 cm = 15 * 100 + 17 = 1500 + 17 = 1517 cm.Step 3: Breadth = 9 m 2 cm = 9 * 100 + 2 = 900 + 2 = 902 cm.Step 4: Side of one square tile = 41 cm, so its area = 41 * 41 sq. cm.Step 5: Check how many tiles fit along the length: 1517 / 41 = 37, because 41 * 37 = 1517.Step 6: Check how many tiles fit along the breadth: 902 / 41 = 22, because 41 * 22 = 902.Step 7: Therefore, total number of tiles = 37 * 22 = 814. Verification / Alternative check: You can compute total ceiling area directly: Area_ceiling = 1517 * 902 sq. cm. Number of tiles should equal Area_ceiling divided by 41^2. Since both 1517 and 902 are exact multiples of 41, the result is an integer, which we already found to be 37 * 22. Performing this multiplication again confirms that the total number of tiles is 814 and no partial tiles are needed. Why Other Options Are Wrong: Values like 656, 738, 902 and 960 do not match the product of integer counts along each dimension. They can appear if someone divides the area roughly or rounds the room dimensions or miscomputes 1517 / 41 or 902 / 41. Since the dimensions divide exactly, any answer other than 814 is inconsistent with the exact tiling requirement. Common Pitfalls: Common mistakes include failing to convert metres to centimetres correctly, leading to wrong multiples, or approximating division instead of checking for exact divisibility. Some students directly divide total area by tile area without verifying that the result is an integer, which could allow a non-whole number of tiles. Careful unit conversion and exact arithmetic are crucial in tiling problems. Final Answer: The least number of square tiles required is 814.

Question 20

O is the circumcentre of triangle ABC. If the distance AO from vertex A to the circumcentre is 8 cm, what is the length of BO?

Accepted Answer

Introduction / Context: This is a conceptual geometry question about the circumcentre of a triangle. The circumcentre is the point that is equidistant from all three vertices of the triangle, and it is the centre of the circumcircle passing through those vertices. The question checks your understanding of this fundamental property. Although no coordinates or side lengths are given, recognising the definition of the circumcentre is enough to answer correctly. Given Data / Assumptions: - Triangle ABC is any triangle (it may be acute, right or obtuse). - O is the circumcentre of triangle ABC, that is, the centre of the circumcircle passing through A, B and C. - AO, the distance from vertex A to the circumcentre, is 8 cm. - We are asked to find BO, the distance from vertex B to the circumcentre. - The standard property of a circumcentre applies: all vertices are equidistant from O. Concept / Approach: By definition, the circumcentre of a triangle is the point where the perpendicular bisectors of the sides meet, and it is equidistant from all three vertices. This means OA = OB = OC, where OA, OB and OC are radii of the circumcircle. Therefore, if you know the distance from the circumcentre to one vertex, you automatically know the distance to the other vertices, because all are equal to the circumradius. Step-by-Step Solution: Step 1: Recall that in any triangle, the circumcentre is the centre of a circle that passes through all three vertices A, B and C.Step 2: By this definition, distances OA, OB and OC are all equal and represent the radius r of the circumcircle.Step 3: The problem gives AO = 8 cm.Step 4: Therefore, OA = OB = OC = 8 cm.Step 5: Hence, BO must equal the same circumradius.Step 6: So, BO = 8 cm.Step 7: The length of BO is directly determined from the fundamental property of the circumcentre without any further calculation. Verification / Alternative check: You can imagine drawing any triangle and constructing its perpendicular bisectors, which meet at the circumcentre O. If you then draw the circle centred at O and passing through A, the radius OA fixes the size of the circle. Because B and C lie on the same circle, the distances OB and OC must also be equal to the radius. Changing the shape of the triangle while keeping the same circumcentre and circumradius does not alter this equality. This conceptual picture reconfirms that OB equals OA. Why Other Options Are Wrong: Values like 3 cm, 6 cm and 12 cm imply that different vertices are at different distances from O, which contradicts the definition of a circumcentre. The option Cannot be determined is also incorrect because the property OA = OB = OC is universally true for any triangle with a defined circumcentre, so no extra information such as side lengths or angles is needed to find BO once AO is known. Common Pitfalls: Some students confuse the circumcentre with other centres of a triangle such as the centroid, incenter or orthocentre. For these other centres, distances to the vertices are not necessarily equal, which may make them think that BO cannot be determined. Remember that only the circumcentre is equidistant from all vertices. Mixing up these centres is a common conceptual error in geometry problems. Final Answer: Therefore, the length of BO is 8 cm.

Volume and Surface Area Questions