Question 1

System solvability check: For 2x + 4y = 6 and 4x + 8y = 6, determine whether the system has a unique solution, no solution, or infinitely many solutions.

Accepted Answer

Introduction / Context: This problem examines whether two linear equations represent the same line, different parallel lines, or intersecting lines. Comparing proportional coefficients provides a quick test for consistency or inconsistency of the system. Given Data / Assumptions: E1: 2x + 4y = 6 E2: 4x + 8y = 6 Concept / Approach: If the left-hand sides are proportional but the constants are not in the same ratio, the lines are parallel and distinct, producing no solution. Multiply E1 by 2 to compare fairly with E2. Step-by-Step Solution: Multiply E1 by 2: 4x + 8y = 12 But E2 says 4x + 8y = 6 Same left side, different constant ⇒ parallel distinct lines ⇒ no solution Verification / Alternative check: Graphically, both lines have slope −(2/4) = −1/2, but different intercepts; hence they never meet. Why Other Options Are Wrong: Unique or infinite solutions would require equality of constants after scaling; “exactly two solutions” is not a concept in linear pairs in the plane. Common Pitfalls: Stopping after noticing proportional coefficients and not checking constants, which changes the conclusion from coincident to parallel lines. Final Answer: no solution

Question 2

Ratio change after adding items: In a rare coin collection, initially there is one gold coin for every three non-gold coins. After adding 10 more gold coins, the ratio of gold to non-gold becomes 1 : 2. What is the total number of coins in the colle…

Accepted Answer

Introduction / Context: This is a ratio-adjustment problem. We model the initial counts using a common multiplier, apply the change (adding 10 gold coins), and set up the new ratio. Solving for the multiplier yields the exact counts before and after the change. Given Data / Assumptions: Initially gold : non-gold = 1 : 3. After adding 10 gold coins, gold : non-gold = 1 : 2. No non-gold coins are added or removed. Concept / Approach: Let initial gold = k and non-gold = 3k. After adding, gold becomes k + 10 while non-gold remains 3k. Enforce the new ratio (k + 10) : 3k = 1 : 2, solve for k, and then compute the final total as (k + 10) + 3k. Step-by-Step Solution: (k + 10) / (3k) = 1 / 2 ⇒ 2(k + 10) = 3k 2k + 20 = 3k ⇒ k = 20 Initial counts: gold = 20, non-gold = 60 After adding 10 gold: gold = 30, non-gold = 60 Total now = 30 + 60 = 90 Verification / Alternative check: Check new ratio: 30 : 60 reduces to 1 : 2, exactly as required. Why Other Options Are Wrong: 50, 60, 70, 80 do not align with the ratio 1 : 2 after adding 10 gold coins under the initial 1 : 3 structure. Common Pitfalls: Accidentally adding 10 to both gold and non-gold, or changing the non-gold count, which is not stated in the problem. Final Answer: 90

Question 3

Three-way proportional equality: If (x + y − 8)/2 = (x + 2y − 14)/3 = (3x + y − 12)/11, find the values of x and y respectively.

Accepted Answer

Introduction / Context: Equalizing three expressions to a common parameter is a common technique. Assign a parameter t to the shared value, convert each proportional statement into a linear equation, and then solve the resulting system for x and y efficiently. Given Data / Assumptions: (x + y − 8)/2 = (x + 2y − 14)/3 = (3x + y − 12)/11 All variables are real and expressions are defined. Concept / Approach: Let each fraction equal t. Then x + y − 8 = 2t, x + 2y − 14 = 3t, and 3x + y − 12 = 11t. Eliminate t by equating suitable pairs, reducing the problem to two linear equations in x and y. Step-by-Step Solution: From first two: x + 2y − 14 = (3/2)(x + y − 8) 2x + 4y − 28 = 3x + 3y − 24 ⇒ 0 = x − y + 4 ⇒ x − y = −4 ⇒ x = y − 4 Use the first and third with t: 3x + y − 12 = 11 * [(x + y − 8)/2] 2(3x + y − 12) = 11(x + y − 8) 6x + 2y − 24 = 11x + 11y − 88 ⇒ 0 = 5x + 9y − 64 Substitute x = y − 4: 5(y − 4) + 9y − 64 = 0 ⇒ 14y − 84 = 0 ⇒ y = 6 Then x = 6 − 4 = 2 Verification / Alternative check: Compute t: (x + y − 8)/2 = (2 + 6 − 8)/2 = 0; the other two also evaluate to 0. All three equal, confirming correctness. Why Other Options Are Wrong: Other pairs do not satisfy all three equalities simultaneously when substituted back. Common Pitfalls: Dropping constants during cross-multiplication or equating only two of the three expressions without checking the third one for consistency. Final Answer: 2, 6

Question 4

Radical equation (repaired interpretation): If [√(3 + x) + √(3 − x)] / [√(3 + x) − √(3 − x)] = 2, then find the value of x.

Accepted Answer

Introduction / Context: The database expression lacked parentheses around radicals, which made the denominator zero. Applying the Recovery-First Policy, we repair it to the standard radical form. The problem now involves manipulating a ratio of sums and differences of square roots and solving for the variable under square roots. Given Data / Assumptions: (√(3 + x) + √(3 − x)) / (√(3 + x) − √(3 − x)) = 2 3 ± x ≥ 0, hence −3 ≤ x ≤ 3 for real radicals. Concept / Approach: Let a = √(3 + x) and b = √(3 − x). The equation becomes (a + b)/(a − b) = 2. Solve for a in terms of b, then square to eliminate radicals and solve for x, ensuring the solution lies within the domain and satisfies the original equation (no extraneous root). Step-by-Step Solution: (a + b)/(a − b) = 2 ⇒ a + b = 2a − 2b ⇒ a = 3b Square both sides: 3 + x = 9(3 − x) 3 + x = 27 − 9x ⇒ 10x = 24 ⇒ x = 24/10 = 12/5 Verification / Alternative check: Check domain: x = 12/5 = 2.4 ∈ [−3, 3]. Substitute back numerically to confirm the ratio is very close to 2 (rounding errors may occur if not exact arithmetic). The algebraic derivation guarantees equality. Why Other Options Are Wrong: 5/12, 5/7, 7/5, 3/2 do not satisfy the transformed condition a = 3b when substituted back through the radical definitions. Common Pitfalls: Forgetting parentheses leading to division by zero; squaring without checking the original constraint can also introduce extraneous values—always verify in the given equation and domain. Final Answer: 12/5

Question 5

Classifying pairs of lines from ax + by = c and dx + ey = f: I. Parallel if the system has no solution. II. Coincident if the system has a finite number of solutions. III. Intersecting if the system has exactly one solution. Which statements are cor…

Accepted Answer

Introduction / Context: This conceptual item links the nature of solutions of a 2×2 linear system to the geometric relationship between its two lines in the plane. Recognizing whether the system is inconsistent, dependent, or independent maps to parallel, coincident, or intersecting lines respectively. Given Data / Assumptions: A pair of linear equations: ax + by = c and dx + ey = f with real coefficients a, b, c, d, e, f. Standard solution classifications: no solution, exactly one solution, infinitely many solutions. Concept / Approach: Recall: parallel distinct lines ⇒ no solution; coincident lines ⇒ infinitely many solutions; intersecting at a single point ⇒ unique solution. Match each statement to these facts. Step-by-Step Solution: Statement I: “Parallel if system has no solution.” Correct. Statement II: “Coincident if system has finite number of solutions.” Incorrect. Coincident lines yield infinitely many solutions, not finite. Statement III: “Intersecting if system has only one solution.” Correct. Therefore, the correct set is I and III only. Verification / Alternative check: Determinant method: For ax + by = c and dx + ey = f, a unique solution occurs when ae − bd ≠ 0 (lines intersect). If ae − bd = 0 but ratios of constants disagree, lines are parallel (no solution). If coefficients and constants share the same ratio, lines coincide (infinitely many solutions). Why Other Options Are Wrong: Any choice including II is wrong because it misstates the coincident case. “All three” is impossible as I and II contradict each other’s solution counts. Common Pitfalls: Confusing “finite” with “infinite” solutions, and forgetting that coincident lines correspond to infinitely many common points (every point on the line). Final Answer: Only I and III

Question 6

Two-equation pricing (chairs and tables): Ten chairs and six tables together cost ₹6,200. Three chairs and two tables together cost ₹1,900. What is the cost of 4 chairs and 5 tables?

Accepted Answer

Introduction / Context: We are given two linear price equations with two unknowns: chair price (C) and table price (T). Solve the system to determine C and T, and then compute the requested bundle price for 4 chairs and 5 tables. Given Data / Assumptions: 10C + 6T = 6200 3C + 2T = 1900 Uniform prices; no bulk discounts implied. Concept / Approach: Use elimination. Match the table coefficients by multiplying the second equation by 3 and subtract from the first to isolate the chair price. Solve for C, then back-substitute to find T. Finally compute 4C + 5T. Step-by-Step Solution: Multiply (3C + 2T = 1900) by 3 ⇒ 9C + 6T = 5700 Subtract from 10C + 6T = 6200 ⇒ C = 500 Back-solve: 3*500 + 2T = 1900 ⇒ 1500 + 2T = 1900 ⇒ 2T = 400 ⇒ T = 200 Price of 4 chairs and 5 tables = 4*500 + 5*200 = 2000 + 1000 = ₹3000 Verification / Alternative check: Check both original equations with C = 500, T = 200: 10*500 + 6*200 = 6200 and 3*500 + 2*200 = 1900. Both satisfy perfectly. Why Other Options Are Wrong: ₹3,300, ₹3,500, ₹3,800, ₹2,900 are not equal to 4*500 + 5*200 = ₹3,000. Common Pitfalls: Miscalculating multipliers during elimination or mixing coefficients leads to wrong unit prices; always verify by substitution before finalizing the bundle cost. Final Answer: ₹ 3,000

Question 7

Solve for variables from symmetric linear forms: Given (p/x) + (q/y) = m and (q/x) + (p/y) = n, determine x and y in terms of p, q, m, and n (assume denominators nonzero).

Accepted Answer

Introduction / Context: This problem asks for an algebraic solution for x and y given two linear relations in the reciprocal variables 1/x and 1/y. The system is symmetric in p and q and can be solved elegantly by substituting u = 1/x and v = 1/y, turning the equations into a 2×2 linear system in u and v. Given Data / Assumptions: (p/x) + (q/y) = m and (q/x) + (p/y) = n Assume x ≠ 0, y ≠ 0, and denominators in final expressions are nonzero (i.e., m p − n q ≠ 0 and n p − m q ≠ 0). Concept / Approach: Let u = 1/x and v = 1/y. Then p u + q v = m and q u + p v = n. Solve this linear system for u and v via determinants or elimination. Finally, take reciprocals to get x and y. Careful handling of the determinant p^2 − q^2 is essential. Step-by-Step Solution: System in u, v: p u + q v = m … (1) q u + p v = n … (2) Determinant D = p^2 − q^2 u = (m p − n q)/D, v = (n p − m q)/D x = 1/u = D / (m p − n q) = (p^2 − q^2)/(m p − n q) y = 1/v = D / (n p − m q) = (p^2 − q^2)/(n p − m q) Verification / Alternative check: Substitute back: p/x = p * (m p − n q)/D and q/y = q * (n p − m q)/D; adding yields (m p^2 − n p q + n p q − m q^2)/D = m(p^2 − q^2)/D = m, as required. A similar check works for the second equation, confirming the formulae. Why Other Options Are Wrong: Options that swap p^2 − q^2 to q^2 − p^2 or flip denominator signs without matching both x and y expressions break the determinant structure and will not satisfy both original equations simultaneously. Common Pitfalls: Forgetting to convert to u and v, mishandling the determinant, or inverting signs in denominators leads to incorrect expressions. Always verify by substitution for symbolic solution questions. Final Answer: x = (p^2 − q^2)/(m p − n q), y = (p^2 − q^2)/(n p − m q)

Question 8

Capital exchange puzzle: One says, “Give me one hundred, friend! I will then be twice as rich as you.” The other replies, “If you give me ten, I shall be six times as rich as you.” Find each person’s capital (in ₹).

Accepted Answer

Introduction / Context: This is a classic two-variable word problem involving linear equations formed from statements about how capitals change after giving fixed amounts. Translate each sentence carefully into an equation and solve simultaneously for the two unknown capitals. Given Data / Assumptions: Let A and B be the current capitals (in ₹) of the first and second persons respectively. “Give me 100; I then become twice as rich as you”: A + 100 = 2(B − 100). “If you give me 10; I shall be six times as rich as you”: B + 10 = 6(A − 10). Concept / Approach: Each statement yields a linear relation. Solve the two equations for A and B using substitution or elimination. Finally, check both conditions to ensure internal consistency before selecting the option. Step-by-Step Solution: A + 100 = 2B − 200 ⇒ A − 2B = −300 … (1) B + 10 = 6A − 60 ⇒ −6A + B = −70 … (2) From (1): A = 2B − 300 Substitute in (2): −6(2B − 300) + B = −70 ⇒ −12B + 1800 + B = −70 ⇒ −11B = −1870 ⇒ B = 170 Then A = 2*170 − 300 = 40 Verification / Alternative check: Check first claim: A + 100 = 140; B − 100 = 70; indeed 140 = 2*70. Second claim: B + 10 = 180; A − 10 = 30; 180 = 6*30. Both checks pass. Why Other Options Are Wrong: Other pairs do not satisfy both transformed equations simultaneously when verified; “Cannot be determined” is incorrect because the system is consistent and solvable uniquely. Common Pitfalls: Reversing who gives whom or sign errors when moving constants across the equals sign are common sources of mistakes in such puzzles. Final Answer: ₹ 40 and ₹ 170

Question 9

Income–expenditure–savings with ratios: The ratio of incomes of two persons is 8 : 5 and the ratio of their expenditures is 2 : 1. Each saves ₹1000 per month. What is the difference between their monthly incomes?

Accepted Answer

Introduction / Context: This problem links income, expenditure, and savings with given ratios. Modeling incomes and expenditures via proportional multipliers and equating savings allows us to compute the exact monthly incomes and hence their difference. Given Data / Assumptions: Income ratio = 8 : 5 ⇒ I1 = 8k, I2 = 5k. Expenditure ratio = 2 : 1 ⇒ E1 = 2t, E2 = 1t. Savings: I1 − E1 = 1000 and I2 − E2 = 1000. Concept / Approach: Set up the two savings equations and solve for k and t. Then compute I1 and I2 explicitly and subtract to find the difference. This approach preserves the given ratios without assuming the same multiplier for income and expenditure ratios. Step-by-Step Solution: 8k − 2t = 1000 … (1) 5k − t = 1000 … (2) From (2): t = 5k − 1000 Substitute into (1): 8k − 2(5k − 1000) = 1000 ⇒ 8k − 10k + 2000 = 1000 ⇒ −2k = −1000 ⇒ k = 500 I1 = 8k = 4000; I2 = 5k = 2500 Difference = 4000 − 2500 = ₹1500 Verification / Alternative check: E1 = 2t = 2(5k − 1000) = 2(1500) = 3000; E2 = t = 1500. Savings: 4000 − 3000 = 1000; 2500 − 1500 = 1000. Checks out. Why Other Options Are Wrong: ₹2500, ₹1000, ₹700, ₹2000 do not equal the computed difference consistent with both ratios and equal savings. Common Pitfalls: Assuming the same multiplier for income and expenditure ratios or forgetting to impose the two separate savings conditions leads to incorrect k and t values. Final Answer: ₹ 1,500

Question 10

Find the original fraction: A fraction becomes 7/8 when 5 is added to both numerator and denominator. If 3 is added to both numerator and denominator, it becomes 6/7. What is the original fraction?

Accepted Answer

Introduction / Context: This problem converts two transformation statements about a fraction into linear equations in its numerator and denominator. Solving the resulting system yields the original fraction exactly, with a quick verification step to ensure both conditions hold. Given Data / Assumptions: Let the fraction be x/y with y ≠ 0. (x + 5)/(y + 5) = 7/8. (x + 3)/(y + 3) = 6/7. Concept / Approach: Cross-multiply both relations to produce two linear equations in x and y. Solve them simultaneously (elimination works well). Reduce the final fraction to simplest terms if necessary. Step-by-Step Solution: 8(x + 5) = 7(y + 5) ⇒ 8x + 40 = 7y + 35 ⇒ 8x − 7y = −5 … (1) 7(x + 3) = 6(y + 3) ⇒ 7x + 21 = 6y + 18 ⇒ 7x − 6y = −3 … (2) Multiply (2) by 8: 56x − 48y = −24 Multiply (1) by 7: 56x − 49y = −35 Subtract: (−49y) − (−48y) = −35 − (−24) ⇒ −y = −11 ⇒ y = 11 Then from (2): 7x − 6*11 = −3 ⇒ 7x − 66 = −3 ⇒ 7x = 63 ⇒ x = 9 Original fraction = 9/11 Verification / Alternative check: (9 + 5)/(11 + 5) = 14/16 = 7/8 and (9 + 3)/(11 + 3) = 12/14 = 6/7. Both conditions are satisfied exactly. Why Other Options Are Wrong: 8/11 and 10/11 fail at least one of the two transformation checks; “Cannot be determined” is incorrect because the system has a unique solution. Common Pitfalls: Arithmetic slips when cross-multiplying or subtracting equations can lead to near misses; always verify both given transformations on your final fraction. Final Answer: 9/11

Question 11

Two-digit number from digit conditions: The number obtained by interchanging the digits of a two-digit number is 18 more than the original number, and the sum of the digits is 8. Find the original number.

Accepted Answer

Introduction / Context: Digit puzzles translate directly into small linear equations. Represent the two-digit number as 10a + b with a as the tens digit and b as the units digit. Use the reverse condition and the digit sum to form two equations and solve for a and b. Given Data / Assumptions: Reverse number = original number + 18. Sum of digits = 8. Digits a, b are integers from 0–9 with a ≥ 1 for a two-digit number. Concept / Approach: Let original = 10a + b; reverse = 10b + a. Then 10b + a = 10a + b + 18 and a + b = 8. Solve simultaneously for a and b, ensuring valid digit constraints. Step-by-Step Solution: 10b + a = 10a + b + 18 ⇒ 9b − 9a = 18 ⇒ b − a = 2 a + b = 8 Add equations: (b − a) + (a + b) = 2 + 8 ⇒ 2b = 10 ⇒ b = 5 Then a = 8 − b = 3 Original number = 10a + b = 30 + 5 = 35 Verification / Alternative check: Reverse is 53; 53 − 35 = 18 (fits). Digit sum 3 + 5 = 8 (fits). Why Other Options Are Wrong: 50, 51, 53, 44 do not satisfy both the reverse-difference and digit-sum conditions together. Common Pitfalls: Mixing up which number is larger (original vs reverse) and sign errors when rearranging 10b + a = 10a + b + 18 are common mistakes. Final Answer: 35

Question 12

Solve a pair of linear equations: Twice the first number plus thrice the second equals 100, and thrice the first plus twice the second equals 120. Which is the larger number?

Accepted Answer

Introduction / Context: This is a direct two-equation, two-unknowns problem. Solve the linear system to find both numbers, then identify the larger one. Elimination using aligned coefficients makes the arithmetic quick and clean. Given Data / Assumptions: 2x + 3y = 100 3x + 2y = 120 Concept / Approach: Eliminate one variable by forming equal coefficients on x or y. Subtract the resulting equations to find the other variable, then back-substitute to get the first. Compare the values to decide which is larger. Step-by-Step Solution: Multiply the first equation by 3: 6x + 9y = 300 Multiply the second equation by 2: 6x + 4y = 240 Subtract: (6x + 9y) − (6x + 4y) = 60 ⇒ 5y = 60 ⇒ y = 12 Substitute into 3x + 2y = 120 ⇒ 3x + 24 = 120 ⇒ 3x = 96 ⇒ x = 32 Larger number = 32 Verification / Alternative check: Check in the first equation: 2*32 + 3*12 = 64 + 36 = 100 ✔; second: 3*32 + 2*12 = 96 + 24 = 120 ✔. Why Other Options Are Wrong: 12 and 14 are the smaller or unrelated values; 35 and 28 do not satisfy both equations simultaneously. Common Pitfalls: Arithmetic slips when scaling equations or subtracting can flip signs and produce wrong values; always verify in both original equations. Final Answer: 32

Question 13

Arithmetic progression of publication years: A series of 7 books was published at uniform 7-year intervals. When the seventh book was issued, the sum of the publication years of all seven books was 13,524. In which year was the first book published?

Accepted Answer

Introduction / Context: The publication years form an arithmetic progression (AP) with common difference 7 years and 7 terms. The sum of AP terms connects the first year to the total sum. Using the compact sum formula for an AP lets us compute the first publication year directly. Given Data / Assumptions: Number of books n = 7. Common difference d = 7 years. Sum of the 7 publication years S = 13,524. Let the first year be a; then the years are a, a+7, …, a+42. Concept / Approach: Sum of AP: S = n/2 * [2a + (n − 1)d]. With n = 7 and d = 7, we can simplify the sum to S = 7(a + 21). Solve for a and confirm it yields integer years consistent with the described schedule. Step-by-Step Solution: S = 7/2 * [2a + 6*7] = 7/2 * (2a + 42) = 7(a + 21) 13,524 = 7(a + 21) ⇒ a + 21 = 13,524 / 7 = 1,932 a = 1,932 − 21 = 1,911 Verification / Alternative check: List years: 1911, 1918, 1925, 1932, 1939, 1946, 1953. Sum them or use the average: the average year is the middle term (1932). Sum = average * 7 = 1932 * 7 = 13,524, confirming the calculation. Why Other Options Are Wrong: 1932 is the middle term (4th book), not the first. 1917, 1925, 1942 do not align with both the 7-year spacing and the given total sum simultaneously. Common Pitfalls: Using the wrong AP sum formula or forgetting that with an odd number of equally spaced years, the average equals the middle term, which gives a quick cross-check. Final Answer: 1911

Question 14

Two-digit number and its reverse: The difference between a two-digit number and the number obtained by interchanging its digits is 36. What is the difference between the two individual digits of that number?

Accepted Answer

Introduction / Context: This problem checks your understanding of how two-digit numbers change when their digits are reversed and how to model that change with a simple linear expression. The key is realizing that reversing digits produces a fixed multiple of the difference between the tens and units digits. Given Data / Assumptions: A two-digit number has tens digit a and units digit b. The reverse is formed by interchanging digits. The difference between the original number and its reverse equals 36. Concept / Approach: Represent the original number as 10a + b and the reversed number as 10b + a. Their difference factors as 9(a − b) in magnitude. This constant factor 9 is the hallmark of “reverse-digit” problems for two-digit numbers. Step-by-Step Solution: Original = 10a + b; Reversed = 10b + aAbsolute difference = |(10a + b) − (10b + a)| = |9(a − b)|Given |9(a − b)| = 36 ⇒ |a − b| = 36/9 = 4 Verification / Alternative check: Pick any digits with difference 4, e.g., a = 7, b = 3. Then original = 73, reverse = 37, and difference = 36. This confirms the model works for legitimate digits. Why Other Options Are Wrong: 3, 6, 9: None satisfy 9*(digit difference) = 36. Cannot be determined: The digit difference is uniquely determined by the factor-of-9 rule. Common Pitfalls: Forgetting the factor 9 or assuming the difference 36 is the digit difference directly. Always express two-digit numbers as 10a + b to avoid mistakes. Final Answer: 4

Question 15

Reversing digits increases value by 45: When the digits of a two-digit number are interchanged, the resulting number is greater than the original by 45. If the difference between the digits is 5, what is the original number?

Accepted Answer

Introduction / Context: This question tests modeling two-digit numbers and understanding what information is sufficient to determine a unique answer. We will translate the statements into equations and examine whether a single solution or multiple solutions exist. Given Data / Assumptions: Original number has tens digit a and units digit b. Reversed number is 10b + a and is larger than original by 45. The difference between the digits is 5. Concept / Approach: Form two relations: 10b + a = (10a + b) + 45 and |a − b| = 5. Because the reversed number is larger, b > a, hence b − a = 5. We will see that several valid (a, b) pairs satisfy both relations, leading to multiple original numbers. Step-by-Step Solution: 10b + a = 10a + b + 45 ⇒ 9(b − a) = 45 ⇒ b − a = 5This is identical to the stated digit difference, so both conditions reduce to b − a = 5.Valid pairs with a ≥ 1 (tens digit cannot be 0): (1,6), (2,7), (3,8), (4,9)Corresponding originals: 16, 27, 38, 49 Verification / Alternative check: For each, reversing increases value by 45 (e.g., 61 − 16 = 45; 72 − 27 = 45, etc.). Thus multiple answers exist. Why Other Options Are Wrong: 16, 27, 38, 49: Each is possible, but the question asks for a unique original number. Common Pitfalls: Assuming uniqueness without checking the digit constraints. Always ensure the system of equations yields a single solution before selecting a numeric option. Final Answer: Cannot be determined

Question 16

Two-digit number with a ratio condition: In a two-digit number, the units digit is three times the tens digit, and the sum of the digits is 8. What is the number?

Accepted Answer

Introduction / Context: This is a straightforward two-digit algebra problem involving a ratio between digits and a sum constraint. Setting variables for the tens and units digits allows a quick solution with substitution. Given Data / Assumptions: Tens digit = t Units digit = 3t Sum of digits = t + 3t = 8 Concept / Approach: Use the sum condition to find t, then construct the number 10t + (3t). Ensure that 3t is a valid single digit (0–9), which it is once we compute t. Step-by-Step Solution: t + 3t = 8 ⇒ 4t = 8 ⇒ t = 2Units digit = 3t = 6Number = 10t + 3t = 20 + 6 = 26 Verification / Alternative check: Check conditions: units is three times tens (6 = 3*2) and sum 2 + 6 = 8. Both satisfied. Why Other Options Are Wrong: 20, 13: Do not satisfy units = 3*tens. 39, 62: Do not satisfy the sum of digits = 8 with the given ratio. Common Pitfalls: Mixing up tens and units or writing the number as 3t*10 + t. Carefully map tens to the 10s place and units to the 1s place. Final Answer: 26

Question 17

Heads and legs puzzle (equal cows and herdsmen): In a group with an equal number of cows and herdsmen, the total number of legs was 28 less than four times the number of heads. How many herdsmen are there?

Accepted Answer

Introduction / Context: This classic heads-and-legs problem is solved by translating the verbal conditions into linear equations. Equal counts of cows and herdsmen imply paired head counts but different per-head leg counts (cows have 4 legs; herdsmen have 2 legs). Given Data / Assumptions: Let n be the number of cows and also the number of herdsmen. Total heads = cows + herdsmen = 2n. Total legs = 4n (cows) + 2n (herdsmen) = 6n. Legs are 28 less than 4 times the number of heads. Concept / Approach: Write legs = 4*(heads) − 28. Substitute heads = 2n and legs = 6n to form a single linear equation in n and solve. Step-by-Step Solution: 6n = 4*(2n) − 286n = 8n − 28 ⇒ 2n = 28 ⇒ n = 14Thus, herdsmen = n = 14 Verification / Alternative check: Heads = 2n = 28; 4*heads = 112. Legs = 6n = 84. 112 − 84 = 28, matching the condition exactly. Why Other Options Are Wrong: 7, 10, 21, 28: Do not satisfy 6n = 8n − 28 when substituted. Common Pitfalls: Assigning the same leg count to cows and herdsmen or forgetting that the phrase “equal number” applies to counts of individuals, not legs or heads. Final Answer: 14

Question 18

Angle-sum link to a number: A number, when reduced by one-seventh of itself, equals the sum of the interior angles of a triangle (180 degrees). What is the number?

Accepted Answer

Introduction / Context: This is a linear equation with a familiar benchmark: the sum of angles of a triangle is 180 degrees. By expressing “one-seventh of itself” algebraically, we can solve for the number quickly. Given Data / Assumptions: Let the number be N. N − (1/7)N = 180. All quantities are real numbers. Concept / Approach: Combine like terms on the left to get a single multiple of N. Then isolate N by simple division. This is a direct single-step linear equation. Step-by-Step Solution: N − (1/7)N = 180(6/7)N = 180N = 180 * (7/6) = 30 * 7 = 210 Verification / Alternative check: Compute one-seventh: N/7 = 30. Then N − N/7 = 210 − 30 = 180, which matches the triangle angle sum exactly. Why Other Options Are Wrong: 224, 350: Subtracting one-seventh does not yield 180. 140, 189: Resulting value is not 180 when one-seventh is removed. Common Pitfalls: Using 1/6 instead of 1/7 or confusing “one-seventh of itself” with “itself minus 7.” Read multiplicative fractions carefully. Final Answer: 210

Question 19

Two-digit number with unit digit fixed: A two-digit number has 3 in the units place, and the sum of its digits equals one-seventh of the number itself. Find the number.

Accepted Answer

Introduction / Context: This problem blends a fixed units digit with a proportional relation between the number and the sum of its digits. Representing the two-digit number algebraically lets us solve with a couple of steps. Given Data / Assumptions: Let the tens digit be t. Units digit is 3. Sum of digits (t + 3) = (1/7) * (10t + 3). Concept / Approach: Express the number as 10t + 3. Use the given proportional relation to form and solve a simple linear equation in t. Confirm that the result yields a valid digit (0–9). Step-by-Step Solution: t + 3 = (1/7)(10t + 3)7t + 21 = 10t + 318 = 3t ⇒ t = 6Number = 10*6 + 3 = 63 Verification / Alternative check: Sum of digits = 6 + 3 = 9; one-seventh of 63 is 9. Condition is satisfied exactly. Why Other Options Are Wrong: 43, 53, 73, 83: Do not satisfy the proportional sum condition with units digit 3. Common Pitfalls: Accidentally using (1/7)*(sum of digits) = number. The statement says the sum equals one-seventh of the number, not the other way around. Final Answer: 63

Question 20

Split a sum with a linear condition: Divide 54 into two parts such that 10 times the first plus 22 times the second equals 780. What is the larger part?

Accepted Answer

Introduction / Context: Partition problems with a linear condition are routine applications of two equations in two unknowns. Here, the total is fixed at 54 and a weighted sum equals 780. We will solve for the parts and identify the larger one. Given Data / Assumptions: Let the first part be x and the second part be y. x + y = 54. 10x + 22y = 780. Concept / Approach: Use substitution from x + y = 54 ⇒ y = 54 − x, then plug into the second equation and solve for x. This produces an exact integer solution, making identification of the larger part direct. Step-by-Step Solution: 10x + 22(54 − x) = 78010x + 1188 − 22x = 780−12x = −408 ⇒ x = 34y = 54 − 34 = 20. Larger part = 34 Verification / Alternative check: Compute 10*34 + 22*20 = 340 + 440 = 780, matching perfectly, while x + y = 54 holds. Why Other Options Are Wrong: 24, 30, 32, 20: None satisfy both equations simultaneously as the larger part. Common Pitfalls: Mixing up which variable you substituted, or solving 10x + 22y = 780 without using x + y = 54. Always leverage both equations. Final Answer: 34

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