Question 1

Polynomials with the same remainder (Recovery applied) If the polynomials ax^3 + 4x^2 − 3x − 4 and x^3 − 4x + a leave the same remainder when divided by (x + 3), find the value of a.

Accepted Answer

Introduction / Context:Two polynomials leave the same remainder upon division by a linear factor if and only if they take the same value at the corresponding root. Recovery-First: the original stem with (x − 3) yields a non-integer a not in options; replacing by (x + 3) corrects the standard intent and aligns with the answer set. Given Data / Assumptions: P1(x) = a x^3 + 4x^2 − 3x − 4. P2(x) = x^3 − 4x + a. Same remainder on division by (x + 3) ⇒ evaluate both at x = −3. Concept / Approach:Remainder Theorem: remainder of division by (x − r) equals P(r). Equal remainders ⇒ P1(r) = P2(r). Step-by-Step Solution:Compute P1(−3) = a(−27) + 4·9 − 3(−3) − 4 = −27a + 36 + 9 − 4 = −27a + 41.Compute P2(−3) = (−27) − 4(−3) + a = −27 + 12 + a = −15 + a.Set equal: −27a + 41 = −15 + a ⇒ −28a = −56 ⇒ a = 2. Verification / Alternative check:At x = −3, both polynomials evaluate to −13, confirming the equal remainder. Why Other Options Are Wrong:−1, 3, 4, 0 do not satisfy the equality at x = −3. Common Pitfalls:Using x = 3 (from the uncorrected stem) yields a non-listed fractional a; always tie the evaluation point to the divisor's root. Final Answer:2

Question 2

Error analysis on fractions: “fifth of a number” vs multiplying by its reciprocal A student should have found (2/5) of a number but instead multiplied the number by (5/2). His answer exceeded the correct one by 5208. What was the number?

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Introduction / Context:This tests proportional reasoning and careful reading. The intended operation is 2/5 of N; the mistaken operation is 5/2 times N. Their difference equals 5208. Given Data / Assumptions: Correct value: (2/5)N. Wrong value: (5/2)N. Wrong − Correct = 5208. Concept / Approach:Set up a single linear equation in N using the two fractions and solve cleanly. Keep everything symbolic until the last step. Step-by-Step Solution:(5/2)N − (2/5)N = 5208.Compute the difference factor: 5/2 − 2/5 = (25 − 4)/10 = 21/10.Thus (21/10)N = 5208 ⇒ N = 5208 * 10 / 21.Since 21 × 248 = 5208, N = 248 × 10 = 2480. Verification / Alternative check:Correct value: (2/5)×2480 = 992; wrong value: (5/2)×2480 = 6200; difference = 5208 as required. Why Other Options Are Wrong:2455, 2456, 2485, 2460 do not satisfy the exact fractional equation. Common Pitfalls:Adding instead of subtracting the fractions; miscomputing 5/2 − 2/5; cancellation errors. Final Answer:2480

Question 3

Mixed-speed travel over fixed time A man covers 18 km in 5 hours partly walking at 2 km/h and partly running at 6 km/h. How many kilometres does he cover by running?

Accepted Answer

Introduction / Context:We have a time split between two constant speeds. With total time and total distance known, set up linear equations in walking time and running time, then compute the running distance. Given Data / Assumptions: Total time = 5 h. Walking speed = 2 km/h; running speed = 6 km/h. Total distance = 18 km. Concept / Approach:Let r = hours of running and w = hours of walking. Then r + w = 5 and 6r + 2w = 18. Solve for r, then compute running distance 6r. Step-by-Step Solution:r + w = 5 ⇒ w = 5 − r.6r + 2w = 18 ⇒ 6r + 2(5 − r) = 18 ⇒ 6r + 10 − 2r = 18.4r = 8 ⇒ r = 2 hours.Running distance = 6 × r = 12 km. Verification / Alternative check:Walking time = 3 h ⇒ walking distance = 2 × 3 = 6 km; total = 6 + 12 = 18 km, consistent. Why Other Options Are Wrong:15 km or 10 km would violate either distance or time totals; 9 km corresponds to r = 1.5 h, not meeting the 18 km total. Common Pitfalls:Confusing time-weighted and distance-weighted averages; forgetting that times (not distances) must sum to 5 hours. Final Answer:12 km

Question 4

Solve for initial counts from two “give 5 oranges” scenarios Ramesh and Ganesh have some oranges. If Ramesh gives 5 to Ganesh, Ganesh will have thrice as many as Ramesh. If Ganesh gives 5 to Ramesh instead, they will then have equal numbers. What is…

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Introduction / Context:Two hypothetical transfers create two linear conditions on the initial counts. Solving them yields the initial numbers and hence their ratio. Given Data / Assumptions: Let initial counts be R (Ramesh) and G (Ganesh). Scenario 1: R gives 5 to G ⇒ (G + 5) = 3(R − 5). Scenario 2: G gives 5 to R ⇒ (R + 5) = (G − 5). Concept / Approach:Translate each scenario into an equation, then solve the pair for R and G. Finally reduce R:G to simplest form. Step-by-Step Solution:From scenario 2: R + 5 = G − 5 ⇒ G = R + 10.Substitute in scenario 1: (R + 10 + 5) = 3(R − 5) ⇒ R + 15 = 3R − 15.30 = 2R ⇒ R = 15; hence G = R + 10 = 25.Initial ratio R:G = 15:25 = 3:5. Verification / Alternative check:Check scenario 1: after transfer, R = 10, G = 30 ⇒ G is thrice R. Scenario 2: after transfer, both have 20 ⇒ equal. Why Other Options Are Wrong:1:3, 5:7, 7:9, 2:3 do not satisfy both transfer conditions simultaneously. Common Pitfalls:Placing the “give 5” on the wrong side; forgetting to update both people's counts in each scenario. Final Answer:3 : 5

Question 5

Two-digit number from digit constraints and a swap-based condition A two-digit number has a ten's digit that exceeds twice its unit's digit by 2. The number formed by interchanging its digits is 5 more than three times the sum of the digits. Find th…

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Introduction / Context:This two-digit number problem couples a digit relation with a condition on the swapped number. We form equations in the ten's digit t and unit's digit u to solve cleanly. Given Data / Assumptions: Original number N = 10t + u. Digit relation: t = 2u + 2. Swapped number N' = 10u + t. N' = 3(t + u) + 5. Concept / Approach:Use the swap condition to link t and u, then substitute the digit relation to obtain a simple linear equation in u. Compute t and build the number. Step-by-Step Solution:N': 10u + t = 3(t + u) + 5 ⇒ 10u + t = 3t + 3u + 5.Rearrange: 7u − 2t = 5.With t = 2u + 2, substitute: 7u − 2(2u + 2) = 5 ⇒ 7u − 4u − 4 = 5 ⇒ 3u = 9 ⇒ u = 3.Then t = 2u + 2 = 8 ⇒ number = 83. Verification / Alternative check:Sum of digits = 8 + 3 = 11; 3×sum + 5 = 38; the swapped number is indeed 38, which equals 10u + t, confirming consistency. Why Other Options Are Wrong:38 is the swapped number, not the original; 56, 65, 74 do not satisfy both constraints. Common Pitfalls:Misplacing “exceeds twice the unit’s digit by 2”; mixing up which number is original vs. swapped. Final Answer:83

Question 6

An exam contains 60 questions. For each correct answer, a student earns +3 marks; for each wrong answer, 1 mark is deducted. A student attempts all 60 questions and secures a total of 120 marks. How many questions did the student answer correctly?

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Introduction / Context:This problem tests linear equation setup in a negative marking scheme. Each response contributes either a positive score (correct) or a penalty (wrong). Because all questions were attempted, the total number of correct and wrong answers must add up to the total questions. Given Data / Assumptions: Total questions = 60. Marking: +3 for correct; −1 for wrong. All questions attempted ⇒ correct + wrong = 60. Total score obtained = 120. Concept / Approach:Let c be the number of correct answers. Then wrong answers are (60 − c). The total score is computed as score = 3*c − 1*(60 − c). Solve the resulting linear equation for c to find the number of correct answers. Step-by-Step Solution: Let c = number of correct answers.Wrong answers = 60 − c.Total score = 3*c − 1*(60 − c) = 3c − 60 + c = 4c − 60.Given total score = 120 ⇒ 4c − 60 = 120.4c = 180 ⇒ c = 180 / 4 = 45. Verification / Alternative check:With c = 45, wrong = 15. Score = 45*3 − 15*1 = 135 − 15 = 120, which matches the given total, confirming correctness. Why Other Options Are Wrong: 42, 18, 15, 48: Substituting any of these into 3*c − (60 − c) does not yield 120. They fail the score equation. Common Pitfalls: Writing the score as 3c − (60) instead of 3c − (60 − c). Forgetting that all questions were attempted, so wrong = 60 − c. Final Answer:45

Question 7

A person has only ₹1 and ₹2 coins. The total number of coins is 50 and the total amount is ₹75. Find the number of ₹1 coins and ₹2 coins, respectively.

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Introduction / Context:Classic coin-count problems involve two variables: the number of lower-denomination coins and higher-denomination coins. You are given both the total count of coins and the total monetary value, which leads to a solvable system of linear equations. Given Data / Assumptions: Total coins = 50. Total value = ₹75. Only denominations used are ₹1 and ₹2 coins. Concept / Approach:Let x be the count of ₹1 coins and y be the count of ₹2 coins. Then x + y = 50 (count equation) and x + 2y = 75 (value equation). Solve simultaneously to get x and y in integers consistent with the constraints. Step-by-Step Solution: x + y = 50 … (1)x + 2y = 75 … (2)Subtract (1) from (2): (x + 2y) − (x + y) = 75 − 50 ⇒ y = 25.From (1): x + 25 = 50 ⇒ x = 25. Verification / Alternative check:Total amount = 25*₹1 + 25*₹2 = 25 + 50 = ₹75; total coins = 25 + 25 = 50. Both conditions satisfied. Why Other Options Are Wrong: 15 and 35; 35 and 15; 30 and 20; 20 and 30: Each pair violates either the total coin count of 50 or the total value of ₹75 when tested. Common Pitfalls: Mistakenly using 2x + y = 75 instead of x + 2y = 75 (mixing denominations). Not checking that both equations are satisfied simultaneously. Final Answer:25 and 25

Question 8

A rare coin collection initially has a ratio of gold to non-gold coins equal to 1:3. After adding 10 more gold coins, the ratio becomes 1:2. What is the total number of coins in the collection after the addition?

Accepted Answer

Introduction / Context:Ratio problems often require translating a verbal description into algebraic relationships. When a quantity changes and a new ratio is specified, you can set up an equation to solve for the original numbers, then recompute the updated totals. Given Data / Assumptions: Initial ratio gold : non-gold = 1 : 3. 10 gold coins are added. New ratio becomes 1 : 2 (gold : non-gold). Concept / Approach:Let initial gold coins = g; then non-gold = 3g. After adding 10 gold coins, the new ratio is (g + 10) : 3g = 1 : 2. Solve for g, then compute the final total after addition. Step-by-Step Solution: (g + 10) / (3g) = 1 / 2Cross-multiply: 2(g + 10) = 3g2g + 20 = 3g ⇒ g = 20Initial total = g + 3g = 4g = 80After adding 10 gold coins: total = 80 + 10 = 90 Verification / Alternative check:New gold = 20 + 10 = 30; non-gold = 60. Ratio = 30 : 60 = 1 : 2, consistent with the condition. Why Other Options Are Wrong: 80, 60, 50, 70: These do not match the updated ratio when the 10 gold coins are added and checked against non-gold counts. Common Pitfalls: Interpreting “1:2” as non-gold : gold instead of gold : non-gold. Forgetting to add 10 only to gold coins, not to both categories. Final Answer:90

Question 9

If the price of a book is reduced by ₹5, a buyer can purchase 2 more books for the same budget of ₹200. What was the original price per book?

Accepted Answer

Introduction / Context:This is a linear relation between price and quantity purchased under a fixed budget. When the unit price decreases, the number of items bought increases. The relationship can be modeled with a simple rational equation and solved algebraically. Given Data / Assumptions: Budget = ₹200. Original price per book = p (₹). Reduced price per book = p − 5 (₹). At the reduced price, the buyer gets 2 more books than before. Concept / Approach:Number of books originally = 200 / p. Number after reduction = 200 / (p − 5). The statement “2 more” translates to 200/(p − 5) = 200/p + 2. Solve this for p to find an admissible positive solution. Step-by-Step Solution: 200/(p − 5) = 200/p + 2200/(p − 5) − 200/p = 2200( p − (p − 5) ) / (p(p − 5)) = 2 ⇒ 200*5 / (p(p − 5)) = 21000 = 2p(p − 5) ⇒ p(p − 5) = 500p^2 − 5p − 500 = 0 ⇒ Discriminant = 25 + 2000 = 2025 ⇒ √2025 = 45p = (5 + 45)/2 = 25 (positive root; negative root discarded) Verification / Alternative check:Original price ₹25 ⇒ can buy 200/25 = 8 books. Reduced price ₹20 ⇒ can buy 200/20 = 10 books. Increase = 2 as required. Why Other Options Are Wrong: ₹30, ₹20, ₹15, ₹35: Substitution contradicts the “2 more books” condition for a ₹200 budget. Common Pitfalls: Mistakes in setting up the equation (e.g., swapping p and p − 5 in denominators). Accepting a negative price solution from the quadratic (which is inadmissible). Final Answer:₹ 25

Question 10

The sum of the squares of two numbers is 234, and the square of their difference is 144. Find the product of the two numbers.

Accepted Answer

Introduction / Context:This problem uses algebraic identities that relate sums and differences to products. Specifically, knowing a^2 + b^2 and (a − b)^2 enables solving for ab via identity manipulations without finding a and b individually. Given Data / Assumptions: a^2 + b^2 = 234. (a − b)^2 = 144. Need ab. Concept / Approach:Use the identity (a − b)^2 = a^2 + b^2 − 2ab. Rearranging gives 2ab = (a^2 + b^2) − (a − b)^2. Substitute the given values to compute ab directly. Step-by-Step Solution: (a − b)^2 = a^2 + b^2 − 2ab144 = 234 − 2ab2ab = 234 − 144 = 90ab = 90 / 2 = 45 Verification / Alternative check:If desired, we can choose numbers with product 45 that fit the constraints (not necessary here). The identity-based computation is sufficient and exact. Why Other Options Are Wrong: 28, 52, 36, 40: These do not satisfy 2ab = 90; substituting back fails the identity relation. Common Pitfalls: Using (a + b)^2 identity mistakenly instead of (a − b)^2. Sign errors when isolating 2ab. Final Answer:45

Question 11

Consider the polynomial p(x) = x^3 + 2x + 1. (The earlier garbled symbols are corrected here for clarity.) What is the degree of this polynomial?

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Introduction / Context:The degree of a polynomial is the highest power of the variable x with a nonzero coefficient. The original text contained extraneous symbols making it non-polynomial. Under the Recovery-First Policy, we repair it to the standard polynomial p(x) = x^3 + 2x + 1 so the question remains meaningful and solvable without changing the core intent (finding degree). Given Data / Assumptions: Polynomial: p(x) = x^3 + 2x + 1. All terms are powers of x with nonnegative integer exponents. Concept / Approach:Identify the term with the largest exponent of x. The exponent associated with that term is the degree. Constants have degree 0, linear terms degree 1, quadratic terms degree 2, cubic terms degree 3, and so on. Step-by-Step Solution: List term degrees: degree(x^3) = 3; degree(2x) = 1; degree(1) = 0.Highest degree among the terms is 3.Therefore, degree(p) = 3. Verification / Alternative check:No like terms combine to raise the degree; the cubic term remains dominant. Hence 3 is final. Why Other Options Are Wrong: 2, 4, 5, 1: They misread the highest power or treat the corrupted original text as valid; degree is set solely by the x^3 term here. Common Pitfalls: Treating non-polynomial expressions (e.g., 1/x) as part of a polynomial; polynomials cannot have negative exponents. Ignoring the highest-power term when constants or linear terms are present. Final Answer:3

Question 12

A piece of cloth costs ₹120 in total. If the piece had been 2 metres longer and the price per metre ₹2 less, the total cost would remain ₹120 unchanged. Find the original rate per metre of the cloth.

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Introduction / Context:This question mixes two unknowns—original length and original price per metre—under a constant total cost. Changing both length and rate but keeping the product constant leads to a system that can be solved using algebraic manipulation and a simple quadratic equation. Given Data / Assumptions: Total cost (original) = ₹120. Original length = L metres; original rate = r ₹/m. After change: length = L + 2, rate = r − 2. Total cost remains ₹120 after the change. Concept / Approach:Model the total cost as rate * length. Then rL = 120 and (r − 2)(L + 2) = 120. Subtract the two equal-cost expressions to eliminate rL and get a relation between r and L. Use that relation with rL = 120 to solve for r (the required rate). Step-by-Step Solution: rL = 120 … (1)(r − 2)(L + 2) = 120 … (2)Expand (2): rL + 2r − 2L − 4 = 120.Using (1), replace rL with 120 ⇒ 120 + 2r − 2L − 4 = 120.2r − 2L − 4 = 0 ⇒ r − L = 2 ⇒ r = L + 2.From (1): (L + 2)L = 120 ⇒ L^2 + 2L − 120 = 0.Solve: Discriminant = 4 + 480 = 484; √484 = 22.L = (−2 + 22)/2 = 10 (reject negative root). Then r = L + 2 = 12. Verification / Alternative check:Original: 10 m × ₹12/m = ₹120. New: 12 m × ₹10/m = ₹120. Unchanged total cost confirmed. Why Other Options Are Wrong: ₹15, ₹10, ₹8, ₹18: Do not satisfy both equations simultaneously when paired with corresponding lengths; only ₹12 works. Common Pitfalls: Setting up (r − 2)(L + 2) = rL but then incorrectly expanding or cancelling terms. Forgetting to reject negative length solutions. Final Answer:₹ 12

Question 13

Find the exact area (in square units) of the triangle formed by the line 2x − 3y + 6 = 0 together with the coordinate axes.

Accepted Answer

Introduction / Context:Lines that intersect the coordinate axes form a right triangle with the axes. The area of this triangle is 1/2 * |x-intercept| * |y-intercept|. We will find intercepts of 2x − 3y + 6 = 0 and compute the area using absolute values of those intercept lengths. Given Data / Assumptions: Line: 2x − 3y + 6 = 0. Intercepts are taken with respect to the x- and y-axes. Concept / Approach:Set y = 0 to get the x-intercept, and set x = 0 to get the y-intercept. The triangle with axes has legs equal to the magnitudes of these intercepts. Then area = (1/2) * |x-intercept| * |y-intercept|. Step-by-Step Solution: For x-intercept: set y = 0 ⇒ 2x + 6 = 0 ⇒ x = −3 ⇒ |x-intercept| = 3.For y-intercept: set x = 0 ⇒ −3y + 6 = 0 ⇒ y = 2 ⇒ |y-intercept| = 2.Area = (1/2) * 3 * 2 = 3 square units. Verification / Alternative check:Convert to intercept form: x/(−3) + y/2 = 1. Using absolute intercepts, area = (1/2)*3*2 = 3, consistent with the above method. Why Other Options Are Wrong: 1.5 sq. units, 6 sq. units, 9 sq. units: These do not match the product of legs divided by 2 for the given intercepts. “3 sq. units 2” is malformed. Common Pitfalls: Using the signed intercept (−3) directly without taking absolute value for length. Confusing slope-intercept form with intercept computation. Final Answer:3 sq. units

Question 14

The distance between the points (0, −5) and (x, 0) is 13 units. Find the possible value(s) of x.

Accepted Answer

Introduction / Context:The two-dimensional distance formula connects coordinate differences to the straight-line distance between points. Here, one point lies on the y-axis and the other on the x-axis, which simplifies computation because one coordinate in each point is zero. Given Data / Assumptions: Point A = (0, −5). Point B = (x, 0). Distance AB = 13 units. Concept / Approach:Use distance formula: For A(x1, y1) and B(x2, y2), distance d satisfies d^2 = (x2 − x1)^2 + (y2 − y1)^2. Substitute the known values, square the distance, and solve for x, remembering that squaring leads to two symmetric solutions (±) when isolating x. Step-by-Step Solution: 13^2 = (x − 0)^2 + (0 − (−5))^2169 = x^2 + 5^2 = x^2 + 25x^2 = 169 − 25 = 144x = ±√144 = ±12 Verification / Alternative check:Check x = 12: distance^2 = 12^2 + 5^2 = 144 + 25 = 169 ⇒ distance = 13. Similarly for x = −12 because x^2 is the same; both satisfy the condition. Why Other Options Are Wrong: 10, ±10, 12, 13: These do not satisfy x^2 = 144 when substituted into the formula; only ±12 works. Common Pitfalls: Dropping the negative solution and reporting only +12. Using 13 directly as x rather than applying the distance relation. Final Answer:± 12

Question 15

Evaluate the symmetric expression for nearby integers: If x = 997, y = 998, and z = 999, compute the value of x^2 + y^2 + z^2 − xy − yz − zx.

Accepted Answer

Introduction / Context:This problem tests recognition of a standard symmetric identity that simplifies expressions of the form x^2 + y^2 + z^2 − xy − yz − zx. Using the identity avoids large-number squaring and reduces the task to simple differences. Given Data / Assumptions: x = 997 y = 998 z = 999 Target expression: x^2 + y^2 + z^2 − xy − yz − zx Concept / Approach:Use the identity: x^2 + y^2 + z^2 − xy − yz − zx = 1/2 * [ (x − y)^2 + (y − z)^2 + (z − x)^2 ]. This identity is derived by expanding the squares and collecting terms, and it dramatically simplifies computation when the numbers are close together. Step-by-Step Solution: Compute pairwise differences: x − y = 997 − 998 = −1.y − z = 998 − 999 = −1.z − x = 999 − 997 = 2.Squares: (−1)^2 + (−1)^2 + (2)^2 = 1 + 1 + 4 = 6.Apply the identity: 1/2 * 6 = 3. Verification / Alternative check: Direct computation with large squares would be tedious; the identity provides a clean shortcut and the result is a small integer, which is consistent with the tiny gaps between x, y, and z. Why Other Options Are Wrong: 9, 16, 4: These do not match the identity-based calculation using true differences −1, −1, and 2. Common Pitfalls: Forgetting the 1/2 factor in the identity. Sign mistakes in differences, though squares remove signs; the only risk is miscomputing 2^2. Final Answer: 3

Question 16

Algebra under the condition a + b + c = 0: Find the multiplier k such that a^2 + b^2 + c^2 = k * (ab + bc + ca).

Accepted Answer

Introduction / Context:This question checks a well-known identity that holds when the linear sum a + b + c equals zero. It ties the sum of squares to the sum of pairwise products, a relationship that often appears in symmetric algebra and factorization problems. Given Data / Assumptions: a + b + c = 0 We seek k such that a^2 + b^2 + c^2 = k * (ab + bc + ca). Concept / Approach:Start with (a + b + c)^2. Expanding yields a^2 + b^2 + c^2 + 2(ab + bc + ca). Under the given condition, the left side is 0, which directly relates the two target sums and allows k to be read off immediately after rearrangement. Step-by-Step Solution: (a + b + c)^2 = a^2 + b^2 + c^2 + 2(ab + bc + ca).Given a + b + c = 0 ⇒ 0 = a^2 + b^2 + c^2 + 2(ab + bc + ca).Thus a^2 + b^2 + c^2 = −2(ab + bc + ca).Therefore k = −2. Verification / Alternative check: Pick a simple triple with sum zero, e.g., a = 1, b = 1, c = −2. Then a^2 + b^2 + c^2 = 1 + 1 + 4 = 6 and ab + bc + ca = 1 − 2 − 2 = −3. Indeed 6 = (−2) * (−3). Why Other Options Are Wrong: 2, 1, 0: None satisfy the derived identity for generic nonzero triples with a + b + c = 0. Common Pitfalls: Losing the sign on the 2(ab + bc + ca) term during rearrangement. Assuming k must be positive; here it is negative. Final Answer: -2

Question 17

Algebraic word problem with staged usage: Arjun carries some arrows. He uses half of them to cut incoming arrows, uses 6 more arrows to kill the charioteer, then uses 1 arrow each to knock down the chariot, the flag, and the bow (3 arrows). Finally,…

Accepted Answer

Introduction / Context:This is a classic number puzzle modeled as an algebraic equation. It uses staged consumption (half of the arrows, then fixed amounts, then an amount depending on the square root of the original total) to form a solvable equation in the original number of arrows, n. Given Data / Assumptions: Half of n are used first: n/2. Six more arrows are used next: +6. Then 1 arrow each for chariot, flag, bow: +3 total. Finally, 4*sqrt(n) + 1 arrows are used. All arrows are exhausted by these uses. Concept / Approach:Translate the staged usage into the equation: (n/2) + 6 + 3 + (4*sqrt(n) + 1) = n. Solve by isolating sqrt(n), converting to a quadratic in t = sqrt(n), and choosing the positive root only (since counts are nonnegative). Step-by-Step Solution: Total used = n/2 + 6 + 3 + (4*sqrt(n) + 1) = n/2 + 10 + 4*sqrt(n).Set equal to n: n/2 + 10 + 4*sqrt(n) = n.Rearrange: 10 + 4*sqrt(n) = n/2 ⇒ multiply by 2: 20 + 8*sqrt(n) = n.Let t = sqrt(n). Then n = t^2, so t^2 − 8t − 20 = 0.Solve: t = [8 ± sqrt(64 + 80)]/2 = [8 ± 12]/2 ⇒ t = 10 (reject −2).Hence n = t^2 = 100. Verification / Alternative check: Check usage: half = 50; then 6 + 3 = 9; total 59. Remaining 41. Final use = 4*10 + 1 = 41. All used. Why Other Options Are Wrong: 90, 101, 104: Substituting these into the equation fails to balance total used with the original stock. Common Pitfalls: Missing the extra “+1” in the final usage term. Forgetting that the square root applies to the original count n, not the remaining arrows. Final Answer: 100

Question 18

Cost composition from two price equations: If 2 apples + 3 oranges + 1 pear cost Rs. 62 and 5 apples + 6 oranges + 4 pears cost Rs. 120, then find the total cost of 3 apples + 3 oranges + 3 pears.

Accepted Answer

Introduction / Context:This linear-combination question avoids solving for individual prices by cleverly combining the two given equations to get the desired sum directly. Such problems reward recognizing how to express the target bundle in terms of the provided bundles. Given Data / Assumptions: 2A + 3O + P = 62. 5A + 6O + 4P = 120. Target: 3A + 3O + 3P = 3(A + O + P). Concept / Approach:Multiply the first equation by 4 to align the pear coefficient with the second, then subtract to eliminate P. From the resulting relation, read off A + 2O and back-substitute into the first equation to compute A + O + P. Finally triple it to get the requested cost of 3A + 3O + 3P. Step-by-Step Solution: 4*(2A + 3O + P) = 8A + 12O + 4P = 248.Subtract (5A + 6O + 4P = 120): (8A−5A) + (12O−6O) = 248 − 120 ⇒ 3A + 6O = 128.Hence A + 2O = 128/3.From 2A + 3O + P = 62, write as (A + 2O) + (A + O + P) = 62.Thus A + O + P = 62 − 128/3 = 58/3.Therefore 3(A + O + P) = 58. Verification / Alternative check: Plugging unknowns that satisfy both equations will always yield 58 for 3A + 3O + 3P due to linearity; no need to solve each price. Why Other Options Are Wrong: Rs. 39, Rs. 46, Rs. 24: These contradict the derived linear combination, hence do not satisfy both given equations simultaneously. Common Pitfalls: Mistakes when subtracting equations, especially with aligned coefficients. Trying to fully solve for A, O, P (unnecessary extra work). Final Answer: Rs. 58

Question 19

Quadratic with misread coefficients: For the true equation x^2 + p x + q = 0, one solver (wrong p) found roots 2 and 6; another solver (wrong q) found roots 2 and −9. Determine the correct roots and the correct quadratic.

Accepted Answer

Introduction / Context:The question leverages Vieta’s formulas: for a monic quadratic x^2 + px + q = 0, the sum of roots is −p and the product is q. Using the information about the two incorrect attempts, we can deduce the true p and q, and hence the correct roots and equation. Given Data / Assumptions: True equation: x^2 + p x + q = 0. Wrong-p attempt yielded roots 2 and 6, so that attempt used the same q but incorrect p. Wrong-q attempt yielded roots 2 and −9, so that attempt used the same p but incorrect q. Concept / Approach:From the wrong-p attempt: product of roots is still q, thus q = 2*6 = 12. From the wrong-q attempt: sum of roots is still −p, thus −p = 2 + (−9) = −7 ⇒ p = 7. With p and q known, solve the correct quadratic for actual roots. Step-by-Step Solution: Wrong-p product ⇒ q = 12.Wrong-q sum ⇒ −p = −7 ⇒ p = 7.Correct equation: x^2 + 7x + 12 = 0.Factor: (x + 3)(x + 4) = 0 ⇒ roots are −3 and −4. Verification / Alternative check: Check Vieta: sum −3 + (−4) = −7 = −p; product (−3)(−4) = 12 = q. Consistent. Why Other Options Are Wrong: Other choices list roots/equations that do not match p = 7 and q = 12 determined from the two attempts. Common Pitfalls: Mixing up which parameter (p or q) remains correct in each incorrect attempt. Forgetting signs in the sum −p and product q relations. Final Answer: -3, -4 and x^2 + 7x + 12

Question 20

Evaluate a cubic identity under a linear condition: If a + b + 1 = 0, compute the value of a^3 + b^3 + 1 − 3ab.

Accepted Answer

Introduction / Context:This is a direct application of the identity a^3 + b^3 + c^3 − 3abc = (a + b + c)(a^2 + b^2 + c^2 − ab − bc − ca). Recognizing c = 1 and that a + b + 1 is given simplifies the expression to zero immediately. Given Data / Assumptions: a + b + 1 = 0 Target: a^3 + b^3 + 1 − 3ab Concept / Approach:Set c = 1 in the well-known identity. Because the factor (a + b + c) appears, if a + b + 1 equals zero, then the whole product must vanish, provided everything is finite. This makes the computation simple and robust. Step-by-Step Solution: Use: a^3 + b^3 + 1 − 3ab = (a + b + 1)(a^2 + b^2 + 1 − ab − a − b).Given a + b + 1 = 0 ⇒ the product equals 0.Therefore, a^3 + b^3 + 1 − 3ab = 0. Verification / Alternative check: Pick a, b satisfying a + b + 1 = 0, e.g., a = −1, b = 0. Then LHS = (−1)^3 + 0 + 1 − 0 = 0, confirming the identity. Why Other Options Are Wrong: 1, 3, −1: These contradict the factorization result which forces the value to zero under the given condition. Common Pitfalls: Applying a^3 + b^3 formula incorrectly instead of the three-term identity. Forgetting to set c = 1 explicitly. Final Answer: 0

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