Question 1

A solid right circular cylinder has radius 4 centimetres and height 8 centimetres. It is melted and recast into a right circular cone of the same radius. What will be the height of the cone, in centimetres, assuming no loss of material?

Accepted Answer

Introduction / Context: This question involves solid geometry and volume conservation. A cylinder is melted and recast into a cone with the same radius. Such problems appear frequently in aptitude exams to test whether you can apply volume formulas and the idea that volume is conserved when a solid is melted and reshaped without loss of material. Given Data / Assumptions: Radius of the original cylinder r = 4 centimetres. Height of the cylinder h1 = 8 centimetres. The cylinder is melted and recast into a cone of radius r = 4 centimetres. Height of the cone h2 is unknown and must be found. No material is lost during melting and recasting, so volumes are equal. pi cancels out in the calculations, so we do not need a numeric value. Concept / Approach: The volume of a right circular cylinder is Vcyl = pi * r^2 * h1. The volume of a right circular cone is Vcone = (1/3) * pi * r^2 * h2. Because the cylinder is melted and recast into the cone without loss of material, we set these volumes equal: pi * r^2 * h1 = (1/3) * pi * r^2 * h2. The common factors pi and r^2 cancel out, leaving a simple relationship between the heights. We then solve for h2. Step-by-Step Solution: Step 1: Write volume of cylinder: Vcyl = pi * r^2 * h1. With r = 4 and h1 = 8, Vcyl = pi * 4^2 * 8 = pi * 16 * 8 = 128 * pi. Step 2: Write volume of cone: Vcone = (1/3) * pi * r^2 * h2. With r = 4, Vcone = (1/3) * pi * 16 * h2 = (16/3) * pi * h2. Step 3: Set volumes equal: 128 * pi = (16/3) * pi * h2. Cancel pi from both sides to get 128 = (16/3) * h2. Step 4: Multiply both sides by 3: 128 * 3 = 16 * h2. So 384 = 16 * h2. Step 5: Divide both sides by 16: h2 = 384 / 16 = 24 centimetres. Verification / Alternative check: We can verify by plugging h2 = 24 back into the cone volume formula. Vcone = (1/3) * pi * 4^2 * 24 = (1/3) * pi * 16 * 24 = (1/3) * pi * 384 = 128 * pi. This matches the original cylinder volume Vcyl = 128 * pi. Since the volumes are equal, our computed height of 24 centimetres is correct. Why Other Options Are Wrong: Option A (48 cm): This would double the correct height and give cone volume (1/3) * pi * 16 * 48 = 256 * pi, which is too large. Option C (36 cm): This leads to a cone volume larger than 128 * pi and violates volume conservation. Option D (12 cm): This gives cone volume (1/3) * pi * 16 * 12 = 64 * pi, which is only half of the required volume. Option E (18 cm): This also produces a volume different from 128 * pi, so it cannot be correct. Common Pitfalls: A common mistake is forgetting to use the factor 1/3 in the cone volume formula or mixing up cylinder and cone formulas. Another issue is trying to plug in pi numerically when it is not necessary, which can introduce rounding errors. Some students also try to conserve surface area instead of volume, which is incorrect for melting and recasting problems. Always remember that when a solid is melted and reshaped, it is the volume that remains constant. Final Answer: The height of the cone is 24 cm (24 centimetres).

Question 2

In a circle of radius 21 centimetres, an arc subtends a central angle of 30 degrees at the centre. Using the arc length formula, find the length of the arc in centimetres, taking pi = 22/7.

Accepted Answer

Introduction / Context: This problem checks your understanding of the relationship between the central angle of a circle and the length of the corresponding arc. Arc length questions are standard in geometry and aptitude tests. The idea is to use the proportion of the angle to the full circle to find the fraction of the circumference that the arc represents. Given Data / Assumptions: Radius of the circle r = 21 centimetres. Central angle theta = 30 degrees. We must find the length of the arc corresponding to this angle. Take pi = 22/7. The full circle has 360 degrees. Concept / Approach: The length of an arc is proportional to the angle it subtends at the centre. For a full circle, circumference = 2 * pi * r and angle = 360 degrees. For an arc with central angle theta, the arc length L is given by: L = (theta / 360) * 2 * pi * r. We substitute theta = 30 degrees, r = 21 centimetres, and pi = 22/7 into this formula and simplify step by step to obtain the arc length in centimetres. Step-by-Step Solution: Step 1: Use formula L = (theta / 360) * 2 * pi * r. With theta = 30 degrees, r = 21 cm, and pi = 22/7, we have L = (30 / 360) * 2 * (22/7) * 21. Step 2: Simplify the fraction 30 / 360 to 1 / 12. So L = (1 / 12) * 2 * (22/7) * 21. Step 3: Compute the full circumference term: 2 * (22/7) * 21. First, 21 divided by 7 is 3, so 2 * 22 * 3 = 132. Thus 2 * pi * r = 132 centimetres. Step 4: Now compute the fraction of the circumference: L = (1 / 12) * 132. 132 divided by 12 is 11. Therefore, the length of the arc is 11 centimetres. Verification / Alternative check: We can cross check by noting that 30 degrees is one twelfth of 360 degrees. So the arc length must be one twelfth of the full circumference. We already found the full circumference as 132 centimetres, so one twelfth of 132 is 11. The answer 11 cm is therefore perfectly consistent with both the formula and the proportional reasoning approach. Why Other Options Are Wrong: Option A (22 cm): This is exactly one sixth of the circumference and would correspond to a 60 degree central angle, not 30 degrees. Option B (16.5 cm): This value corresponds to an angle larger than 30 degrees and does not match the fraction 30/360. Option D (28 cm): This is too large and has no direct proportional link with the given angle. Option E (14 cm): This might tempt students who roughly guess, but it does not match the exact calculation. Common Pitfalls: Students sometimes forget to convert the angle to a fraction of 360 degrees and instead use 30 directly in the circumference formula. Another common mistake is to use pi * r instead of 2 * pi * r for the full circumference, or to mis-handle the fraction 30/360. Taking time to write the formula and simplify the angle ratio first makes the arithmetic straightforward and reduces errors. Final Answer: The length of the arc is 11 cm (11 centimetres).

Question 3

The length and breadth of a rectangle are increased by 40% and 70% respectively. By what percentage does the area of the rectangle increase as a result of these simultaneous changes?

Accepted Answer

Introduction / Context: This problem is a classic example of percentage increase in area when the dimensions of a rectangle change. It highlights that percentage changes in length and breadth combine multiplicatively, not additively. Such questions are frequently used to test understanding of compound percentage ideas in quantitative aptitude exams. Given Data / Assumptions: Original length of the rectangle = L. Original breadth of the rectangle = B. Length is increased by 40%, so new length = 1.4 * L. Breadth is increased by 70%, so new breadth = 1.7 * B. We must find the percentage increase in the area. Concept / Approach: The original area A1 of the rectangle is L * B. After the increases, the new area A2 is (1.4 * L) * (1.7 * B) = 1.4 * 1.7 * L * B. So A2 = 2.38 * A1. The factor 2.38 means the new area is 2.38 times the old area. The percentage increase is computed as ((A2 - A1) / A1) * 100%. That simplifies to (2.38 - 1) * 100% = 1.38 * 100% = 138%. Step-by-Step Solution: Step 1: Let original area A1 = L * B. Step 2: New length = L increased by 40%, so new length = L * (1 + 40/100) = 1.4 * L. Step 3: New breadth = B increased by 70%, so new breadth = B * (1 + 70/100) = 1.7 * B. Step 4: New area A2 = (1.4 * L) * (1.7 * B) = 1.4 * 1.7 * L * B. Step 5: Multiply 1.4 by 1.7: 1.4 * 1.7 = 2.38. So A2 = 2.38 * A1. Step 6: Percentage increase = ((A2 - A1) / A1) * 100%. Substitute A2 = 2.38 * A1: ((2.38 * A1 - A1) / A1) * 100% = (1.38 * A1 / A1) * 100%. This simplifies to 1.38 * 100% = 138%. Verification / Alternative check: We can test with simple numbers. Assume L = 10 and B = 10, so original area A1 = 100 square units. New length = 1.4 * 10 = 14, new breadth = 1.7 * 10 = 17. New area A2 = 14 * 17 = 238 square units. The increase in area is 238 - 100 = 138 square units, which relative to the original area 100 is a 138% increase. This numerical check confirms the algebraic result. Why Other Options Are Wrong: Option A (118%): This might come from partly correct but incomplete multiplication of factors or from adding incorrect adjustment. Option B (110%): This is much smaller and does not align with the product 1.4 * 1.7. Option D (128%): This may result from adding 40% and 70% and then subtracting something, but it is not accurate. Option E (90%): This is clearly too low relative to the large increases in both dimensions. Common Pitfalls: A common mistake is to simply add the percentage changes (40% + 70% = 110%) and assume that is the percentage increase in area. This ignores the multiplicative effect of changing both dimensions. Remember that area scales with the product of the scale factors, not their sum. Another pitfall is arithmetic errors when multiplying decimals like 1.4 and 1.7. Writing them as fractions, such as 14/10 and 17/10, can sometimes help avoid mistakes. Final Answer: The area of the rectangle increases by 138%.

Question 4

Find the slope of a straight line that is parallel to the line passing through the points (4, -2) and (-3, 5).

Accepted Answer

Introduction / Context: This question checks your understanding of the concept of slope in coordinate geometry and the fact that parallel lines have equal slopes. Given two points, you must compute the slope of the line joining them, and then use that slope for any line parallel to it. This is a very common pattern in algebra and analytic geometry questions. Given Data / Assumptions: Two points on the given line: (4, -2) and (-3, 5). We must find the slope of a line parallel to this given line. In the coordinate plane, slope m = (change in y) / (change in x). Concept / Approach: The slope m of a line passing through points (x1, y1) and (x2, y2) is given by: m = (y2 - y1) / (x2 - x1). Parallel lines have equal slopes. So once we compute the slope of the line through the given points, that same value is the slope of any line parallel to it. The rest is straightforward substitution and simplification. Step-by-Step Solution: Step 1: Label the points as (x1, y1) = (4, -2) and (x2, y2) = (-3, 5). Step 2: Use the slope formula m = (y2 - y1) / (x2 - x1). Compute y2 - y1 = 5 - (-2) = 5 + 2 = 7. Compute x2 - x1 = -3 - 4 = -7. Step 3: Therefore m = 7 / (-7) = -1. Step 4: Any line parallel to this line must have the same slope m = -1. Verification / Alternative check: We can check by interchanging the roles of the points. If we take (x1, y1) = (-3, 5) and (x2, y2) = (4, -2), then m = (y2 - y1) / (x2 - x1) = (-2 - 5) / (4 - (-3)) = (-7) / 7 = -1 again. The slope is unchanged, confirming that our computation is consistent. Since parallel lines share the same slope, the required slope must be -1. Why Other Options Are Wrong: Option A (3/7): This could arise from incorrect subtraction or partial simplification but does not match the correct ratio of changes. Option B (1): This would correspond to a line rising one unit for every one unit of x, which is not the case here. Option C (-3/7): This suggests mixing up the differences in x and y. Option E (0): This would imply a horizontal line, which is not supported by the coordinates given. Common Pitfalls: A frequent mistake is to reverse the order of subtraction inconsistently between numerator and denominator, which changes the sign of the slope. Another issue is misreading the coordinates, especially the negative signs. As long as you subtract in the same order for x and y coordinates, the sign will be correct. Also remember that parallel lines share the same slope, so once the slope is correctly computed, that is the final answer. Final Answer: The slope of any line parallel to the given line is -1.

Question 5

Find the volume, in cubic centimetres, of a right circular cylinder whose diameter is 21 centimetres and whose height is 10 centimetres. Take pi = 22/7.

Accepted Answer

Introduction / Context: This question tests your ability to compute the volume of a right circular cylinder using the standard formula. It also checks whether you correctly handle radius when only the diameter is given. Volume questions like this are common in quantitative aptitude and basic geometry sections. Given Data / Assumptions: Diameter of the cylinder = 21 centimetres. Radius r = diameter / 2 = 21 / 2 = 10.5 centimetres. Height h = 10 centimetres. We must find the volume in cubic centimetres. Use pi = 22/7. Concept / Approach: The volume V of a right circular cylinder is given by: V = pi * r^2 * h. We first compute the radius from the diameter, then square the radius, multiply by pi and the height. Because the radius is a decimal (10.5), it is convenient to convert it to a fraction to simplify the algebra. Once we compute the exact value, we compare it with the options to select the correct one. Step-by-Step Solution: Step 1: Radius r = 21 / 2 = 10.5 centimetres. Step 2: Express r as a fraction: r = 21 / 2, so r^2 = (21 / 2)^2 = 441 / 4. Step 3: Use the formula V = pi * r^2 * h. V = (22/7) * (441/4) * 10. Step 4: Simplify (22/7) * (441/4). First, 441 / 7 = 63, so this becomes 22 * 63 / 4. 22 * 63 = 1386, so we have 1386 / 4. 1386 / 4 = 346.5. Step 5: Now multiply by the height h = 10: V = 346.5 * 10 = 3465 cubic centimetres. Verification / Alternative check: We can do a quick approximation by using pi as 3.14. Then V is approximately 3.14 * (10.5)^2 * 10. Since (10.5)^2 is about 110.25, the volume is roughly 3.14 * 110.25 * 10, which is about 3463.85 cubic centimetres. This is very close to 3465, confirming that 3465 cm3 is a reasonable and correct choice given pi = 22/7. Why Other Options Are Wrong: Option A (3118 cm3): This is too low and does not match the exact computation using the correct formula. Option B (3275 cm3): This is also lower than the correct value and may arise from rounding pi incorrectly. Option C (3621 cm3): This is somewhat higher than the correct result and does not align with the exact calculation. Option E (3300 cm3): This is a rounded looking number but still does not match the correctly derived volume. Common Pitfalls: Common errors include using the diameter directly in place of the radius, forgetting to divide by 2 to find the radius, or using the surface area formula instead of the volume formula. Some students also forget to square the radius and only multiply by r once. Carefully writing the formula V = pi * r^2 * h and substituting r as 21 / 2 helps avoid these mistakes. Final Answer: The volume of the cylinder is 3465 cm3 (cubic centimetres).

Question 6

A right pentagonal prism has 15 edges in total. Using basic properties of prisms, determine how many vertices the prism has.

Accepted Answer

Introduction / Context: This question involves basic solid geometry and properties of prisms. It asks you to relate the number of edges of a pentagonal prism to its number of vertices. Understanding how edges, faces, and vertices relate for prisms is useful in many combinatorial and geometry problems. Given Data / Assumptions: The solid is a pentagonal prism, so the base is a pentagon. Total number of edges = 15. We must find the number of vertices. The prism is assumed to be a standard prism with two congruent parallel polygonal faces and rectangular lateral faces. Concept / Approach: For an n sided prism (where the base is an n sided polygon): Number of vertices = 2 * n (n on the top face and n on the bottom face). Number of edges = 3 * n (n edges on the top base, n edges on the bottom base, and n vertical lateral edges). Here, we are told that the prism has 15 edges, so 3 * n = 15. From this we can find n, and then use 2 * n to get the number of vertices. Step-by-Step Solution: Step 1: Use the relation for edges of a prism: number of edges = 3 * n. Given that number of edges = 15, we have 3 * n = 15. Step 2: Solve for n: n = 15 / 3 = 5. Step 3: This means the base is a pentagon (5 sided polygon), which is consistent with a pentagonal prism. Step 4: Number of vertices for an n sided prism = 2 * n. So number of vertices = 2 * 5 = 10. Verification / Alternative check: We can visualise the prism. A pentagon has 5 vertices. A pentagonal prism has one pentagon on the top and one on the bottom, so there are 5 + 5 = 10 vertices. The edges consist of 5 edges on the top pentagon, 5 on the bottom pentagon, and 5 vertical edges joining corresponding vertices. That is 5 + 5 + 5 = 15 edges, which matches the given information. This confirms that our reasoning is consistent and the number of vertices is indeed 10. Why Other Options Are Wrong: Option A (12): This would be the number of edges for some other solids but not the number of vertices for this prism. Option C (15): This repeats the number of edges, not the vertices, and confuses two different counts. Option D (20): This number is too large and does not fit the formula for vertices in an n sided prism. Option E (8): This is fewer than the vertices of even a simple cube, so it is not logical for a pentagonal prism. Common Pitfalls: Students sometimes confuse the formulas for faces, edges, and vertices, or they try to apply Euler's formula without first identifying n. Another mistake is to treat the pentagonal prism as if it were a pyramid, which has a different pattern of edges and vertices. Remember that for a prism with n sided bases, the basic relationships are very regular: vertices = 2 * n and edges = 3 * n. Final Answer: The pentagonal prism has 10 vertices.

Question 7

At what point does the straight line 4x - 3y = -6 intersect the y-axis? Give your answer as an ordered pair (x, y).

Accepted Answer

Introduction / Context: This question tests your understanding of intercepts of a straight line on the coordinate axes. Finding the y intercept from an equation in standard form is a basic and important skill in coordinate geometry. It is often used to sketch graphs quickly and to interpret linear relationships. Given Data / Assumptions: Equation of the line: 4x - 3y = -6. We need the point where the line meets the y axis. On the y axis, the x coordinate is always 0. We must present the answer as an ordered pair (x, y). Concept / Approach: The y axis consists of all points where x = 0. To find the y intercept of a line, we set x = 0 in the line's equation and solve for y. The resulting value of y, together with x = 0, gives the intercept coordinate. This process is simple substitution followed by solving a linear equation in one variable. Step-by-Step Solution: Step 1: Start with the line equation 4x - 3y = -6. Step 2: At the y axis, x = 0. Substitute x = 0 into the equation. This gives 4 * 0 - 3y = -6, so -3y = -6. Step 3: Solve for y: divide both sides by -3 to get y = (-6) / (-3) = 2. Step 4: Therefore the y intercept is the point (0, 2). Verification / Alternative check: We can check by substituting the point (0, 2) back into the original equation. Substitute x = 0 and y = 2: 4 * 0 - 3 * 2 = 0 - 6 = -6, which matches the right hand side of the equation. This confirms that (0, 2) lies on the line and is indeed the point where the line crosses the y axis. Why Other Options Are Wrong: Option B (0, 3/2): Substituting y = 3/2 gives 4 * 0 - 3 * (3/2) = -9/2, which is not equal to -6. Option C (2, 0): This is a point on the x axis, not the y axis, since its x coordinate is 2. Option D (3/2, 0): This is also on the x axis and does not represent a y intercept. Option E (0, -2): Substituting y = -2 gives 4 * 0 - 3 * (-2) = 6, which does not equal -6. Common Pitfalls: A common mistake is to confuse x intercept and y intercept, substituting y = 0 instead of x = 0. Another issue is sign errors when solving the equation, especially with negative coefficients. Always remember that the y intercept has x = 0, and carefully substitute and solve to avoid sign mistakes. Final Answer: The line intercepts the y axis at the point (0, 2).

Question 8

Calculate the area, in square centimetres, of a circle whose radius is 10.5 centimetres. Use pi = 22/7.

Accepted Answer

Introduction / Context: This question checks your ability to use the standard area formula for a circle when the radius is a decimal number. Circles with radius 10.5 centimetres are common in exam problems, because 10.5 can be written as 21/2 and simplifies nicely with pi = 22/7. Given Data / Assumptions: Radius r = 10.5 centimetres. We must find the area A in square centimetres. Use pi = 22/7. Concept / Approach: The area A of a circle is given by: A = pi * r^2. Because r = 10.5, which is 21/2 as a fraction, squaring the radius gives a convenient rational number. We substitute r^2 into the area formula, multiply by pi, and then simplify to get the final area value in square centimetres. Step-by-Step Solution: Step 1: Express the radius as a fraction: r = 10.5 = 21 / 2. Step 2: Compute r^2: r^2 = (21 / 2)^2 = 441 / 4. Step 3: Use the formula A = pi * r^2. A = (22/7) * (441/4). Step 4: Simplify (22/7) * (441/4). First, 441 / 7 = 63, so we get 22 * 63 / 4. Compute 22 * 63 = 1386. So A = 1386 / 4 square centimetres. Step 5: Divide 1386 by 4 to get 346.5. Therefore, the area of the circle is 346.5 square centimetres. Verification / Alternative check: As a rough check, we can approximate pi as 3.14 and compute A approximately as 3.14 * (10.5)^2. Since (10.5)^2 is about 110.25, the approximate area is 3.14 * 110.25 which is close to 346.2 square centimetres. This is very close to 346.5, so the exact answer with pi = 22/7 is consistent with the approximate method. Why Other Options Are Wrong: Option A (693 cm2): This is exactly double the correct area and may come from mistakenly doubling the result. Option B (157.5 cm2): This is far too small and might arise from using radius instead of radius squared. Option C (315 cm2): This still underestimates the area and does not result from the correct formula. Option E (330 cm2): This is a rounded guess but does not match the exact computation. Common Pitfalls: Typical errors include using the diameter instead of the radius, forgetting to square the radius, or inserting pi as 3 instead of 22/7 or 3.14. When the radius is a decimal like 10.5, many students also make mistakes in squaring it. Converting 10.5 to 21/2 and then squaring makes the algebra more systematic and reduces calculation errors. Final Answer: The area of the circle is 346.5 cm2 (square centimetres).

Question 9

What is the area, in square centimetres, of a rhombus if the lengths of its diagonals are 12 centimetres and 14 centimetres?

Accepted Answer

Introduction / Context: This problem focuses on the area formula for a rhombus in terms of its diagonals. Rhombus questions appear frequently because they give practice with a special quadrilateral whose diagonals intersect at right angles and bisect each other. Knowing the diagonal based area formula allows a quick solution without needing side lengths. Given Data / Assumptions: Length of first diagonal d1 = 12 centimetres. Length of second diagonal d2 = 14 centimetres. We must find the area in square centimetres. The figure is a standard rhombus with the usual diagonal properties. Concept / Approach: For a rhombus, if d1 and d2 are the lengths of its diagonals, then its area A is given by: A = (d1 * d2) / 2. This formula arises because the diagonals of a rhombus are perpendicular and divide it into four right angled triangles. We use the given diagonals directly in this formula to compute the area. Step-by-Step Solution: Step 1: Identify the diagonals: d1 = 12 cm and d2 = 14 cm. Step 2: Use the formula A = (d1 * d2) / 2. Compute the product d1 * d2 = 12 * 14. 12 * 14 = 168. Step 3: Now divide by 2: A = 168 / 2 = 84. So the area of the rhombus is 84 square centimetres. Verification / Alternative check: We can think of the rhombus as being composed of four right angled triangles, each with legs of length 12/2 = 6 cm and 14/2 = 7 cm. The area of each triangle is (1/2) * 6 * 7 = 21 square centimetres. Four such triangles have total area 4 * 21 = 84 square centimetres. This alternative reasoning matches the result from the diagonal formula, confirming the answer. Why Other Options Are Wrong: Option A (42 sq cm): This is exactly half of the correct area and would come from forgetting to multiply the diagonals before dividing by 2. Option B (168 sq cm): This is the product d1 * d2 without dividing by 2, so it overestimates the area. Option D (63 sq cm): This does not match any correct intermediate value and is simply incorrect. Option E (56 sq cm): This could result from a mistaken combination of half of one diagonal with the full other diagonal. Common Pitfalls: Students sometimes confuse the formula for the area of a rhombus with that of a parallelogram and use base times height instead. Others may forget the division by 2 in the diagonal based formula. Miscalculating the product 12 * 14 can also lead to errors. Carefully remembering A = (d1 * d2) / 2 and checking multiplication helps avoid these problems. Final Answer: The area of the rhombus is 84 sq cm (square centimetres).

Question 10

The radius of a circle is decreased to 25% of its original value. By what percentage does the area of the circle decrease as a result of this change?

Accepted Answer

Introduction / Context: This question checks your understanding of how the area of a circle depends on its radius. It is not enough to say that area decreases in the same proportion as radius, because area is proportional to the square of the radius. Percentage change problems of this type are common in aptitude tests and help reinforce the concept of square relationships. Given Data / Assumptions: Original radius = R. New radius = 25% of original radius = 0.25 * R. We must find the percentage decrease in area. Area of a circle A is proportional to r^2. Concept / Approach: The area of a circle is given by A = pi * r^2. If the radius changes from R to k * R, then the new area becomes pi * (k * R)^2 = pi * k^2 * R^2. Thus the area scales by a factor of k^2. Here k = 0.25, so k^2 = 0.25^2. Once we know the factor by which the area changes, we can find the percentage decrease by comparing new area to original area. Step-by-Step Solution: Step 1: Original area A1 = pi * R^2. Step 2: New radius = 0.25 * R. Step 3: New area A2 = pi * (0.25 * R)^2 = pi * 0.25^2 * R^2. Compute 0.25^2 = 0.0625. So A2 = 0.0625 * pi * R^2. Step 4: The ratio of new area to original area = A2 / A1 = 0.0625. This means the new area is 6.25% of the original area. Step 5: Percentage decrease = 100% - 6.25% = 93.75%. Verification / Alternative check: To visualise, assume the original radius R = 4 units. Then original area A1 = pi * 4^2 = 16 * pi. New radius is 25% of 4, which is 1. New area A2 = pi * 1^2 = pi. The ratio A2 / A1 = pi / (16 * pi) = 1 / 16 = 0.0625. This corresponds to 6.25% of the original area. Therefore, the area decreased by 100% - 6.25% = 93.75%, which confirms our result. Why Other Options Are Wrong: Option A (25%): This wrongly assumes the area decreases in the same proportion as the radius, ignoring the square law. Option B (43.75%): This is neither 6.25% nor 93.75% and does not derive from any correct ratio. Option C (50%): This suggests a simple halving and would correspond to a radius multiplied by 1 / sqrt(2), not by 0.25. Option E (75%): This comes from thinking that the remaining area is 25%, but that confuses radius and area relationships. Common Pitfalls: Many students forget that area is proportional to the square of the radius and assume a linear relationship. Others may compute 25% of the area and treat that as the decrease rather than the new value. To avoid such mistakes, always express the new radius as a fraction of the old radius, square that fraction, and then interpret the result as a percentage of the original area. Final Answer: The area of the circle decreases by 93.75%.

Question 11

The perimeter of a rectangle is 60 centimetres and its breadth is 14 centimetres. Find the area of the rectangle in square centimetres.

Accepted Answer

Introduction / Context: This problem involves finding the dimensions and area of a rectangle when its perimeter and one side are known. It checks your ability to use the perimeter formula and then compute area once both length and breadth are determined. Rectangular geometry of this type is very common in aptitude tests. Given Data / Assumptions: Perimeter of rectangle P = 60 centimetres. Breadth B = 14 centimetres. Length L is unknown. We must find the area A = L * B in square centimetres. Concept / Approach: The perimeter P of a rectangle with length L and breadth B is given by: P = 2 * (L + B). We can rearrange this to express L in terms of P and B: L + B = P / 2, so L = P / 2 - B. Once we find L, we multiply L and B to obtain the area. This is a straightforward application of perimeter and area formulas. Step-by-Step Solution: Step 1: Use the perimeter formula P = 2 * (L + B). Given P = 60 and B = 14, we have 60 = 2 * (L + 14). Step 2: Divide both sides by 2: 60 / 2 = L + 14. So 30 = L + 14. Step 3: Solve for L: L = 30 - 14 = 16 centimetres. Step 4: Area A = L * B = 16 * 14. 16 * 14 = 224 square centimetres. Verification / Alternative check: We can check by recomputing the perimeter from L and B. With L = 16 cm and B = 14 cm, perimeter P = 2 * (16 + 14) = 2 * 30 = 60 centimetres. This matches the given perimeter, confirming that our length and area are correct. The area of 224 cm2 is therefore consistent with both the given data and the rectangle formulas. Why Other Options Are Wrong: Option A (112 cm2): This might come from mistakenly using half the correct length or breadth in the area formula. Option B (448 cm2): This is exactly double the correct area and could result from an incorrect multiplication or unit mistake. Option D (336 cm2): This suggests a wrong length value was used with correct breadth. Option E (196 cm2): This equals 14 * 14, which assumes the rectangle is a square with side 14, which does not satisfy the given perimeter. Common Pitfalls: Students sometimes confuse the formula for perimeter with that for area, or they forget to divide the perimeter by 2 when solving for L + B. Another mistake is to miscalculate 16 * 14. Writing each algebraic step clearly and checking the final perimeter helps prevent errors. Final Answer: The area of the rectangle is 224 cm2 (square centimetres).

Question 12

The area of a square is 100 square centimetres. Find the length of its diagonal in centimetres.

Accepted Answer

Introduction / Context: This problem examines the relationship between the side and the diagonal of a square. It uses the area of the square to find the side length, and then applies the Pythagorean theorem to get the diagonal. Such questions are standard in geometry and aptitude topics involving squares and right triangles. Given Data / Assumptions: Area of the square A = 100 square centimetres. We must find the length of the diagonal in centimetres. All angles of a square are right angles, and all sides are equal. Concept / Approach: If the side of the square is s, then its area is s^2. The diagonal of a square can be found using the Pythagorean theorem on the right triangle formed by two sides and the diagonal. In such a triangle, diagonal d satisfies d^2 = s^2 + s^2 = 2 * s^2, so d = s * sqrt(2). We first find s from the area, then find d using this relationship. Step-by-Step Solution: Step 1: Let side length be s. Area A = s^2. Given A = 100, so s^2 = 100. Step 2: Solve for s: s = sqrt(100) = 10 centimetres. Step 3: For a square, diagonal d = s * sqrt(2). Substitute s = 10: d = 10 * sqrt(2) centimetres. Verification / Alternative check: We can verify numerically by approximating sqrt(2) as 1.414. Then d is approximately 10 * 1.414 = 14.14 centimetres. If we square this approximate diagonal, we get about 200, which equals 2 * s^2, confirming that the diagonal length formula d = s * sqrt(2) is applied correctly. Since s = 10, the exact form 10 * sqrt(2) is correct. Why Other Options Are Wrong: Option B (10 cm): This is the side length, not the diagonal, so it is too short. Option C (20 cm): This would be twice the side length and is longer than the diagonal should be. Option D (20*sqrt(2) cm): This is even larger and corresponds to a square with side 20, not side 10. Option E (5*sqrt(2) cm): This is too small and would correspond to a square with side 5, not area 100. Common Pitfalls: A common error is to think that the diagonal equals side times 2 instead of side times sqrt(2). Another mistake is to miscompute the side length from the area, especially when the area is not a perfect square. In this case, the area 100 makes the side length easy, but the idea remains the same: take the square root of the area to get the side, then multiply by sqrt(2) to get the diagonal. Final Answer: The length of the diagonal of the square is 10*sqrt(2) cm.

Question 13

If the radius of a circle is increased by 14%, by approximately what percentage does the area of the circle increase?

Accepted Answer

Introduction / Context: This question tests understanding of how percentage changes in the radius of a circle affect its area. Because area depends on the square of the radius, the percentage change in area is not the same as the percentage change in radius. Recognising this square relationship is important in many quantitative reasoning problems. Given Data / Assumptions: Original radius = R. Radius is increased by 14%. New radius = 1.14 * R. We must find the percentage increase in area. Area is proportional to the square of the radius. Concept / Approach: The area A of a circle is A = pi * r^2. If the radius changes from R to k * R, the new area becomes pi * (k * R)^2 = pi * k^2 * R^2, which is k^2 times the original area. Here k = 1.14. We compute k^2, subtract 1, and convert to a percentage to find the percentage increase in area. Step-by-Step Solution: Step 1: Original area A1 = pi * R^2. Step 2: New radius r2 = 1.14 * R. Step 3: New area A2 = pi * (1.14 * R)^2 = pi * 1.14^2 * R^2. Compute 1.14^2: 1.14 * 1.14 = 1.2996. So A2 = 1.2996 * pi * R^2. Step 4: Ratio of new area to old area = A2 / A1 = 1.2996. This means the new area is 129.96% of the old area. Step 5: Percentage increase in area = (1.2996 - 1) * 100% = 0.2996 * 100% = 29.96%. Verification / Alternative check: We can use an example. Let R = 10 units, so original area A1 = pi * 100. New radius is 1.14 * 10 = 11.4 units, so new area A2 = pi * (11.4)^2 = pi * 129.96. The factor by which area has increased is 129.96 / 100 = 1.2996, meaning an increase of 29.96%. This numerical check matches the algebraic result. Why Other Options Are Wrong: Option A (28%): This is close but underestimates the true increase and may come from approximating 1.14^2 poorly. Option C (14%): This assumes a linear relationship between radius and area and ignores the square dependence. Option D (14.98%): This is roughly half the correct increase and has no proper basis. Option E (19.6%): This may arise from incorrectly multiplying 14% by 1.4 or some similar shortcut. Common Pitfalls: A typical error is to think that if the radius increases by 14%, the area also increases by 14%, which is incorrect because of the square relationship. Another mistake is to approximate 1.14^2 as 1.28 or 1.26 instead of 1.2996. To avoid mistakes, write out the squared factor explicitly and do the multiplication carefully or use fraction based approximations when possible. Final Answer: The area of the circle increases by approximately 29.96%.

Question 14

In an equilateral triangle of side 24 centimetres, a circle is inscribed so that it touches all three sides. Find the area, in square centimetres, of the remaining portion of the triangle that lies outside the circle but inside the triangle.

Accepted Answer

Introduction / Context: This question combines properties of an equilateral triangle with those of its incircle. You are asked to find the area of the region inside the triangle but outside the circle, which is a common type of shaded region problem. It tests your ability to compute triangle area, find the inradius, and compute circle area, then subtract. Given Data / Assumptions: The triangle is equilateral. Side length a = 24 centimetres. A circle is inscribed, touching all three sides (incircle). We must find area of triangle minus area of circle. Use the standard formulas for equilateral triangle area and inradius. Use pi = 22/7 for circle area approximation. Concept / Approach: For an equilateral triangle of side a, the area Atriangle is (sqrt(3) / 4) * a^2. The inradius r of an equilateral triangle is given by r = a * sqrt(3) / 6. The area of the incircle is Acircle = pi * r^2. The required shaded area is Atriangle - Acircle. We compute both areas using a = 24, then subtract, and finally match the numerical result to the closest option. Step-by-Step Solution: Step 1: Compute area of the equilateral triangle. Atriangle = (sqrt(3) / 4) * a^2. Here a = 24, so a^2 = 576. Atriangle = (sqrt(3) / 4) * 576 = 144 * sqrt(3) square centimetres. Step 2: Find the inradius r = a * sqrt(3) / 6. Substitute a = 24: r = 24 * sqrt(3) / 6 = 4 * sqrt(3) centimetres. Step 3: Compute area of incircle: Acircle = pi * r^2. r^2 = (4 * sqrt(3))^2 = 16 * 3 = 48. Using pi = 22/7, Acircle = (22/7) * 48 = 1056 / 7 square centimetres. 1056 / 7 is approximately 150.857 square centimetres. Step 4: Approximate the triangle area numerically. Take sqrt(3) ≈ 1.732. Atriangle ≈ 144 * 1.732 ≈ 249.408 square centimetres. Step 5: Shaded area = Atriangle - Acircle ≈ 249.408 - 150.857 ≈ 98.551 square centimetres. Rounded to one decimal place, the shaded area is about 98.5 sq cm. Verification / Alternative check: An approximate check is to see that the circle should occupy a bit more than half of the triangle's area. Our triangle area is about 249.4 sq cm, and the circle area is about 150.9 sq cm. The ratio 150.9 / 249.4 is a bit over 0.6, which seems reasonable, since the incircle is fairly large inside an equilateral triangle. The remaining area of about 98.5 sq cm is therefore realistic and consistent with these approximate calculations. Why Other Options Are Wrong: Option A (36.6 sq cm): This is far too small and would imply the circle covers almost the entire triangle, which is not possible. Option B (54.2 sq cm): This still underestimates the shaded region and does not match the triangle and circle areas. Option C (72.8 sq cm): This is closer than the smaller options but still significantly below the calculated value. Option E (120 sq cm): This is larger than the computed shaded area and would require the circle to be much smaller relative to the triangle than it actually is. Common Pitfalls: Students may confuse the formulas for inradius and circumradius or misremember the area formula for an equilateral triangle. Another common issue is to use radius a/2 instead of a * sqrt(3)/6 for the incircle. Rounding errors in sqrt(3) and pi can also affect the final result. Writing each formula clearly and doing the algebra symbolically before substituting approximate values helps minimise mistakes. Final Answer: The area of the remaining portion of the triangle outside the circle is approximately 98.5 sq cm.

Question 15

One interior angle of a rhombus is 60° and the length of its shorter diagonal is 8 cm. What is the area of the rhombus in square centimetres?

Accepted Answer

Introduction / Context: This problem tests your understanding of the geometry of a rhombus when one interior angle and one diagonal are known. A rhombus has four equal sides, and its diagonals intersect at right angles. When one angle is 60 degrees, there are special relationships between the side length and the diagonals which allow us to compute the area efficiently without first finding all interior details separately. Given Data / Assumptions: The quadrilateral is a rhombus. One interior angle of the rhombus is 60 degrees. The shorter diagonal has length 8 cm. All sides of the rhombus are equal in length. Area of a rhombus can be found using the formula Area = (d1 * d2) / 2, where d1 and d2 are the diagonals. Concept / Approach: In a rhombus, if the interior angle between two sides is 60 degrees, the diagonals are related to the side length a by special formulas. For such a rhombus, one diagonal equals a and the other diagonal equals a * square root of 3. In this question the shorter diagonal is given as 8 cm, so that diagonal must equal a. Once we know the side a, we immediately know the other diagonal and can calculate the area using the diagonal product formula. Step-by-Step Solution: Let the side length of the rhombus be a cm. For a rhombus with interior angle 60 degrees, the diagonals are a and a * square root of 3. The shorter diagonal is given as 8 cm, so a = 8 cm. Longer diagonal d2 = a * square root of 3 = 8 * square root of 3. Area of rhombus = (d1 * d2) / 2. Substitute d1 = 8 and d2 = 8 * square root of 3. Area = (8 * 8 * square root of 3) / 2 = (64 * square root of 3) / 2 = 32 * square root of 3 sq cm. Verification / Alternative check: Another way is to split the rhombus into four congruent right triangles by its diagonals. Each triangle has sides related to the side of the rhombus, and using trigonometry you again get one diagonal equal to a and the other equal to a * square root of 3. This confirms that using a = 8 cm gives the correct area 32 * square root of 3 square centimetres. Why Other Options Are Wrong: 64√3 sq cm: This treats the product of the diagonals as the area without dividing by 2. 32√2 sq cm and 64√2 sq cm: These use square root of 2 instead of square root of 3 and do not match the geometry of a 60 degree rhombus. 48√3 sq cm: This comes from an incorrect assumption about diagonal lengths or an arithmetic slip. Common Pitfalls: Forgetting that the area formula for a rhombus with diagonals uses division by 2. Using generic formulas that relate diagonals to sides without accounting for the specific 60 degree angle. Assuming the given diagonal is the longer one rather than the shorter one. Final Answer: The area of the rhombus is 32√3 sq cm

Question 16

A triangle is inscribed in a semicircle of radius 6 m. What is the area of the largest possible triangle that can be inscribed in this semicircle?

Accepted Answer

Introduction / Context: This question explores the classic result about the largest triangle that can be inscribed in a semicircle. It builds on two key ideas: first, that any angle subtended by the diameter in a semicircle is a right angle, and second, that the area of a triangle depends on its base and corresponding height. The problem asks for the maximum area, which occurs for a specific configuration of the triangle inside the semicircle. Given Data / Assumptions: Radius of the semicircle r = 6 m. Diameter of the semicircle is 2r = 12 m. The triangle is completely contained in the semicircle and one side lies along the diameter for the maximal area case. The largest area occurs when the triangle is right angled with the hypotenuse along the diameter. Concept / Approach: A key theorem states that any angle in a semicircle is a right angle. When a triangle is inscribed in a semicircle with the diameter as one side, the vertex opposite the diameter lies on the semicircle, and the triangle is right angled at that vertex. Among all such triangles, the one with the largest area has its right angle at the point of the semicircle such that the triangle height equals the radius. Thus, the base is the diameter and the maximum height is the radius, giving a simple formula for the maximum area. Step-by-Step Solution: Radius r = 6 m, so diameter = 2r = 12 m. For the largest triangle, take the diameter as base and the radius as height. Base b = 12 m, height h = 6 m. Area of a triangle = (1/2) * base * height. Area = (1/2) * 12 * 6 = 6 * 6 = 36 sq m. Verification / Alternative check: If you attempt any other triangle in the semicircle whose base is shorter than the diameter or whose height is less than the radius, the product base * height becomes smaller, and hence the area decreases. Therefore, the combination base = 12 m and height = 6 m indeed yields the maximum possible area, confirming 36 square metres as the correct answer. Why Other Options Are Wrong: 72 sq m: This is double the correct value and would require base or height values not compatible with a semicircle of radius 6 m. 18 sq m and 12 sq m: These correspond to smaller base or height values and do not represent the maximum area triangle. 24 sq m: Also smaller than 36 sq m and results from taking a reduced effective height or base. Common Pitfalls: Forgetting that the largest triangle is right angled with the diameter as hypotenuse. Using the radius incorrectly as the base rather than as the height. Misapplying area formulas for sectors or segments instead of the triangle formula. Final Answer: The maximum possible area of the triangle is 36 sq m

Question 17

The area of a rectangle is 60 sq cm and its perimeter is 34 cm. What is the length of the diagonal of the rectangle?

Accepted Answer

Introduction / Context: This problem combines two basic properties of rectangles: area and perimeter. By using both pieces of information together, you can find the side lengths and then use the Pythagoras theorem to determine the diagonal. It is a good example of how algebra and geometry work hand in hand in mensuration questions. Given Data / Assumptions: Area of the rectangle = 60 sq cm. Perimeter of the rectangle = 34 cm. Let length = l cm and breadth = b cm. Area formula: l * b = 60. Perimeter formula: 2 * (l + b) = 34, so l + b = 17. Diagonal d satisfies d^2 = l^2 + b^2 for a rectangle. Concept / Approach: We are given the sum of length and breadth and their product. From algebra, we know (l + b)^2 = l^2 + b^2 + 2lb. Using this identity and the given numbers, we can compute l^2 + b^2 without finding l and b individually. Since for a rectangle the diagonal is the hypotenuse of a right triangle with legs l and b, we apply the Pythagoras theorem to find the diagonal directly from l^2 + b^2. Step-by-Step Solution: Given l * b = 60 and l + b = 17. Compute (l + b)^2 = 17^2 = 289. Use identity: (l + b)^2 = l^2 + b^2 + 2lb. So l^2 + b^2 = 289 - 2 * 60 = 289 - 120 = 169. For a rectangle, diagonal d satisfies d^2 = l^2 + b^2. Hence d^2 = 169, so d = square root of 169 = 13 cm. Verification / Alternative check: We can also find individual side lengths. Solve l + b = 17 and l * b = 60. The pair (12, 5) satisfies both. Then diagonal d = square root of (12^2 + 5^2) = square root of (144 + 25) = square root of 169 = 13 cm, which confirms the earlier calculation and the correctness of the answer. Why Other Options Are Wrong: 11 cm and 9 cm: These correspond to smaller sums of squares of sides and do not satisfy the given area and perimeter simultaneously. 15 cm and 17 cm: These diagonals would require side lengths that do not match both the stated area and perimeter conditions. Common Pitfalls: Trying to guess side lengths without using the algebraic identity and making arithmetic errors. Forgetting that for a rectangle, the diagonal is the hypotenuse of a right triangle formed by length and breadth. Miscomputing (l + b)^2 or the subtraction step that produces l^2 + b^2. Final Answer: The diagonal of the rectangle is 13 cm long

Question 18

The diagonal of a rectangle is 17 cm and its breadth is 8 cm. Find the perimeter of the rectangle in centimetres.

Accepted Answer

Introduction / Context: This question checks your understanding of how the Pythagoras theorem applies to rectangles. The diagonal of a rectangle acts as the hypotenuse of a right triangle formed by its length and breadth. Once the missing side is computed, calculating the perimeter is straightforward using basic perimeter formulas. Given Data / Assumptions: Diagonal of the rectangle d = 17 cm. Breadth b = 8 cm. Let length be l cm. In a rectangle, diagonal^2 = length^2 + breadth^2. Perimeter P = 2 * (l + b). Concept / Approach: Use the Pythagoras theorem on the right triangle formed by the length, breadth, and diagonal. Knowing two sides (breadth and diagonal), you can solve for the third side (length). Then substitute length and breadth into the perimeter formula to find the total boundary length of the rectangle. Step-by-Step Solution: Given diagonal d = 17 cm and breadth b = 8 cm. Use the relation d^2 = l^2 + b^2. So 17^2 = l^2 + 8^2. 289 = l^2 + 64. l^2 = 289 - 64 = 225. l = square root of 225 = 15 cm. Now find perimeter P = 2 * (l + b) = 2 * (15 + 8) = 2 * 23 = 46 cm. Verification / Alternative check: You may quickly verify by checking the Pythagoras triple 8, 15, 17. Since 8^2 + 15^2 = 64 + 225 = 289 and 17^2 = 289, this matches exactly, confirming that the right triangle assumption is correct. So length 15 cm and breadth 8 cm give a consistent rectangle with diagonal 17 cm and perimeter 46 cm. Why Other Options Are Wrong: 92 and 84: These correspond to larger perimeters that would require larger side lengths than those allowed by the diagonal. 42 and 34: These correspond to smaller perimeters and would not match a diagonal of 17 cm when using the Pythagoras theorem. Common Pitfalls: Forgetting to square the given diagonal and breadth when applying the Pythagoras theorem. Using perimeter formula incorrectly, such as l + b instead of 2 * (l + b). Making arithmetic mistakes in subtracting 64 from 289 or in taking the square root of 225. Final Answer: The perimeter of the rectangle is 46 cm

Question 19

The base of an isosceles triangle has length 2x + 2y + 4z and its perimeter is 4x + 2y + 6z. What is the length of each of the equal sides of the triangle, expressed in terms of x, y and z?

Accepted Answer

Introduction / Context: This problem is an algebraic application of the perimeter formula for an isosceles triangle. You are given symbolic expressions for the base and the perimeter and asked to find the length of each equal side in terms of x, y and z. It tests your ability to manipulate algebraic expressions and apply the basic perimeter concept for special triangles. Given Data / Assumptions: The triangle is isosceles. Base length B = 2x + 2y + 4z. Perimeter P = 4x + 2y + 6z. Let each equal side have length S. Perimeter of any triangle = sum of all three sides = B + 2S. Concept / Approach: In an isosceles triangle, two sides are equal and the third side is the base. If we know the total perimeter and the base length, we can subtract the base from the perimeter to obtain the combined length of the two equal sides. Dividing this result by 2 gives the length of each equal side. This logic applies equally well when the lengths are given as algebraic expressions rather than numeric values. Step-by-Step Solution: Let the equal side length be S. Perimeter P = B + 2S. Given P = 4x + 2y + 6z and B = 2x + 2y + 4z. So 4x + 2y + 6z = (2x + 2y + 4z) + 2S. Subtract base from both sides: (4x + 2y + 6z) - (2x + 2y + 4z) = 2S. This simplifies to 2x + 0y + 2z = 2S. So 2S = 2x + 2z, hence S = x + z. Verification / Alternative check: You can verify by reconstructing the perimeter using S = x + z and B = 2x + 2y + 4z. Then P = B + 2S = (2x + 2y + 4z) + 2(x + z) = 2x + 2y + 4z + 2x + 2z = 4x + 2y + 6z, which matches the given perimeter exactly, confirming that S = x + z is correct. Why Other Options Are Wrong: x + y and x + y + z: These values do not satisfy the perimeter equation when substituted back. 2(x + y): Leads to a perimeter expression that does not match 4x + 2y + 6z. 2x + z: This mis-allocates the z term and fails the check against the given perimeter. Common Pitfalls: Forgetting that there are two equal sides, and using P = B + S instead of P = B + 2S. Making sign errors when subtracting algebraic expressions. Not simplifying terms correctly, especially when combining x, y and z terms. Final Answer: Each of the equal sides has length x + z

Question 20

What is the least number that must be subtracted from 1294 so that, when the result is divided by 9, 11 and 13, the remainder in each case is 6?

Accepted Answer

Introduction / Context: This number theory problem checks your understanding of common remainders and the use of the least common multiple for several divisors. You are given a starting number and asked to adjust it so that dividing by 9, 11 and 13 always leaves the same remainder 6. The goal is to find the smallest possible subtraction to achieve this uniform remainder condition. Given Data / Assumptions: Original number = 1294. Divisors: 9, 11 and 13. Desired remainder when divided by each divisor = 6. Let the new number be N = 1294 - k, where k is the least number to subtract. N leaves remainder 6 when divided by 9, 11 and 13. Concept / Approach: If N leaves remainder 6 when divided by each of 9, 11 and 13, then N - 6 is divisible by all three of these numbers. Therefore, N - 6 must be a multiple of the least common multiple of 9, 11 and 13. Once we compute this least common multiple, we look for multiples of it that are close to 1294 and then choose the one that gives the largest N less than or equal to 1294. The difference between 1294 and that N gives the required k. Step-by-Step Solution: Let N = 1294 - k. We want N ≡ 6 (mod 9), N ≡ 6 (mod 11), N ≡ 6 (mod 13). This means N - 6 is divisible by 9, 11 and 13. Compute least common multiple of 9, 11 and 13. Since they are pairwise coprime, LCM = 9 * 11 * 13 = 1287. So N - 6 must equal 1287 * t for some integer t. We need N as large as possible but not greater than 1294, so try t = 1. Then N - 6 = 1287, so N = 1287 + 6 = 1293. Now 1293 ≤ 1294, which is valid. Therefore k = 1294 - 1293 = 1. Verification / Alternative check: Check that 1293 leaves remainder 6 with each divisor. 1293 ÷ 9 = 143 with remainder 6, 1293 ÷ 11 = 117 with remainder 6, and 1293 ÷ 13 = 99 with remainder 6. All three checks confirm that 1293 satisfies the condition, so subtracting 1 from 1294 is both necessary and sufficient. Why Other Options Are Wrong: 2, 3, 4 and 5: Subtracting any of these values gives numbers that do not simultaneously leave remainder 6 when divided by 9, 11 and 13. They fail at least one of the divisibility checks for N - 6. Common Pitfalls: Trying to test each option only against one divisor and not checking all three. Confusing common remainder problems with problems that require remainder zero, that is pure divisibility. Forgetting to account for the remainder shift, that is working with N instead of N - 6 when applying the least common multiple idea. Final Answer: The least number that must be subtracted is 1

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