Question 1

What is the area (in square centimetres) of an equilateral triangle whose side length is 14 cm?

Accepted Answer

Introduction / Context: Equilateral triangles are triangles with all sides equal and all interior angles equal to 60°. There is a standard formula for the area of an equilateral triangle in terms of its side length. This formula is frequently used in aptitude and competitive exams because it allows quick calculation without needing to derive the height separately each time. Given Data / Assumptions: The triangle is equilateral. Each side of the triangle is 14 cm long. We need to calculate the area in square centimetres. Concept / Approach: The standard area formula for an equilateral triangle with side length a is: Area = (√3 / 4) * a^2 This formula is derived from drawing the altitude, which splits the equilateral triangle into two right triangles, and then using the Pythagorean theorem to express the height in terms of the side length. Once we know the formula, we can substitute the given side length directly. Step-by-Step Solution: Step 1: Identify the side length a = 14 cm. Step 2: Use the formula Area = (√3 / 4) * a^2. Step 3: Compute a^2: 14^2 = 196. Step 4: Substitute into the formula: Area = (√3 / 4) * 196. Step 5: Simplify 196 / 4 = 49. Step 6: Therefore, Area = 49√3 square centimetres. Verification / Alternative check: We can verify by computing the height first. In an equilateral triangle of side a, the height h is given by h = (√3 / 2) * a. For a = 14, h = (√3 / 2) * 14 = 7√3. The area is then (1/2) * base * height = (1/2) * 14 * 7√3 = 7 * 7√3 = 49√3. This matches the formula-based result and confirms the area. Why Other Options Are Wrong: 98√3: This is double the correct area, possibly from forgetting to divide by 4 in the formula. (49/2)√3 and (49/4)√3: These values underestimate the area, coming from misuse of the denominator factor. 196: This is simply a^2 and ignores the √3/4 factor entirely. Common Pitfalls: Some students mistakenly treat an equilateral triangle like a right triangle and miscalculate the height. Others forget the exact formula and either miss the √3 factor or the 1/4 factor. Memorising the compact formula Area = (√3 / 4) * a^2 is very helpful and avoids repeated derivations during timed exams. Final Answer: Thus, the area of the equilateral triangle is 49√3 sq cm.

Question 2

Point P is the midpoint of segment AB. The coordinates of P are (3, 1) and the coordinates of B are (5, -4). What are the coordinates of point A?

Accepted Answer

Introduction / Context: The midpoint formula is a fundamental concept in coordinate geometry. It gives the coordinates of the point exactly halfway between two given points. This question reverses that idea: you are given the midpoint and one endpoint and must find the other endpoint. Understanding how to manipulate the midpoint formula is essential for solving such problems efficiently. Given Data / Assumptions: Segment AB has midpoint P. Coordinates of P are (3, 1). Coordinates of B are (5, -4). We must determine the coordinates of A(x, y). Concept / Approach: If A(x1, y1) and B(x2, y2) are endpoints of a segment, the midpoint P has coordinates: P = ( (x1 + x2) / 2 , (y1 + y2) / 2 ) Here P and B are known, and we need A. So we can set up two equations by equating the x and y coordinates of the midpoint to the given values and then solve for the unknown coordinates of A. Step-by-Step Solution: Step 1: Let A have coordinates (x, y). Step 2: Use the midpoint formula for the x coordinate: (x + 5) / 2 = 3. Step 3: Multiply both sides by 2: x + 5 = 6, so x = 6 - 5 = 1. Step 4: Use the midpoint formula for the y coordinate: (y + (-4)) / 2 = 1. Step 5: Simplify the numerator: (y - 4) / 2 = 1. Step 6: Multiply both sides by 2: y - 4 = 2, so y = 2 + 4 = 6. Step 7: Therefore, A has coordinates (1, 6). Verification / Alternative check: Check that P really is the midpoint of A(1, 6) and B(5, -4). The average of the x coordinates is (1 + 5) / 2 = 6 / 2 = 3. The average of the y coordinates is (6 + (-4)) / 2 = 2 / 2 = 1. This matches the given midpoint (3, 1), so our solution is consistent and correct. Why Other Options Are Wrong: (-1, 7), (1, -7), and (-1, -7): These do not produce a midpoint of (3, 1) when paired with B(5, -4). (7, -1): This would give a midpoint with x coordinate (7 + 5) / 2 = 6, not 3. Common Pitfalls: Some learners mistakenly subtract instead of averaging or mis-handle negative coordinates. Another issue is mixing x and y equations when solving for the unknown point. Always apply the midpoint formula component wise: solve separately for the x coordinate and the y coordinate using the average method. Final Answer: Thus, the coordinates of point A are (1, 6).

Question 3

What is the slope of a line that is perpendicular to the line passing through the points (-5, 1) and (-2, 0)?

Accepted Answer

Introduction / Context: In coordinate geometry, slopes help describe the orientation of lines. Two lines are perpendicular if the angle between them is 90°, and there is a specific relationship between their slopes. This question tests your ability to compute the slope of a line from two points and then use the negative reciprocal rule to find the slope of a perpendicular line. Given Data / Assumptions: A line passes through points A(-5, 1) and B(-2, 0). We need the slope of any line that is perpendicular to this line. The coordinate plane is standard with x horizontal and y vertical. Concept / Approach: First, find the slope m of the given line using the formula: m = (y2 - y1) / (x2 - x1) If a line has slope m, then any line perpendicular to it has slope m_perp given by: m_perp = -1 / m This is known as the negative reciprocal relationship between slopes of perpendicular lines (provided neither line is vertical or horizontal in a conflicting way). Step-by-Step Solution: Step 1: Label the points A(-5, 1) and B(-2, 0). Step 2: Compute the slope m of line AB using m = (y2 - y1) / (x2 - x1). Step 3: Substitute the coordinates: m = (0 - 1) / (-2 - (-5)). Step 4: Simplify the numerator: 0 - 1 = -1. Step 5: Simplify the denominator: -2 - (-5) = -2 + 5 = 3. Step 6: Therefore, m = -1 / 3. Step 7: The slope of a perpendicular line is the negative reciprocal of -1 / 3, so m_perp = -1 / (-1 / 3) = 3. Verification / Alternative check: If two slopes m1 and m2 correspond to perpendicular lines, then their product should be -1. Here, m1 = -1/3 and m2 = 3. The product is (-1/3) * 3 = -1, which satisfies the perpendicular condition exactly. This confirms that the slope 3 is correct for a perpendicular line. Why Other Options Are Wrong: -3: This would make the product of slopes (-1/3) * (-3) = 1, indicating parallel but not perpendicular lines. -1/3: This is the slope of the original line, so it gives parallel lines, not perpendicular ones. 1/3: This gives a product of slopes (-1/3) * (1/3) = -1/9, which does not indicate perpendicularity. 0: This slope would give a horizontal line, which is perpendicular only to vertical lines, not to the given line with slope -1/3. Common Pitfalls: Students frequently confuse "negative reciprocal" with simply "negative" or merely "reciprocal". Another error is making sign mistakes while subtracting coordinates. Always calculate the original slope carefully, then take the negative reciprocal accurately to find the slope of a line perpendicular to it. Final Answer: Therefore, the slope of a line perpendicular to the line through (-5, 1) and (-2, 0) is 3.

Question 4

What is the area (in square centimetres) of a rectangle whose diagonal is 25 cm and one of its sides is 24 cm long?

Accepted Answer

Introduction / Context: This problem combines the Pythagorean theorem with the area formula for a rectangle. A rectangle has opposite sides equal and all angles equal to 90°, so its diagonal forms a right triangle with the two adjacent sides as legs. By using the diagonal and one side, we can find the other side using Pythagoras, and then compute the area as length multiplied by breadth. Given Data / Assumptions: The shape is a rectangle. The length of the diagonal is 25 cm. One side of the rectangle is 24 cm. We must find the area of the rectangle in square centimetres. Concept / Approach: Let the sides of the rectangle be a and b, and the diagonal be d. The Pythagorean theorem gives: d^2 = a^2 + b^2 Once both side lengths are known, the area is given by: Area = a * b We are given d and one side, so we solve for the other side and then compute the area. Step-by-Step Solution: Step 1: Let one side of the rectangle be a = 24 cm and the other side be b (unknown). Step 2: The diagonal is given as d = 25 cm. Step 3: Apply the Pythagorean theorem: d^2 = a^2 + b^2. Step 4: Substitute the values: 25^2 = 24^2 + b^2. Step 5: Compute squares: 625 = 576 + b^2. Step 6: Rearrange to find b^2: b^2 = 625 - 576 = 49. Step 7: Take the square root: b = sqrt(49) = 7 cm. Step 8: Compute the area: Area = a * b = 24 * 7. Step 9: Multiply: 24 * 7 = 168 sq cm. Verification / Alternative check: We can check the diagonal using the side lengths 24 cm and 7 cm. The diagonal squared should equal 24^2 + 7^2 = 576 + 49 = 625, so the diagonal is sqrt(625) = 25 cm, which matches the given value. The area is then clearly 24 * 7 = 168 sq cm, confirming that the computed area is consistent with both the Pythagorean theorem and the given dimensions. Why Other Options Are Wrong: 186, 144, 132, and 150: These values come from incorrect combinations or arithmetic errors. None of them correspond to 24 multiplied by the correctly derived second side of 7 cm. Common Pitfalls: Some students mistakenly add or subtract the diagonal and given side instead of using the Pythagorean theorem. Others might use the wrong formula for area. Remember that a rectangle's diagonal, together with its sides, always satisfies a^2 + b^2 = d^2, and area is simply the product of the two side lengths once they are known. Final Answer: Therefore, the area of the rectangle is 168 sq cm.

Question 5

The area of the four walls of a cuboid room (lateral surface area) is 57 sq m. If its length is 5.5 m and its height is 3 m, what is the breadth of the cuboid (in metres)?

Accepted Answer

Introduction / Context: This question deals with the lateral surface area of a cuboid, which corresponds to the area of its four vertical walls. For practical applications, such as painting or wallpapering a room, this lateral area is more important than the total surface area. Knowing the relationship between the lateral surface area and the dimensions of a cuboid allows us to find unknown dimensions when sufficient information is provided. Given Data / Assumptions: The solid is a cuboid (rectangular box). The area of the four walls (lateral surface area) is 57 square metres. Length (l) of the cuboid is 5.5 m. Height (h) of the cuboid is 3 m. We must find the breadth (b) in metres. Concept / Approach: The lateral surface area (area of the four walls) of a cuboid of length l, breadth b, and height h is given by: Lateral area = 2h(l + b) This formula comes from adding the areas of the four rectangles: two of size l * h and two of size b * h. We substitute the known values of lateral area, length, and height and then solve for breadth b. Step-by-Step Solution: Step 1: Write the formula for lateral surface area: L = 2h(l + b). Step 2: Substitute L = 57 sq m, h = 3 m, and l = 5.5 m. Step 3: The equation becomes 57 = 2 * 3 * (5.5 + b). Step 4: Simplify the constant product: 2 * 3 = 6, so 57 = 6 * (5.5 + b). Step 5: Divide both sides by 6: (5.5 + b) = 57 / 6. Step 6: Compute 57 / 6 = 9.5. Step 7: So 5.5 + b = 9.5. Step 8: Solve for b: b = 9.5 - 5.5 = 4. Step 9: Hence, the breadth is 4 metres. Verification / Alternative check: We can quickly verify by plugging b = 4 back into the formula. Then l + b = 5.5 + 4 = 9.5. Lateral area becomes 2 * 3 * 9.5 = 6 * 9.5 = 57 sq m, which matches the given value. Thus the breadth of 4 m is consistent with all the information provided. Why Other Options Are Wrong: 4.5, 3, 3.5, and 2.5: Substituting any of these into l + b would not produce 9.5, and therefore the lateral surface area would not equal 57 sq m. These are distractors arising from algebraic or arithmetic mistakes. Common Pitfalls: One common error is to use the total surface area formula instead of the lateral surface area formula, leading to extra terms such as 2lb that are not needed. Another pitfall is miscalculating the division or the product 2h(l + b). Carefully substituting values and solving the simple linear equation step by step helps avoid such mistakes. Final Answer: Therefore, the breadth of the cuboid is 4 m.

Question 6

In triangle ABC, points D and E lie on sides AB and AC respectively, with DE parallel to BC. If AD:DB = 1:4 and the area of triangle ADE is 6 sq cm, what is the ratio of area of triangle ADE to the area of quadrilateral BDEC?

Accepted Answer

Introduction / Context: This problem again uses similar triangles formed by a line drawn parallel to the base of a triangle. When such a line is drawn, it produces a smaller, similar triangle and a trapezoidal region (here the quadrilateral BDEC). Understanding how the areas of the similar triangles scale with the square of the side ratio allows us to compute area ratios between the small triangle and the remaining quadrilateral. Given Data / Assumptions: D lies on AB and E lies on AC of triangle ABC. DE is parallel to BC, so triangle ADE is similar to triangle ABC. The ratio AD : DB is 1 : 4. The area of triangle ADE is 6 square centimetres. We must find the ratio area(ADE) : area(BDEC). Concept / Approach: Because DE is parallel to BC, triangles ADE and ABC are similar. If AD : AB is the linear ratio, then the area ratio of the smaller to the larger triangle is the square of that ratio. Once we know the areas of the two triangles, we can find the area of the quadrilateral by subtraction, and then form the required ratio of the small triangle to the quadrilateral. Step-by-Step Solution: Step 1: From AD : DB = 1 : 4, we find AB = AD + DB = 1 + 4 = 5 parts. Step 2: Therefore AD / AB = 1 / 5. Step 3: The area ratio of triangle ADE to triangle ABC is (AD / AB)^2 = (1 / 5)^2 = 1 / 25. Step 4: Let the area of triangle ABC be A. Then area(ADE) = (1 / 25) * A. Step 5: We are given area(ADE) = 6 sq cm, so (1 / 25) * A = 6. Step 6: Solve for A: A = 6 * 25 = 150 sq cm. Step 7: The area of quadrilateral BDEC is area(ABC) - area(ADE) = 150 - 6 = 144 sq cm. Step 8: Now form the ratio area(ADE) : area(BDEC) = 6 : 144. Step 9: Simplify 6 : 144 by dividing both numbers by 6 to get 1 : 24. Verification / Alternative check: We can confirm the area ratio between the two triangles is 1 : 25. That means the trapezoidal region (quadrilateral BDEC) should have 24 times the area of the small triangle, since 1 part belongs to triangle ADE and 24 parts to the remaining region. Since area(ADE) = 6, the quadrilateral should have area 6 * 24 = 144, matching the computed value. This confirms that the ratio 1 : 24 is correct. Why Other Options Are Wrong: 1:12, 1:6, 1:16, and 1:20: These ratios do not match the squared side scaling. They would imply different relationships between AD and AB or an incorrect area of the entire triangle, contradicting the similarity conditions and given area. Common Pitfalls: Some students mistakenly take the area ratio as 1 : 5 instead of 1 : 25, forgetting that area scales with the square of the side ratio. Others mix up which region is the quadrilateral and which is the whole triangle. Clearly separating the full triangle area from the small similar triangle area and then subtracting is the key to obtaining the correct ratio. Final Answer: Therefore, the required ratio of the area of triangle ADE to the area of quadrilateral BDEC is 1:24.

Question 7

The height of an equilateral triangle is 18 cm. What is the area of the triangle (in square centimetres)?

Accepted Answer

Introduction / Context: This question asks you to find the area of an equilateral triangle when its height is given. Equilateral triangles have all sides equal and all angles equal to 60°, and there is a specific relationship between the side length and the height. By using this relationship and the standard area formula, we can compute the area from the given height. Given Data / Assumptions: The triangle is equilateral. The height h of the triangle is 18 cm. We must find the area in square centimetres. Concept / Approach: For an equilateral triangle of side length a, the height h is: h = (√3 / 2) * a From this, we can express a in terms of h: a = (2h) / √3 The area of an equilateral triangle is: Area = (√3 / 4) * a^2 By substituting the expression for a into the area formula, or by first finding a numerically and then using the area formula, we can find the required area. Step-by-Step Solution: Step 1: Use the height relation h = (√3 / 2) * a with h = 18. Step 2: Rearrange to find a: a = (2h) / √3 = (2 * 18) / √3 = 36 / √3. Step 3: Rationalise a if desired: a = (36 * √3) / 3 = 12√3. Step 4: Now apply the area formula Area = (√3 / 4) * a^2. Step 5: Compute a^2: (12√3)^2 = 144 * 3 = 432. Step 6: Substitute into the area formula: Area = (√3 / 4) * 432. Step 7: Compute 432 / 4 = 108, giving Area = 108√3 square centimetres. Verification / Alternative check: Alternatively, using base b and height h, area can be written as Area = (1/2) * b * h. Here, the base b is equal to the side length a = 12√3. Then: Area = (1/2) * 12√3 * 18 = 6√3 * 18 = 108√3 sq cm. This matches the area obtained from the equilateral triangle formula, so the result is confirmed. Why Other Options Are Wrong: 36√3 sq m and 96√3 sq m: These are in square metres instead of square centimetres and also have incorrect numerical values. 108 sq cm: This omits the √3 factor and therefore underestimates the area. 54√3 sq cm: This is exactly half of the correct area and may result from improperly using base or height. Common Pitfalls: Students often confuse side length and height in equilateral triangles, or they forget the exact relationships involving √3. Another common error is mixing units or miscomputing squares of expressions with √3. Carefully using the formulas h = (√3 / 2) * a and Area = (√3 / 4) * a^2, along with neat algebra, helps avoid these mistakes. Final Answer: Thus, the area of the equilateral triangle is 108√3 sq. cm.

Question 8

The sides of a triangle are in the ratio 1/2 : 1/3 : 1/4 and its perimeter is 104 cm. What is the length of the longest side (in cm)?

Accepted Answer

Introduction / Context: This question involves the concept of side ratios in a triangle and the perimeter. When the sides are given in a certain ratio, we can introduce a common multiplying factor to convert the ratio into actual side lengths. Using the perimeter, which is the sum of all sides, we can determine that factor and hence each side. This is a common style of question in aptitude tests that combines ratios and basic algebra. Given Data / Assumptions: The three sides of the triangle are in the ratio 1/2 : 1/3 : 1/4. The perimeter of the triangle is 104 cm. We must find the length of the longest side. Concept / Approach: If the side lengths are proportional to given ratios, we can write them as a common factor k multiplied by each ratio term. Here, the sides will be: side 1 = (1/2) * k side 2 = (1/3) * k side 3 = (1/4) * k The perimeter is the sum of all three sides, so we can set up an equation involving k and solve for it. Once k is found, we can compute each side and identify the longest one. Step-by-Step Solution: Step 1: Let the common factor be k. Step 2: Express the sides in terms of k: s1 = k/2, s2 = k/3, s3 = k/4. Step 3: Write the perimeter equation: k/2 + k/3 + k/4 = 104. Step 4: Find a common denominator, which is 12. Step 5: Convert each fraction: k/2 = 6k/12, k/3 = 4k/12, k/4 = 3k/12. Step 6: Add these: (6k + 4k + 3k) / 12 = 13k / 12. Step 7: Set 13k / 12 = 104. Step 8: Multiply both sides by 12: 13k = 104 * 12. Step 9: Compute 104 * 12 = 1248, so 13k = 1248. Step 10: Divide by 13: k = 1248 / 13 = 96. Step 11: Now compute the sides: s1 = 96 / 2 = 48, s2 = 96 / 3 = 32, s3 = 96 / 4 = 24. Step 12: The longest side is therefore 48 cm. Verification / Alternative check: Verify the perimeter: 48 + 32 + 24 = 80 + 24 = 104 cm, which matches the given perimeter. The ratio check also works: 48 : 32 : 24 simplifies by dividing each term by 96 to (1/2) : (1/3) : (1/4) in fractional form, matching the given ratio when written with a common denominator. This confirms that the values are correct and consistent. Why Other Options Are Wrong: 52, 32, 26, and 24: None of these match the largest side once the perimeter and side ratio are applied correctly. They come from incorrect arithmetic or misinterpretation of the ratio. Common Pitfalls: Common errors include adding the ratio terms incorrectly, forgetting to use a common denominator, or equating the sum directly to the perimeter without multiplying by a common factor. Students may also incorrectly assume the largest ratio term corresponds to the largest integer without computing the actual side lengths. Working with a clear algebraic representation and checking both the perimeter and the ratio avoids these mistakes. Final Answer: Hence, the length of the longest side of the triangle is 48 cm.

Question 9

The four walls and the ceiling of a rectangular room have to be painted. The room has length 25 m, breadth 12 m and height 10 m. Painter A can paint 200 m^2 in 5 days, while painter B can paint 250 m^2 in 2 days. If painters A and B work together an…

Accepted Answer

Introduction / Context: This question combines the concept of surface area of a room with the idea of combined rates of work. We first need to calculate the total area that must be painted, and then determine how long two painters working together will take to complete that area, based on their individual speeds. Such questions are common in quantitative aptitude tests where geometry and time-and-work concepts are blended. Given Data / Assumptions: Length of the room = 25 m. Breadth of the room = 12 m. Height of the room = 10 m. Four walls and the ceiling are to be painted, the floor is not painted. Painter A paints 200 m^2 in 5 days. Painter B paints 250 m^2 in 2 days. Both painters work together at constant rates until the entire job is finished. Concept / Approach: The solution has two main parts. First, we compute the total area to be painted, which is the sum of the areas of the four walls and the ceiling. For a rectangular room, the total area of four walls is 2 * height * (length + breadth), and the area of the ceiling is length * breadth. Second, we convert the work done by each painter into a daily rate of painting in square metres per day and then add these rates to get their combined rate. Finally, total time equals total area divided by the combined daily rate. Step-by-Step Solution: Area of four walls = 2 * height * (length + breadth)= 2 * 10 * (25 + 12)= 20 * 37 = 740 m^2.Area of ceiling = length * breadth = 25 * 12 = 300 m^2.Total area to be painted = 740 + 300 = 1040 m^2.Painter A paints 200 m^2 in 5 days, so A's rate = 200 / 5 = 40 m^2 per day.Painter B paints 250 m^2 in 2 days, so B's rate = 250 / 2 = 125 m^2 per day.Combined rate of A and B = 40 + 125 = 165 m^2 per day.Time taken together = total area / combined rate = 1040 / 165 days.Compute 1040 / 165 = 6 + 50/165 = 6 + 10/33.Therefore, time taken = 6 10/33 days. Verification / Alternative check: We can check the reasonableness of the answer by estimating. If the combined rate is 165 m^2 per day, in 6 days they would paint 6 * 165 = 990 m^2. The remaining area is 1040 - 990 = 50 m^2. At 165 m^2 per day, the time for the remaining 50 m^2 is 50 / 165 = 10/33 days. Adding 6 and 10/33 gives 6 10/33 days, which matches the exact calculation and confirms that the result is consistent. Why Other Options Are Wrong: 6 days: At 165 m^2 per day, 6 days cover only 990 m^2, which is less than 1040 m^2, so the painting would not be complete. 7 10/33 days: This is more than required; it would imply extra time after the work is already finished, which is not correct for the minimum completion time. 8 days: Clearly larger than 6 10/33 days, so this would overestimate the actual required time. 5 5/11 days: This is less than 6 days, and the total work finished in such a short time at 165 m^2 per day would be less than 1040 m^2, so it cannot be correct. Common Pitfalls: Candidates often forget to include the ceiling area and calculate only the wall area. Another common mistake is miscomputing the daily painting rates, especially for painter B, or adding times instead of rates. Mixing up perimeter formulas or misusing area of walls as 2 * (length + breadth) without multiplying by height is another typical error. Careful attention to the formula for four walls and the distinction between rate and time helps avoid these issues. Final Answer: The correct time required for painters A and B working together to paint the four walls and ceiling of the room is 6 10/33 days.

Question 10

The side of a square and the diameter of a circle are equal in length. If the side of the square is taken equal to the diameter of the circle and π is taken as 22/7, what is the ratio between the area of the square and the area of the circle?

Accepted Answer

Introduction / Context: This problem compares the areas of two basic plane figures, a square and a circle, when their key dimensions are linked. The key relationship is that the side of the square is equal to the diameter of the circle. Such questions check understanding of area formulas and the ability to manipulate algebraic expressions and ratios, especially when a specific value for π is provided for easy calculation. Given Data / Assumptions: Side of the square = s. Diameter of the circle = s (same as the square side). Radius of the circle = s / 2. Area of square = s^2. Area of circle = π * (radius)^2. Use π = 22/7 as instructed. Concept / Approach: The idea is to express the area of both figures in terms of the single variable s, since side of the square and diameter of the circle are equal. Once both areas are expressed in the same variable, we can form a ratio of square area to circle area and simplify that ratio. Because the variable s appears in both numerator and denominator, it cancels out, leaving a pure numerical ratio that can be simplified using the given value of π. Step-by-Step Solution: Let the side of the square be s.Then the diameter of the circle is also s, so the radius r = s / 2.Area of the square = s^2.Area of the circle = π * r^2 = π * (s / 2)^2 = π * s^2 / 4.Ratio of areas (square : circle) = s^2 : (π * s^2 / 4).Cancel s^2 from both sides to get 1 : (π / 4).This is equivalent to 4 : π.Using π = 22/7, the ratio becomes 4 : 22/7.Multiply both terms by 7 to clear the fraction: 4 * 7 : 22 = 28 : 22.Simplify 28 : 22 by dividing both by 2 to get 14 : 11. Verification / Alternative check: We can verify with a numerical example. Let s = 14 units. Then radius r = 7 units. Area of square = 14^2 = 196 square units. Area of circle = (22/7) * 7^2 = (22/7) * 49 = 22 * 7 = 154 square units. Ratio 196 : 154 simplifies on dividing by 14 to 14 : 11, which confirms the derived ratio is correct and consistent with the formula-based solution. Why Other Options Are Wrong: 28 : 11 is too large and results from not simplifying the intermediate ratio properly. 7 : 22 and 22 : 7 invert or distort the correct relationship between square and circle area. 11 : 14 is the reverse of the correct ratio and would correspond to area of circle : area of square, not the other way around. Therefore, only 14 : 11 matches the correct derivation. Common Pitfalls: One mistake is to forget that the diameter of the circle equals the side of the square, and incorrectly take the radius as s instead of s / 2. Another common error is to leave the ratio in terms of π without substituting π = 22/7 or to fail to simplify the fraction completely. Students may also accidentally calculate the ratio of circle area to square area instead of square to circle. Careful attention to the wording of the problem prevents such errors. Final Answer: The ratio of the area of the square to the area of the circle is 14 : 11.

Question 11

A piece of wire 132 cm long is bent successively to form three different closed plane figures: an equilateral triangle, a square, and a circle. In which of these three shapes will the enclosed area be the greatest when the entire wire is used each t…

Accepted Answer

Introduction / Context: This question examines the relationship between perimeter and area for different plane figures. When we use the same length of wire to form different shapes, each shape will have the same perimeter but a different enclosed area. The goal is to know which shape maximizes the area for a given fixed perimeter. This concept is very important in optimization problems in geometry and appears frequently in aptitude tests. Given Data / Assumptions: Length of the wire = 132 cm. The wire is bent into three shapes: an equilateral triangle, a square, and a circle. In each case, the entire 132 cm is used as the perimeter of that shape. We compare only the areas enclosed by these shapes, not their perimeters (which are the same). No material is lost or added when changing from one shape to another. Concept / Approach: A well-known geometric result states that among all closed plane figures with the same perimeter, the circle encloses the maximum possible area. Among polygons with a fixed perimeter, regular polygons (all sides equal and all angles equal) have larger areas than irregular ones, and as the number of sides increases, the area approaches that of the circle. Here, all three shapes are regular (equilateral triangle, square, circle), but the circle is known to give the largest area for the same perimeter, so we can answer the question using this theoretical fact without performing detailed numerical calculations. Step-by-Step Solution: Perimeter of each shape = 132 cm (by construction).For an equilateral triangle with perimeter 132 cm, each side is 132 / 3 = 44 cm.For a square with perimeter 132 cm, each side is 132 / 4 = 33 cm.For a circle with circumference 132 cm, radius r is found from 2 * π * r = 132.Although we could calculate the exact areas of each shape, we can rely on the theorem that for a given perimeter, the circle maximizes the enclosed area.Thus, the circle made from the wire will enclose more area than the equilateral triangle and the square formed from the same wire. Verification / Alternative check: If we wish, we can verify this numerically. For the square, area = side^2 = 33^2 = 1089 square centimetres. For the equilateral triangle with side 44 cm, area = (sqrt(3) / 4) * 44^2 ≈ 0.433 * 1936 ≈ 838.7 square centimetres, which is less than the area of the square. For the circle, using an approximate value of π, the radius r is about 132 / (2 * π) ≈ 21.0 cm. Then area of the circle is approximately π * r^2 ≈ 3.14 * 441 ≈ 1384 square centimetres, which is clearly greater than both 1089 and 838.7, confirming that the circle has the largest area. Why Other Options Are Wrong: The equilateral triangle and the square both enclose less area than the circle for the same perimeter, as shown by theory and numerical calculation. The option claiming equal areas in all shapes is incorrect because area depends on how the perimeter is distributed in the shape. The option stating that the result cannot be determined is also incorrect, since classical geometry directly answers this question and even numeric calculations confirm it. Common Pitfalls: Some learners mistakenly think that the shape with more sides always has more area, or that all shapes with the same perimeter have the same area. Others might incorrectly assume that a square and a circle give nearly the same area and might guess rather than reason. Remember that the circle is uniquely optimal for maximum area at a fixed perimeter, and this fact is widely used in design and optimization problems. Final Answer: The shape in which the wire encloses the greatest area is the circle.

Question 12

In an isosceles triangle, the two equal sides are each twice as long as the third side. An equilateral triangle is constructed so that its perimeter is the same as that of this isosceles triangle. What is the ratio of the area of the isosceles trian…

Accepted Answer

Introduction / Context: This problem connects properties of two special triangles: an isosceles triangle and an equilateral triangle. Both triangles share the same perimeter, but their side lengths and shapes differ. The aim is to compare their areas by forming a ratio. The question tests knowledge of triangle area formulas, handling of square roots, and careful algebraic manipulation while keeping track of the same perimeter condition. Given Data / Assumptions: In the isosceles triangle, the two equal sides have length 2x each. The third side (base) of the isosceles triangle has length x. Perimeter of the isosceles triangle = 2x + 2x + x = 5x. An equilateral triangle is constructed with the same perimeter, so its side length is 5x / 3. We compare area of isosceles triangle to area of equilateral triangle. We ignore any units, since they cancel in the ratio. Concept / Approach: First, we express the area of the isosceles triangle using base and height. The equal sides and base are known in terms of x, so the height can be found via the Pythagoras theorem. Second, we write the area of the equilateral triangle using the standard formula (sqrt(3) / 4) * side^2. Both areas will be written fully in terms of x, allowing us to take their ratio. The variable x will cancel out, leaving a numeric ratio involving square roots. We then simplify this ratio to match the given options. Step-by-Step Solution: Let the isosceles triangle have equal sides 2x and base x.Perimeter = 2x + 2x + x = 5x.For the isosceles triangle, drop a perpendicular from the vertex opposite the base to the base, splitting the base x into two segments of length x / 2 each.Height h of the isosceles triangle satisfies h^2 + (x / 2)^2 = (2x)^2.So h^2 = 4x^2 - x^2 / 4 = (16x^2 - x^2) / 4 = 15x^2 / 4.Thus h = (x / 2) * sqrt(15).Area of isosceles triangle = (1 / 2) * base * height = (1 / 2) * x * (x / 2 * sqrt(15)) = (x^2 * sqrt(15)) / 4.For the equilateral triangle with the same perimeter, each side a = 5x / 3.Area of equilateral triangle = (sqrt(3) / 4) * a^2 = (sqrt(3) / 4) * (25x^2 / 9) = (25x^2 * sqrt(3)) / 36.Ratio of areas (isosceles : equilateral) = (x^2 * sqrt(15) / 4) : (25x^2 * sqrt(3) / 36).Cancel x^2 and simplify: = (sqrt(15) / 4) * (36 / (25 * sqrt(3))) = (36 * sqrt(15)) / (100 * sqrt(3)).Now sqrt(15) / sqrt(3) = sqrt(5), so the ratio becomes (36 * sqrt(5)) / 100.Thus ratio of areas = 36√5 : 100. Verification / Alternative check: We can choose a simple value, such as x = 1, and compute numerical areas. For x = 1, the isosceles triangle has sides 2, 2, and 1. Using the formula derived, area is approximately (1 * 3.873) / 4 ≈ 0.968. The equilateral triangle has sides 5/3 and area approximately (1.732 / 4) * (25 / 9) ≈ 1.201. Then the ratio 0.968 : 1.201 ≈ 0.806. The numeric value of 36√5 / 100 ≈ 36 * 2.236 / 100 ≈ 80.5 / 100 ≈ 0.805, which is essentially the same, confirming our algebraic work. Why Other Options Are Wrong: Options 30√5 : 100, 32√5 : 100, and 42√5 : 100 represent different numeric values that do not match the ratio derived via clean algebra. The option 9√5 : 25 looks related to the simplified form but is not the same as 36√5 : 100 when converted to a common base. Only 36√5 : 100 exactly matches the simplified ratio obtained from the correct area formulas. Common Pitfalls: Errors often occur while computing the height of the isosceles triangle or when squaring and simplifying fractions. Another typical mistake is to forget that both triangles must have the same perimeter, leading to incorrect side lengths for the equilateral triangle. Some students also cancel x incorrectly or mishandle the square roots when simplifying the final ratio. Careful stepwise manipulation and checking each algebraic step prevents these problems. Final Answer: The ratio of the area of the isosceles triangle to that of the equilateral triangle with the same perimeter is 36√5 : 100.

Question 13

In an equilateral triangle ABC, the centroid is denoted by G. Each side of the triangle is 6 cm long. What is the length (in cm) of the segment AG drawn from vertex A to the centroid G?

Accepted Answer

Introduction / Context: This question involves a standard result from triangle geometry, specifically for an equilateral triangle. The centroid G of a triangle is the common intersection point of its medians, and in any triangle it divides each median in the ratio 2 : 1 from vertex to midpoint. For an equilateral triangle, due to its symmetry, the centroid, circumcentre, incentre, and orthocentre all coincide, and the medians have a simple relation to the side length. The problem asks for the distance from a vertex to the centroid in such a triangle. Given Data / Assumptions: Triangle ABC is equilateral. Each side of the triangle is 6 cm. G is the centroid of triangle ABC. The centroid divides each median in the ratio 2 : 1, measured from the vertex. We need the length of AG in centimetres. Concept / Approach: In any triangle, a median is a line segment joining a vertex to the midpoint of the opposite side. The centroid is the intersection point of the three medians, and it divides each median in the ratio 2 : 1. For an equilateral triangle of side a, the length of any median is equal to the altitude, which can be found using the Pythagoras theorem. Once we know the median length, we use the centroid property to find the distance from the vertex to the centroid, which is two-thirds of the median. Step-by-Step Solution: Let the side length a = 6 cm.Consider triangle ABC with median from vertex A to midpoint of BC.The midpoint of BC divides BC into two segments of length 3 cm each.The median from A is also the altitude in an equilateral triangle, so we can use the Pythagoras theorem.Let the median (and altitude) have length m. Then m^2 + 3^2 = 6^2.So m^2 = 36 - 9 = 27.Thus m = √27 = 3√3 cm.The centroid G divides the median in the ratio 2 : 1 from the vertex, so AG = (2 / 3) * m.Therefore, AG = (2 / 3) * 3√3 = 2√3 cm. Verification / Alternative check: We can also recall the direct formula for the distance from a vertex to the centroid in any equilateral triangle: AG = (2 / 3) * height, and the height of an equilateral triangle with side a is (√3 / 2) * a. For a = 6, height = (√3 / 2) * 6 = 3√3. Then AG = (2 / 3) * 3√3 = 2√3 cm, which perfectly matches the computed value. This alternative approach confirms the result without re-deriving the median with coordinates. Why Other Options Are Wrong: Values such as 2√2 cm, 3√2 cm, and 3√3 cm correspond to incorrect uses of geometric formulas or misapplication of the Pythagoras theorem. For example, 3√3 cm is the full median length, not the distance from the vertex to the centroid. The option 4 cm is a simple but incorrect guess that does not correspond to any standard ratio in an equilateral triangle. Only 2√3 cm correctly represents the length from vertex A to the centroid G. Common Pitfalls: Students often confuse the full altitude or median with the vertex-to-centroid distance and choose 3√3 cm instead of 2√3 cm. Another common error is using half the side incorrectly in the Pythagoras calculation. A few learners may also forget that the centroid divides the median in the ratio 2 : 1 and assume that AG is half the median. Remembering the standard ratios and understanding the geometry of special triangles is essential to avoid these mistakes. Final Answer: The length of AG in the equilateral triangle of side 6 cm is 2√3 cm.

Question 14

In a circle with radius 5 cm, a diameter AB and a chord PQ (which is not a diameter) intersect at point X such that PQ is perpendicular to AB at X. If AX : XB = 3 : 2 along the diameter AB, what is the length (in cm) of the chord PQ?

Accepted Answer

Introduction / Context: This problem deals with the geometry of a circle involving a diameter, a chord perpendicular to that diameter, and a given ratio on the diameter. It tests understanding of how to use basic coordinate geometry or Pythagoras theorem inside a circle. By using the given ratio on the diameter and the fact that the chord is perpendicular to the diameter, we can find the exact position of the intersection point and then compute the chord length using the circle equation or distance formula. Given Data / Assumptions: Circle radius r = 5 cm. AB is a diameter of the circle. PQ is a chord that is not a diameter. PQ is perpendicular to AB at their intersection point X. AX : XB = 3 : 2 along the diameter AB. We need the length of chord PQ. Concept / Approach: A convenient way to solve this is to use a coordinate system. We can place the centre of the circle at the origin and take the diameter AB along the horizontal axis. Once we know the coordinates of point X on AB using the given ratio, we exploit the fact that PQ is perpendicular to AB, so it must be a vertical line through X. The endpoints of the chord PQ are then the intersection of this vertical line with the circle. Using the circle equation x^2 + y^2 = r^2, we can find the y-coordinates of the chord endpoints and thus the length of PQ. Step-by-Step Solution: Let the centre of the circle O be at (0, 0) with radius 5 cm.Take diameter AB along the x-axis, so A = (−5, 0) and B = (5, 0).Given AX : XB = 3 : 2, and AB = 10 cm, so AX = 6 cm and XB = 4 cm.Starting from A at x = −5, point X is 6 cm to the right, so X = (−5 + 6, 0) = (1, 0).Chord PQ is perpendicular to AB at X, so PQ lies on the vertical line x = 1.Points P and Q are where this line meets the circle x^2 + y^2 = 25.Substitute x = 1: 1^2 + y^2 = 25 so y^2 = 24.Thus y = ±√24 = ±2√6.So P = (1, 2√6) and Q = (1, −2√6).Length of chord PQ = difference in y-coordinates = 2√6 − (−2√6) = 4√6 cm. Verification / Alternative check: We can verify using the chord length formula. The distance from the centre to the chord is the x-coordinate of X, which is 1 cm. For a circle of radius r, a chord at distance d from the centre has length L = 2 * √(r^2 − d^2). Here, r = 5 and d = 1, so L = 2 * √(25 − 1) = 2 * √24 = 4√6 cm. This matches our coordinate geometry result and confirms the correctness of the calculation. Why Other Options Are Wrong: The options 2√13 cm, 5√3 cm, and 6√5 cm all correspond to chord lengths that would require different distances from the centre or different radii. None satisfy the exact relation with a radius of 5 cm and a chord at distance 1 cm from the centre. The option 8 cm is a simple rational guess and does not match the exact geometric requirements. Only 4√6 cm is consistent with the circle equation and perpendicular chord construction. Common Pitfalls: Some learners may misplace the centre or choose the wrong orientation for the diameter, leading to incorrect coordinates for X. Another common error is not using the ratio AX : XB correctly along the full length of 10 cm. Mistakes in solving 1^2 + y^2 = 25 or incorrectly calculating the chord length from the two y-values also occur. Being systematic about coordinate placement and carefully applying the chord length formula helps avoid these issues. Final Answer: The length of the chord PQ is 4√6 cm.

Question 15

Let x and y be two acute angles such that x + y < 90° and sin(2x − 20°) = cos(2y + 20°). Under these conditions, what is the value of tan(x + y)?

Accepted Answer

Introduction / Context: This problem is a trigonometric identity question involving acute angles and relationships between sine and cosine functions. It combines transformations of angles, the co-function identity between sine and cosine, and the use of angle sum restrictions to find the value of tan(x + y). Such questions are standard in trigonometry sections of aptitude tests and require both algebraic manipulation and a good grasp of basic trigonometric identities. Given Data / Assumptions: x and y are acute angles (0° < x < 90°, 0° < y < 90°). x + y < 90°. sin(2x − 20°) = cos(2y + 20°). We need to determine tan(x + y). We assume the principal acute angle relationships, consistent with typical aptitude-style questions. Concept / Approach: The key idea is to use the co-function identity: cos θ = sin(90° − θ). The equation sin(2x − 20°) = cos(2y + 20°) can be rewritten entirely in terms of sine. Once we express cos(2y + 20°) as sin(90° − (2y + 20°)), we can equate the arguments of the sine functions. Under the standard principal-angle assumption for acute angles in such problems, we set the angles equal, solve for a relation between x and y, and then find the value of x + y and its tangent. Step-by-Step Solution: Start with sin(2x − 20°) = cos(2y + 20°).Use the co-function identity cos θ = sin(90° − θ).So cos(2y + 20°) = sin(90° − (2y + 20°)) = sin(70° − 2y).Thus sin(2x − 20°) = sin(70° − 2y).For acute-angle based aptitude questions, we typically take the principal solution, equating the arguments:2x − 20° = 70° − 2y.Rearrange: 2x + 2y = 90°.This gives x + y = 45°.Now tan(x + y) = tan(45°) = 1. Verification / Alternative check: We can choose a concrete pair of angles satisfying x + y = 45° and x, y acute, for example x = 25° and y = 20°. Then 2x − 20° = 2 * 25° − 20° = 30°, and 2y + 20° = 2 * 20° + 20° = 60°. We have sin(30°) = 1/2, cos(60°) = 1/2, so the given equation sin(2x − 20°) = cos(2y + 20°) holds. For this pair, x + y = 45°, and tan(45°) = 1, which is consistent with the derived answer. This kind of numerical test supports the algebraic reasoning. Why Other Options Are Wrong: √3 and 1/√3 correspond to tan(60°) and tan(30°) respectively, which would require x + y to be 60° or 30°, contradicting the equation obtained from the given relation. The option 2 + √2 does not correspond to a standard tangent value for simple angles and is not supported by any identity here. The option 0 would imply x + y = 0° or 180°, which is impossible for acute x and y with x + y < 90°. Only the value 1 matches tan(45°), which arises naturally from the relationship between x and y. Common Pitfalls: One frequent error is to forget to convert cos to sin using the co-function identity and instead attempt more complicated transformations. Another is to mishandle the equation sin A = sin B and ignore the principal acute-angle behaviour, leading to extraneous angle relations. Students may also incorrectly assume that x + y = 90°, which directly contradicts the given condition x + y < 90°. Careful attention to the identities and the conditions on the angles avoids these mistakes. Final Answer: The value of tan(x + y) under the given conditions is 1.

Question 16

Two metal pipes have lengths 1.5 m and 1.2 m respectively. Both pipes are to be cut into smaller pieces of equal length, with no leftover material from either pipe. What is the greatest possible length (in metres) of each piece so that all the piece…

Accepted Answer

Introduction / Context: This is a classic application of the highest common factor (HCF) or greatest common divisor (GCD) concept in a geometric setting. The question asks for the greatest possible length of equal pieces into which two given pipe lengths can be cut without leaving any remainder. This is equivalent to finding the greatest length that divides both lengths exactly, which is precisely the idea of finding the HCF of the two measurements when expressed in a common unit. Given Data / Assumptions: Length of first pipe = 1.5 m. Length of second pipe = 1.2 m. Both pipes are to be cut into equal pieces. No leftover length (no wastage) is allowed from either pipe. All pieces must have the same length. Concept / Approach: To find the greatest possible common piece length, we need a length that divides both 1.5 m and 1.2 m exactly. This is a greatest common divisor in the context of measurements. It is easier to work in centimetres to avoid decimals. We convert both lengths to centimetres, find their HCF, and then convert the result back to metres. The HCF ensures that the pieces are as long as possible while still dividing both original lengths exactly. Step-by-Step Solution: Convert lengths to centimetres: 1.5 m = 150 cm, 1.2 m = 120 cm.Now we need the HCF of 150 and 120.List prime factors: 150 = 2 * 3 * 5 * 5, and 120 = 2 * 2 * 2 * 3 * 5.Common prime factors are 2, 3, and 5.Their product is 2 * 3 * 5 = 30.So the HCF of 150 and 120 is 30.This means the greatest common length in centimetres is 30 cm.Convert back to metres: 30 cm = 30 / 100 m = 0.3 m.Therefore, each piece can be at most 0.3 m long. Verification / Alternative check: We can verify by dividing each original length by 0.3 m directly. For the first pipe: 1.5 / 0.3 = 5 pieces, which is an integer. For the second pipe: 1.2 / 0.3 = 4 pieces, also an integer. No leftover length remains in either case. If we tried a larger value, such as 0.4 m, then 1.2 / 0.4 = 3 pieces works, but 1.5 / 0.4 is not an integer, so 0.4 m is too large and not acceptable. This confirms that 0.3 m is indeed the greatest possible common length. Why Other Options Are Wrong: 0.13 m and 0.2 m both divide the lengths, but they are smaller than 0.3 m and hence do not represent the greatest possible common length. The values 0.4 m and 0.41 m fail to divide one or both of the original lengths exactly, leaving remainders. Therefore, these cannot be used to cut the pipes into equal pieces without wastage. Only 0.3 m satisfies both conditions: exact division and maximal possible length. Common Pitfalls: Some candidates try to work directly with decimals and make rounding errors. Others may find a common divisor but forget to check whether it is the greatest. Mistakes can also occur when converting between metres and centimetres. A systematic approach using integer measurements and prime factorization or the Euclidean algorithm helps avoid these issues and quickly leads to the correct HCF. Final Answer: The greatest possible length of each piece is 0.3 m.

Question 17

Two concentric circles (circles with the same centre) have radii 68 cm and 22 cm respectively. What is the area, in terms of π, of the closed region (ring-shaped area) bounded between the two circular boundaries?

Accepted Answer

Introduction / Context: This question focuses on the area of a ring-shaped region (annulus) formed by two concentric circles. The region of interest is the area between the outer circle and the inner circle. To find this, we subtract the area of the smaller inner circle from the area of the larger outer circle. This is a very standard geometry problem that reinforces understanding of the circle area formula and basic algebraic subtraction. Given Data / Assumptions: Radius of the outer circle, R = 68 cm. Radius of the inner circle, r = 22 cm. Both circles share the same centre (they are concentric). We need the area of the region between the two circles in square centimetres, in terms of π. Concept / Approach: The area of a circle with radius r is π * r^2. For an annulus formed by two concentric circles, the area of the ring is equal to the area of the larger circle minus the area of the smaller circle. So we use the formula: area of annulus = π * R^2 − π * r^2 = π * (R^2 − r^2). The problem is then reduced to computing R^2 and r^2, subtracting them, and multiplying the result by π. Step-by-Step Solution: Outer radius R = 68 cm, inner radius r = 22 cm.Area of outer circle = π * R^2 = π * 68^2.Area of inner circle = π * r^2 = π * 22^2.Compute 68^2: 68 * 68 = 4624.Compute 22^2: 22 * 22 = 484.Difference R^2 − r^2 = 4624 − 484 = 4140.Area of annulus = π * (R^2 − r^2) = π * 4140.Thus the required area = 4140π square centimetres. Verification / Alternative check: We can quickly verify the arithmetic. The inner area is much smaller because r is significantly less than R. The outer circle area is 4624π and the inner circle area is 484π, so the difference is (4624 − 484)π = 4140π, confirming the subtraction is correct. The difference 4624 − 4000 = 624 and 624 − 484 = 140, so the calculation step by step is consistent. This check reassures us that the numeric difference is accurate. Why Other Options Are Wrong: Options like 4110π, 4080π, 4050π, and 4000π all result from incorrect subtraction or miscalculated squares of 68 or 22. For example, if someone mistakenly used 67 or 21 instead, or miscalculated 68^2, they might arrive at one of those wrong values. Only 4140π sq.cm. corresponds to the correct computation of π * (68^2 − 22^2). Common Pitfalls: A common mistake is to add the areas of the two circles instead of subtracting, forgetting that the inner circle region is removed from the ring. Another frequent error is miscalculating the squares of 68 or 22. Sometimes students forget to keep π as a common factor and instead approximate it early, which can lead to rounding errors. Keeping π symbolically and focusing on accurate integer arithmetic avoids these problems. Final Answer: The area of the ring-shaped region between the two concentric circles is 4140π sq.cm.

Question 18

A solid sphere of radius 6 cm is completely melted and recast into a long, thin wire in the shape of a right circular cylinder with radius 0.2 cm. Assuming no loss of material during melting and recasting, what is the length of the wire (in metres)?

Accepted Answer

Introduction / Context: This problem is about conservation of volume when a solid is melted and recast into another shape. A solid sphere is melted to form a cylindrical wire. Since there is no loss of material, the volume of the sphere must be equal to the volume of the cylinder. By using the standard formulas for the volumes of a sphere and a cylinder, we can equate these volumes and solve for the unknown length of the wire. This is a common type of question in geometry and mensuration sections of aptitude exams. Given Data / Assumptions: Radius of the sphere, R = 6 cm. The sphere is melted and recast into a cylindrical wire. Radius of the cylindrical wire, r = 0.2 cm. Length of the wire = L cm (to be found). No loss of material during melting and recasting, so volumes are equal. Concept / Approach: The volume of a sphere is given by (4 / 3) * π * R^3. The volume of a right circular cylinder is π * r^2 * L. Because the material is conserved, the volume of the original sphere must equal the volume of the resulting cylinder. We set (4 / 3) * π * R^3 = π * r^2 * L, cancel π, and solve for L. After finding L in centimetres, we convert the result to metres as required by the question. Step-by-Step Solution: Volume of the sphere = (4 / 3) * π * R^3.Here R = 6 cm, so R^3 = 6^3 = 216.Thus volume of sphere = (4 / 3) * π * 216 = 4 * 72 * π = 288π cubic centimetres.Volume of the cylindrical wire = π * r^2 * L.Radius of cylinder r = 0.2 cm, so r^2 = (0.2)^2 = 0.04.Thus volume of cylinder = π * 0.04 * L.Equate volumes: 288π = π * 0.04 * L.Cancel π from both sides: 288 = 0.04 * L.Solve for L: L = 288 / 0.04.Compute 288 / 0.04 = 288 * (1 / 0.04) = 288 * 25 = 7200 cm.Convert to metres: 7200 cm = 7200 / 100 m = 72 m. Verification / Alternative check: We can quickly check if the numbers are reasonable. The radius of the wire is very small compared to the radius of the sphere, so the wire must be quite long. A length of 72 m (7200 cm) is much larger than the original diameter of the sphere (12 cm), which is intuitive. If we incorrectly used r = 2 cm instead of 0.2 cm, we would get a much smaller length, which would not match the thin-wire description. Our calculations with r = 0.2 cm and 288 / 0.04 = 7200 are consistent and error free. Why Other Options Are Wrong: The options 81 m, 80 m, 75 m, and 90 m do not satisfy the exact volume equality condition when used in the formula π * 0.04 * L. Substituting any of these values for L into the cylinder volume formula gives volumes that are not equal to 288π cubic centimetres. Only L = 72 m gives the correct cylindrical volume of 288π cubic centimetres, matching the sphere's volume. Common Pitfalls: Students often forget to convert the final answer from centimetres to metres or mistakenly use the diameter instead of the radius in the cylinder volume formula. Another common error is squaring 0.2 incorrectly or mishandling the division by 0.04. Some may also forget to cancel π and carry it through unnecessarily, complicating the arithmetic. Careful unit handling and correct use of the volume formulas ensure a straightforward solution. Final Answer: The length of the wire formed by melting the sphere is 72 m.

Question 19

In trapezium ABCD, sides AB and CD are parallel, and ∠ADC = 90°. The lengths of the parallel sides are AB = 15 cm and CD = 40 cm, and the diagonal AC has length 41 cm. What is the area (in cm²) of trapezium ABCD?

Accepted Answer

Introduction / Context: This problem deals with a trapezium (trapezoid) in which one non-parallel side and one parallel side meet at a right angle. We are given the lengths of the two parallel sides and one diagonal. The goal is to find the area of the trapezium. To solve this, we interpret the trapezium as a figure that contains right triangles and a rectangle-like structure, and we use the Pythagoras theorem to find the height, which is the perpendicular distance between the parallel sides. Once the height is known, we can use the standard area formula for a trapezium. Given Data / Assumptions: AB and CD are parallel sides of trapezium ABCD. ∠ADC = 90°, so AD is perpendicular to CD. AB = 15 cm. CD = 40 cm. Diagonal AC = 41 cm. We must find the area of ABCD in square centimetres. Concept / Approach: Place the trapezium on a coordinate plane for clarity. Because ∠ADC = 90°, we can treat AD as a vertical segment (height) and CD as horizontal. Then AB will be parallel to CD at the top. By using the coordinates of A, D, and C, and the length of diagonal AC, we apply the Pythagoras theorem to find the height AD. Once the height is known, we use the area formula for a trapezium: Area = (1 / 2) * (sum of parallel sides) * height. Step-by-Step Solution: Let D be at (0, 0) and place CD along the x-axis, so C = (40, 0).Since ∠ADC = 90°, take AD along the y-axis. Let A = (0, h), where h is the height.AB is parallel to CD and has length 15 cm, so B has coordinates (15, h).We know the diagonal AC has length 41 cm.Coordinates: A = (0, h) and C = (40, 0).Use the distance formula: AC^2 = (40 − 0)^2 + (0 − h)^2 = 40^2 + h^2.Given AC = 41 cm, so 41^2 = 1600 + h^2.Compute 41^2 = 1681, so 1681 = 1600 + h^2.Thus h^2 = 1681 − 1600 = 81.So h = 9 cm (height of the trapezium).Area of trapezium = (1 / 2) * (AB + CD) * height = (1 / 2) * (15 + 40) * 9.Compute: (15 + 40) = 55, so area = (1 / 2) * 55 * 9 = 27.5 * 9 = 247.5 cm². Verification / Alternative check: It is worth checking the internal right triangle dimensions. With height h = 9 cm, AD = 9 and CD = 40, triangle ADC is a right triangle with legs 9 and 40, giving AC as √(9^2 + 40^2) = √(81 + 1600) = √1681 = 41, which matches the given diagonal. This confirms that the coordinate placement and height calculation are consistent with the problem statement and supports the correctness of the final area value. Why Other Options Are Wrong: An area of 245 cm² or 240 cm² would result from using an incorrect height approximation or miscomputing the Pythagoras step. An area of 250 cm² would come from slightly miscalculating (55 * 9) or rounding. The option 225 cm² is too low and does not match the correct height. Only 247.5 cm² exactly matches the precise area obtained from the correct height and trapezium area formula. Common Pitfalls: One frequent error is to forget that the diagonal AC connects points (0, h) and (40, 0) and so misapply the distance formula. Another mistake is to assume AD equals 41 or to treat the trapezium as a rectangle. Some candidates also forget to take the half factor in the area formula for a trapezium or incorrectly add the side lengths. Carefully setting up the coordinate geometry and applying Pythagoras leads to a reliable solution. Final Answer: The area of trapezium ABCD is 247.5 cm².

Question 20

A rhombus has all its sides equal to 10 cm, and one of its diagonals is 12 cm long. Using these measurements, what is the area (in cm²) of the rhombus?

Accepted Answer

Introduction / Context: This problem concerns a rhombus, a special type of quadrilateral in which all four sides are equal and the diagonals intersect at right angles and bisect each other. We are given the length of each side and one diagonal and must find the area of the rhombus. The key relationships here are between the side length, the diagonals, and the right triangles formed when the diagonals intersect. Using these relationships, we can find the second diagonal and then apply the standard area formula for a rhombus. Given Data / Assumptions: Each side of the rhombus = 10 cm. One diagonal (say d1) = 12 cm. The diagonals of a rhombus are perpendicular and bisect each other. We must find the area of the rhombus in square centimetres. Concept / Approach: In a rhombus, the diagonals intersect at right angles and bisect each other. So each half of a diagonal forms a right triangle with half of the other diagonal and the side of the rhombus. If d1 and d2 are the diagonals, then (d1 / 2)^2 + (d2 / 2)^2 = side^2. We can use this relation to find the unknown diagonal d2. Once both diagonals are known, the area of the rhombus is given by (1 / 2) * d1 * d2. Step-by-Step Solution: Let the side of the rhombus be a = 10 cm.Let the diagonals be d1 and d2, with d1 = 12 cm and d2 unknown.Half of d1 is d1 / 2 = 12 / 2 = 6 cm.Half of d2 is d2 / 2.In one of the right triangles formed by the diagonals, the hypotenuse is the side a and the legs are d1 / 2 and d2 / 2.So (d1 / 2)^2 + (d2 / 2)^2 = a^2.Substitute values: 6^2 + (d2 / 2)^2 = 10^2.That is 36 + (d2^2 / 4) = 100.Rearrange: d2^2 / 4 = 100 − 36 = 64.So d2^2 = 64 * 4 = 256, giving d2 = 16 cm.Area of the rhombus = (1 / 2) * d1 * d2.So area = (1 / 2) * 12 * 16 = 6 * 16 = 96 cm². Verification / Alternative check: We can verify by checking the right triangle formed by half diagonals. With d1 = 12 and d2 = 16, half lengths are 6 and 8. Then the side of the rhombus should be √(6^2 + 8^2) = √(36 + 64) = √100 = 10 cm, which matches the given side. This confirms that d2 = 16 cm is correct and ensures that the area calculation based on these diagonals is accurate. Why Other Options Are Wrong: The value 48 cm² would result from mistakenly multiplying 12 and 8 without using the half factor correctly. The values 144 cm² and 192 cm² are too large and typically arise from misapplying formulas or ignoring the halving when computing diagonals. The value 120 cm² does not follow from any consistent application of the rhombus area formulas. Only 96 cm² is consistent with both the diagonal relationship and the correct area formula (1 / 2) * d1 * d2. Common Pitfalls: Many students forget that the diagonals are bisected, and therefore use full diagonal lengths in the Pythagoras theorem rather than half lengths, leading to incorrect values. Another mistake is to mix up the formulas for the area of a rhombus and a general quadrilateral. Some also bypass the diagonal relation and attempt to use base-height formulas without a clear known height, which complicates the problem unnecessarily. Sticking to the diagonal properties makes this problem straightforward. Final Answer: The area of the rhombus is 96 cm².

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