Q: The area of a circle is 346.5 sq cm. Find its circumference (in cm), assuming pi = 22/7.

Introduction / Context: This problem connects a circle’s area to its circumference. The area formula is A = pi*r^2, and circumference is C = 2*pi*r. Since the area is known, we first solve for r by rearranging the area formula: r^2 = A/pi, then r = sqrt(r^2). After finding r, we substitute into C = 2*pi*r. The value of pi is specified as 22/7, which usually indicates the numbers were chosen to make r a clean value. The final circumference should be a reasonable size: since area is about 346.5, r will be around 10, making circumference around 60 to 70, which helps validate the result. Given Data / Assumptions: Area A = 346.5 sq cm pi = 22/7 A = pi*r^2 C = 2*pi*r Concept / Approach: Solve for r from A = pi*r^2, then compute circumference using C = 2*pi*r. Keep all units in cm and sq cm consistently. Step-by-Step Solution: r^2 = A / pi = 346.5 / (22/7) = 346.5 * 7 / 22 346.5 / 22 = 15.75, so r^2 = 15.75 * 7 = 110.25 r = sqrt(110.25) = 10.5 cm C = 2*pi*r = 2*(22/7)*10.5 (22/7)*10.5 = 22*1.5 = 33 C = 2*33 = 66 cm Verification / Alternative check: Check area using r=10.5: A = (22/7)*10.5^2 = (22/7)*110.25 = 22*15.75 = 346.5 sq cm, exactly matching the given area. So the circumference computed from this r is consistent. Why Other Options Are Wrong: 38 is too small for a circle with this area. 76 and 88 would require a larger r, which would increase area beyond 346.5. 132 is exactly double of 66 and can come from mistakenly doubling twice. Common Pitfalls: Using diameter instead of radius in formulas, forgetting to take square root after finding r^2, or mixing pi approximations inconsistently. Final Answer: The circumference is 66 cm.

Question 1

The diagonals of two squares are in the ratio 2:5. Find the ratio of their areas.

Accepted Answer

Introduction / Context: This question tests scaling laws for similar shapes. Squares are always similar, so any linear measurement (like side, diagonal, perimeter) scales by the same factor between two squares. The diagonal of a square is proportional to its side because diagonal = side*sqrt(2). Therefore, if diagonals are in the ratio 2:5, the sides are also in the ratio 2:5. Area depends on the square of a linear dimension, so the ratio of areas becomes the square of the ratio of sides (or diagonals). This is a common aptitude shortcut: linear ratio k leads to area ratio k^2. Given Data / Assumptions: Diagonal ratio d1:d2 = 2:5 For a square, diagonal is proportional to side Area of a square is proportional to side^2 Concept / Approach: Convert diagonal ratio to side ratio (same), then square it to get area ratio: 2^2:5^2 = 4:25. Step-by-Step Solution: d1:d2 = 2:5 Since d = s*sqrt(2), s1:s2 = d1:d2 = 2:5 Area ratio = s1^2:s2^2 = 2^2:5^2 Area ratio = 4:25 Verification / Alternative check: Assume diagonals are 2 and 5. Then sides are 2/sqrt(2) and 5/sqrt(2). Areas become (2^2/2) and (5^2/2), i.e., 2 and 12.5. Ratio 2:12.5 simplifies to 4:25, confirming the same result. Why Other Options Are Wrong: 2:25 and 10:25 come from incorrect squaring or multiplying ratios. 3:25 and 6:25 are random-looking distractors that do not match squaring 2:5. Common Pitfalls: Using the diagonal ratio directly as the area ratio (not squaring), or mixing up which ratio to square. Final Answer: The ratio of their areas is 4:25.

Question 2

The area of a square with side a is equal to the area of a triangle with base a. What is the altitude (height) of the triangle?

Accepted Answer

Introduction / Context: This question tests equating areas of two shapes and solving for an unknown dimension. A square of side a has a fixed area a*a = a^2. A triangle with base a and altitude h has area (1/2)*a*h. If these two areas are equal, we set a^2 = (1/2)*a*h and solve for h. The key is careful cancellation of a (assuming a is not zero) and keeping the triangle area formula correct with the 1/2 factor. This is a straightforward substitution and algebra simplification problem commonly used in aptitude tests for quick concept checking. Given Data / Assumptions: Square side = a Area of square = a^2 Triangle base = a Triangle altitude = h Area of triangle = (1/2)*a*h Concept / Approach: Equate the two areas and solve for h: a^2 = (1/2)*a*h. Cancel a and isolate h. Step-by-Step Solution: Area(square) = a^2 Area(triangle) = (1/2)*a*h Given they are equal: a^2 = (1/2)*a*h Cancel a from both sides: a = (1/2)*h Multiply both sides by 2: h = 2a Verification / Alternative check: If h = 2a, triangle area becomes (1/2)*a*(2a) = a^2, exactly matching the square’s area. This direct substitution confirms the result without extra steps. Why Other Options Are Wrong: a would make triangle area (1/2)*a*a = a^2/2, only half of the square. 3a and 5a make triangle area larger than the square. a/2 makes triangle area far smaller than the square. Common Pitfalls: Forgetting the 1/2 in triangle area, canceling incorrectly, or assuming altitude equals base without using the equality condition. Final Answer: The altitude of the triangle is 2a.

Question 3

A square has a side length of 14 feet. What is the area of the square (in square feet)?

Accepted Answer

Introduction / Context: This question checks the basic area formula for a square. A square is a special rectangle where all sides are equal. The area of a square is found by multiplying the side by itself: Area = side * side. Since the side is given in feet, the computed area will be in square feet. This is a direct substitution problem, but it is important not to confuse area with perimeter. Perimeter would be 4*side, while area is side^2. Many options in such questions are designed to trap students who multiply by 4 or add numbers instead of squaring. Given Data / Assumptions: Side of square = 14 feet Area of square = side * side Units: feet for length, square feet for area Concept / Approach: Compute area by squaring the side length: 14^2. Keep the result as a numeric value in square feet. Step-by-Step Solution: Area = 14 * 14 Area = 196 square feet Verification / Alternative check: Since 14 is close to 15, the area should be close to 15^2 = 225. 196 is a bit smaller, which makes sense. Also, if you compute 14^2 as (10+4)^2 = 100 + 80 + 16 = 196, you confirm the multiplication cleanly. Why Other Options Are Wrong: 176 can come from 14*12.5 or a partial mistake. 186 and 206 are near-miss distractors from arithmetic slips. 224 can come from perimeter-like thinking (4*56) confusion or wrong operation. Common Pitfalls: Computing perimeter instead of area, doubling instead of squaring, or making multiplication errors in 14*14. Final Answer: The area of the square is 196 square feet.

Question 4

A parallelogram has area P. Two triangles have areas R and T. All three figures are constructed on the same base and have the same altitude. Which relation must be true?

Accepted Answer

Introduction / Context: This question tests a fundamental area relationship between a triangle and a parallelogram built on the same base and with the same altitude (height). The area of a parallelogram is base * height. The area of a triangle on the same base and height is (1/2) * base * height. So, for any triangle sharing the same base and altitude as the parallelogram, its area is exactly half of the parallelogram’s area. That immediately implies the parallelogram’s area is twice the triangle’s area. The variables R and T represent areas of triangles (possibly different triangles), but the key point is that each such triangle has half the parallelogram’s area under the same base and altitude condition. Therefore, P = 2R must hold (and also P = 2T, and R = T, but the question asks for one relation). Given Data / Assumptions: All figures share the same base All figures share the same altitude (height) Area(parallelogram) = base * height Area(triangle) = (1/2) * base * height P is parallelogram area, R and T are triangle areas Concept / Approach: Compare formulas using the same base and height: P = base*height and R = (1/2)*base*height, so P = 2R. Step-by-Step Solution: Let the common base be b and common altitude be h Parallelogram area P = b*h Triangle area R = (1/2)*b*h So P / R = (b*h) / ((1/2)*b*h) = 2 Therefore, P = 2R Verification / Alternative check: If b=10 and h=6, parallelogram area is 60. Any triangle with the same base 10 and height 6 has area (1/2)*10*6 = 30. Indeed, 60 = 2*30, matching P = 2R regardless of the triangle’s slant or shape as long as base and height are the same. Why Other Options Are Wrong: P = R ignores the 1/2 factor in triangle area. P = R/2 reverses the correct relationship. R = 2P is the opposite of the truth. P = R + T is not guaranteed by the formulas and depends on which triangles are chosen. Common Pitfalls: Forgetting the 1/2 in the triangle area formula, mixing up base and altitude, or assuming shapes must be congruent rather than just sharing base and height. Final Answer: The relation that must be true is P = 2R.

Question 5

A man walks from one corner of a square plot to the opposite corner. Approximately what percent distance is saved by walking diagonally instead of walking along two sides (along the edges)?

Accepted Answer

Introduction / Context: This question tests understanding of a square’s diagonal and comparing two travel paths. If the square has side s, walking along the edges from one corner to the opposite corner requires walking two sides: distance = s + s = 2s. Walking diagonally uses the square’s diagonal, which is s*sqrt(2) by the Pythagoras theorem. The “percent saved” is computed as: (saved distance / original distance) * 100, where “original distance” is the along-the-edges distance 2s, and “saved distance” is (2s - s*sqrt(2)). Because sqrt(2) is approximately 1.414, the savings becomes a percentage close to 29.3%, which is usually rounded to about 30% in aptitude options. Given Data / Assumptions: Square side length = s Edge path distance (two sides) = 2s Diagonal distance = s*sqrt(2) Percent saved = ((2s - s*sqrt(2)) / (2s)) * 100 Use sqrt(2) ≈ 1.414 for approximation Concept / Approach: Compute the fractional saving: (2 - sqrt(2)) / 2, then convert to percent and round to the nearest option. Step-by-Step Solution: Edge distance = 2s Diagonal distance = s*sqrt(2) Saved distance = 2s - s*sqrt(2) = s*(2 - sqrt(2)) Fraction saved = (s*(2 - sqrt(2))) / (2s) = (2 - sqrt(2)) / 2 Using sqrt(2) ≈ 1.414: (2 - 1.414) / 2 = 0.586 / 2 = 0.293 Percent saved ≈ 0.293 * 100 = 29.3% ≈ 30% Verification / Alternative check: Try s=100 m. Edge distance = 200 m. Diagonal ≈ 141.4 m. Saved ≈ 58.6 m. Percent saved ≈ 58.6/200*100 = 29.3%, confirming the same result numerically. Why Other Options Are Wrong: 10% and 20% underestimate the diagonal advantage. 40% overestimates savings; diagonal is not half of 2s. 25% is a common guess but does not match the sqrt(2) based computation. Common Pitfalls: Comparing diagonal to one side (s) instead of two sides (2s), forgetting to divide by the original distance when converting to percent, or using an incorrect value for sqrt(2). Final Answer: Approximately 30% distance is saved by walking diagonally.

Question 6

The area of a rectangle is 80 sq cm. If its length is 16 cm, what is its width (in cm)?

Accepted Answer

Introduction / Context: This question checks the basic area relationship of a rectangle: Area = length * width. When area and one dimension are known, the other dimension is found by division: width = area / length. Since the units are in centimetres and square centimetres, the division returns a value in centimetres. This is a direct single-formula substitution problem, but accuracy depends on not mixing up which number to divide by and not confusing perimeter with area. Given Data / Assumptions: Area = 80 sq cm Length = 16 cm Area formula: Area = length * width Concept / Approach: Rearrange the rectangle area formula: width = Area / length. Substitute the given values and compute the division. Step-by-Step Solution: Area = length * width 80 = 16 * width width = 80 / 16 width = 5 cm Verification / Alternative check: Multiply back to confirm: 16 * 5 = 80 sq cm, which matches the given area. This quick check ensures no division mistake was made. Why Other Options Are Wrong: 10 cm would give area 16*10=160, too large. 15 cm and 20 cm produce even larger areas. 8 cm gives 128, which does not match 80. Common Pitfalls: Dividing by the wrong number, writing width = length/area, or mixing units (cm vs mm). Final Answer: The width of the rectangle is 5 cm.

Question 7

The area of a square is 49 sq mm. What is the length of each side of the square (in mm)?

Accepted Answer

Introduction / Context: This question tests the inverse relationship between the area of a square and its side length. The area of a square is side^2. If the area is known, the side is found by taking the square root of the area. Since the area here is 49 sq mm, the side will be sqrt(49) mm. This is a simple recall and calculation question, but it is important to keep units straight: area is in square millimetres, so the side length will be in millimetres. Given Data / Assumptions: Area of square = 49 sq mm Area formula: Area = side^2 Side length = sqrt(Area) Concept / Approach: Compute side = sqrt(49). Since 49 is a perfect square (7*7), the answer is an integer in mm. Step-by-Step Solution: Area = side^2 49 = side^2 side = sqrt(49) = 7 So the side length is 7 mm Verification / Alternative check: Square the result to verify: 7^2 = 49. That matches the given area, confirming the computed side is correct. Why Other Options Are Wrong: 5 mm gives area 25 sq mm, too small. 6 mm gives area 36 sq mm, still too small. 8 mm gives area 64 sq mm, too large. 9 mm gives area 81 sq mm, too large. Common Pitfalls: Forgetting to take square root, confusing area units with length units, or mistakenly dividing by 2 instead of applying sqrt. Final Answer: The side length of the square is 7 mm.

Question 8

Find the area of a square with side length s = 10 feet. What is the area (in square feet)?

Accepted Answer

Introduction / Context: This question checks basic area calculation for a square. When the side length is given, the area is found by squaring the side: Area = s*s. Since s is measured in feet, the area will be measured in square feet. The options are simple round numbers designed to catch common errors like multiplying by 4 (perimeter thinking) or multiplying by 2. The correct operation is squaring, not doubling or quadrupling. Given Data / Assumptions: Side length s = 10 feet Area of square = s^2 Units: feet for side, square feet for area Concept / Approach: Compute s^2: Area = 10^2. This is a direct formula substitution problem. Step-by-Step Solution: Area = s^2 Area = 10^2 Area = 10 * 10 = 100 square feet Verification / Alternative check: Since 10 is a base-10 number, squaring it ends with two zeros: 10^2 = 100. Also, if you imagine a 10-by-10 grid of 1 sq ft tiles, there are 100 tiles, which matches the computed area. Why Other Options Are Wrong: 200 and 300 can come from doubling or tripling 100 without reason. 400 can come from confusing with perimeter squared or other incorrect operations. 500 is unrelated and is typically a distractor for careless guessing. Common Pitfalls: Computing perimeter (4*10=40) instead of area, or using 2*s instead of s*s. Final Answer: The area is 100 square feet.

Question 9

A rectangular wheelchair slalom course has length 30 m and breadth 16 m. Find the area of the course (in sq m).

Accepted Answer

Introduction / Context: This question tests the area formula for a rectangle in a real-world context (a race course). The area of a rectangle is found by multiplying its length and breadth: Area = length * breadth. The dimensions are given in metres, so the resulting area will be in square metres. This is a direct calculation problem, and the options are close enough that a small multiplication mistake could lead you to a wrong choice. The safest method is to multiply carefully and keep units consistent. Given Data / Assumptions: Length = 30 m Breadth = 16 m Area of rectangle = length * breadth Concept / Approach: Use rectangle area formula and multiply 30 by 16. Keep the result in sq m. Step-by-Step Solution: Area = 30 * 16 30 * 16 = (3 * 16) * 10 = 48 * 10 Area = 480 sq m Verification / Alternative check: Estimate: 30*16 is close to 30*15=450, plus one more 30 gives 480. This quick check matches the computed answer and confirms there is no arithmetic slip. Why Other Options Are Wrong: 560 is 35*16 or 28*20 style confusion, not 30*16. 180 is too small and may come from adding (30+16) and squaring incorrectly. 460 and 500 are near distractors for sloppy multiplication. Common Pitfalls: Adding length and breadth instead of multiplying, dropping a zero (getting 48 instead of 480), or mixing up dimensions from a diagram. Final Answer: The area of the course is 480 sq m.

Question 10

In trapezium ABCD, AB is parallel to CD, AB < CD, CD = 6 cm, and the distance between the parallel sides is 4 cm. If the area of trapezium ABCD is 16 sq cm, find AB (in cm).

Accepted Answer

Introduction / Context: This question tests the area formula of a trapezium (trapezoid). A trapezium has one pair of parallel sides (here AB and CD). The area is the average of the parallel sides multiplied by the perpendicular distance (height) between them: Area = (1/2) * (AB + CD) * height. We are given CD and height and the area, so we can solve for AB using a simple linear equation. Since AB is stated to be less than CD, the solution should come out less than 6 cm, which provides a useful reasonableness check at the end. Given Data / Assumptions: AB || CD CD = 6 cm Height (distance between parallel sides) = 4 cm Area = 16 sq cm Area formula: (1/2) * (AB + CD) * height Concept / Approach: Substitute into the trapezium area formula and solve for AB: 16 = (1/2) * (AB + 6) * 4. Step-by-Step Solution: Area = (1/2) * (AB + CD) * height 16 = (1/2) * (AB + 6) * 4 (1/2) * 4 = 2, so 16 = 2 * (AB + 6) Divide by 2: 8 = AB + 6 AB = 8 - 6 = 2 cm Verification / Alternative check: Plug AB=2 back: area = (1/2)*(2+6)*4 = (1/2)*8*4 = 16 sq cm. It matches exactly, and AB=2 is indeed less than CD=6, consistent with the condition AB < CD. Why Other Options Are Wrong: 1 cm and 3 cm produce areas 14 or 18 sq cm, not 16. 8 cm violates AB < CD and would also increase area a lot. 4 cm would give area (1/2)*(10)*4 = 20 sq cm. Common Pitfalls: Forgetting the 1/2 factor, using CD-AB instead of CD+AB, or using the slant side instead of height as the perpendicular distance. Final Answer: AB = 2 cm.

Question 11

The area of a circle is 346.5 sq cm. Find its circumference (in cm), assuming pi = 22/7.

Accepted Answer

Introduction / Context: This problem connects a circle’s area to its circumference. The area formula is A = pi*r^2, and circumference is C = 2*pi*r. Since the area is known, we first solve for r by rearranging the area formula: r^2 = A/pi, then r = sqrt(r^2). After finding r, we substitute into C = 2*pi*r. The value of pi is specified as 22/7, which usually indicates the numbers were chosen to make r a clean value. The final circumference should be a reasonable size: since area is about 346.5, r will be around 10, making circumference around 60 to 70, which helps validate the result. Given Data / Assumptions: Area A = 346.5 sq cm pi = 22/7 A = pi*r^2 C = 2*pi*r Concept / Approach: Solve for r from A = pi*r^2, then compute circumference using C = 2*pi*r. Keep all units in cm and sq cm consistently. Step-by-Step Solution: r^2 = A / pi = 346.5 / (22/7) = 346.5 * 7 / 22 346.5 / 22 = 15.75, so r^2 = 15.75 * 7 = 110.25 r = sqrt(110.25) = 10.5 cm C = 2*pi*r = 2*(22/7)*10.5 (22/7)*10.5 = 22*1.5 = 33 C = 2*33 = 66 cm Verification / Alternative check: Check area using r=10.5: A = (22/7)*10.5^2 = (22/7)*110.25 = 22*15.75 = 346.5 sq cm, exactly matching the given area. So the circumference computed from this r is consistent. Why Other Options Are Wrong: 38 is too small for a circle with this area. 76 and 88 would require a larger r, which would increase area beyond 346.5. 132 is exactly double of 66 and can come from mistakenly doubling twice. Common Pitfalls: Using diameter instead of radius in formulas, forgetting to take square root after finding r^2, or mixing pi approximations inconsistently. Final Answer: The circumference is 66 cm.

Question 12

Find the total surface area of a cube with side length 7.5 cm. What is the total surface area (in sq cm)?

Accepted Answer

Introduction / Context: This question tests the total surface area of a cube. A cube has 6 identical square faces. If the side length is a, the area of one face is a^2, and total surface area is 6*a^2. Since a is given as 7.5 cm, we square 7.5 to get the area of one face in sq cm, then multiply by 6. The arithmetic involves decimals, so it helps to compute 7.5^2 carefully. The final answer should be in square centimetres because the input length is in centimetres. Given Data / Assumptions: Cube side length a = 7.5 cm Area of one face = a^2 Total surface area = 6*a^2 Concept / Approach: Compute a^2 and multiply by 6: TSA = 6*(7.5^2). Keep units as sq cm. Step-by-Step Solution: a^2 = 7.5 * 7.5 = 56.25 Total surface area = 6 * 56.25 6 * 56.25 = 337.5 So TSA = 337.5 sq cm Verification / Alternative check: Since 7.5 is close to 8, the surface area should be close to 6*8^2 = 6*64 = 384. Our value 337.5 is smaller, which makes sense because 7.5 is smaller than 8. This sanity check supports the computed result. Why Other Options Are Wrong: 364.5 and 446 are inconsistent with 6*a^2 for a=7.5. 684 is roughly double 337.5 and can come from multiplying by 12 instead of 6. 450 is an unrelated distractor for rounding errors. Common Pitfalls: Using volume formula a^3 instead of surface area, forgetting there are 6 faces, or squaring 7.5 incorrectly. Final Answer: The total surface area is 337.5 sq cm.

Question 13

The area of a circle is equal to the area of a rectangle. The rectangle has perimeter 50 cm, and its length is 3 cm more than its breadth. What is the diameter of the circle (in cm), assuming pi = 22/7?

Accepted Answer

Introduction / Context: This problem links two shapes by equating their areas. First, we must find the rectangle’s area using its perimeter and the relation between length and breadth. Perimeter gives L + B, and the “length is 3 cm more” condition gives L in terms of B. Solving these yields exact L and B, and therefore the rectangle area L*B. Because the circle’s area equals this rectangle area, we set pi*r^2 equal to the rectangle area and solve for r. Finally, diameter is 2*r. The value pi = 22/7 is provided to keep the radius and diameter neat and avoid messy decimals. Given Data / Assumptions: Rectangle perimeter = 50 cm Rectangle length L = B + 3 cm Perimeter formula: 2*(L + B) = 50 Area rectangle = L*B Area circle = pi*r^2, pi = 22/7 Diameter = 2*r Concept / Approach: Find L and B from perimeter and difference, compute rectangle area, set it equal to pi*r^2, solve r, then compute diameter. Step-by-Step Solution: 2*(L + B) = 50 => L + B = 25 L = B + 3 Substitute: (B + 3) + B = 25 => 2B = 22 => B = 11 Then L = 11 + 3 = 14 Rectangle area = L*B = 14*11 = 154 sq cm Circle area = 154 = (22/7)*r^2 r^2 = 154 * 7 / 22 = 49 => r = 7 Diameter = 2*r = 14 cm Verification / Alternative check: Check circle area with r=7: (22/7)*49 = 22*7 = 154 sq cm, exactly equal to rectangle area. This confirms the computed diameter is correct. Why Other Options Are Wrong: 7 cm is the radius, not the diameter. 21 cm, 28 cm are too large and would produce areas much bigger than 154 sq cm. 10 cm is a common guess but does not match the derived radius. Common Pitfalls: Forgetting to divide perimeter by 2, mixing up length and breadth, forgetting to double the radius to get diameter, or using pi inconsistently. Final Answer: The diameter of the circle is 14 cm.

Question 14

The diagonal of a square is 14 cm. What is the area of the square (in sq cm)?

Accepted Answer

Introduction / Context: This question tests the relationship between a square’s diagonal and its area. In a square, the diagonal forms a right triangle with the two sides. By the Pythagoras theorem, diagonal d and side s satisfy d = s*sqrt(2). Instead of calculating s with square roots and then squaring again, a useful shortcut is: Area = s^2 = (d^2)/2. This works because s = d/sqrt(2), so s^2 = d^2/2. With diagonal 14 cm, d^2 is 196, and dividing by 2 gives a clean area value. Units remain consistent: diagonal is in cm, so area is in sq cm. Given Data / Assumptions: Diagonal d = 14 cm For square: d = s*sqrt(2) Area = s^2 Shortcut: Area = d^2 / 2 Concept / Approach: Compute area using Area = d^2/2. This avoids dealing with sqrt(2) directly and reduces rounding errors. Step-by-Step Solution: Area = d^2 / 2 d^2 = 14 * 14 = 196 Area = 196 / 2 = 98 sq cm Verification / Alternative check: Find side: s = 14 / sqrt(2) = 7*sqrt(2). Then area s^2 = (7*sqrt(2))^2 = 49*2 = 98 sq cm. This matches the shortcut result and confirms correctness. Why Other Options Are Wrong: 196 sq cm happens if you wrongly treat diagonal as side and compute 14^2. 49 sq cm can come from using half the diagonal as side (7^2), which is incorrect. 77 and 100 are distractors from arithmetic or rounding errors. Common Pitfalls: Confusing diagonal with side, forgetting the division by 2 in d^2/2, or applying Pythagoras incorrectly. Final Answer: The area of the square is 98 sq cm.

Question 15

Two circles touch externally, so the distance between their centres equals the sum of their radii. The distance between their centres is 12 cm. The radius of one circle is 7 cm. Find the diameter of the other circle (in cm).

Accepted Answer

Introduction / Context: This question tests the basic property of externally tangent circles. When two circles touch externally, they meet at exactly one point, and the line joining their centres passes through that point of tangency. In this case, the distance between their centres equals the sum of their radii: d = r1 + r2. If you know the centre distance and one radius, you can find the other radius by subtraction: r2 = d - r1. Finally, the diameter of the second circle is 2*r2. This is a direct substitution and arithmetic problem with a clear geometry meaning. Given Data / Assumptions: Circles touch externally Distance between centres d = 12 cm Radius of one circle r1 = 7 cm For external touching: d = r1 + r2 Diameter of other circle = 2*r2 Concept / Approach: Use d = r1 + r2 to find r2. Then compute diameter = 2*r2. Step-by-Step Solution: d = r1 + r2 12 = 7 + r2 r2 = 12 - 7 = 5 cm Diameter of other circle = 2*r2 = 2*5 = 10 cm Verification / Alternative check: If the other circle’s radius were 5 cm, then the sum of radii would be 7 + 5 = 12 cm, exactly matching the given centre distance. This is precisely the condition for external tangency, so the computed radius and diameter are consistent with the geometry. Why Other Options Are Wrong: 5 is the radius, not the diameter. 8 and 12 imply radii 4 and 6, which would not sum with 7 to make 12. 14 is the diameter of the first circle (2*7), not the second. Common Pitfalls: Adding instead of subtracting to get r2, confusing radius with diameter, or mixing up internal vs external tangency rules (difference of radii vs sum of radii). Final Answer: The diameter of the other circle is 10 cm.

Question 16

The area of a rhombus is 84 square centimetres and the length of one of its diagonals is 7 centimetres. Find the length, in centimetres, of the other diagonal of the rhombus.

Accepted Answer

Introduction / Context: This question tests the relationship between the area of a rhombus and its diagonals. Rhombus based questions are common in aptitude tests because they combine geometry formulas with simple algebra. If you know how the diagonals of a rhombus determine its area, you can easily find an unknown diagonal when the area and one diagonal are given. Given Data / Assumptions: Area of the rhombus = 84 square centimetres. Length of one diagonal = 7 centimetres. The other diagonal is unknown and must be found. The figure is a standard rhombus, so the usual formula for area applies. Concept / Approach: The area of a rhombus can be expressed in terms of its diagonals. If d1 and d2 are the lengths of the diagonals, then the area is given by: Area = (d1 * d2) / 2. Here, one diagonal and the area are known, so we rearrange the formula to solve for the second diagonal. This turns the problem into a straightforward algebraic equation in one variable. Step-by-Step Solution: Let d1 = 7 cm and let d2 be the unknown diagonal in centimetres. Area = (d1 * d2) / 2. So 84 = (7 * d2) / 2. Multiply both sides by 2 to clear the denominator: 84 * 2 = 7 * d2. 168 = 7 * d2. Now divide both sides by 7: d2 = 168 / 7. d2 = 24 centimetres. Verification / Alternative check: We can substitute d1 = 7 and d2 = 24 back into the area formula to verify the result. Area = (7 * 24) / 2. First compute 7 * 24 = 168. Then divide by 2 to get 84. This matches the given area, so the computed value of the second diagonal is correct and consistent with the formula. Why Other Options Are Wrong: Option A (12): If d2 = 12, area would be (7 * 12) / 2 = 42 square centimetres, which is only half of the required area. Option C (48): If d2 = 48, area would be (7 * 48) / 2 = 168 square centimetres, which is exactly double the required area. Option D (36): If d2 = 36, area would be (7 * 36) / 2 = 126 square centimetres, which is greater than the given area. Option E (18): If d2 = 18, area would be (7 * 18) / 2 = 63 square centimetres, which is less than 84. Common Pitfalls: Many learners mistakenly use the formula for the area of a rectangle or for base times height instead of the diagonal based formula for a rhombus. Another common error is to forget the division by 2 in the formula Area = (d1 * d2) / 2. Some students also reverse operations when solving the equation and may multiply instead of dividing, which leads to incorrect diagonal lengths. Writing the formula clearly and performing algebraic steps slowly helps avoid these mistakes. Final Answer: The length of the other diagonal of the rhombus is 24 centimetres.

Question 17

A semicircle has a radius of 28 centimetres. Using the standard formula for the perimeter of a semicircle, which includes the curved part and the diameter, find its total perimeter in centimetres. Take pi = 22/7.

Accepted Answer

Introduction / Context: This problem checks your understanding of the perimeter of a semicircle, which is a very common shape in aptitude and geometry questions. The key idea is that the perimeter of a semicircle is not only the curved arc, but also includes the straight diameter. Forgetting the diameter is one of the most common mistakes in such problems. Given Data / Assumptions: Radius of the semicircle r = 28 centimetres. We must find the total perimeter of the semicircle. Use pi = 22/7 for calculations. Perimeter means curved part plus the straight diameter. Concept / Approach: For a full circle, circumference = 2 * pi * r. For a semicircle, the curved part is half of the full circumference, which is pi * r. However, the perimeter of a semicircle also includes the diameter, which is 2 * r. Therefore, the perimeter P of a semicircle is given by: P = pi * r + 2 * r. We simply substitute the given radius and the given value of pi into this formula and compute the numerical value step by step. Step-by-Step Solution: Given r = 28 cm and pi = 22/7. Perimeter of semicircle P = pi * r + 2 * r. First compute pi * r: pi * r = (22/7) * 28. 28 divided by 7 is 4, so pi * r = 22 * 4 = 88. Now compute the diameter term: 2 * r = 2 * 28 = 56. Add both contributions: P = 88 + 56 = 144. Therefore, the perimeter of the semicircle is 144 centimetres. Verification / Alternative check: We can quickly check the reasonableness of the answer. A full circle of radius 28 cm would have circumference 2 * pi * r = 2 * (22/7) * 28 = 176 cm. A semicircle will have slightly less than this value as its total perimeter, since it has half the circumference (88) plus one diameter (56), giving 144, which is less than 176 but more than the half circumference 88. The number 144 therefore makes sense in this context. Why Other Options Are Wrong: Option A (242): This is larger than the circumference of the full circle, so it cannot be the perimeter of a semicircle. Option C (121): This does not match pi * r + 2 * r and comes from no standard combination of radius and pi. Option D (288): This is exactly double the correct value and might result from mistakenly using 2 * (pi * r + 2 * r). Option E (200): This is an arbitrary incorrect value and does not match the correct formula. Common Pitfalls: The biggest pitfall is using only the curved part of the semicircle and forgetting to add the diameter, which would give 88 instead of 144. Another common issue is using the wrong formula, such as 2 * pi * r for a semicircle. Some students also make arithmetic errors when multiplying by 22/7. Writing out each step clearly and remembering that a semicircle has both a curved boundary and a straight boundary helps avoid errors. Final Answer: The perimeter of the semicircle is 144 centimetres.

Question 18

ABCD is a quadrilateral inscribed in a circle of radius r. The bisectors of angles DAB and BCD meet the circle again at points X and Y respectively. What is the length of the chord XY in terms of r?

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Introduction / Context: This is a conceptual geometry question involving a cyclic quadrilateral, angle bisectors, and a circle. It tests your understanding of special properties of angles in a circle and how angle bisectors of opposite angles of a cyclic quadrilateral behave. Instead of heavy calculation, the key is to recall a known geometric result that connects these angle bisectors to a diameter of the circle. Given Data / Assumptions: ABCD is a quadrilateral inscribed in a circle of radius r. The bisector of angle DAB meets the circle again at point X. The bisector of angle BCD meets the circle again at point Y. We must find the length of XY in terms of r. All usual circle theorems for cyclic quadrilaterals apply. Concept / Approach: In a cyclic quadrilateral, opposite angles are supplementary. That is, angle DAB plus angle DCB equals 180 degrees. When we take the bisectors of these opposite angles, the resulting half angles add up to 90 degrees. The rays along these bisectors, when extended to meet the circle, mark points that form a right angle at the centre. In a circle, any angle subtending a diameter is a right angle. Therefore, the chord joining the intersection points of these bisectors with the circle is actually a diameter of the circle. Once we know that XY is a diameter, its length is simply 2 * r. Step-by-Step Solution: Step 1: In a cyclic quadrilateral ABCD, angle DAB + angle DCB = 180 degrees. Step 2: Let angle DAB = alpha and angle DCB = beta, so alpha + beta = 180 degrees. Step 3: The bisector of angle DAB makes an angle alpha/2 with each side of that angle, and the bisector of angle BCD makes an angle beta/2 with each side of that angle. Step 4: Because alpha + beta = 180 degrees, we have alpha/2 + beta/2 = 90 degrees. Step 5: The two bisectors, when extended to the circle, define points X and Y such that the central angle subtended by chord XY is 180 degrees, meaning XY is a diameter. Step 6: The length of any diameter of a circle of radius r is 2 * r. Verification / Alternative check: Another way to see this is by recalling that in a circle, if two points on the circle subtend a right angle at any point on the circumference, then the segment joining those two points is a diameter. The angle between the two bisectors is 90 degrees, which ensures that they subtend a right angle somewhere on the circle. This confirms that XY must be a diameter, and therefore its length is 2 * r. This matches the result obtained from the angle sum reasoning. Why Other Options Are Wrong: Option A (pi*r^2): This represents the area of a circle, not a length, so it is dimensionally incorrect. Option C (r + 2): This mixes a length r with a pure number 2 without any geometric justification. Option D ((pi*r^2)/2): This is half the area of a circle and again is an area, not a linear measure. Option E (r/2): This is only a quarter of the diameter and has no special geometric significance in this configuration. Common Pitfalls: A common mistake is to try to apply chord length formulas or coordinate geometry directly without recognising that the chord is a diameter. Another pitfall is confusing area formulas with length formulas, which leads to expressions involving pi*r^2 instead of simple multiples of r. Students may also forget that opposite angles of a cyclic quadrilateral are supplementary, which is the key starting point. Remembering and using this property quickly reduces the question to a simple fact about diameters. Final Answer: The length of chord XY is 2*r units.

Question 19

A tent is to be built in the form of a right circular cylinder of radius 10 metres surmounted by a right circular cone of the same radius. The height of the cylindrical part is 5 metres and the slant height of the conical part is 15 metres. If only …

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Introduction / Context: This question combines the curved surface area of a cylinder and a cone, then adds a percentage extra for stitching and folding. Such problems model real life applications like making tents or covers and require you to carefully select only the relevant surfaces. You must know and apply surface area formulas correctly and then handle the percentage increase accurately. Given Data / Assumptions: Radius r of both cylinder and cone = 10 metres. Height of cylindrical part h = 5 metres. Slant height of conical part l = 15 metres. Only curved surfaces need canvas, so bases are not covered. Extra canvas allowed for folding and stitching = 20% of the total curved area. Use pi = 22/7. Concept / Approach: We first compute the curved surface area of the cylinder, which is given by 2 * pi * r * h. Next, we compute the curved surface area of the cone, which is pi * r * l. Adding these gives the total curved area that would be needed without any extra allowance. Then we increase this total by 20% to account for folding and stitching. Mathematically, that means multiplying the total curved area by 1.2. The final result is rounded to a reasonable whole number of square metres for practical use. Step-by-Step Solution: Step 1: Curved surface area of cylinder = 2 * pi * r * h. Substitute values: 2 * (22/7) * 10 * 5. Compute: 2 * 10 * 5 = 100, so area = (22/7) * 100 = 2200/7 square metres, which is about 314.2857 sq m. Step 2: Curved surface area of cone = pi * r * l = (22/7) * 10 * 15. Compute: 10 * 15 = 150, so area = (22/7) * 150 = 3300/7 square metres, which is about 471.4286 sq m. Step 3: Total curved area without extra = 2200/7 + 3300/7 = 5500/7 sq m. 5500/7 is approximately 785.7143 square metres. Step 4: Allow 20% extra: required canvas = total area * 1.2. Required canvas = (5500/7) * 1.2 = (5500 * 1.2) / 7 = 6600/7 sq m. 6600/7 is approximately 942.8571 square metres, which we round to about 943 sq m. Verification / Alternative check: A quick check is to approximate pi as 3.14 and repeat the calculation. Curved area of cylinder becomes roughly 2 * 3.14 * 10 * 5 = 314 sq m. Curved area of cone becomes approximately 3.14 * 10 * 15 = 471 sq m. Their sum is about 785 sq m. Adding 20% gives 785 * 1.2 which is again about 942 sq m. This aligns well with our earlier, more exact calculation and supports the rounded answer of approximately 943 square metres of canvas. Why Other Options Are Wrong: Option B (786 sq m): This is the total curved area before adding the 20% extra, so it ignores the stitching allowance. Option C (1100 sq m): This is much larger than the correctly computed value and cannot be justified by the given dimensions. Option D (628 sq m): This is significantly lower than the basic curved area itself, so it is not realistic. Option E (1570 sq m): This is more than one and a half times the correct value and does not follow from any correct step. Common Pitfalls: Students often forget that only curved surfaces need to be counted, mistakenly including the area of the base. Another frequent error is to forget to add the 20% extra, or to add 20 as a fixed number instead of 20% as a proportion. Sometimes the formula for surface area of a cone is misremembered as involving the vertical height instead of the slant height. Carefully identifying each formula and applying the percentage as a multiplier of 1.2 prevents these mistakes. Final Answer: The total canvas required is approximately 943 sq m (square metres).

Question 20

A square field has an area of X square metres. A cylindrical ditch of radius 7 metres and depth 2 metres is dug in the field, and all the earth removed from the ditch is spread uniformly over the remaining part of the field, so that the ground level…

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Introduction / Context: This question links volume and area concepts. Earth removed from a cylindrical ditch is used to raise the level of the remaining part of a square field. Such problems are typical in aptitude tests to check your ability to equate volumes and handle percentage like increases in height. Converting the geometric description into a simple volume balance equation is the main skill here. Given Data / Assumptions: Square field area = X square metres. Radius of cylindrical ditch r = 7 metres. Depth (height) of ditch h = 2 metres. Earth from ditch is spread uniformly over the remaining field. The rise in level of the remaining field = 0.77 metres. The cylindrical ditch footprint removes area equal to pi * r^2 from the field. Use pi = 22/7. Concept / Approach: The volume of earth dug out from the cylindrical ditch is volume of cylinder = pi * r^2 * h. This volume is spread over the remaining area of the field, which is total field area minus the circular area of the ditch. If the rise in level over this remaining area is t, then the added volume is t times the remaining area. So we set volume dug out equal to volume spread: pi * r^2 * h = t * (X - pi * r^2). Then we solve this equation for X. Finally, we compare our result with the options to pick the correct value. Step-by-Step Solution: Step 1: Volume of cylindrical ditch = pi * r^2 * h. With r = 7 and h = 2, r^2 = 49, so volume = pi * 49 * 2 = 98 * pi. Using pi = 22/7, volume = 98 * (22/7) = 98 * 22 / 7 = 14 * 22 = 308 cubic metres. Step 2: Remaining area of field after ditch is dug = X - pi * r^2. Compute pi * r^2 = (22/7) * 49 = 22 * 7 = 154 square metres. So remaining area = X - 154. Step 3: Volume added to remaining area = rise * remaining area = 0.77 * (X - 154). Equate volumes: 0.77 * (X - 154) = 308. Step 4: Write 0.77 as 77/100. So (77/100) * (X - 154) = 308. Multiply both sides by 100: 77 * (X - 154) = 30800. Step 5: Divide both sides by 77: X - 154 = 30800 / 77. Since 77 * 400 = 30800, we get X - 154 = 400. Step 6: So X = 400 + 154 = 554 square metres. Verification / Alternative check: We can check quickly by reversing the logic. If X = 554, then remaining area after removing the circular ditch is 554 - 154 = 400 square metres. When this area is raised by 0.77 metres, the volume gained is 400 * 0.77 = 308 cubic metres. This matches perfectly with the volume of earth dug from the ditch, which we computed as 308 cubic metres. Therefore, the value X = 554 sq m is fully consistent and correct. Why Other Options Are Wrong: Option A (548 sq m): This would give remaining area 394 sq m, and 394 * 0.77 is not equal to 308, so volume does not match. Option B (524 sq m): Remaining area becomes 370 sq m; multiplying by 0.77 again fails to produce 308 cubic metres. Option C (518 sq m): This leads to an even smaller remaining area and an incorrect volume balance. Option E (540 sq m): This value also fails the volume equality when you compute 0.77 * (540 - 154). Common Pitfalls: One frequent mistake is forgetting to subtract the area of the circular ditch from the square field before applying the rise in height. Another is using diameter instead of radius in the formula for the area or volume of the cylinder. Students may also mis-handle the decimal 0.77, so converting it to a fraction like 77/100 can make algebra cleaner. Always ensure that volume removed equals volume added when dealing with earthwork type problems. Final Answer: The area of the square field is 554 sq m (square metres).

Area Questions