Question 1

Surd decomposition and parameter sum Given that √28 − 6√3 is attempted to be written in the form √3·a + b with a, b rational numbers, determine the value of (a + b).

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Introduction / Context:This question explores expressing a radical combination in a restricted “linear in √3” form. We test whether √28 − 6√3 can be written as √3·a + b where a and b are rational numbers, and then interpret what (a + b) would be. The core idea is to compare like radicals and rational parts consistently. Given Data / Assumptions: Expression: √28 − 6√3. Target form: √3·a + b with a, b ∈ rationals. We treat √7 and √3 as linearly independent over the rationals (they cannot be converted into each other with rational coefficients). Concept / Approach:Break √28 as 2√7. Then √28 − 6√3 = 2√7 − 6√3. If the target is √3·a + b, then any √7 term must vanish or be representable using √3 and rationals, which is not possible since √7 is irrational and independent from √3 over Q. Step-by-Step Solution:Rewrite √28 = 2√7, so the expression is 2√7 − 6√3.Target form is a√3 + b with a, b rational.There is no way to express 2√7 as q√3 + r for rational q, r, because √7 and √3 are distinct quadratic irrationals.Therefore, the proposed representation does not exist with rational a and b. Verification / Alternative check:Assume, for contradiction, 2√7 = a√3 + b. Squaring gives 28 = 3a^2 + b^2 + 2ab√3. The left side has no √3 term; hence 2ab must be 0. If ab = 0, then either a = 0 or b = 0, which still cannot yield 28 from 3a^2 + b^2 without contradicting the absence of √3 and the presence of √7 upon taking square roots. Hence impossible. Why Other Options Are Wrong:Numerical choices like −2, 2, 1, −1 assume a valid (a, b) exists; but no rational pair exists, so those sums are meaningless for this constraint. Common Pitfalls:Confusing √28 with 2√7 and then trying to blend √7 into a combination of √3 and rationals; these radicals cannot be interchanged with rational coefficients. Final Answer:Not possible (no such rational a, b)

Question 2

Cube–reciprocal identity with a = (√2 − 1)^(1/3) Evaluate (a − a^−1)^3 + 3(a − a^−1).

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Introduction / Context:This problem tests recognition of a compact algebraic identity that relates (a − 1/a)^3 + 3(a − 1/a) to a^3 − 1/a^3. Using the given a = (√2 − 1)^(1/3), we can evaluate the expression exactly without messy decimals. Given Data / Assumptions: a = (√2 − 1)^(1/3). Target: (a − a^−1)^3 + 3(a − a^−1). All algebra performed with exact radicals. Concept / Approach:Identity: (t)^3 + 3t = a^3 − 1/a^3 where t = a − 1/a. Expand (a − 1/a)^3 to verify this, or use the well-known result to avoid expansion. Then compute a^3 and a^−3 directly from the definition of a. Step-by-Step Solution:Let t = a − 1/a. Then t^3 + 3t = a^3 − 1/a^3.Given a = (√2 − 1)^(1/3) ⇒ a^3 = √2 − 1.Compute a^−3 = 1/(√2 − 1) = (√2 + 1)/[(√2 − 1)(√2 + 1)] = √2 + 1.Hence a^3 − a^−3 = (√2 − 1) − (√2 + 1) = −2. Verification / Alternative check:Expand (a − 1/a)^3 = a^3 − 3a + 3/a − 1/a^3. Adding 3(a − 1/a) gives a^3 − 1/a^3 indeed, confirming the identity. Why Other Options Are Wrong:2, 2√2, and √2 result from sign or reciprocal mistakes; 0 would require a^3 = 1/a^3, which is false here. Common Pitfalls:Forgetting to rationalize 1/(√2 − 1), or misapplying the identity by using plus instead of minus in the reciprocal term. Final Answer:-2

Question 3

Surd ratios and symmetric powers Let x = (√3 + 1)/(√3 − 1) and y = (√3 − 1)/(√3 + 1). Evaluate x^2/y + y^2/x.

Accepted Answer

Introduction / Context:Expressions featuring x and its reciprocal often reduce via symmetry. Here, y is the reciprocal of x, enabling a simplification from mixed fractions to a sum of cubes of x and 1/x. Given Data / Assumptions: x = (√3 + 1)/(√3 − 1), y = (√3 − 1)/(√3 + 1). We need S = x^2/y + y^2/x. Concept / Approach:Notice y = 1/x. Then S = x^2 * x + (1/x^2) * (1/x) = x^3 + 1/x^3. Next compute x in simplest form, then use the identity x^3 + 1/x^3 = (x + 1/x)^3 − 3(x + 1/x). Step-by-Step Solution:Simplify x: (√3 + 1)/(√3 − 1) = ((√3 + 1)^2)/(3 − 1) = (3 + 1 + 2√3)/2 = 2 + √3.Thus 1/x = 2 − √3 (since (2 + √3)(2 − √3) = 1).Then x + 1/x = (2 + √3) + (2 − √3) = 4.Compute x^3 + 1/x^3 = (x + 1/x)^3 − 3(x + 1/x) = 4^3 − 3*4 = 64 − 12 = 52. Verification / Alternative check:Directly cube x and 1/x numerically using √3 ≈ 1.732; sums confirm the exact value 52. Why Other Options Are Wrong:76 and 64 come from squaring 4 or forgetting the subtraction term; 4 corresponds to x + 1/x, not the cube sum; 28 is a rough, incorrect estimate. Common Pitfalls:Not noticing y = 1/x; failing to rationalize x first, which hides the elegant cancellation. Final Answer:52

Question 4

Use x + a/x = b to simplify a rational expression If x + a/x = b, evaluate (x^2 + b x + a) / (b x^2 − x^3).

Accepted Answer

Introduction / Context:This problem relies on manipulating the relation x + a/x = b. Converting it to a quadratic in x lets us replace x^2 terms and reduce a complicated rational expression to a simple ratio. Given Data / Assumptions: x + a/x = b, with x ≠ 0 and a ≠ 0. Expression: E = (x^2 + b x + a) / (b x^2 − x^3). Concept / Approach:From x + a/x = b, multiply by x to get x^2 + a = b x ⇒ x^2 − b x + a = 0. Use this to replace x^2 and also to simplify b − x in terms of a/x, which collapses the denominator neatly. Step-by-Step Solution:From x^2 − b x + a = 0, we have x^2 = b x − a.Numerator: x^2 + b x + a = (b x − a) + b x + a = 2 b x.Also from x + a/x = b ⇒ b − x = a/x.Denominator: b x^2 − x^3 = x^2 (b − x) = x^2 * (a/x) = a x.Therefore, E = (2 b x) / (a x) = 2b/a. Verification / Alternative check:Pick sample values a = 2, x satisfying x + 2/x = b; compute both sides numerically to confirm E = 2b/a regardless of the particular root x. Why Other Options Are Wrong:a + b, b/a, and ab ignore the key transformations; a/(2b) flips the correct ratio. Common Pitfalls:Losing track of signs when substituting x^2 = b x − a; not noticing that b − x = a/x. Final Answer:2b/a

Question 5

Triangles with a line parallel to base – Using similarity: In ∆ABC, D and E lie on AB and AC respectively such that AD = 8 cm, DB = 12 cm (so AB = 20 cm), AE = 6 cm, EC = 9 cm (so AC = 15 cm). Show that DE ∥ BC and find BC in terms of DE.

Accepted Answer

Introduction / Context:When a segment joins points on two sides of a triangle in the same ratio, it is parallel to the third side (Converse of Basic Proportionality Theorem), leading to similar triangles and a fixed scale factor. Given Data / Assumptions: AD/DB = 8/12 = 2/3. AE/EC = 6/9 = 2/3. Therefore DE ∥ BC. Concept / Approach:Similarity of ∆ADE and ∆ABC gives scale factor AD/AB = 8/20 = 2/5. Corresponding sides satisfy DE/BC = 2/5 ⇒ BC = (5/2)DE. Step-by-Step Solution: AD/DB = AE/EC ⇒ DE ∥ BC.AD/AB = 8/20 = 2/5 ⇒ DE/BC = 2/5 ⇒ BC = (5/2)DE. Final Answer:5/2 DE

Question 6

A tailor has 37.5 metres of cloth. He can cut 8 equal pieces from each metre of cloth. How many pieces can he make from exactly half of the cloth he has?

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Introduction / Context: This is a straightforward application of fractions and unitary method. We are given the total length of cloth and the number of pieces obtainable from one metre. The problem then asks for the number of pieces that can be cut from half of the total cloth. Such questions are standard in aptitude tests involving proportional reasoning. Given Data / Assumptions: Total length of cloth = 37.5 metres. From 1 metre of cloth, the tailor can cut 8 equal pieces. We are interested in half of the total cloth, that is 37.5 / 2 metres. All pieces are of equal size. Concept / Approach: First, determine the length of cloth in half the total amount. Next, use the rate of 8 pieces per metre to find the number of pieces from that half length. This is essentially multiplying the number of metres in half by the number of pieces per metre. Step-by-Step Solution: Step 1: Compute half of 37.5 metres: 37.5 / 2 = 18.75 metres. Step 2: From 1 metre, the tailor can make 8 pieces. Step 3: From 18.75 metres, total pieces = 18.75 × 8. Step 4: Calculate 18.75 × 8 = (18 × 8) + (0.75 × 8) = 144 + 6 = 150. Step 5: Therefore, the tailor can make 150 pieces from half of the cloth. Verification / Alternative check: We can find the total pieces from the full 37.5 metres and then divide by 2. From 37.5 metres, total pieces = 37.5 × 8 = 300. Half of 300 is 150, which matches the earlier result. Thus the answer is consistent. Why Other Options Are Wrong: 300: This is the number of pieces from the full 37.5 metres, not from half of it. 175 and 200: These are higher than the correct count for 18.75 metres and do not match any logical calculation based on 8 pieces per metre. 250: This corresponds to 31.25 metres of cloth, which is more than half of 37.5 metres, so it is not correct. Common Pitfalls: Some learners mistakenly multiply the total 37.5 metres by 8 and then forget to halve the number of pieces. Others miscalculate half of 37.5. Always handle the fraction step carefully and then apply the pieces-per-metre rate. Final Answer: The tailor can make 150 pieces from half of the cloth he has.

Question 7

In triangle ABC, points D and E lie on sides AB and AC respectively, and segment DE is drawn parallel to side BC. The segments on side AB satisfy the ratio AD : DB = 2 : 5. The area of the smaller triangle ADE is 8 cm². Using properties of similar t…

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Introduction: This geometry question tests the concept of similar triangles formed by a line drawn parallel to the base of a triangle, and how their areas are related to the ratios of corresponding sides. Once similarity is understood, the area of the quadrilateral can be found by subtracting the smaller triangle area from the larger triangle area. Given Data / Assumptions: In triangle ABC, D lies on AB and E lies on AC. DE is parallel to BC. AD : DB = 2 : 5, so AB is divided internally in this ratio. Area of triangle ADE = 8 cm². We must find the area of quadrilateral BDEC. Concept / Approach: When a line segment is drawn parallel to one side of a triangle and cutting the other two sides, the smaller triangle formed at the vertex is similar to the original triangle. The ratio of similarity is the ratio of corresponding sides, and the ratio of their areas is the square of the ratio of their corresponding sides. After obtaining the area of the whole triangle ABC, the quadrilateral area is simply the difference between the total area and the area of the smaller triangle. Step-by-Step Solution: Let AB = AD + DB. Given AD : DB = 2 : 5, so AD : AB = 2 : (2 + 5) = 2 : 7. Thus, the ratio of similarity (ADE to ABC) is AD / AB = 2 / 7. Area scale factor = (2 / 7)² = 4 / 49. Let area of triangle ABC = K. Then area(ADE) = (4 / 49) * K. Given area(ADE) = 8, so 8 = (4 / 49) * K. Solve for K: K = 8 * (49 / 4) = 2 * 49 = 98 cm². Quadrilateral BDEC is the remaining part: area(BDEC) = area(ABC) − area(ADE) = 98 − 8 = 90 cm². Verification / Alternative check: As a ratio, area(ADE) : area(ABC) = 4 : 49. That means area(BDEC) : area(ADE) = (49 − 4) : 4 = 45 : 4. If area(ADE) is 8, then area(BDEC) = (45 / 4) * 8 = 45 * 2 = 90 cm², which confirms our previous result. Why Other Options Are Wrong: Values 86, 94, and 98 cm² do not satisfy the required area ratio. In particular, 98 cm² is the total area of triangle ABC, not the quadrilateral. Any value other than 90 would break the fixed similarity ratio implied by AD : AB. Common Pitfalls: Learners sometimes assume that area ratios are directly proportional to side ratios instead of the square of side ratios. Another error is to forget that AB is the sum AD + DB when forming the 2 : 7 ratio. Being precise with ratios and squaring for area is essential. Final Answer: The area of quadrilateral BDEC is 90 cm².

Question 8

Consider the straight line that passes through the points (8, 2) and (3, 1). What is the slope of the line that is perpendicular to this line?

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Introduction: This coordinate geometry question checks your understanding of slopes of lines and the relationship between the slopes of perpendicular lines. Once you find the slope of the given line, you can use the negative reciprocal to obtain the slope of any line perpendicular to it. Given Data / Assumptions: The original line passes through points (8, 2) and (3, 1). We need the slope of a line that is perpendicular to this line. All work is in the standard Cartesian coordinate system. Concept / Approach: The slope m of a line passing through (x₁, y₁) and (x₂, y₂) is: m = (y₂ − y₁) / (x₂ − x₁). If two lines are perpendicular and neither is vertical, then the product of their slopes is −1. Thus, if one line has slope m, the perpendicular line has slope m⊥ = −1 / m. This negative reciprocal relationship is the key to the solution. Step-by-Step Solution: First find the slope of the given line through (8, 2) and (3, 1). m = (1 − 2) / (3 − 8) = (−1) / (−5) = 1/5. So the slope of the given line is 1/5. For a line perpendicular to this, we need a slope m⊥ such that m * m⊥ = −1. So (1/5) * m⊥ = −1. Therefore m⊥ = −1 / (1/5) = −5. Verification / Alternative check: You can verify by checking the product: (1/5) * (−5) = −1, which confirms perpendicularity. This is the standard condition for two non vertical lines to be perpendicular in the plane. Why Other Options Are Wrong: The slope 5 has product 5 * (1/5) = 1, indicating parallelism in a different sense, not perpendicularity. The slopes 1/5 and −1/5 are either the original slope or a line that is not perpendicular because their product with 1/5 is not −1. Only −5 satisfies the negative reciprocal condition. Common Pitfalls: A common error is to change only the sign or only invert the slope instead of doing both. Another mistake is miscalculating the original slope by mixing up x and y differences. Always compute the initial slope carefully, then take its negative reciprocal for a perpendicular line. Final Answer: The slope of the line perpendicular to the given line is −5.

Question 9

In an examination scoring system, a student scores 4 marks for every correct answer and loses 1 mark for every wrong answer. If the student attempts all 60 questions and secures a total of 130 marks, how many questions were answered correctly?

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Introduction / Context: This question tests basic algebra and logical reasoning using a simple exam scoring rule. The student earns marks for correct answers and loses marks for wrong ones while attempting every question. Problems like this appear frequently in quantitative aptitude and reasoning sections, and they help evaluate your ability to set up equations based on word descriptions of scoring systems. Given Data / Assumptions: Total number of questions attempted = 60. Marks for each correct answer = +4. Marks lost for each wrong answer = -1. Total marks obtained by the student = 130. Let the number of correct answers be C and wrong answers be W. Since all questions are attempted, C + W = 60. Concept / Approach: We convert the scoring rule into an algebraic expression. The total marks T are: T = 4 * C - 1 * W. Because C + W = 60, we can express W as 60 - C and substitute into the total marks expression. This yields a single linear equation in C, which we solve to determine the number of correct answers. These linear equations are foundational in many aptitude topics. Step-by-Step Solution: Step 1: Write the total questions equation: C + W = 60. Step 2: Express W in terms of C: W = 60 - C. Step 3: Write the total marks equation: T = 4C - W. Step 4: Given total marks T = 130, so 130 = 4C - W. Step 5: Substitute W = 60 - C into the marks equation: 130 = 4C - (60 - C). Step 6: Simplify: 130 = 4C - 60 + C = 5C - 60. Step 7: Rearrange to solve for C: 5C = 130 + 60 = 190. Step 8: Divide by 5: C = 190 / 5 = 38. Step 9: Therefore, the student answered 38 questions correctly. Verification / Alternative check: Compute W and total marks to confirm. Wrong answers: W = 60 - C = 60 - 38 = 22. Marks from correct answers: Marks_correct = 4 * 38 = 152. Penalty from wrong answers: Marks_wrong = -1 * 22 = -22. Total marks: T = 152 - 22 = 130. This matches the given total, which verifies that 38 correct answers is the right count. Why Other Options Are Wrong: 35, 40, 42, and 36 all lead to different totals when you compute 4 * correct minus 1 * wrong with 60 questions. None of those totals equals 130, so those options do not satisfy the scoring conditions. Common Pitfalls: Students sometimes forget that all questions are attempted and mistakenly use only one equation, or they mix up the signs in the total marks formula. Others try trial and error with the options rather than forming the correct linear equation. Writing both the total questions equation and the total marks equation clearly, then substituting and solving systematically, is the safest method. Final Answer: The student answered 38 questions correctly.

Question 10

In aptitude (Linear Equations and Word Problems), 2 tables and 3 chairs together cost ₹3,500, while 3 tables and 2 chairs together cost ₹4,000. Assuming each table has the same price and each chair has the same price, what is the cost of one table (…

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Introduction / Context: This question is a classic linear equations word problem where item prices must be deduced from combined totals. It tests your ability to translate a real world purchasing scenario into algebraic equations and then solve simultaneous equations to find the unknown prices. Such reasoning is important for both quantitative aptitude and everyday financial decision making. Given Data / Assumptions: 2 tables plus 3 chairs cost ₹3,500. 3 tables plus 2 chairs cost ₹4,000. Each table has a fixed price, say T rupees. Each chair has a fixed price, say C rupees. We must find T, the cost of one table. Concept / Approach: We form two linear equations based on the given total costs: 2T + 3C = 3,500 3T + 2C = 4,000 These two equations in two unknowns can be solved by elimination or substitution. The elimination method is usually faster for exam settings. Once we solve for T and C, we pick the value of T that matches one of the answer options. Step-by-Step Solution: Step 1: Let the price of one table be T and the price of one chair be C. Step 2: From the first condition, write 2T + 3C = 3,500. Step 3: From the second condition, write 3T + 2C = 4,000. Step 4: Multiply the first equation by 3 to get 6T + 9C = 10,500. Step 5: Multiply the second equation by 2 to get 6T + 4C = 8,000. Step 6: Subtract the second new equation from the first new equation: (6T + 9C) - (6T + 4C) = 10,500 - 8,000. Step 7: This simplifies to 5C = 2,500. Step 8: Solve for C: C = 2,500 / 5 = 500. Step 9: Substitute C = 500 into 2T + 3C = 3,500. Step 10: So 2T + 3 * 500 = 3,500, which gives 2T + 1,500 = 3,500. Step 11: Therefore, 2T = 2,000 and T = 1,000. Step 12: Thus the cost of one table is ₹1,000. Verification / Alternative check: Check with the second original equation. With T = 1,000 and C = 500: 3T + 2C = 3 * 1,000 + 2 * 500 = 3,000 + 1,000 = 4,000. This matches the given total for the second combination, confirming that the values are consistent. The first combination also works, since 2 * 1,000 + 3 * 500 = 2,000 + 1,500 = 3,500. Why Other Options Are Wrong: ₹500, ₹1,500, ₹2,000, and ₹750 do not satisfy both equations when matched with any nonnegative chair price. Only T = ₹1,000, C = ₹500 makes both combined totals correct, so the other options are invalid. Common Pitfalls: Some students guess values from the options without checking both equations, or they make arithmetic errors while eliminating variables. Misaligning signs during subtraction is another frequent issue. A good practice is to rewrite the equations carefully, align like terms, and check the final values in both original equations to confirm accuracy. Final Answer: The cost of one table is ₹1,000.

Question 11

In aptitude (Ratio and Probability style word problems), A fires 5 shots for every 3 shots fired by B. A hits once in every 3 shots, while B hits once in every 2 shots. When B has missed 27 times, how many birds has A killed?

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Introduction / Context: This question blends ideas of ratios and success rates. Two shooters fire at birds with different accuracy and frequency. One shooter has a higher hit rate, but both follow fixed ratios of shots and outcomes. The question uses the number of misses for one shooter to determine how many hits the other shooter makes. This type of reasoning problem appears frequently in aptitude exams to test logical and quantitative skills together. Given Data / Assumptions: A fires 5 shots for every 3 shots fired by B, so the ratio of shots A : B is 5 : 3. A hits once in every 3 shots, so A kills 1 bird per 3 shots. B hits once in every 2 shots, so B misses 1 shot for every 2 shots fired. B has missed 27 times in total. We must find how many birds A has killed, assuming the pattern remains consistent. Concept / Approach: For B, in every pair of 2 shots: 1 hit and 1 miss occur. So misses and hits for B are equal, and the number of misses tells us the number of such 2 shot cycles. If B has 27 misses, there must be 27 hits as well, and total shots for B are twice the number of cycles. Once we know B's shots, we use the 5 : 3 ratio to find A's number of shots, and then use A's accuracy (1 hit per 3 shots) to determine how many birds A kills. Step-by-Step Solution: Step 1: For B, each 2 shots contain 1 hit and 1 miss. Step 2: Therefore, the number of misses equals the number of hits for B. Step 3: Given B has missed 27 times, B has also hit 27 times. Step 4: Each pair of 2 shots produces 1 miss and 1 hit, so the number of pairs is 27. Step 5: Total shots fired by B = 2 * 27 = 54. Step 6: Using the ratio A : B = 5 : 3 for shots, if B fires 54 shots, the corresponding shots for A are: A's shots = 54 * (5 / 3) = 54 * 5 / 3 = 18 * 5 = 90. Step 7: A hits once in every 3 shots, so A's number of hits is 90 / 3 = 30. Step 8: Thus A has killed 30 birds. Verification / Alternative check: We can verify by summarizing the outcomes. B fired 54 shots and, by design, hit 27 birds and missed 27 birds. A fired 90 shots and hit 30 birds, missing 60. The ratio of shots A : B remains 90 : 54, which simplifies to 5 : 3, consistent with the given condition. The number of misses for B is exactly 27 as required, confirming that the calculations are correct. Why Other Options Are Wrong: 60 birds, 72 birds, 90 birds, and 45 birds all correspond to different relationships between A's shots and B's misses and would break either the 5 : 3 shot ratio or A's success rate of 1 hit per 3 shots. Only 30 birds aligns with all constraints in the question. Common Pitfalls: A frequent mistake is to misinterpret the phrase hits once in every 2 shots for B and think this means 2 hits in 1 shot cycle. Others ignore the ratio of shots and try to work with absolute numbers prematurely. Always translate verbal conditions into simple ratios and success rates, and then link them step by step. Tracking shots, hits, and misses separately for each shooter avoids confusion. Final Answer: When B has missed 27 times, A has killed 30 birds.

Question 12

In aptitude (Algebra and Custom Operations), a custom binary operation is defined as a * b = 2a - 3b + ab. Using this definition, what is the value of (3 * 5) + (5 * 3)?

Accepted Answer

Introduction / Context: This problem introduces a custom binary operation defined on real numbers. The question is not about standard arithmetic but about correctly applying the given rule. In many aptitude and reasoning tests, such custom operations are used to check whether you read definitions carefully and substitute values accurately instead of assuming usual multiplication or addition rules. Given Data / Assumptions: The operation is defined as a * b = 2a - 3b + ab. We need to compute (3 * 5) + (5 * 3). All calculations are done with real numbers using this definition. There is no assumption that * is commutative; we must evaluate both orders separately. Concept / Approach: The key idea is to treat the definition a * b = 2a - 3b + ab as a formula. To find 3 * 5, substitute a = 3 and b = 5. To find 5 * 3, substitute a = 5 and b = 3. Then add the two results using ordinary addition. It is important not to swap a and b casually because the expression uses a and b in different ways, which generally makes the operation non commutative. Step-by-Step Solution: Step 1: Compute 3 * 5 using the definition a * b = 2a - 3b + ab. Step 2: For 3 * 5, take a = 3 and b = 5. Step 3: Substitute: 3 * 5 = 2 * 3 - 3 * 5 + 3 * 5. Step 4: Calculate: 2 * 3 = 6, 3 * 5 = 15. Step 5: So 3 * 5 = 6 - 15 + 15 = 6. Step 6: Now compute 5 * 3 with a = 5 and b = 3. Step 7: Substitute: 5 * 3 = 2 * 5 - 3 * 3 + 5 * 3. Step 8: Calculate: 2 * 5 = 10, 3 * 3 = 9, and 5 * 3 = 15. Step 9: So 5 * 3 = 10 - 9 + 15 = 16. Step 10: Finally, compute (3 * 5) + (5 * 3) = 6 + 16 = 22. Verification / Alternative check: We can double check each step carefully. For 3 * 5, plug directly into the formula: 2a - 3b + ab = 2 * 3 - 3 * 5 + 3 * 5. The last two terms cancel, leaving 6. For 5 * 3, apply 2a - 3b + ab = 2 * 5 - 3 * 3 + 5 * 3 = 10 - 9 + 15 = 16. There is no simplification mistake, so the sum 6 + 16 = 22 is confirmed. Why Other Options Are Wrong: 23, 24, 25, and 28 all arise from possible arithmetic or substitution errors such as mixing up a and b, miscomputing products, or using 3 * 5 as standard multiplication instead of the defined operation. Once the definition is applied correctly to both pairs, only the value 22 fits the calculation. Common Pitfalls: A frequent mistake is to treat * as ordinary multiplication, quickly computing 3 * 5 = 15 and 5 * 3 = 15 and then adding to get 30. Others reverse the roles of a and b or forget to distribute the minus sign correctly in the expression 2a - 3b + ab. Always rewrite the formula and substitute step by step for each pair to avoid confusion in custom operation problems. Final Answer: The value of (3 * 5) + (5 * 3) under the given custom operation is 22.

Question 13

In aptitude (Algebraic Identities), if a + b + c = 13 and a^2 + b^2 + c^2 = 69, what is the value of the expression (ab + bc + ca)?

Accepted Answer

Introduction / Context: This problem checks your familiarity with standard algebraic identities involving sums of variables and sums of their squares. Rather than finding individual values of a, b, and c, the question asks for a combined symmetric expression ab + bc + ca. Such questions are frequently used in aptitude and algebra sections to test conceptual understanding of identities and manipulation skills. Given Data / Assumptions: a + b + c = 13. a^2 + b^2 + c^2 = 69. We must find the value of ab + bc + ca. a, b, and c are real numbers, but their exact values are not required. Concept / Approach: The key identity relating the square of a sum to sums of squares and pairwise products is: (a + b + c)^2 = a^2 + b^2 + c^2 + 2(ab + bc + ca). We are given the value of a + b + c and a^2 + b^2 + c^2, so we can substitute them into the identity and solve for ab + bc + ca. This avoids the need to find the individual variables, which may not even be unique. Step-by-Step Solution: Step 1: Start from the identity (a + b + c)^2 = a^2 + b^2 + c^2 + 2(ab + bc + ca). Step 2: Substitute a + b + c = 13, so (a + b + c)^2 = 13^2 = 169. Step 3: Substitute a^2 + b^2 + c^2 = 69 into the identity. Step 4: The equation becomes 169 = 69 + 2(ab + bc + ca). Step 5: Subtract 69 from both sides: 169 - 69 = 2(ab + bc + ca). Step 6: Compute 169 - 69 = 100. Step 7: So 2(ab + bc + ca) = 100. Step 8: Divide by 2: ab + bc + ca = 100 / 2 = 50. Step 9: Hence, the required value is 50. Verification / Alternative check: To check consistency, we can attempt constructing one possible triple (a, b, c) that fits the conditions, though it is not necessary. However, the algebraic identity is standard and holds for all real numbers, so once we have substituted correct numeric values, the result is reliable. This kind of cross check reassures us that there is no arithmetic mistake in squaring 13 or subtracting 69 from 169. Why Other Options Are Wrong: 30, 10, 70, and 40 all fail to satisfy the identity when plugged back into 69 + 2(ab + bc + ca). Only ab + bc + ca = 50 makes a^2 + b^2 + c^2 + 2(ab + bc + ca) equal to 169, which is the square of 13. Common Pitfalls: Students sometimes use the wrong identity or forget the factor 2 in front of ab + bc + ca. Another typical mistake is to compute 13^2 incorrectly or to make subtraction errors with 169 and 69. Carefully writing the identity and each arithmetic step significantly reduces these errors in exam settings. Final Answer: The value of the expression ab + bc + ca is 50.

Question 14

In aptitude (Fractions and Reciprocals), find the difference between the number 18 and its reciprocal. That is, compute 18 - (1 / 18) and express the result as a simplified improper fraction.

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Introduction / Context: This question involves basic operations with fractions and reciprocals. You must subtract a small fraction, which is the reciprocal of a whole number, from that whole number and express the result as a simplified fraction. Such problems test your comfort with expressing integers as fractions and performing accurate subtraction using a common denominator. Given Data / Assumptions: The whole number is 18. The reciprocal of 18 is 1 / 18. We must find 18 - (1 / 18). The final answer must be written as a simplified improper fraction. Concept / Approach: To subtract a fraction from a whole number, it is convenient to express the whole number as a fraction with the same denominator. Since the reciprocal has denominator 18, we convert 18 into a fraction with denominator 18: 18 = 18 * (18 / 18) = 324 / 18. Then the subtraction 18 - (1 / 18) becomes: 324 / 18 - 1 / 18. Now we subtract the numerators while keeping the common denominator, and simplify if possible. Step-by-Step Solution: Step 1: Write 18 as a fraction with denominator 18: 18 = 324 / 18. Step 2: The expression becomes 324 / 18 - 1 / 18. Step 3: Combine the fractions with common denominator 18. Step 4: Subtract numerators: 324 - 1 = 323. Step 5: So the result is 323 / 18. Step 6: Check if 323 and 18 share any common factor. 323 is 17 * 19, and 18 is 2 * 3 * 3, so they have no common factors. Step 7: Therefore 323 / 18 is already in simplest form. Verification / Alternative check: We can convert 323 / 18 into a mixed fraction to verify that it is slightly less than 18. Divide 323 by 18: 18 * 17 = 306, and 323 - 306 = 17. So 323 / 18 = 17 and 17 / 18. This is 17 17/18, which is indeed a bit less than 18 as expected, because we subtracted a positive amount 1 / 18 from 18. The direction and magnitude of the change make sense. Why Other Options Are Wrong: 305/18: This would correspond to subtracting 19 / 18 instead of 1 / 18 from 18. 323/16 and 324/18: These have incorrect denominators or numerators and do not match the operation 18 - 1 / 18. 17 1/18: As a mixed fraction this equals (17 * 18 + 1) / 18 = 307 / 18, which is not equal to 323 / 18 and therefore not correct. Common Pitfalls: A common mistake is to forget to convert the integer into a fraction with the same denominator, leading to invalid operations on numerators and denominators. Another frequent issue is arithmetic slips in large numerators like 324 or 323. Writing the intermediate fraction clearly and checking the mixed number equivalent helps catch such errors. Final Answer: The difference between 18 and its reciprocal is 323/18.

Question 15

In aptitude (Ratio and Work Problems), 25 buckets of water are required to fill a tank using the current bucket size. If the bucket capacity is reduced to two fifths of its present capacity, how many buckets will be required to fill the same tank?

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Introduction / Context: This question involves proportional reasoning about capacity and quantity. It describes a tank filled using a bucket of a certain size, and then asks how many buckets are needed if each bucket becomes smaller. Problems of this type are common in aptitude tests and help you practice direct inverse proportionality between capacity and number of units required to perform the same job. Given Data / Assumptions: Initially, 25 buckets of water are needed to fill the tank. Each original bucket has some capacity, say V units. The new bucket has capacity equal to two fifths of the original capacity, that is (2 / 5) * V. The tank volume remains the same in both scenarios. We must find the number of new buckets required. Concept / Approach: The total volume of the tank can be expressed in terms of the original bucket as: Tank volume = 25 * V. With the smaller bucket, each bucket carries only two fifths of V, but the tank volume is unchanged. If N new buckets are needed, then: N * (2 / 5) * V = 25 * V. Since V is nonzero, it cancels out, and we can solve the resulting equation for N. This is a simple example of inverse proportional reasoning: reducing the capacity by a factor causes the number of buckets to increase by the reciprocal factor. Step-by-Step Solution: Step 1: Let the original bucket capacity be V units. Step 2: Total tank volume in terms of V is 25 * V. Step 3: New bucket capacity is (2 / 5) * V. Step 4: Let N be the number of new buckets required, so N * (2 / 5) * V = 25 * V. Step 5: Cancel V from both sides: N * (2 / 5) = 25. Step 6: Solve for N: N = 25 * (5 / 2) = 125 / 2. Step 7: Compute 125 / 2 = 62.5. Step 8: Therefore, 62.5 new buckets are required to fill the tank. Verification / Alternative check: We can verify with actual units. Assume V = 10 litres for convenience. Then the tank volume is 25 * 10 = 250 litres. New bucket capacity is (2 / 5) * 10 = 4 litres. To supply 250 litres with 4 litre buckets, we need 250 / 4 = 62.5 buckets. This matches our earlier calculation and is consistent with the idea that a smaller bucket means more trips. Why Other Options Are Wrong: 52.5, 72.5, 82.5, and 50 do not satisfy the basic volume equation N * (2 / 5) = 25. Only N = 62.5 yields the required total of 25V units of water. Any other choice either under fills or overfills the tank based on the assumed capacity. Common Pitfalls: Students sometimes scale the number of buckets incorrectly, for example by multiplying by two fifths instead of dividing, or they confuse whether capacity and number vary directly or inversely. Remember that when the size of each unit decreases, you need more units to cover the same total volume. Writing out the equation with an abstract capacity V helps keep the relationships clear. Final Answer: When the bucket capacity is reduced to two fifths of its original size, the number of buckets required to fill the tank becomes 62.5.

Question 16

A water tank is initially 2/5 full of water. When 16 litres of water are added to the tank, the level rises so that it becomes 6/7 full of its total capacity. What is the full capacity of the water tank, in litres?

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Introduction / Context: This aptitude question tests your understanding of fractions, linear equations, and the idea of comparing two different fill levels of the same water tank. Problems like this appear frequently in bank exams, placement tests, and general competitive exams because they check whether you can translate a word problem into a simple algebraic equation. The key idea is that the capacity of the tank remains fixed, while the amount of water inside changes when more water is added. By carefully relating the fractional levels of 2/5 and 6/7 and the added 16 litres, we can determine the full capacity of the tank. Given Data / Assumptions: The water tank is initially 2/5 full. After adding 16 litres of water, the tank becomes 6/7 full. The capacity of the tank is the same in both situations. We assume no leakage or loss of water. Concept / Approach: The concept used is simple linear algebra based on fractions of the same whole. Let the full capacity of the tank be C litres. Initially the volume of water is (2/5) * C. After adding 16 litres, the volume becomes (6/7) * C. Since the only change between the two situations is the addition of 16 litres, the difference between the later amount and the initial amount must be exactly 16 litres. This gives us a single linear equation in the unknown C, which can then be solved easily. Step-by-Step Solution: Let the full capacity of the tank be C litres. Initial quantity of water = (2/5) * C litres. Final quantity of water after adding 16 litres = (6/7) * C litres. According to the problem, initial quantity + 16 = final quantity. So, (2/5) * C + 16 = (6/7) * C. Bring the fractional terms to one side: 16 = (6/7) * C - (2/5) * C. Compute the difference using a common denominator of 35: (6/7) * C = (30/35) * C and (2/5) * C = (14/35) * C. Therefore, (6/7) * C - (2/5) * C = (30/35 - 14/35) * C = (16/35) * C. So, 16 = (16/35) * C. Solve for C by multiplying both sides by 35 and dividing by 16. C = 16 * 35 / 16 = 35 litres. Verification / Alternative check: We can quickly check the result by substituting C = 35 litres back into the original conditions. Initially, the tank has (2/5) * 35 = 14 litres of water. After adding 16 litres, the volume becomes 14 + 16 = 30 litres. Now calculate 6/7 of the full capacity: (6/7) * 35 = 30 litres. Since the final amount of water matches 6/7 of the capacity exactly, the value C = 35 litres is fully consistent with the given data, confirming that the calculation is correct. Why Other Options Are Wrong: 28 litres: For capacity 28 litres, 2/5 of 28 is 11.2 litres, and adding 16 gives 27.2 litres, which is not 6/7 of 28 (which is 24 litres). 32 litres: For capacity 32 litres, 2/5 of 32 is 12.8 litres. Adding 16 gives 28.8 litres, while 6/7 of 32 is 27.428571 litres, so it does not match. 42 litres: For capacity 42 litres, 2/5 of 42 is 16.8 litres, and adding 16 gives 32.8 litres, but 6/7 of 42 is 36 litres, so this is also inconsistent. 30 litres: For capacity 30 litres, 2/5 of 30 is 12 litres, and adding 16 gives 28 litres, whereas 6/7 of 30 is 25.714285 litres, so this option fails too. Common Pitfalls: Many learners mistakenly set up the equation using 2/5 of the difference between levels instead of applying the fractions to the full capacity. Another common mistake is to treat the denominators 5 and 7 casually and introduce arithmetic errors when finding a common denominator. Some students also incorrectly add fractions of different denominators directly without converting them to a common base. Careful algebraic manipulation and checking the equation after writing it are essential to avoid these errors. Final Answer: Thus, the full capacity of the water tank is 35 litres.

Question 17

In a regular week there are 5 working days, each with 8 regular working hours. A man is paid Rs 2.40 per hour for regular hours and Rs 3.20 per hour for overtime. In 4 such weeks he earns a total of Rs 432. For how many hours in total, including bot…

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Introduction / Context: This problem is a classic wages and work question from quantitative aptitude. It combines the concept of regular working hours with overtime and different hourly rates. Such problems are very common in bank exams, SSC exams, and placement tests, and they are meant to check whether you can construct a linear equation from a word description involving time, rate of pay, and total earnings. The key idea is to separate regular hours, which are fixed for 4 weeks, from overtime hours, which are unknown, and then use the total income to find the extra hours worked. Given Data / Assumptions: There are 5 working days in a regular week. Each working day has 8 regular working hours. The worker is paid Rs 2.40 per regular hour. The worker is paid Rs 3.20 per overtime hour. The period considered is 4 such weeks. Total earnings in 4 weeks are Rs 432. Concept / Approach: The approach is to first calculate the total regular working hours in 4 weeks and the corresponding income from those hours. Then, one introduces a variable for the overtime hours, calculates the overtime earnings in terms of that variable, and sets the sum equal to the known total income. This results in a simple linear equation with one unknown, namely the overtime hours. After solving for the overtime hours, we add them to the fixed regular hours to obtain the total hours worked in 4 weeks. Step-by-Step Solution: Regular hours in one week = 5 days * 8 hours per day = 40 hours. Regular hours in 4 weeks = 4 * 40 = 160 hours. Let the total overtime hours in 4 weeks be x hours. Regular pay for 160 hours = 160 * 2.40 = Rs 384. Overtime pay for x hours = x * 3.20. Total earnings are given as 432, so: 384 + 3.20 * x = 432. Subtract 384 from both sides: 3.20 * x = 432 - 384 = 48. So, x = 48 / 3.20 = 15 hours. Total hours worked in 4 weeks = regular hours + overtime hours = 160 + 15 = 175 hours. Verification / Alternative check: To verify, compute the income from 175 total hours, with the first 160 counted at the regular rate and the remaining 15 as overtime. Regular income remains 160 * 2.40 = 384. Overtime income is 15 * 3.20 = 48. The total income is therefore 384 + 48 = 432, which matches the given total earning. This confirms that the computed overtime hours and thus the total working hours are correct. Why Other Options Are Wrong: 160 hours: This would correspond to no overtime at all, giving only 160 * 2.40 = Rs 384, which is less than 432. 180 hours: This implies 20 overtime hours, giving extra 20 * 3.20 = 64, so total earnings would be 384 + 64 = 448, which is too high. 195 hours: This means 35 overtime hours and earnings 384 + 35 * 3.20 = 496, again not 432. 200 hours: This would mean 40 overtime hours and total income 384 + 40 * 3.20 = 512, far from the given value. Common Pitfalls: A common mistake is to treat all hours as if they are paid at a single rate, which ignores the concept of overtime and leads to a wrong equation. Another frequent error is miscounting the number of regular hours by forgetting that there are 4 weeks or miscalculating 5 days * 8 hours. Some students also incorrectly divide by 2.4 instead of 3.2 when solving for the overtime hours. Careful reading of the rates and correct identification of the unknown variable are essential. Final Answer: The man works a total of 175 hours in the 4 weeks, including both regular and overtime hours.

Question 18

A bag contains a total of 324 coins consisting only of 20 paise coins and 25 paise coins. The total value of all the coins together is Rs 71. How many 20 paise coins are there in the bag?

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Introduction / Context: This question is a standard mixture and coins problem from quantitative aptitude. It involves two denominations of coins, their total count, and their combined value. Such problems test your ability to convert a verbal description into algebraic equations, to handle paise and rupees correctly, and to solve simultaneous linear equations. These are very common in competitive exams, where candidates must quickly set up equations and find the correct number of each type of coin. Given Data / Assumptions: Total number of coins in the bag: 324. The bag contains only 20 paise coins and 25 paise coins. Total value of all coins = Rs 71. 1 rupee = 100 paise. We assume every 20 paise coin and every 25 paise coin is genuine and identical in value within its type. Concept / Approach: The core approach is to let the number of 20 paise coins and 25 paise coins be variables and then write two equations. The first equation is based on the total number of coins. The second equation is based on the total value, expressed in paise to avoid decimals. After that, we solve the two simultaneous equations to determine the number of each type of coin. Finally, we read off how many 20 paise coins are present, which is what the problem asks us to find. Step-by-Step Solution: Let the number of 20 paise coins be x. Let the number of 25 paise coins be y. Total number of coins: x + y = 324. Convert the total value to paise: Rs 71 = 7100 paise. Value equation in paise: 20 * x + 25 * y = 7100. From x + y = 324, express y as y = 324 - x. Substitute into the value equation: 20 * x + 25 * (324 - x) = 7100. Simplify: 20 * x + 25 * 324 - 25 * x = 7100. This gives: 20x - 25x + 8100 = 7100, so -5x + 8100 = 7100. Rearrange: -5x = 7100 - 8100 = -1000, so x = 200. Therefore, the number of 20 paise coins is 200. Verification / Alternative check: We can verify the numbers by computing the value again. If x = 200, then y = 324 - 200 = 124. Value of 20 paise coins = 200 * 20 = 4000 paise. Value of 25 paise coins = 124 * 25 = 3100 paise. Total value = 4000 + 3100 = 7100 paise, which equals Rs 71. This matches the given total value, so the solution is correct. Why Other Options Are Wrong: 50 coins: If there were only 50 coins of 20 paise, the remaining 274 coins would be 25 paise; the resulting value would not match Rs 71. 100 coins: With 100 coins of 20 paise and 224 coins of 25 paise, the total value exceeds Rs 71. 150 coins: With 150 and 174 coins of each type, the total value again does not equal Rs 71. 175 coins: Similar checking shows that using 175 coins of 20 paise does not give the correct total value either. Only 200 fits both equations exactly. Common Pitfalls: A very common mistake is to mix rupees and paise in the same equation without converting to one unit, which leads to incorrect arithmetic. Another error is to misinterpret the total number of coins as the total value, or to incorrectly set up the second equation. Some students also forget that the total value must be exactly Rs 71, not approximately. Using paise throughout and proceeding carefully with algebra avoids these pitfalls. Final Answer: The bag contains 200 coins of 20 paise each.

Question 19

Simplify the expression (1.5^3 + 4.7^3 + 3.8^3 - 3 * 1.5 * 4.7 * 3.8) divided by (1.5^2 + 4.7^2 + 3.8^2 - 1.5 * 4.7 - 4.7 * 3.8 - 1.5 * 3.8) by using an algebraic identity and find its numerical value.

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Introduction / Context: This is an algebraic simplification problem that checks whether you recognise and correctly apply a standard identity for the sum of cubes. Instead of expanding all powers and computing a very long numerical expression, the problem becomes easy if you see the pattern. This type of question is common in aptitude and algebra sections because it rewards conceptual understanding of identities rather than brute force calculation. Given Data / Assumptions: The expression in the numerator is 1.5^3 + 4.7^3 + 3.8^3 - 3 * 1.5 * 4.7 * 3.8. The expression in the denominator is 1.5^2 + 4.7^2 + 3.8^2 - 1.5 * 4.7 - 4.7 * 3.8 - 1.5 * 3.8. All numbers are real and standard arithmetic rules apply. We assume the denominator is not zero so the division is valid. Concept / Approach: The numerator and denominator match the well known algebraic identity involving three variables a, b, and c. The identity is: a^3 + b^3 + c^3 - 3abc = (a + b + c) * (a^2 + b^2 + c^2 - ab - bc - ca). This means that the given numerator is exactly equal to (a + b + c) multiplied by the denominator. Therefore, when we divide the numerator by the denominator, the complicated factor cancels out and the result is simply a + b + c. We only need to identify a, b, and c and then compute their sum. Step-by-Step Solution: Compare the numerator with a^3 + b^3 + c^3 - 3abc. We see that a = 1.5, b = 4.7, and c = 3.8. The denominator matches a^2 + b^2 + c^2 - ab - bc - ca. By the identity, a^3 + b^3 + c^3 - 3abc = (a + b + c) * (a^2 + b^2 + c^2 - ab - bc - ca). Therefore, the given fraction is: [(a^3 + b^3 + c^3 - 3abc)] / [a^2 + b^2 + c^2 - ab - bc - ca] = a + b + c. Compute the sum a + b + c = 1.5 + 4.7 + 3.8. First, 1.5 + 4.7 = 6.2. Then, 6.2 + 3.8 = 10. So, the value of the expression is 10. Verification / Alternative check: A direct verification method is to approximate both numerator and denominator numerically and then divide. Although it is tedious to compute each cube and each product exactly by hand, a calculator would confirm that the numerator is equal to 10 times the denominator. This agrees perfectly with the identity based solution. The identity method is preferred in exams, because it is much faster and avoids arithmetic errors with decimals. Why Other Options Are Wrong: 6 and 8: These values are smaller than the true sum a + b + c and would correspond to an incorrect or partial application of the identity, or to miscalculation in adding the decimal numbers. 12 and 14: These are larger than the correct sum and might be guessed by students who think that the identity produces something like 2(a + b + c) or a more complicated factor. Correct use of the identity gives exactly a + b + c, which is 10. Common Pitfalls: A very common mistake is to not recognise the identity and start expanding each cube and product manually. This is time consuming and increases the chance of making arithmetic errors, especially with decimal numbers. Another pitfall is to misremember the identity, for example writing the denominator incorrectly. Also, some students add 1.5, 4.7, and 3.8 incorrectly by misplacing the decimal points. Careful use of the identity and correct addition of decimals avoids these issues. Final Answer: The simplified value of the given expression is 10.

Question 20

In a cricket innings, the highest individual score is equal to 3/11 of the total team score. The next highest individual score is equal to 3/11 of the remaining runs after subtracting the highest score from the total. If the difference between the h…

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Introduction / Context: This question comes from sports based arithmetic, often used to make ratio and fraction topics more interesting. It involves two players scores that are given as fractions of the total team score and of the remaining runs. By converting these statements into algebra, we can find the total score. This type of problem tests your skill in writing expressions with fractions and solving a linear equation based on a given difference between two quantities. Given Data / Assumptions: Total team score is T runs. Highest individual score = 3/11 of T. Remaining runs after removing the highest score = T - 3/11 T. Next highest score = 3/11 of the remaining runs. Difference between highest and next highest scores is 9 runs. All scores are assumed to be whole numbers of runs. Concept / Approach: We introduce a variable T for the total team score and express both the highest and next highest scores in terms of T. Since the next highest is taken as a fraction of the remaining runs, we must carefully subtract the highest score before applying the fraction. The relationship that their difference is 9 runs gives us a straightforward equation in T. Solving that equation provides the total team score. The fractions and arithmetic must be handled with care to avoid mistakes. Step-by-Step Solution: Let the total team score be T runs. Highest score H = (3/11) * T. Remaining runs after removing H = T - (3/11) * T. Compute remaining runs: T - 3T/11 = (11T/11 - 3T/11) = 8T/11. Next highest score S = (3/11) of remaining runs = (3/11) * (8T/11) = 24T/121. Given that H - S = 9 runs. So, (3/11) * T - 24T/121 = 9. Convert 3/11 to a fraction with denominator 121: 3/11 = 33/121. Thus, H - S = (33T/121) - (24T/121) = 9T/121. So, 9T/121 = 9. Multiply both sides by 121/9 to get T = 9 * 121 / 9 = 121. Therefore, the total team score is 121 runs. Verification / Alternative check: Check the internal values to be sure. If T = 121, then the highest score H = 3/11 * 121 = 33 runs. Remaining runs = 121 - 33 = 88. Next highest score S = 3/11 of 88 = 3 * 8 = 24 runs. Difference between highest and next highest = 33 - 24 = 9 runs, which matches the given information exactly. This confirms that the total team score of 121 runs is correct. Why Other Options Are Wrong: 110 runs: If T = 110, then H = 30 and the remaining runs are 80. S would be 3/11 of 80, which is not an integer equal to 21, and the difference is not 9. 132 runs: For T = 132, H = 36 and remaining runs are 96, so S = 3/11 of 96 which is not 27, and the difference does not become 9. 143 and 154 runs: Similar substitution shows that the difference between highest and next highest does not equal 9 for these totals either. Common Pitfalls: Many learners mistakenly take both scores directly as fractions of the total score, ignoring that the second one is a fraction of the remaining runs. This changes the equation and leads to a wrong total value. Another pitfall is mismanaging the fraction arithmetic, especially the conversion of 3/11 to 33/121. Some students also forget to apply the correct order of operations when subtracting. Writing out each step clearly with a common denominator helps avoid these mistakes. Final Answer: The total team score in the innings was 121 runs.

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