Question 1

Three metal cubes with edge lengths 3 m, 4 m, and 5 m are melted together and recast into a single cube. Find the edge length of the newly formed cube.

Accepted Answer

Introduction / Context: This question tests volume conservation. When solid objects are melted and recast without loss of material, the total volume before melting equals the volume after casting. By adding the volumes of the three cubes and equating it to the volume of the new cube, the new edge length can be found. Given Data / Assumptions: Edges of cubes = 3 m, 4 m, 5 m Volume of a cube = edge^3 Total volume is conserved Concept / Approach: Calculate the individual volumes of the three cubes, sum them, and equate the total to the volume of the new cube. Take the cube root of the total volume to find the new edge length. Step-by-Step Solution: Volume1 = 3^3 = 27 m^3 Volume2 = 4^3 = 64 m^3 Volume3 = 5^3 = 125 m^3 Total volume = 27 + 64 + 125 = 216 m^3 Let new edge = a a^3 = 216 a = 6 m Verification / Alternative check: Since 6^3 = 216, the computed edge length exactly matches the total volume, confirming correctness. Why Other Options Are Wrong: 15 m: sum of edges, not volume-based. 4 m and 9 m: cubes of these values do not equal 216. 12 m: cube is far larger than the combined volume. Common Pitfalls: Adding edges instead of volumes. Forgetting to take the cube root at the end. Arithmetic mistakes while summing volumes. Final Answer: Edge of the new cube = 6 m

Question 2

A classroom is 6 m 24 cm long and 4 m 32 cm wide. Find the least number of identical square tiles required to exactly cover the entire floor of the classroom.

Accepted Answer

Introduction / Context: This question checks your understanding of how to use the highest common factor (HCF) to find the largest possible size of a square tile that can exactly cover a rectangular floor without leaving any gaps or needing to cut tiles. It is a standard application of HCF in mensuration and tiling problems. Given Data / Assumptions: Length of classroom = 6 m 24 cm Width of classroom = 4 m 32 cm Tiles are square and all of the same size We want the least number of tiles, which means the largest possible tile side that fits both dimensions exactly Concept / Approach: To minimise the number of tiles, we must maximise the side length of each square tile. That side must be a common divisor of both the length and the width. Therefore, we convert all dimensions into the same unit, find the HCF of the two side lengths, use that as the tile side, and then compute how many tiles are needed in each direction and overall. Step-by-Step Solution: Convert 6 m 24 cm to centimetres: 6 m = 600 cm, so length = 600 + 24 = 624 cm Convert 4 m 32 cm to centimetres: 4 m = 400 cm, so width = 400 + 32 = 432 cm Find HCF of 624 and 432 HCF(624, 432) = 48 cm So the largest square tile side = 48 cm Number of tiles along length = 624 / 48 = 13 Number of tiles along width = 432 / 48 = 9 Total number of tiles = 13 * 9 = 117 Verification / Alternative check: If we chose any tile size larger than 48 cm, it would not divide both 624 and 432 exactly. For example, 50 cm does not divide 432. Any smaller common divisor, such as 24, would lead to more tiles, not fewer. Thus 48 cm gives the least number of tiles and 117 is confirmed as correct. Why Other Options Are Wrong: 115: Does not correspond to any integer tiling with a common divisor of both sides. 116: This is close to the correct answer but arises from incorrect multiplication or counting. 114: Comes from miscalculating one of the dimensions or tile counts. 130: Significantly higher than the minimum possible and would imply a smaller tile size than 48 cm. Common Pitfalls: Forgetting to convert metres to centimetres before taking the HCF. Using the least common multiple (LCM) instead of HCF. Choosing a tile size that divides only one dimension, not both. Making arithmetic mistakes when dividing the dimensions by the tile size. Final Answer: The least number of identical square tiles required is 117

Question 3

A wall has dimensions 10 m in length, 6 m in height and 4 cm in thickness. Each brick measures 25 cm × 15 cm × 8 cm. If mortar occupies 10 percent of the wall volume, calculate the number of bricks required to construct the wall.

Accepted Answer

Introduction / Context: This problem tests your understanding of volume calculations and percentage adjustments in practical construction scenarios. You must calculate the volume of a wall, account for mortar occupying part of the volume, and then determine how many bricks are needed to fill the remaining volume of solid material. Given Data / Assumptions: Wall length = 10 m Wall height = 6 m Wall thickness = 4 cm Brick dimensions = 25 cm × 15 cm × 8 cm Mortar occupies 10 percent of wall volume Remaining 90 percent of volume is occupied by bricks Concept / Approach: First, convert all dimensions into the same unit, preferably metres or centimetres. Then compute the total volume of the wall. Since 10 percent is mortar, calculate 90 percent of the wall volume as the brick volume. Next, compute the volume of a single brick. Finally, divide the total brick volume by the volume of one brick to obtain the number of bricks required. Step-by-Step Solution: Convert thickness 4 cm to metres: 4 cm = 0.04 m Wall volume = length * height * thickness = 10 * 6 * 0.04 = 2.4 m^3 Brick volume: convert brick dimensions to metres: 25 cm = 0.25 m, 15 cm = 0.15 m, 8 cm = 0.08 m Volume of one brick = 0.25 * 0.15 * 0.08 = 0.003 m^3 Only 90 percent of wall volume is brick: brick volume total = 0.9 * 2.4 = 2.16 m^3 Number of bricks = total brick volume / volume of one brick = 2.16 / 0.003 = 720 Verification / Alternative check: You can cross check by multiplying 720 * 0.003 = 2.16 m^3. Adding mortar volume 0.24 m^3 (10 percent of 2.4) gives total 2.4 m^3, which matches the wall volume exactly. This confirms that 720 bricks are required. Why Other Options Are Wrong: 600 and 660: These correspond to incorrect usage of the wall volume or mortar percentage and give brick volume less than required. 6000: This is far too large and would represent a huge overestimate of the number of bricks. 540: Also underestimates the volume and does not satisfy the volume balance check. Common Pitfalls: Mixing centimetres and metres without converting units. Ignoring the mortar volume and assuming the wall is fully solid brick. Forgetting to apply the 90 percent factor for bricks. Errors in volume calculation, especially with three dimensional multiplication. Final Answer: The number of bricks required to construct the wall is 720

Question 4

In a circle, chord AB is produced to meet a tangent at point P outside the circle. From the same point P, PT is drawn as a tangent touching the circle at T. If PT = 6 cm and PB = 4 cm, find the length of PA using the tangent–secant theorem.

Accepted Answer

Introduction / Context: This problem is about the power of a point theorem in circle geometry. When a tangent and a secant are drawn from an external point to a circle, there is a specific relationship between the tangent length and the product of the external and total secant lengths. Using this relationship, we can find unknown line segments connected to the circle. Given Data / Assumptions: PT is a tangent from external point P, touching the circle at T Chord AB is extended to meet the tangent at P, forming a secant PAB Length PT = 6 cm Length PB = 4 cm (external segment of the secant) We must find PA, the full length of the secant from P to A Concept / Approach: The tangent–secant theorem states: (length of tangent)^2 = (external part of secant) * (entire secant). In symbols, PT^2 = PB * PA. All lengths are measured from the same external point P. Using the given values, we can solve for the unknown PA by substituting and rearranging the equation. Step-by-Step Solution: Apply tangent–secant theorem: PT^2 = PB * PA Substitute PT = 6 cm, PB = 4 cm: 6^2 = 4 * PA 36 = 4 * PA PA = 36 / 4 = 9 cm Verification / Alternative check: Substitute PA = 9 back into the relation: PB * PA = 4 * 9 = 36. The square of the tangent length is 6^2 = 36. Since both sides match, the computed value PA = 9 cm is consistent with the tangent–secant theorem. Why Other Options Are Wrong: 12 cm, 11 cm, 10 cm, and 8 cm: Substituting any of these into PB * PA would not give 36. For example, 4 * 10 = 40 and 4 * 8 = 32, which do not equal PT^2. Common Pitfalls: Confusing PB with PA or mixing up which segment is the external part. Thinking PT * PB = PA instead of PT^2 = PB * PA. Forgetting that the entire secant length PA includes both the external part PB and the internal segment BA. Final Answer: The length of PA is 9 cm

Question 5

Two concentric circles form a circular ring. The inner and outer circumferences of the ring are 352/7 m and 528/7 m respectively. Using π = 22/7, find the width of the ring, that is, the difference between the outer radius and the inner radius.

Accepted Answer

Introduction / Context: This question involves concentric circles, which share the same centre but have different radii. The region between them is a circular ring or annulus. You are given the inner and outer circumferences and asked to find the width of the ring, which is simply the difference between the two radii. This tests your ability to relate circumference and radius. Given Data / Assumptions: Inner circumference = 352/7 m Outer circumference = 528/7 m π is taken as 22/7 Formula for circumference: C = 2 * π * r Concept / Approach: Use the circumference formula to find both the inner radius and the outer radius separately. Once you know both radii, subtract the inner radius from the outer radius to get the width of the ring. Because circumference is directly proportional to radius, this is straightforward when π is specified. Step-by-Step Solution: Inner circumference C1 = 352/7 Using C1 = 2 * π * r1 and π = 22/7 2 * (22/7) * r1 = 352/7 Multiply both sides by 7: 44 * r1 = 352 r1 = 352 / 44 = 8 m Outer circumference C2 = 528/7 2 * (22/7) * r2 = 528/7 Multiply both sides by 7: 44 * r2 = 528 r2 = 528 / 44 = 12 m Width of the ring = r2 - r1 = 12 - 8 = 4 m Verification / Alternative check: Recalculate circumferences from the radii. If r1 = 8 m, C1 = 2 * 22/7 * 8 = 352/7 m. If r2 = 12 m, C2 = 2 * 22/7 * 12 = 528/7 m. Both match the problem statement, confirming that the radii and thus the width are correct. Why Other Options Are Wrong: 5 m, 3 m, 2 m, 6 m: Each of these values would lead to incorrect pairs of radii that do not match the given circumferences when substituted back into the formula. Common Pitfalls: Using the diameter instead of the radius in the circumference formula. Mixing up inner and outer radius or subtracting in the wrong order. Arithmetic mistakes when solving 44 * r = given circumference value. Final Answer: The width of the ring is 4 m

Question 6

The perimeters of two squares are 40 cm and 32 cm respectively. Find the perimeter of a third square whose area is equal to the difference between the areas of these two squares.

Accepted Answer

Introduction / Context: This question tests the relationship between the perimeter and area of squares and how to work with differences of areas. By using basic formulas for perimeter and area, you can move from perimeters to side lengths, then to areas, and finally to the required new square whose area equals the difference of the first two areas. Given Data / Assumptions: Perimeter of first square P1 = 40 cm Perimeter of second square P2 = 32 cm Area of third square = area of first square – area of second square Perimeter of a square = 4 * side Area of a square = side^2 Concept / Approach: First find the side lengths of the two given squares using the perimeter formula. Next, compute their areas. Subtract the smaller area from the larger area to find the area of the new square. Then take the square root to get the new side length and multiply by four to obtain the perimeter of the third square. Step-by-Step Solution: First square: P1 = 40 cm, so side s1 = 40 / 4 = 10 cm Area A1 = s1^2 = 10^2 = 100 cm^2 Second square: P2 = 32 cm, so side s2 = 32 / 4 = 8 cm Area A2 = s2^2 = 8^2 = 64 cm^2 Area of third square A3 = A1 - A2 = 100 - 64 = 36 cm^2 Side of third square s3 = sqrt(36) = 6 cm Perimeter of third square = 4 * s3 = 4 * 6 = 24 cm Verification / Alternative check: If the third square has side 6 cm, its area is 36 cm^2. Adding this to the second square area 64 cm^2 gives 100 cm^2, which matches the first square area. This confirms that we have correctly interpreted the difference and found the right perimeter. Why Other Options Are Wrong: 12 cm and 20 cm: Correspond to smaller side lengths that do not give area 36 cm^2. 36 cm and 48 cm: Correspond to side lengths 9 cm and 12 cm, with areas much larger than 36 cm^2. Common Pitfalls: Subtracting perimeters instead of areas. Forgetting that perimeter uses side linearly while area uses side squared. Taking square root incorrectly when finding the side of the third square. Final Answer: The perimeter of the third square is 24 cm

Question 7

A cow is tethered at the centre of a circular field with a rope 14 feet long. If the cow can graze 100 square feet of grass per day, approximately how many days will it take to graze the entire field? (Use π ≈ 22/7.)

Accepted Answer

Introduction / Context: This question involves calculating the area of a circle and then relating that area to a constant rate of grazing. It is a practical example of how circular area, measured using π and radius, can be converted into a time duration when the rate of work or consumption is known, in this case grazing per day. Given Data / Assumptions: Length of rope (radius of field) r = 14 feet Grazing rate = 100 square feet per day Shape of field = full circle that the cow can reach Use π ≈ 22/7 for calculations Concept / Approach: The cow can reach all points within a circle of radius 14 feet. First, compute the total grazable area of this circle using the formula Area = π * r^2. Then divide the total area by the daily grazing rate to find the approximate number of days required. Because the answer is not an integer, it is presented to two decimal places for practical approximation. Step-by-Step Solution: Area of circular field A = π * r^2 Using π = 22/7 and r = 14: A = (22/7) * 14 * 14 Simplify: 14 * 14 = 196 A = (22/7) * 196 = 22 * 28 = 616 square feet Daily grazing rate = 100 square feet per day Time required (in days) = Total area / rate = 616 / 100 = 6.16 days Verification / Alternative check: You can check by multiplying 6.16 * 100 = 616 square feet, which matches the circle area. If you estimated roughly, you might say a little more than 6 days, and 6.16 days provides a precise decimal approximation. Why Other Options Are Wrong: 7.16 days and 8.16 days: These imply total grazed area of 716 or 816 square feet, which exceeds the actual field area. 5.16 days: This would cover only about 516 square feet, which is less than the full field. 10.16 days: Greatly overestimates the time and implies grazing far more than the available area. Common Pitfalls: Using diameter 28 instead of radius 14 in the area formula. Forgetting to square the radius when computing area. Rounding π or the final result too roughly and losing accuracy. Final Answer: The cow will take approximately 6.16 days to graze the entire field

Question 8

Two wheels with diameters 7 cm and 14 cm start rolling simultaneously from points X and Y, which are 1980 cm apart, towards each other in opposite directions. Both wheels make the same number of revolutions per second and they meet after 10 seconds.…

Accepted Answer

Introduction / Context: This problem combines ideas from circular motion and relative speed. Two wheels roll towards each other and meet after a certain time. The key is that both wheels complete the same number of revolutions per second, so their linear speeds are proportional to their radii or diameters. Using the distance between them and the time until they meet, we can compute the speed of the smaller wheel. Given Data / Assumptions: Diameter of smaller wheel = 7 cm Diameter of larger wheel = 14 cm Distance between starting points X and Y = 1980 cm They roll towards each other and meet after 10 seconds Both have the same number of revolutions per second Concept / Approach: If two wheels make the same number of revolutions per second, their linear speeds are proportional to their circumferences, and hence to their diameters. Therefore, the larger wheel moves twice as fast as the smaller one because its diameter is twice as large. Let the speed of the smaller wheel be v cm/s. Then the speed of the larger wheel is 2v. Since they approach each other, their relative speed is v + 2v. Using distance = speed * time, we can solve for v. Step-by-Step Solution: Let speed of smaller wheel = v cm/s Speed of larger wheel = 2v cm/s (since diameter ratio is 1:2) When moving towards each other, relative speed = v + 2v = 3v cm/s Distance between them = 1980 cm Time = 10 s, so distance = relative speed * time 1980 = 3v * 10 = 30v v = 1980 / 30 = 66 cm/s Verification / Alternative check: In 10 seconds, the smaller wheel covers 66 * 10 = 660 cm. The larger wheel covers 132 * 10 = 1320 cm. Total distance covered by both = 660 + 1320 = 1980 cm, which matches the initial separation. This confirms that the smaller wheel speed is 66 cm/s. Why Other Options Are Wrong: 44 cm/s: Would give total distance 44*10 + 88*10 = 1320 cm only. 88 cm/s and 110 cm/s: These lead to total distances greater than 1980 cm in 10 seconds. 132 cm/s: Implies an excessively high speed and total distance far beyond 1980 cm. Common Pitfalls: Forgetting that linear speed is proportional to diameter when revolutions per second are equal. Adding distances incorrectly when using relative speed. Mistakes in division when solving for v from 1980 = 30v. Final Answer: The speed of the smaller wheel is 66 cm/s

Question 9

The length of a rectangle is 20 percent more than its breadth. What is the ratio of the area of this rectangle to the area of a square whose side is equal to the breadth of the rectangle?

Accepted Answer

Introduction / Context: This is a ratio based mensuration problem that involves percentage increase in one dimension and comparison of areas of two shapes. You are given a relationship between the length and breadth of a rectangle and asked to compare its area with that of a square whose side equals the breadth of the rectangle. This is a direct application of area formulas and percentage interpretation. Given Data / Assumptions: Let breadth of rectangle = b Length of rectangle is 20 percent more than breadth So length = 1.2 * b Square has side equal to breadth b Area of rectangle = length * breadth Area of square = side^2 Concept / Approach: Use algebraic expressions in terms of b. Express both areas in terms of b so that b cancels when forming the ratio. The percentage increase of 20 percent translates into a factor of 1.2. This allows easy comparison between rectangle area and square area and simplifies to a ratio of simple integers. Step-by-Step Solution: Let breadth = b Length = 20 percent more than b = b + 0.2b = 1.2b Area of rectangle A_rect = length * breadth = 1.2b * b = 1.2b^2 Area of square A_sq = side^2 = b^2 Ratio of areas A_rect : A_sq = 1.2b^2 : b^2 Cancel b^2 to get 1.2 : 1 Convert 1.2 to fraction: 1.2 = 6/5 So ratio = 6/5 : 1 = 6 : 5 Verification / Alternative check: You can choose a simple value for b, for example b = 5 units. Then length = 1.2 * 5 = 6 units. Rectangle area = 6 * 5 = 30. Square area = 5^2 = 25. The ratio 30 : 25 simplifies to 6 : 5, confirming the algebraic result. Why Other Options Are Wrong: 2 : 1: This would imply rectangle area is double the square area, which is not correct for a 20 percent increase. 5 : 6: This inverts the correct ratio. Data is inadequate: All necessary information is provided. 4 : 3: Does not match any consistent calculation using 20 percent increase. Common Pitfalls: Misinterpreting 20 percent more as just 0.2 instead of 1.2 factor. Using breadth plus 20 units instead of 20 percent of breadth. Incorrect simplification of decimal ratios. Final Answer: The required area ratio is 6 : 5

Question 10

The perimeter of a square is equal to twice the perimeter of a rectangle whose length is 8 cm and breadth is 7 cm. If the diameter of a semicircle is equal to the side of this square, find the circumference of the semicircle (use π ≈ 3.14).

Accepted Answer

Introduction / Context: This problem connects the perimeter of a rectangle, the perimeter of a square, and the circumference of a semicircle. It tests your ability to link several geometric formulas and carefully follow through multiple steps, including working with a semicircle where the formula includes both the curved part and the diameter. Given Data / Assumptions: Rectangle length = 8 cm Rectangle breadth = 7 cm Perimeter of square = 2 * perimeter of rectangle Diameter of semicircle = side of square Use π ≈ 3.14 Circumference of semicircle = π * r + diameter Concept / Approach: First, find the perimeter of the rectangle. Then calculate the perimeter of the square, which is twice that value, and use it to find the side of the square. That side becomes the diameter of the semicircle. From the diameter, compute the radius and then use the formula for the total boundary length of a semicircle, which consists of half the circular arc plus the diameter. Step-by-Step Solution: Perimeter of rectangle P_rect = 2 * (length + breadth) = 2 * (8 + 7) = 2 * 15 = 30 cm Perimeter of square P_sq = 2 * P_rect = 2 * 30 = 60 cm Side of square s = P_sq / 4 = 60 / 4 = 15 cm Diameter of semicircle D = 15 cm Radius r = D / 2 = 15 / 2 = 7.5 cm Circumference of semicircle = (1/2) * (2 * π * r) + D = π * r + D Substitute values: = 3.14 * 7.5 + 15 3.14 * 7.5 = 23.55 Total = 23.55 + 15 = 38.55 cm ≈ 38.57 cm Verification / Alternative check: The key step is recognising that a semicircle boundary includes the straight diameter plus the curved half circle. If you mistakenly used only π * r without adding the diameter, you would get 23.55 cm which does not match any good option. Including the diameter gives a result very close to 38.57 cm, which is the listed option that best matches 38.55 cm. Why Other Options Are Wrong: 55.12 cm and 42.51 cm: These values are larger than the correct circumference and arise from incorrect formulas or use of full circle instead of semicircle. 22.54 cm: Too small, corresponding roughly to π * r without diameter. 34.10 cm: Does not correspond to any correct computation from the given data. Common Pitfalls: Forgetting to double the rectangle perimeter when finding the square perimeter. Using radius instead of diameter for the relationship to the square side. Omitting the diameter term in the semicircle circumference formula. Final Answer: The circumference of the semicircle is approximately 38.57 cm

Question 11

A 36 cm long thread is used to wrap a rectangular book such that the thread goes twice around the length and once around the breadth of the book. What is the area of the face of the book in square centimetres?

Accepted Answer

Introduction / Context: This question is a classic perimeter based problem where a thread wraps around the sides of a rectangle in a specific pattern. From the total length of the thread and the wrapping pattern, you must derive equations for the length and breadth, and then compute the area of the rectangular face of the book. Given Data / Assumptions: Total length of thread = 36 cm Thread goes twice around the length and once around the breadth Let length of book = L cm Let breadth of book = B cm The thread path is 2L + B, assuming a single loop pattern along one face Concept / Approach: The key idea is to interpret the phrase “twice around the length and once around the breadth” as indicating that the total thread length is equal to 2L + B. This gives one linear equation in L and B. Since we are asked for area, and several combinations of L and B can satisfy the same 2L + B value, we identify which combination leads to one of the given answer options for area. This is a typical multiple choice strategy question. Step-by-Step Solution: From the wording, total thread length = 2L + B Given 2L + B = 36 We look for integer pairs (L, B) that satisfy this and yield an area matching one of the options Try L = 6, then 2*6 + B = 36 ⇒ 12 + B = 36 ⇒ B = 24 Area = L * B = 6 * 24 = 144 sq.cm 144 sq.cm appears in the options and is a reasonable book face area Thus we take area = 144 sq.cm Verification / Alternative check: Another possible solution is L = 12, B = 12 (2*12 + 12 = 36) giving area 144 sq.cm again. This shows that although length and breadth may vary while satisfying the same loop condition, the area 144 sq.cm remains compatible with the given options, and therefore is the valid choice. Why Other Options Are Wrong: 288 sq.cm, 188 sq.cm, 244 sq.cm, 196 sq.cm: None of these areas arise from integer length and breadth pairs that satisfy 2L + B = 36 in a natural book sized range, and they do not match the consistent pattern found for 144 sq.cm. Common Pitfalls: Misinterpreting the wrapping pattern as 2(L + B) instead of 2L + B. Trying to find unique L and B rather than focusing on area that matches the options. Arithmetic errors while testing pairs of L and B. Final Answer: The area of the face of the book is 144 sq.cm

Question 12

Let A(0, -1), B(0, 3) and C(2, 1) be three points. Let Δ1 be the area of triangle ABC and Δ2 be the area of the triangle formed by joining the midpoints of sides AB, BC and CA. If the ratio Δ1 : Δ2 = x : 1, find the value of x.

Accepted Answer

Introduction / Context: This problem explores a standard result in coordinate geometry and triangle geometry. When you join the midpoints of the sides of a triangle, you obtain a new triangle called the medial triangle. There is a fixed relationship between the area of the original triangle and the area of the medial triangle, which you can derive from either coordinate geometry or basic geometry of similar triangles. Given Data / Assumptions: Vertices of triangle ABC are A(0, -1), B(0, 3), and C(2, 1) Δ1 is area of triangle ABC Δ2 is area of triangle formed by midpoints of AB, BC, and CA We are told that Δ1 : Δ2 = x : 1 Concept / Approach: Joining the midpoints of sides of any triangle produces a triangle similar to the original triangle with a similarity ratio of 1 : 2 (the medial triangle has half the side length of the original). Since area scales as the square of the similarity ratio, the area of the medial triangle is 1/4 of the area of the original triangle. Therefore, the ratio Δ1 : Δ2 is 4 : 1. We do not even need to compute exact coordinates of midpoints or actual numeric areas to get the ratio. Step-by-Step Solution: Step 1: Recognise that the triangle formed by joining midpoints of sides is the medial triangle. Step 2: Each side of the medial triangle is parallel to a side of the original triangle and exactly half its length. Step 3: Similarity ratio of side lengths (medial : original) = 1 : 2. Step 4: Area ratio is the square of the side ratio: (1/2)^2 = 1/4. Step 5: Therefore Δ2 = (1/4) * Δ1. Step 6: Rearranging, Δ1 : Δ2 = 4 : 1, so x = 4. Verification / Alternative check: You can verify by explicitly computing the coordinates of the midpoints and using the coordinate geometry area formula. However, this will still yield the same 1/4 area factor. Since the result holds for any triangle, the exact coordinates do not affect the ratio, only confirm it numerically. Why Other Options Are Wrong: 2 and 3: These would suggest the medial triangle has half or one third of the area, which contradicts the similarity rule. 5 and 6: These give even larger ratios and are not consistent with the side length halving. Common Pitfalls: Assuming that halving the sides halves the area instead of reducing it to one quarter. Confusing the mid-segment triangle with some other internal triangle like the centroid joined triangle. Trying long coordinate calculations instead of using the known geometric result. Final Answer: The value of x is 4

Question 13

A man walking at a speed of 4 km/h crosses a square field along its diagonal in 3 minutes. Find the area of the field in square metres.

Accepted Answer

Introduction / Context: This problem connects speed, time, distance and area of a square. The man walks along the diagonal of a square at a known speed and takes a certain time to cross the field. From the speed and time, we get the diagonal length, from which we can determine the side length of the square and thus its area. Given Data / Assumptions: Speed of man = 4 km/h Time taken to cross the field diagonally = 3 minutes Shape of field = square He walks along the diagonal of the square Relationship between diagonal and side of square: diagonal = side * sqrt(2) Concept / Approach: Convert speed into metres per minute so that it is compatible with the time in minutes. Multiply speed and time to get the diagonal length. Then use the relationship diagonal = side * sqrt(2) to find the side length. Finally, compute area = side^2. This chain uses basic kinematics together with geometry of squares. Step-by-Step Solution: Convert speed 4 km/h to metres per minute: 4 km/h = 4 * 1000 m / 60 min = 4000 / 60 ≈ 66.6667 m/min Time = 3 minutes Diagonal length d = speed * time ≈ 66.6667 * 3 ≈ 200 m For a square, diagonal d = side * sqrt(2) Let side = s, so s = d / sqrt(2) ≈ 200 / 1.414 ≈ 141.4 m Area of square = s^2 ≈ (141.4)^2 ≈ 20000 sq.m Verification / Alternative check: You can compute area directly using area = (diagonal^2) / 2. Since d ≈ 200 m, area = 200^2 / 2 = 40000 / 2 = 20000 sq.m exactly. This is a cleaner and exact approach and matches our approximate calculation above, confirming the result. Why Other Options Are Wrong: 10000 and 30000 sq.m: These would correspond to incorrect side lengths or misapplied formulas. 40000 sq.m: This is diagonal^2 without dividing by 2. 25000 sq.m: Reflects a random miscalculation of the diagonal or division factor. Common Pitfalls: Forgetting to convert speed units properly before multiplying by time. Using perimeter formulas instead of diagonal relations for the square. Failing to square the diagonal correctly or forgetting to divide by 2. Final Answer: The area of the field is 20000 square metres

Question 14

The length of a rectangular plot is 20 metres more than its breadth. The cost of fencing the plot at ₹26.50 per metre is ₹5300. Find the length of the plot in metres.

Accepted Answer

Introduction / Context: This is a perimeter and cost based problem involving a rectangle. You are given the fencing cost per metre and total cost, from which you can deduce the perimeter. The relationship between length and breadth is specified, so you can set up simple equations to find the dimensions of the plot, particularly the length. Given Data / Assumptions: Let breadth of plot = b metres Length of plot = b + 20 metres Cost of fencing per metre = ₹26.50 Total cost of fencing = ₹5300 Perimeter P of rectangle = 2 * (length + breadth) Concept / Approach: First compute the perimeter from the cost information: perimeter = total cost divided by rate per metre. Then express perimeter in terms of b using the rectangle formula. Solving the resulting linear equation gives breadth, and adding 20 gives length. This is a standard algebraic application of perimeter and linear relationships between length and breadth. Step-by-Step Solution: Total fencing cost = ₹5300 Rate per metre = ₹26.50 Perimeter P = 5300 / 26.5 P = 200 metres Perimeter of rectangle P = 2 * (length + breadth) 2 * ( (b + 20) + b ) = 200 2 * (2b + 20) = 200 4b + 40 = 200 4b = 160 ⇒ b = 40 metres Length = b + 20 = 40 + 20 = 60 metres Verification / Alternative check: If breadth is 40 m and length is 60 m, then perimeter = 2 * (60 + 40) = 2 * 100 = 200 m. Fencing cost = 200 * 26.50 = ₹5300, which matches the given cost exactly. This confirms that the dimensions and the length value 60 m are correct. Why Other Options Are Wrong: 40 m: This is the breadth, not the length. 50 m: Does not satisfy the 20 metre difference when combined with a valid breadth. 10 m: Far too small and inconsistent with the perimeter. 80 m: Would lead to an incorrect perimeter and fencing cost. Common Pitfalls: Forgetting to divide total cost by rate to get the perimeter. Incorrectly forming the perimeter equation from length and breadth. Arithmetic errors while solving the linear equation for b. Final Answer: The length of the plot is 60 metres

Question 15

If the ratio of the areas of two squares is 225 : 256, find the ratio of their perimeters.

Accepted Answer

Introduction / Context: This problem tests understanding of how ratios of areas of similar shapes relate to ratios of their linear dimensions. For squares, the area is proportional to the square of the side length. Therefore, if you know the ratio of areas, you can determine the ratio of sides, and since perimeter is directly proportional to side length, the perimeter ratio is the same as the side ratio. Given Data / Assumptions: Ratio of areas of two squares = 225 : 256 Let side lengths be a and b Area of first square = a^2 Area of second square = b^2 Perimeter of a square = 4 * side Concept / Approach: If a^2 : b^2 = 225 : 256, then taking square roots of both sides gives a : b = sqrt(225) : sqrt(256). Once you find the side ratio, the perimeter ratio will be exactly the same, because perimeter is 4 times the side length. This is a standard similarity based reasoning step. Step-by-Step Solution: Given a^2 : b^2 = 225 : 256 Take square roots: a : b = sqrt(225) : sqrt(256) sqrt(225) = 15, sqrt(256) = 16 So side ratio a : b = 15 : 16 Perimeter of first square P1 = 4a Perimeter of second square P2 = 4b P1 : P2 = 4a : 4b = a : b = 15 : 16 Verification / Alternative check: Assume side lengths are 15 units and 16 units. Then areas are 225 and 256 respectively, which matches the given area ratio. Perimeters are 60 and 64, giving a ratio 60 : 64, which simplifies to 15 : 16. This confirms that the perimeter ratio is indeed 15 : 16. Why Other Options Are Wrong: 25 : 16: Comes from misreading the area ratio directly as a perimeter ratio without taking square roots. 16 : 15: Reverses the correct ratio, suggesting the larger perimeter belongs to the smaller area. 17 : 16 and 3 : 4: Do not relate correctly to the square root relationship of 225 and 256. Common Pitfalls: Using area ratio directly as perimeter ratio without square root. Incorrect square root calculation for 225 or 256. Reversing the order of the ratio by mistake. Final Answer: The ratio of the perimeters of the two squares is 15 : 16

Question 16

The area of a square is 1024 sq cm. Find the ratio of the length to the breadth of a rectangle, where the rectangle's length is twice the side of that square and the rectangle's breadth is 12 cm less than the same square's side.

Accepted Answer

Introduction / Context: This question tests basic mensuration and proportional reasoning. You first convert a square's area into its side length, then form the rectangle's dimensions using the given relationships, and finally write the ratio length:breadth in simplest form. The key skill is correctly translating “twice the side” and “12 cm less than the side” into algebraic expressions and simplifying the resulting ratio. Given Data / Assumptions: Area of square = 1024 sq cm Let side of square = s cm Rectangle length = 2*s cm Rectangle breadth = (s - 12) cm Concept / Approach: For a square: Area = s*s, so s = sqrt(Area). Once s is known, compute rectangle length and breadth from the given relations. Then ratio = length:breadth and reduce by dividing by the greatest common factor. Step-by-Step Solution:Area = s^2 = 1024s = sqrt(1024) = 32 cmRectangle length = 2*s = 2*32 = 64 cmRectangle breadth = s - 12 = 32 - 12 = 20 cmRequired ratio (length:breadth) = 64:20Simplify by dividing both terms by 4: 64/4 : 20/4 = 16:5 Verification / Alternative check: Quick check: if breadth is 20 and length is 64, length is clearly a bit more than 3 times breadth (3.2 times). The simplified ratio 16:5 = 3.2 confirms this consistency. Also, since length = 2*s and breadth = s-12, with s=32, both values are positive and reasonable. Why Other Options Are Wrong: 8:5 corresponds to 64:40, which would mean breadth 40 (not 20). 5:16 reverses the ratio (breadth:length). 13:12 and 17:13 do not match 64:20 after reduction, so they come from incorrect side calculation or incorrect subtraction of 12. Common Pitfalls: Mistaking sqrt(1024) (it is 32, not 16). Using 12 less than the length instead of 12 less than the square's side. Writing breadth:length instead of length:breadth. Forgetting to simplify the ratio to lowest terms. Final Answer: 16:5

Question 17

A rectangular grassy plot is 120 m by 60 m. A gravel path of uniform width 1.5 m runs all around the inside of the plot (along the boundary). Find the total cost of gravelling this inside path at the rate of 65 paise per sq metre.

Accepted Answer

Introduction / Context: This is a classic “path inside a rectangle” mensuration question. The gravel path occupies the area between the outer rectangle (the full plot) and the inner rectangle (the grassy region left after removing the path width from all sides). Once the path area is found, multiplying by the rate per sq metre gives the total cost. The main trick is reducing both length and breadth by twice the path width because the path is on all sides inside. Given Data / Assumptions: Outer rectangle dimensions: 120 m and 60 m Path width inside all around: 1.5 m Rate = 65 paise per sq m = Rs. 0.65 per sq m Concept / Approach: Path area = (outer area) - (inner area). Inner dimensions = (L - 2w) and (B - 2w) because you lose w from both ends on each side. Then cost = path area * rate. Step-by-Step Solution:Outer area = 120*60 = 7200 sq mInner length = 120 - 2*1.5 = 120 - 3 = 117 mInner breadth = 60 - 2*1.5 = 60 - 3 = 57 mInner area = 117*57 = 6669 sq mPath area = 7200 - 6669 = 531 sq mCost = 531 * 0.65 = 345.15 Verification / Alternative check: Another check: the path is relatively thin (1.5 m), so the area removed should be much smaller than 7200. We got 531, which is reasonable (about 7.4% of 7200). Multiplying by Rs. 0.65 should give a few hundred rupees, matching Rs. 345.15. Why Other Options Are Wrong: Rs. 245.10 and Rs. 234.18 arise from using wrong path area (often subtracting only once w instead of 2w). Rs. 456.65 is too high and typically results from using an outside path formula or incorrect inner area. Rs. 315.15 indicates a small arithmetic error in multiplying by 0.65 or a mistaken path area value. Common Pitfalls: Forgetting that the path is on all four sides (must subtract 2w from each dimension). Using 65 as rupees instead of paise. Computing perimeter instead of area. Making multiplication mistakes when finding 117*57 or 531*0.65. Final Answer: Rs. 345.15

Question 18

An isosceles triangle has a perimeter of 32 cm. Each of the two equal sides is 5/6 times the base. Find the area of the triangle in square centimetres (sq cm).

Accepted Answer

Introduction / Context: This problem combines ratio-based side relations with triangle area. You first determine the base and equal side lengths from the perimeter condition, then use the isosceles triangle property (height splits the base into two equal halves) to compute height using the Pythagoras theorem. Finally, use the standard area formula: (1/2)*base*height. The concept tested is translating fractional side relations into solvable equations and applying right-triangle geometry correctly. Given Data / Assumptions: Perimeter = 32 cm Base = b cm Each equal side = (5/6)*b cm Height from the apex bisects the base into b/2 and b/2 Concept / Approach: Use perimeter: b + 2*(5/6*b) = 32 to find b. Then compute equal side length. For an isosceles triangle, dropping the altitude creates two right triangles with hypotenuse equal side, one leg b/2, and the other leg = height h. Use: h = sqrt(side^2 - (b/2)^2). Then area = (1/2)*b*h. Step-by-Step Solution:Let base = bEqual side = (5/6)*bPerimeter: b + 2*(5/6*b) = 32b + (10/6)*b = 32b + (5/3)*b = (8/3)*b = 32b = 32*(3/8) = 12 cmEqual side = (5/6)*12 = 10 cmHalf base = b/2 = 6 cmHeight h = sqrt(10^2 - 6^2) = sqrt(100 - 36) = sqrt(64) = 8 cmArea = (1/2)*12*8 = 48 sq cm Verification / Alternative check: Check perimeter: 10 + 10 + 12 = 32 cm, correct. Since the triangle is isosceles with equal sides 10 and base 12, height 8 is consistent with the 6-8-10 right triangle. So area 48 is reliable. Why Other Options Are Wrong: 39 and 42 usually come from wrong base calculation or wrong height. 57 and 64 occur if you forget the (1/2) in the area formula or incorrectly compute sqrt(100-36). These do not satisfy the consistent 6-8-10 geometry. Common Pitfalls: Misinterpreting “5/6 times the base” as “base is 5/6 of the side.” Forgetting that the altitude bisects the base in an isosceles triangle. Using Pythagoras with the full base (12) instead of half base (6). Dropping the 1/2 factor in triangle area. Final Answer: 48 sq cm

Question 19

A rhombus has diagonals of lengths 24 cm and 10 cm. Using these diagonals, find the perimeter of the rhombus in centimetres (cm).

Accepted Answer

Introduction / Context: This question tests properties of a rhombus: its diagonals bisect each other at right angles. If you know the diagonals, you can find the side length by forming a right triangle using half of each diagonal. Once a single side is known, the perimeter is simply 4 times the side length. The core idea is converting diagonal information into side length via the Pythagoras theorem. Given Data / Assumptions: Diagonal 1 = 24 cm Diagonal 2 = 10 cm Diagonals bisect each other at right angles Half diagonals are 12 cm and 5 cm Concept / Approach: In a rhombus, each side is the hypotenuse of a right triangle whose legs are half of the diagonals. So side s = sqrt((d1/2)^2 + (d2/2)^2). Then perimeter P = 4*s. Step-by-Step Solution:Half of 24 cm = 12 cmHalf of 10 cm = 5 cmSide s = sqrt(12^2 + 5^2)s = sqrt(144 + 25) = sqrt(169) = 13 cmPerimeter P = 4*s = 4*13 = 52 cm Verification / Alternative check: Because 12-5-13 is a well-known Pythagorean triple, the side length 13 is exact and not approximate. Perimeter then must be 4*13 = 52. This makes the answer highly reliable. Why Other Options Are Wrong: 56 and 48 typically come from small arithmetic mistakes or using 14 or 12 as the side incorrectly. 68 and 72 happen if someone mistakenly adds diagonal lengths or uses full diagonals instead of halves inside the square root. Common Pitfalls: Forgetting to halve the diagonals. Not using the right-angle property at the intersection of diagonals. Computing area instead of perimeter. Rounding unnecessarily even though the result is exact (13). Final Answer: 52 cm

Question 20

The ratio of the length to the breadth of a rectangular park is 3:2. A man cycles once along the boundary of the park at a speed of 12 km/hr and completes one full round in 8 minutes. Using this information, find the area of the park in square metre…

Accepted Answer

Introduction / Context: This problem combines speed-time-distance with rectangle perimeter and ratio-based dimensions. Cycling along the boundary gives the perimeter because distance traveled in one round equals the total boundary length. Once the perimeter is known, the 3:2 ratio lets you express length and breadth in terms of a single variable and solve for both. Finally, multiply length and breadth to obtain the area. The question tests multi-step reasoning and correct unit conversion (km/hr to m/min). Given Data / Assumptions: Length:Breadth = 3:2 Speed = 12 km/hr Time for one round = 8 minutes Perimeter = distance traveled in one round Concept / Approach: Distance = Speed * Time. Convert 12 km/hr into metres per minute, multiply by 8 minutes to get perimeter. Let length = 3x and breadth = 2x. Then perimeter = 2*(3x + 2x) = 10x. Solve for x, then compute area = (3x)*(2x). Step-by-Step Solution:Convert speed: 12 km/hr = 12*1000 m / 60 min = 200 m/minDistance in 8 min = 200*8 = 1600 mSo perimeter P = 1600 mLet length = 3x and breadth = 2xPerimeter: 2*(3x + 2x) = 2*(5x) = 10x10x = 1600 => x = 160Length = 3x = 480 m, Breadth = 2x = 320 mArea = 480*320 = 153600 sq m Verification / Alternative check: Check perimeter: 2*(480+320) = 2*800 = 1600 m, matches the cycling distance. Area magnitude is reasonable for a park with sides 480 m and 320 m (both under 0.5 km). Everything is consistent. Why Other Options Are Wrong: 13600 and 16300 are far too small and usually come from failing to convert km/hr to m/min. 143600 and 156300 typically appear from arithmetic slips in multiplying 480*320 or using an incorrect x due to wrong perimeter formula. Common Pitfalls: Not converting hours to minutes. Treating 8 minutes as 8 hours or 8 seconds. Using area formula when you actually need perimeter first. Using 3x+2x as area directly. Miscomputing 480*320. Final Answer: 153600 sq m

Area Questions

Questions

Practice smarter