Question 1

Arithmetic progression (AP) constraints: The 4th term of an AP is 37, and the 6th term is 12 more than the 4th term. What is the sum of the 2nd and 8th terms?

Accepted Answer

Introduction / Context: This problem tests basic AP formulas: the k-th term t_k = a + (k − 1)d. Knowing two terms lets us find the common difference and first term, then any other term or combination such as t2 + t8. Given Data / Assumptions: t4 = a + 3d = 37. t6 = a + 5d = t4 + 12 = 49. Find t2 + t8. Concept / Approach: Solve for d from the difference between t6 and t4, then back-solve a. Compute t2 and t8 and add. Alternatively, use an identity to avoid computing a explicitly. Step-by-Step Solution: From t6 − t4 = (a + 5d) − (a + 3d) = 2d = 12 ⇒ d = 6.From t4: a + 3d = 37 ⇒ a + 18 = 37 ⇒ a = 19.t2 = a + d = 19 + 6 = 25.t8 = a + 7d = 19 + 42 = 61.Sum t2 + t8 = 25 + 61 = 86. Verification / Alternative check: Identity: t2 + t8 = (a + d) + (a + 7d) = 2a + 8d = 2(a + 3d) + 2d = 2*37 + 12 = 86. Confirms without computing a separately. Why Other Options Are Wrong: 80, 74, 92: Do not satisfy the AP relationships given by t4 and t6. Common Pitfalls: Misreading “6th is 12 more than 4th” or mixing term indices. Always convert to algebraic equalities and proceed systematically. Final Answer: 86

Question 2

Counting multiples (divisibility by 6): How many three-digit numbers (from 100 to 999) are exactly divisible by 6?

Accepted Answer

Introduction / Context: A number divisible by 6 must be divisible by both 2 and 3, i.e., it must be a multiple of 6. Counting the multiples of a fixed integer in a closed interval can be done using floor division endpoints. Given Data / Assumptions: Range: 100 to 999 inclusive. Divisor: 6. Concept / Approach: The count of multiples of k in [A, B] is floor(B/k) − floor((A−1)/k). Apply with A = 100, B = 999, k = 6. Step-by-Step Solution: floor(999/6) = 166.floor(99/6) = 16.Count = 166 − 16 = 150. Verification / Alternative check: First multiple ≥ 100 is 102 (6 * 17); last multiple ≤ 999 is 996 (6 * 166). Number of terms in the AP 102, 108, …, 996 is (166 − 17 + 1) = 150. Why Other Options Are Wrong: 149 or 151: Off by one due to endpoint mishandling. 166: That is floor(999/6); ignores the 1–99 block. Common Pitfalls: Using B/k − A/k without floors or forgetting to subtract the multiples below A. Final Answer: 150

Question 3

Square triangular savings day: On March 1, 2016, Sherry saves ₹1. From March 2 onward, each day he saves ₹1 more than the previous day. Find the first date after March 1, 2016 on which his total savings becomes a perfect square.

Accepted Answer

Introduction / Context: The daily savings form an arithmetic sequence starting at 1 with common difference 1. The cumulative savings after n days equals the n-th triangular number n(n + 1)/2. The question asks for the earliest n > 1 such that this total is a perfect square. Given Data / Assumptions: Day 1 (March 1): save 1. Day d: save d rupees. Total after n days: T_n = n(n + 1)/2. Concept / Approach: We need T_n to be a square (a square triangular number). The smallest solutions are well known: n = 1 gives T_1 = 1 (but this is not “after March 1”), and the next is n = 8 giving T_8 = 36, a perfect square. Step-by-Step Solution: Compute T_2 = 3, T_3 = 6, …; none square until n = 8.At n = 8: T_8 = 8 * 9 / 2 = 36 = 6^2.The 8th day from March 1, 2016 is March 8, 2016. Verification / Alternative check: Known square triangular sequence starts 1, 36, 1225, … corresponding to n = 1, 8, 49, … So the earliest date after March 1 is March 8. Why Other Options Are Wrong: 17th, 18th, 26th March: These correspond to n = 17, 18, 26; their triangular totals are not perfect squares. Hence “None of these” is correct. Common Pitfalls: Confusing “after March 1” (n > 1) or mistaking daily deposit with cumulative total. Always test triangular numbers explicitly. Final Answer: None of these

Question 4

AP installments to clear a debt: A debt of ₹3600 is to be repaid in 40 yearly installments forming an AP. After paying 30 installments, one-third of the debt remains unpaid. Find the value of the first installment.

Accepted Answer

Introduction / Context: Installments in arithmetic progression allow the use of AP sum formulas. Knowing the total sum (debt) and the partial sum after 30 terms lets us solve for the first term and common difference. Then we can extract the first installment directly. Given Data / Assumptions: Total debt = ₹3600 = S_40. Unpaid after 30 installments = one-third of 3600 ⇒ unpaid = 1200, so S_30 = 2400. Installments are an AP with first term a and common difference d. Concept / Approach: Use S_n = n/2 * [2a + (n − 1)d]. Set up equations for S_30 and S_40 and solve for a and d, then identify a. Step-by-Step Solution: S_30 = 2400 ⇒ 30/2 * (2a + 29d) = 2400 ⇒ 15(2a + 29d) = 2400 ⇒ 2a + 29d = 160.S_40 = 3600 ⇒ 20(2a + 39d) = 3600 ⇒ 2a + 39d = 180.Subtract equations: (2a + 39d) − (2a + 29d) = 20 ⇒ 10d = 20 ⇒ d = 2.Plug back: 2a + 29*2 = 160 ⇒ 2a + 58 = 160 ⇒ 2a = 102 ⇒ a = 51. Verification / Alternative check: Check S_40: 20[2*51 + 39*2] = 20[102 + 78] = 20*180 = 3600. Check S_30: 15[102 + 58] = 15*160 = 2400. Consistent. Why Other Options Are Wrong: 55, 53, 49: Do not satisfy both S_30 = 2400 and S_40 = 3600 simultaneously with any constant d. Common Pitfalls: Forgetting that the unpaid part fixes S_30, or misusing the AP sum formula (e.g., mixing n − 1 and n). Solve the linear system carefully. Final Answer: 51

Question 5

Split a number into three AP parts: The number 15 is divided into three parts in AP, and the sum of their squares is 83. What is the smallest part?

Accepted Answer

Introduction / Context: When three numbers are in AP, they can be written as (a − d), a, (a + d). Their sum and sum of squares yield two equations in a and d. Solving provides the parts and identifies the smallest one. Given Data / Assumptions: Sum = 15 ⇒ (a − d) + a + (a + d) = 3a = 15. Sum of squares = 83 ⇒ (a − d)^2 + a^2 + (a + d)^2 = 83. Concept / Approach: Use 3a = 15 ⇒ a = 5. For the squares, expand and simplify to 3a^2 + 2d^2 = 83, then solve for d. Smallest part is a − d. Step-by-Step Solution: a = 15 / 3 = 5.Sum of squares: (a − d)^2 + a^2 + (a + d)^2 = a^2 + d^2 − 2ad + a^2 + a^2 + d^2 + 2ad = 3a^2 + 2d^2.3*5^2 + 2d^2 = 83 ⇒ 75 + 2d^2 = 83 ⇒ 2d^2 = 8 ⇒ d^2 = 4 ⇒ d = 2 (magnitude).Parts: 3, 5, 7. Smallest = 3. Verification / Alternative check: Check squares: 3^2 + 5^2 + 7^2 = 9 + 25 + 49 = 83. Perfect match. Why Other Options Are Wrong: 5, 6, 8: Not the smallest member of the valid triplet 3, 5, 7. Common Pitfalls: Dropping the 2 in 2d^2 when summing squares, or overlooking that d can be ±2 (we need the smallest, thus a − d). Final Answer: 3

Question 6

Doubling daily wages (geometric sequence): A boy earns ₹1 on day 1, ₹2 on day 2, ₹4 on day 3, doubling each day. If he works from 1st February to 20th February (20 days), how much does he earn in total?

Accepted Answer

Introduction / Context: This is a geometric progression (GP) with first term 1 and common ratio 2. The total for n days equals the sum of the first n powers of 2 starting at 2^0. The standard GP sum formula applies directly here. Given Data / Assumptions: Day k earning = 2^(k−1) rupees (k = 1 to 20). Number of days = 20 (Feb 1 through Feb 20 inclusive). Concept / Approach: Sum of GP: S_n = a * (r^n − 1) / (r − 1). With a = 1 and r = 2, this simplifies to S_n = 2^n − 1. Substitute n = 20 to get the total earnings. Step-by-Step Solution: S_20 = 2^20 − 1.This is the exact closed-form total; numerically it equals 1,048,575. Verification / Alternative check: Quick check for small n: S_1 = 1 = 2^1 − 1; S_2 = 1 + 2 = 3 = 2^2 − 1. Pattern holds; thus S_20 = 2^20 − 1. Why Other Options Are Wrong: 2^19 − 1: Sum for 19 days, not 20. 2^20: Overcounts by 1. 2^19: Not a sum; it is a single day’s wage on day 20−1. Common Pitfalls: Confusing arithmetic and geometric sequences, or forgetting that the first term is 1 (2^0). Always apply the GP sum formula carefully. Final Answer: 2^20 - 1

Question 7

Sum of two-digit numbers with remainder 3 on division by 7: What is the sum of all two-digit numbers that leave a remainder 3 when divided by 7?

Accepted Answer

Introduction / Context: Numbers congruent to 3 modulo 7 form an arithmetic sequence with common difference 7. Restricting to two-digit numbers gives a finite AP, whose sum we can compute with the AP sum formula. Given Data / Assumptions: Two-digit range: 10 to 99 inclusive. Condition: n ≡ 3 (mod 7). Concept / Approach: Find the first and last two-digit numbers congruent to 3 mod 7, count terms, then sum using S_n = n/2 * (first + last). Step-by-Step Solution: First term: 10 (since 10 = 7*1 + 3).Last term: 94 (since 94 = 7*13 + 3 ≤ 99).Number of terms: n = ((94 − 10)/7) + 1 = 12 + 1 = 13.Sum S = 13/2 * (10 + 94) = 13/2 * 104 = 13 * 52 = 676. Verification / Alternative check: Average term equals (first + last)/2 = 52. With 13 terms, total = 13 * 52 = 676. Confirms the AP sum result. Why Other Options Are Wrong: 94: This is the last term, not the sum. 696: Too high; would imply 13 terms averaging about 53.5, which is not correct. None of these: Incorrect because 676 is attainable. Common Pitfalls: Forgetting that 10 is valid (two-digit) and congruent to 3 mod 7, or miscounting terms by dropping the +1 in the count formula. Final Answer: 676

Question 8

AP with reciprocal targets: In an arithmetic progression, the m-th term is 1/n and the n-th term is 1/m. What is the sum of the first mn terms?

Accepted Answer

Introduction / Context: We are given two AP terms at positions m and n that are reciprocals in a symmetric way. Solving for the common difference and first term lets us compute the sum of the first mn terms succinctly. Given Data / Assumptions: t_m = a + (m − 1)d = 1/n. t_n = a + (n − 1)d = 1/m. Concept / Approach: Subtract equations to find d, then back-solve for a. Finally, use S_k = k/2[2a + (k − 1)d] with k = mn, simplifying carefully. Step-by-Step Solution: Subtract: (m − n)d = 1/n − 1/m = (m − n)/(mn) ⇒ d = 1/(mn).From t_m: a = 1/n − (m − 1)d = 1/n − (m − 1)/(mn) = 1/(mn).Now S_{mn} = (mn)/2 [2a + (mn − 1)d] = (mn)/2 [2/(mn) + (mn − 1)/(mn)].Inside bracket = (mn + 1)/(mn). Multiply gives S_{mn} = (mn + 1)/2. Verification / Alternative check: Symmetry suggests both a and d equal 1/(mn), which our algebra confirms, making the final sum compact and elegant. Why Other Options Are Wrong: (mn−1)/4, (mn+1)/4, (mn−1)/2: Do not follow from the derived a and d; algebra leads to (mn+1)/2 specifically. Common Pitfalls: Sign errors when subtracting the two equations, or forgetting to divide by mn consistently across terms. Final Answer: (mn+1)/2

Question 9

AP sums with exclusions: An AP has 10 terms. The sum of all terms except the 1st is 99, and the sum of all terms except the 6th is 89. If t1 + t5 = 10, find the 3rd term.

Accepted Answer

Introduction / Context: This problem combines AP sum identities with partial exclusions. Using S_10 − t_1 and S_10 − t_6 yields two equations. Together with t_1 + t_5 = 10, they allow solving for the first term and common difference, then t_3. Given Data / Assumptions: S_10 − t_1 = 99. S_10 − t_6 = 89. t_1 + t_5 = 10. Concept / Approach: Let t_1 = a; t_k = a + (k − 1)d. Compute S_10 = 10/2[2a + 9d] = 5(2a + 9d). Use exclusions to form a linear system in a, d; then find t_3 = a + 2d. Step-by-Step Solution: S_10 = 5(2a + 9d) = 10a + 45d.S_10 − t_1 = 99 ⇒ (10a + 45d) − a = 99 ⇒ 9a + 45d = 99. (1)S_10 − t_6 = 89 ⇒ (10a + 45d) − (a + 5d) = 89 ⇒ 9a + 40d = 89. (2)Subtract (1) − (2): (9a + 45d) − (9a + 40d) = 10 ⇒ 5d = 10 ⇒ d = 2.From (2): 9a + 40*2 = 89 ⇒ 9a + 80 = 89 ⇒ 9a = 9 ⇒ a = 1.t_3 = a + 2d = 1 + 4 = 5. Verification / Alternative check: t_1 + t_5 = a + (a + 4d) = 2a + 4d = 2*1 + 8 = 10, as given. Consistent. Why Other Options Are Wrong: 15, 8, 10: Do not align with the solved a = 1 and d = 2. Common Pitfalls: Dropping signs or mixing indices when excluding terms. Keep S_10 explicit and carefully subtract the appropriate term. Final Answer: 5

Question 10

Linear equation from a verbal sentence: If 24 is subtracted from a number, the result is four-sevenths of the original number. What is the sum of the digits of that number?

Accepted Answer

Introduction / Context: Translate the statement into an equation, solve for the number, then compute the sum of its digits. This is a straightforward linear equation in one variable with a follow-up digit-sum calculation. Given Data / Assumptions: x − 24 = (4/7) x. x is an integer (implied by a sensible “digits” question). Concept / Approach: Collect x terms on one side. Solve for x by isolating it using standard algebra. Then add the digits of the resulting number. Step-by-Step Solution: x − (4/7)x = 24.(3/7)x = 24 ⇒ x = 24 * 7 / 3 = 56.Sum of digits of 56 = 5 + 6 = 11. Verification / Alternative check: Check the sentence: 56 − 24 = 32 and (4/7)*56 = 32. Equality holds; digit sum 11 is confirmed. Why Other Options Are Wrong: 1, 9: Not the digit sum of 56. Data inadequate: The equation uniquely determines x. Common Pitfalls: Subtracting in the wrong direction or mishandling fractions when moving terms. Multiply both sides as needed to avoid fraction errors. Final Answer: 11

Question 11

Translating a mixed-fraction condition: The sum of one-half and one-fifth of a number exceeds one-third of that number by 7 1/3. What is the number?

Accepted Answer

Introduction / Context: Convert words into an algebraic equation involving fractions, then solve for the unknown number. Mixed fractions should be converted to improper fractions to simplify arithmetic. Given Data / Assumptions: (1/2 + 1/5)x − (1/3)x = 7 1/3. 7 1/3 = 22/3. Concept / Approach: Combine like terms on the left: (1/2 + 1/5 − 1/3)x. Compute the common denominator and simplify to get a single coefficient times x. Then isolate x. Step-by-Step Solution: 1/2 + 1/5 = 7/10.7/10 − 1/3 = 21/30 − 10/30 = 11/30.(11/30)x = 22/3 ⇒ x = (22/3) * (30/11) = (22*30)/(33) = 660/33 = 20. Verification / Alternative check: Check: (1/2 + 1/5) of 20 = (10 + 4) = 14; one-third of 20 = 20/3 ≈ 6.666…; difference 14 − 6.666… = 7.333… = 7 1/3, correct. Why Other Options Are Wrong: 15, 18, 30: Do not satisfy the exact fractional relation; plugging them in fails the difference test. Common Pitfalls: Misconverting 7 1/3 to an improper fraction or adding fractions without a common denominator. Final Answer: 20

Question 12

Income split (salary and tips): A driver's tips in a week were 5/4 of his salary. What fraction of his total income came from tips?

Accepted Answer

Introduction / Context: This is a ratio-to-fraction conversion. Given a part (tips) relative to another part (salary), we first express both as parts of the whole, then compute the fraction for tips over total income. Given Data / Assumptions: Let salary = S. Tips = (5/4)S. Total income = S + tips. Concept / Approach: Compute total = S + (5/4)S. Then tips fraction = tips / total. Simplify by factoring S and cancelling common factors. Step-by-Step Solution: Total = S + (5/4)S = (9/4)S.Fraction from tips = [(5/4)S] / [(9/4)S] = 5/9. Verification / Alternative check: Pick a concrete S (e.g., 4). Tips = 5; total = 9; tips/total = 5/9. The fraction is independent of S, confirming the algebra. Why Other Options Are Wrong: 4/9: Would correspond to tips being 4/5 of salary, not 5/4. 5/8: Ignores the correct total; 5/8 would mean total of 8 units with tips 5, but here total is 9 units (4 salary + 5 tips). 5/4: That is the ratio of tips to salary, not the fraction of total income. Common Pitfalls: Confusing ratio “of salary” with fraction “of total income.” Always compute the whole first, then form the desired part/whole fraction. Final Answer: 5/9

Question 13

Counting guests from dish ratios: At a party there is 1 bowl of rice for every 2 guests, 1 bowl of juice for every 3 guests, and 1 bowl of meat for every 4 guests. If there are 65 bowls in total, how many guests attended?

Accepted Answer

Introduction / Context: The counts of bowls are given per guest ratios. For G guests, the total bowls equal G/2 + G/3 + G/4. Setting this total to 65 yields a simple linear equation in G after taking a common denominator. Given Data / Assumptions: Bowl counts: rice = G/2, juice = G/3, meat = G/4. Total bowls = 65. Guests are a whole number; ratios are exact. Concept / Approach: Solve G/2 + G/3 + G/4 = 65. Use the common denominator 12 to combine terms and isolate G. Step-by-Step Solution: G/2 + G/3 + G/4 = (6G + 4G + 3G)/12 = 13G/12.13G/12 = 65 ⇒ 13G = 780 ⇒ G = 780/13 = 60. Verification / Alternative check: Compute bowls for G = 60: rice = 30, juice = 20, meat = 15; total = 65. Checks out exactly. Why Other Options Are Wrong: 65: Would require bowl/guest ratios summing to 1, which they do not. 24 or 48: Produce totals far from 65 given the bowl ratios. Common Pitfalls: Adding numerators without a common denominator or forgetting that each term depends on the same G. Final Answer: 60

Question 14

Two scenarios, two equations (money transfer): Ram and Mohan have some money. If Ram gives ₹30 to Mohan, Mohan then has twice as much as Ram has left. If instead Mohan gives ₹10 to Ram, Ram then has three times as much as Mohan has left. How much mo…

Accepted Answer

Introduction / Context: This is a pair of linear equations derived from two hypothetical money transfers. Define variables for Ram's and Mohan's current amounts, translate each scenario into an equation, and solve simultaneously. Given Data / Assumptions: Let R = Ram's money, M = Mohan's money. Scenario 1: M + 30 = 2(R − 30). Scenario 2: R + 10 = 3(M − 10). Concept / Approach: Each scenario produces a linear relationship between R and M. Solve the system by substitution or elimination to find integer solutions matching the options. Step-by-Step Solution: From Scenario 1: M + 30 = 2R − 60 ⇒ M = 2R − 90.From Scenario 2: R + 10 = 3M − 30 ⇒ R = 3M − 40.Substitute R into M = 2R − 90: M = 2(3M − 40) − 90 = 6M − 80 − 90 ⇒ −5M = −170 ⇒ M = 34.Then R = 3*34 − 40 = 102 − 40 = 62. Verification / Alternative check: Check Scenario 1: Ram gives 30 ⇒ R − 30 = 32; Mohan gets 30 ⇒ M + 30 = 64 = 2*32. Check Scenario 2: Mohan gives 10 ⇒ M − 10 = 24; Ram gets 10 ⇒ R + 10 = 72 = 3*24. Both conditions satisfied. Why Other Options Are Wrong: ₹6,₹2 or ₹43,₹26 or ₹170,₹124: Do not satisfy both equations simultaneously when tested. Common Pitfalls: Sign mistakes when moving terms across the equals sign, or mixing up which person gives/receives money in each scenario. Final Answer: ₹62,₹34

Question 15

Sum–difference to product (two-number system): The sum of two numbers is 25 and their difference is 13. Keeping both conditions true, determine the product of the two numbers.

Accepted Answer

Introduction / Context: This question tests solving a two-variable linear system built from a given sum and difference, and then using the recovered numbers to compute their product. It reinforces the standard method for reconstructing two unknowns when their sum and difference are known. Given Data / Assumptions: x + y = 25 x − y = 13 x and y are real numbers; no additional constraints are needed. Concept / Approach: With the pair of linear equations, the variables can be found by adding and subtracting the equations. Once x and y are known, compute x * y. This is a classic system solution and product evaluation task. Step-by-Step Solution: Add the equations: (x + y) + (x − y) = 25 + 13 → 2x = 38 → x = 19Substitute in x + y = 25: 19 + y = 25 → y = 6Product = x * y = 19 * 6 = 114 Verification / Alternative check: Check the difference: 19 − 6 = 13 (matches). The product computed from the validated pair is therefore correct. Why Other Options Are Wrong: 104, 315, 325, and 156 are products of other pairs that do not simultaneously satisfy sum 25 and difference 13. Common Pitfalls: Mixing up which variable is larger or trying to guess pairs instead of solving systematically. Always use add/subtract to avoid mistakes. Final Answer: 114

Question 16

Use (a^2 − b^2) = (a − b)(a + b): The difference of the squares of two numbers is 256000 and their sum is 1000. Find the two numbers.

Accepted Answer

Introduction / Context: This problem is a direct application of the identity a^2 − b^2 = (a − b)(a + b). With the sum given, the difference of the squares quickly yields the difference of the numbers, from which the pair is recovered uniquely (assuming a ≥ b). Given Data / Assumptions: a^2 − b^2 = 256000 a + b = 1000 We look for real numbers a ≥ b; integers are suggested by the data. Concept / Approach: Factor the square difference: (a − b)(a + b) = 256000. Since a + b = 1000, it follows that a − b = 256000 / 1000 = 256. Solve the simultaneous equations for a and b. Step-by-Step Solution: a + b = 1000a − b = 256Add to get a: 2a = 1256 → a = 628Then b = 1000 − 628 = 372 Verification / Alternative check: Compute a^2 − b^2 = (a − b)(a + b) = 256 * 1000 = 256000, which matches exactly. Why Other Options Are Wrong: 600,400 gives (a − b) = 200, not 256; 640,360 gives (a − b) = 280; the others do not satisfy both the sum and square-difference conditions. Common Pitfalls: Confusing a^2 − b^2 with (a − b)^2. The correct identity uses the product (a − b)(a + b). Final Answer: 628, 372

Question 17

Fractional composition of a number: Two-fifths of one-fourth of three-sevenths of a number equals 15. What is one-half of that number?

Accepted Answer

Introduction / Context: This is a unit-fractions chaining problem that checks comfort with multiplying fractions and solving for an unknown. We are told that successive fractions of a number produce 15, and we are asked to find half of the original number. Given Data / Assumptions: (2/5) * (1/4) * (3/7) * N = 15 N is a real number Goal: compute (1/2) * N Concept / Approach: Combine the fractional multipliers first to avoid errors. Then solve for N by dividing 15 by the combined factor. Finally, take half of N as required. Step-by-Step Solution: Combined factor = (2 * 1 * 3) / (5 * 4 * 7) = 6 / 140 = 3 / 70(3/70) * N = 15 → N = 15 * (70/3) = 350Half of N = 350 / 2 = 175 Verification / Alternative check: Back-substitute: (2/5)*(1/4)*(3/7)*350 = (3/70)*350 = 15, consistent. Half is then clearly 175. Why Other Options Are Wrong: 168, 196, 210, 240 result from arithmetic slips such as inverting fractions incorrectly or mis-multiplying numerators/denominators. Common Pitfalls: Mixing the fraction order or canceling incorrectly. Keep numerators and denominators aligned and simplify only when safe. Final Answer: 175

Question 18

Squares in a fixed ratio with a linear offset: The square of one number is 224 less than 8 times the square of the other. If the numbers are in the ratio 3:4, find the two numbers.

Accepted Answer

Introduction / Context: The problem pairs a ratio condition with a relation between squares. By parameterizing the numbers using the ratio and substituting into the square relation, we can solve for the scale factor and hence determine the actual numbers. Given Data / Assumptions: Numbers are in the ratio 3:4 → let them be 3k and 4k The square of one is 224 less than 8 times the square of the other Concept / Approach: Test which arrangement satisfies the condition. Assume (4k)^2 = 8*(3k)^2 − 224 and solve for k. If it yields a consistent positive k, we have the valid assignment. Otherwise, try the converse. Step-by-Step Solution: Assume b = 4k, a = 3k and b^2 = 8a^2 − 22416k^2 = 8 * 9k^2 − 224 = 72k^2 − 224Rearrange: 16k^2 − 72k^2 = −224 → −56k^2 = −224 → k^2 = 4 → k = 2Numbers = 3k, 4k = 6, 8 Verification / Alternative check: Check the given relation: 8*(6^2) − 224 = 8*36 − 224 = 288 − 224 = 64, and 8^2 = 64, so it holds exactly. Why Other Options Are Wrong: 3,6 and 4,8 do not satisfy the exact offset of 224 under the 8x square relation. 6,4 flips the ratio order. 9,12 keeps 3:4 but fails the square condition. Common Pitfalls: Interpreting “224 less than” in the wrong direction. Always place the subtraction from the larger expression as stated. Final Answer: 6 , 8

Question 19

Tour-budget linear equation (extend by 4 days, daily spend falls by ₹3): A traveler has a total budget of ₹360 for daily expenses. He extends the tour by 4 days, which reduces his daily spending by ₹3. Find the original planned number of tour days.

Accepted Answer

Introduction / Context: This problem turns a budget scenario into a linear equation in the number of days. The total spend is fixed, so daily expense equals 360 divided by the number of days. Extending the tour reduces daily spend by a known amount, giving an equation to solve. Given Data / Assumptions: Total budget = ₹360 Original days = n → daily spend = 360 / n Extended days = n + 4 → new daily spend = 360 / (n + 4) New daily spend = old daily spend − ₹3 Concept / Approach: Form the equation 360/(n + 4) = 360/n − 3 and solve for n. This is a rational equation leading to a quadratic that factors cleanly. Step-by-Step Solution: 360/(n + 4) = 360/n − 3Multiply by n(n + 4): 360n = (n + 4)(360 − 3n)Expand: 360n = 360n − 3n^2 + 1440 − 12nRearrange: 3n^2 + 12n − 1440 = 0 → n^2 + 4n − 480 = 0Solve: n = [−4 + √(16 + 1920)] / 2 = (−4 + 44)/2 = 20 Verification / Alternative check: Old daily = 360/20 = ₹18; new daily = 360/24 = ₹15; drop is ₹3, which matches. Why Other Options Are Wrong: 15, 16, 18, 24 do not satisfy the exact ₹3 reduction when checked back with 360/n and 360/(n+4). Common Pitfalls: Sign slips while expanding or canceling 360n prematurely. Keep terms aligned before simplifying. Final Answer: 20

Question 20

Three consecutive odd numbers, sum condition: The sum of three consecutive odd numbers is 20 more than the first of these numbers. What is the middle number?

Accepted Answer

Introduction / Context: Problems with consecutive integers are easily modeled using algebra. Represent consecutive odd numbers as n, n + 2, n + 4, and translate the sum condition into an equation. Given Data / Assumptions: First odd = n Second odd = n + 2 Third odd = n + 4 Sum = n + (n + 2) + (n + 4) = 20 + n Concept / Approach: Set up and solve the linear equation from the stated condition. The middle number is then n + 2. Step-by-Step Solution: 3n + 6 = n + 202n = 14 → n = 7Middle number = n + 2 = 9 Verification / Alternative check: Check: 7 + 9 + 11 = 27, which is 20 more than the first (7 + 20 = 27). Condition satisfied. Why Other Options Are Wrong: 7 is the first, not the middle; 11 and 13 do not match the derived middle given the condition; “Data inadequate” is incorrect because the information is sufficient. Common Pitfalls: Using n, n + 1, n + 2 instead of odd steps; forgetting the +2 pattern for odd numbers. Final Answer: 9

Linear Equation Questions