Question 1

Exam qualification threshold and maximum marks It is required to score at least 40% to qualify. A candidate scored 200 marks and failed by 8 marks. What were the maximum marks for the exam?

Accepted Answer

Introduction / Context: This aptitude question tests percentage application in exams. The key idea is that the qualifying cutoff is a fixed percentage of the maximum marks. If a candidate falls short by a given amount, we can back-calculate the maximum possible marks of the test paper using simple algebra with percentages. Given Data / Assumptions: You must score 40% of the maximum marks to qualify. The candidate scored 200 marks. The candidate failed by 8 marks (i.e., needed 8 more marks to just qualify). Let the maximum marks be M (a positive number). Concept / Approach: The qualifying cutoff = 40% of M = 0.40 * M. “Failed by 8 marks” means the candidate’s score is 8 less than the cutoff. So candidate score + 8 = cutoff. From that, solve for M directly. Step-by-Step Solution:Let M be the maximum marks.Cutoff to qualify = 40% of M = 0.40 * M.Given: 200 is 8 less than the cutoff ⇒ 200 + 8 = 0.40 * M.Compute: 208 = 0.40 * M.Therefore M = 208 / 0.40 = 520. Verification / Alternative check: 40% of 520 is 0.40 * 520 = 208. The candidate has 200, which is exactly 8 less than 208 as stated. This confirms the calculation is consistent. Why Other Options Are Wrong:540: 40% is 216, not 208; the shortfall would be 16, not 8.502: 40% is 200.8; shortfall would be 0.8, not 8.Couldn’t be determined: Sufficient data exists to find M uniquely. Common Pitfalls: Confusing “failed by 8” as “scored 8” or using 60% instead of 40%. Another error is subtracting 8 from the cutoff incorrectly or using 8% instead of 8 marks. Keep units straight: marks vs percentages. Final Answer: 520

Question 2

Two-subject marks with linear relations One-third of Rahul's Mathematics marks exceeds one-half of his Hindi marks by 30. If together he scored 480 in these two subjects, how many marks did he get in Hindi?

Accepted Answer

Introduction / Context: This problem converts wording into simultaneous linear equations. It assesses your skill in forming equations from verbal conditions and solving them cleanly by substitution or elimination. Given Data / Assumptions: M = marks in Mathematics; H = marks in Hindi. (1/3)M exceeds (1/2)H by 30 ⇒ (1/3)M − (1/2)H = 30. Total marks: M + H = 480. We assume marks are nonnegative and the algebra is exact. Concept / Approach: Translate the comparison into an equation and combine it with the total. Use elimination by clearing denominators to avoid fractional coefficients, then solve systematically for H first (or M first) as convenient. Step-by-Step Solution:(1/3)M − (1/2)H = 30.Multiply by 6 to clear fractions: 2M − 3H = 180.Also given: M + H = 480.From 2M − 3H = 180 ⇒ 2M = 180 + 3H ⇒ M = 90 + 1.5H.Substitute in M + H = 480: (90 + 1.5H) + H = 480 ⇒ 2.5H = 390 ⇒ H = 156.Thus Hindi marks = 156 and Mathematics marks = 480 − 156 = 324. Verification / Alternative check: Check the condition: (1/3)*324 = 108 and (1/2)*156 = 78; 108 − 78 = 30, which matches the statement. Sum 324 + 156 = 480 also matches. Why Other Options Are Wrong:200, 196, 294: In each case, either the sum with M would not be 480 or the “exceeds by 30” relation would fail. Common Pitfalls: Forgetting to multiply both sides when clearing fractions; mixing up which quantity “exceeds” the other; or adding instead of subtracting in the relation. Carefully convert words to symbols before solving. Final Answer: 156

Question 3

Milk container capacity from fractional fills and transfers A container was 4/5 full. After removing 12 bottles and then adding 8 bottles, it became 3/4 full. How many bottles can the container hold in total?

Accepted Answer

Introduction / Context: This is a classic capacity-and-transfer problem. The container’s capacity in “bottles” is unknown, yet fractional fullness and discrete bottle moves allow determining the total capacity using algebra with fractions and integers. Given Data / Assumptions: Total capacity = B bottles. Initial volume = (4/5) * B. Remove 12 bottles; add 8 bottles. Final volume = (3/4) * B. Concept / Approach: Track the number of bottles through each operation. Since “remove 12” then “add 8” is a net change of −4 bottles, write the final equality in terms of B and solve a simple linear equation in B. Step-by-Step Solution:Initial amount = (4/5)B.After removing 12: (4/5)B − 12.After adding 8: (4/5)B − 12 + 8 = (4/5)B − 4.This equals the final fullness: (3/4)B.Therefore (4/5)B − 4 = (3/4)B ⇒ (4/5 − 3/4)B = 4.Compute the fraction: 4/5 − 3/4 = 16/20 − 15/20 = 1/20.So (1/20)B = 4 ⇒ B = 80. Verification / Alternative check: 4/5 of 80 is 64. Remove 12 ⇒ 52. Add 8 ⇒ 60. 3/4 of 80 is 60. Everything matches exactly. Why Other Options Are Wrong:40, 30, 90: Plugging any of these into the equation fails to satisfy the final fullness condition. Common Pitfalls: Forgetting that remove then add is a net −4; mishandling fractional subtraction (4/5 − 3/4). Always reduce to a single equation in B and solve carefully. Final Answer: 80

Question 4

Two-digit number with digit-sum and digit-swap conditions Interchanging the digits makes the number 27 larger; the sum of the digits is 13. What is the original two-digit number?

Accepted Answer

Introduction / Context: Problems with two-digit numbers often use digit variables for tens and units. Here, the sum of digits and the effect of swapping digits are given, allowing a clean system of linear equations in the digit variables. Given Data / Assumptions: Let the tens digit be t and the units digit be u. Original number = 10t + u. Swapped number = 10u + t. 10u + t = (10t + u) + 27. t + u = 13. Concept / Approach: Use the swap relation and the sum relation. Rearrange to isolate u − t, then combine with t + u to find the unique pair (t, u). Ensure digits are integers from 0 to 9 and tens digit is nonzero. Step-by-Step Solution:From swap relation: 10u + t = 10t + u + 27 ⇒ 9u − 9t = 27 ⇒ u − t = 3.Sum relation: t + u = 13.Add both equations: (u − t) + (t + u) = 3 + 13 ⇒ 2u = 16 ⇒ u = 8.Then t = 13 − u = 5.Original number = 10t + u = 10*5 + 8 = 58. Verification / Alternative check: Swap digits to get 85. Difference 85 − 58 = 27. Sum of digits 5 + 8 = 13. Both conditions hold. Why Other Options Are Wrong:67, 76, 85: Each fails either the digit-sum condition, the swap-difference condition, or both (note 85 is the swapped number, not the original). Common Pitfalls: Mixing the swap direction (thinking original is larger); or writing 10t + u and 10u + t incorrectly. Keep consistent definitions for tens and units throughout. Final Answer: 58

Question 5

Three positive numbers with pairwise products Given ab = 42, bc = 56, and ca = 48 for positive numbers a, b, c, find the value of the third number c.

Accepted Answer

Introduction / Context: Using pairwise products is a common technique to recover individual values of positive numbers. Here, three equations relate the three unknowns through products; careful algebra isolates one variable directly without solving a full system. Given Data / Assumptions: ab = 42 bc = 56 ca = 48 a, b, c are positive real numbers. Concept / Approach: Multiply all three equations to get (ab)(bc)(ca) = (abc)^2. Then use a ratio to isolate c. A simpler method is to divide the product of the two equations involving c by the equation not involving c alone: c^2 = (bc * ca) / (ab). Step-by-Step Solution:Compute c^2 = (bc * ca) / (ab).Plug values: c^2 = (56 * 48) / 42.Simplify: 56/42 = 4/3, so c^2 = (4/3) * 48 = 64.Therefore c = sqrt(64) = 8 (positive branch since numbers are positive). Verification / Alternative check: If c = 8, then from bc = 56 ⇒ b = 7; from ca = 48 ⇒ a = 6. Check ab = 6 * 7 = 42, which matches. All conditions are satisfied. Why Other Options Are Wrong:12, 6, 7: Substituting any of these values breaks at least one of the provided product equations when solving for the remaining variables. Common Pitfalls: Adding products instead of multiplying/dividing; forgetting to take the positive square root; or arithmetic slips while simplifying 56*48/42. Keep factorization simple to avoid errors. Final Answer: 8

Question 6

Income allocation and remaining amount A man spends 2/5 of income on rent, 3/10 on food, and 1/8 on conveyance. If ₹ 2800 remains, find his total spending on food and conveyance together.

Accepted Answer

Introduction / Context: This question combines fractional budgeting with a known leftover amount to deduce total income. Once income is known, targeted expenses can be computed by applying the relevant fractions. Given Data / Assumptions: Fractions of income: rent = 2/5, food = 3/10, conveyance = 1/8. Remaining amount (after all three) = ₹ 2800. Total income = I (unknown). Concept / Approach: Sum the spent fractions, subtract from 1 to get the remaining fraction. Equate that fraction of I to ₹ 2800 and solve for I. Then compute (food + conveyance) as (3/10 + 1/8) * I exactly. Step-by-Step Solution:Spent fraction = 2/5 + 3/10 + 1/8.Compute common denominator 40: 2/5 = 16/40, 3/10 = 12/40, 1/8 = 5/40.Sum = (16 + 12 + 5)/40 = 33/40.Remaining fraction = 1 − 33/40 = 7/40.Given 7/40 * I = 2800 ⇒ I = 2800 * 40 / 7 = 16000.Food + conveyance = (3/10 + 1/8) * I = (24/80 + 10/80) * 16000 = (34/80) * 16000 = 17/40 * 16000 = 6800. Verification / Alternative check: Compute each: rent = 2/5 * 16000 = 6400; food = 4800; conveyance = 2000; remaining = 16000 − (6400 + 4800 + 2000) = 2800. All values are consistent. Why Other Options Are Wrong:₹ 7000, ₹ 5500, ₹ 6700: Each deviates from the exact fraction 17/40 of the derived income. Common Pitfalls: Arithmetic slips in fraction addition; using 1/10 instead of 3/10; or forgetting to convert to a common denominator. Work with exact fractions to avoid rounding errors. Final Answer: ₹ 6800

Question 7

Proportional distribution among sons, daughters, and friends ₹ 17200 is divided among 5 sons, 4 daughters, and 2 friends. Each daughter gets 4 times each friend, and each son gets 5 times each friend. How much does each daughter receive?

Accepted Answer

Introduction / Context: Weighted distribution problems assign different multiples of a base share to different groups. By expressing all shares in terms of a single unit (the friend’s share), the total amount can be partitioned precisely. Given Data / Assumptions: Total amount = ₹ 17200. 5 sons, 4 daughters, 2 friends. Each daughter gets 4 times each friend; each son gets 5 times each friend. Let x = amount received by each friend. Concept / Approach: Convert all recipients’ amounts into multiples of x. Sum all “x” units across recipients to get the total in terms of x, then solve for x. Multiply x by 4 to find each daughter’s amount. Step-by-Step Solution:Friend’s share = x ⇒ total for 2 friends = 2x.Each daughter = 4x ⇒ total for 4 daughters = 16x.Each son = 5x ⇒ total for 5 sons = 25x.Grand total = 2x + 16x + 25x = 43x.43x = 17200 ⇒ x = 17200 / 43 = 400.Each daughter receives 4x = 4 * 400 = ₹ 1600. Verification / Alternative check: Compute all: sons = 25*400 = 10000; daughters = 16*400 = 6400; friends = 2*400 = 800; total = 10000 + 6400 + 800 = 17200. Perfect match. Why Other Options Are Wrong:₹ 800, ₹ 1000, ₹ 1500 are not equal to 4x when x = 400; they either reflect the friend’s share or incorrect multiples. Common Pitfalls: Forgetting to multiply by the count of people in each category; mixing up the multiples (4x vs 5x). Always total the weighted counts first, then back-solve for x. Final Answer: ₹ 1600

Question 8

Men and women ratio changes after people leave After 15 women leave a group, there are 2 men for each woman. Then 45 men leave and the ratio becomes 5 women for each man. Find twice the initial number of women.

Accepted Answer

Introduction / Context: Ratio problems with people entering or leaving convert naturally into linear equations. By expressing final ratios after changes, you can solve for initial counts of men and women, then compute the requested derived quantity. Given Data / Assumptions: Initial men = m; initial women = w. After 15 women leave: women = w − 15; men = m; ratio men : women = 2 : 1 ⇒ m = 2(w − 15). Then 45 men leave: men = m − 45; women unchanged at w − 15; ratio women : men = 5 : 1 ⇒ w − 15 = 5(m − 45). All counts are nonnegative integers with w ≥ 15 and m ≥ 45 (consistent with leaving). Concept / Approach: Use the two ratio equations to eliminate m and solve for w. Then compute the requested double (2w). Keep algebra tidy to avoid sign mistakes. Step-by-Step Solution:From first ratio: m = 2(w − 15) = 2w − 30.From second ratio: w − 15 = 5(m − 45).Substitute m: w − 15 = 5((2w − 30) − 45) = 5(2w − 75) = 10w − 375.Rearrange: w − 15 = 10w − 375 ⇒ −9w = −360 ⇒ w = 40.Therefore twice the initial number of women = 2w = 80. Verification / Alternative check: If w = 40, then m = 2(40 − 15) = 50. After the first change: women = 25, men = 50 (ratio 2:1). After 45 men leave: men = 5, women = 25 (ratio 5:1). Everything matches perfectly. Why Other Options Are Wrong:25, 23, 35: These do not equal 2w with w = 40 and would fail the ratio checks when back-substituted. Common Pitfalls: Mixing the order of changes; flipping the ratios (men:women vs women:men); or missing that women remain the same in the second step. Carefully encode each stage before solving. Final Answer: 80

Question 9

Adult–child fare ratio application An adult fare is six times a child fare. If one adult fare is ₹ 114, what is the total for 4 adults and 5 children traveling the same distance?

Accepted Answer

Introduction / Context: Ratio-based unit pricing questions require converting one known fare into the other using the given multiple, then totaling for a group mix. It checks comfort with proportional reasoning and simple arithmetic. Given Data / Assumptions: Adult fare (A) = 6 * child fare (C). A = ₹ 114 ⇒ C = A / 6. Group: 4 adults and 5 children. Concept / Approach: First determine the child fare from the adult fare using the 6:1 relation. Then compute the sum for the exact group composition by multiplying fares by counts and adding results. Step-by-Step Solution:Given A = ₹ 114 and A = 6C ⇒ C = 114 / 6 = ₹ 19.Cost for 4 adults = 4 * 114 = ₹ 456.Cost for 5 children = 5 * 19 = ₹ 95.Total = 456 + 95 = ₹ 551. Verification / Alternative check: Check the ratio: 114 is indeed 6 times 19. The totals are simple whole-number multiplications, so the sum is reliable. Why Other Options Are Wrong:₹ 505, ₹ 560, ₹ 572: These correspond to incorrect child fare calculations or arithmetic mistakes in the group totals. Common Pitfalls: Taking 114/5 instead of 114/6 for the child fare or miscounting children/adults. Reconfirm the multiple (adult is six times child) before computing. Final Answer: ₹ 551

Question 10

Linear pricing with a difference constraint The total cost of 12 pens and 5 pencils is ₹ 111. One pencil costs ₹ 5 less than one pen. What is the total cost of 8 pens and 9 pencils?

Accepted Answer

Introduction / Context: Price-mix questions use small linear systems. With the total cost of a combination and a per-item difference between two products, you can solve for individual prices and then compute the requested new combination cost. Given Data / Assumptions: Let p = price of one pen; q = price of one pencil. 12p + 5q = 111. q = p − 5 (pencil costs ₹ 5 less than pen). Find cost of 8 pens and 9 pencils = 8p + 9q. Concept / Approach: Substitute q = p − 5 into the total equation to solve for p first. Then find q. Use these to compute the new linear combination 8p + 9q. Keep values exact and in rupees. Step-by-Step Solution:12p + 5q = 111 and q = p − 5.Substitute: 12p + 5(p − 5) = 111 ⇒ 12p + 5p − 25 = 111.17p = 136 ⇒ p = 8.Then q = p − 5 = 3.Compute 8p + 9q = 8*8 + 9*3 = 64 + 27 = 91. Verification / Alternative check: Validate the given total: 12p + 5q = 12*8 + 5*3 = 96 + 15 = 111, matching the statement. Hence, prices are consistent. Why Other Options Are Wrong:₹ 89, ₹ 97, ₹ 78: These come from arithmetic slips (e.g., miscomputing 17p, or miscalculating 8p + 9q after finding p and q). Common Pitfalls: Reversing the difference (setting q = p + 5), or plugging into the wrong target expression. Ensure q is ₹ 5 less than p and double-check the final multiplication and addition. Final Answer: ₹ 91

Question 11

Simplification with equal sharing: When a fixed amount is shared among 14 girls, each receives ₹160 more than she would if the same amount were equally shared among 18 girls. Find the total amount.

Accepted Answer

Introduction / Context:In this quantitative aptitude problem on equal distribution, the same sum of money is shared among two different group sizes. Because the group of 14 is smaller than the group of 18, the per-person share increases by a fixed amount. We use unitary method or simple algebra to compute the total amount. Given Data / Assumptions: Total amount = A (unknown) Share with 14 girls = A/14 Share with 18 girls = A/18 Increase per girl when there are 14 instead of 18 = ₹160 Concept / Approach:When the same amount is divided among fewer people, each person gets more. The difference between the two per-person amounts equals ₹160. Set up a linear equation in A and solve. Step-by-Step Solution: A/14 = A/18 + 160A * (1/14 - 1/18) = 1601/14 - 1/18 = (18 - 14) / 252 = 4/252 = 1/63A * (1/63) = 160A = 160 * 63 = 10080 Verification / Alternative check: Share at 18 = 10080/18 = 560Share at 14 = 10080/14 = 720Difference = 720 - 560 = 160 (matches) Why Other Options Are Wrong: ₹ 5040: Half of the correct total; would not yield a ₹160 difference. ₹ 10070: Close but incorrect; difference would not be exactly ₹160. ₹ 5000: Too small to produce the stated per-person increase. Common Pitfalls: Adding 160 directly to the total rather than to the per-person share. Computing 1/18 - 1/14 instead of 1/14 - 1/18, which flips the sign. Final Answer: ₹ 10080

Question 12

Currency note mix: A man has 170 currency notes consisting of ₹100 and ₹50 denominations totaling ₹10000. How much money is in ₹50 notes?

Accepted Answer

Introduction / Context:This is a standard system-of-equations problem involving two denominations. We know the total number of notes and the total value, and we must determine the value tied to one denomination (₹50 notes). Given Data / Assumptions: Total notes = 170 Denominations: ₹100 and ₹50 Total value = ₹10000 Concept / Approach:Let x be the number of ₹100 notes and y be the number of ₹50 notes. Then x + y = 170 and 100x + 50y = 10000. Solve this linear system to find y and compute 50y. Step-by-Step Solution: x + y = 170 … (1)100x + 50y = 10000Divide value equation by 50: 2x + y = 200 … (2)(2) - (1): (2x + y) - (x + y) = 200 - 170 ⇒ x = 30Then y = 170 - 30 = 140Value in ₹50 notes = 50 * 140 = ₹7000 Verification / Alternative check: Value in ₹100 notes = 30 * 100 = ₹3000Total = 3000 + 7000 = ₹10000 (consistent) Why Other Options Are Wrong: ₹ 4000: Implies only 80 fifty-rupee notes; would not total ₹10000 with 170 notes. ₹ 9000: Exceeds the maximum possible given only 170 notes. ₹ 6000: Leads to inconsistent counts with the total number of notes. Common Pitfalls: Forgetting to divide the value equation by 50 to simplify. Arithmetic slips when subtracting equations. Final Answer: ₹ 7000

Question 13

Attendance-based pay and fines: A worker is engaged over 120 days. He earns ₹20 for each working day and incurs a deduction (fine) of ₹3 for each non-working day. His total remuneration after 120 days is ₹560. How many days was he absent?

Accepted Answer

Introduction / Context:This is a linear equation problem involving separate contributions for working and non-working days. The worker receives a positive amount per working day and a deduction per absence, resulting in a relatively small net amount over a long period, which implies many absences. Given Data / Assumptions: Total calendar days considered = 120 Payment per working day = ₹20 Deduction per non-working day = ₹3 Total remuneration = ₹560 Concept / Approach:Let w = number of working days and a = number of absences. Then w + a = 120. Net pay = 20w - 3a. Solve the simultaneous equations for a. Step-by-Step Solution: w + a = 120 ⇒ w = 120 - aNet pay: 20w - 3a = 56020(120 - a) - 3a = 5602400 - 20a - 3a = 5602400 - 23a = 560 ⇒ 23a = 1840 ⇒ a = 1840 / 23 = 80 Verification / Alternative check: w = 120 - 80 = 40Net = 20*40 - 3*80 = 800 - 240 = 560 (matches) Why Other Options Are Wrong: 60/70/40: Substituting these in w + a = 120 and 20w - 3a fails to produce ₹560. Common Pitfalls: Misinterpreting the deduction as reducing the pay per working day instead of subtracting per absence. Arithmetic mistakes while simplifying 2400 - 23a = 560. Final Answer: 80

Question 14

Quality control with defective rate: A crate has 1 bruised apple for every 40 apples. If 3 out of every 4 bruised apples are unsaleable and there are 9 unsaleable apples, how many apples are in the crate?

Accepted Answer

Introduction / Context:This problem combines ratio interpretation with fractional filtering of defects. We are told the frequency of bruised apples and the fraction of bruised apples considered unsaleable. From the number of unsaleable apples, we back-calculate the total count. Given Data / Assumptions: Bruised frequency: 1 bruised per 40 apples Unsaleable among bruised: 3/4 Unsaleable apples = 9 Concept / Approach:Let total apples be N. Then bruised apples = N/40. Unsaleable apples = (3/4) * (N/40) = 3N/160. Set 3N/160 = 9 and solve for N. Step-by-Step Solution: Unsaleable = 3N/160 = 9Multiply both sides by 160: 3N = 1440N = 1440 / 3 = 480 Verification / Alternative check: Bruised = 480/40 = 12Unsaleable = (3/4)*12 = 9 (consistent) Why Other Options Are Wrong: 500/390/632: None satisfy 3N/160 = 9; they give non-integer or wrong unsaleable counts. Common Pitfalls: Using 1/4 instead of 3/4 of bruised apples as unsaleable. Forgetting that 1 bruised per 40 apples means bruised = N/40, not N*40. Final Answer: 480

Question 15

Linear population change: City A has population 136000 decreasing by 2400 per year, and City B has population 84000 increasing by 1600 per year. After how many years will their populations be equal?

Accepted Answer

Introduction / Context:This problem models two populations changing linearly but in opposite directions. We set both expressions equal and solve for time. It is a straightforward linear equation in t (years). Given Data / Assumptions: City A at t years: A(t) = 136000 - 2400*t City B at t years: B(t) = 84000 + 1600*t Equality A(t) = B(t) Concept / Approach:Equate A(t) and B(t) to find the year where the populations match. Since both changes are linear with constant rates, a single solution exists if rates differ. Step-by-Step Solution: 136000 - 2400t = 84000 + 1600t136000 - 84000 = 1600t + 2400t52000 = 4000tt = 52000 / 4000 = 13 Verification / Alternative check: A(13) = 136000 - 2400*13 = 136000 - 31200 = 104800B(13) = 84000 + 1600*13 = 84000 + 20800 = 104800 (matches) Why Other Options Are Wrong: 15, 18, 19: Substitution yields mismatched populations, so they are incorrect. Common Pitfalls: Sign errors when moving terms (mixing +1600t and -2400t). Arithmetic errors dividing 52000 by 4000. Final Answer: 13

Question 16

Nested fractions and brackets: Compute the value of 5 - [ 3/4 + { 2 1/2 - ( 1/2 + 1/6 - 1/7 ) } ] / 2.

Accepted Answer

Introduction / Context:This calculation tests careful handling of mixed numbers, nested brackets, and fraction operations. Accuracy requires converting mixed numbers to improper fractions, respecting the order of operations, and simplifying step by step. Given Data / Assumptions: Expression: 5 - [ 3/4 + { 2 1/2 - ( 1/2 + 1/6 - 1/7 ) } ] / 2 2 1/2 is a mixed number, equal to 5/2 Standard precedence: parentheses, then division/multiplication, then addition/subtraction Concept / Approach:Work from the innermost parentheses outward. Convert mixed numbers early. Keep everything as exact fractions to avoid rounding errors. Finally, subtract from 5 as an improper fraction with a common denominator. Step-by-Step Solution: Innermost: (1/2 + 1/6 - 1/7) = 1/2 + 1/6 = 2/3; 2/3 - 1/7 = 14/21 - 3/21 = 11/21Now: 2 1/2 - (result) = 5/2 - 11/21 = (105 - 22)/42 = 83/42Add 3/4: 3/4 + 83/42 = (63 + 166)/84 = 229/84Divide by 2: (229/84)/2 = 229/168Finally: 5 - 229/168 = (840 - 229)/168 = 611/168 Verification / Alternative check: Approximate decimals: 611/168 ≈ 3.6369; evaluating the original numerically also gives ≈ 3.6369, confirming. Why Other Options Are Wrong: 423/168, 583/168, 577/168: Each results from a common slip, such as reversing subtraction inside parentheses or forgetting to divide the bracketed sum by 2. Common Pitfalls: Failing to convert 2 1/2 to 5/2 at the start. Not maintaining common denominators when adding/subtracting fractions. Final Answer: 611/168

Question 17

Algebraic identity application: Evaluate (2.247)^3 + (1.730)^3 + (1.023)^3 - 3 * 2.247 * 1.730 * 1.023, divided by [(2.247)^2 + (1.730)^2 + (1.023)^2 - 2.247*1.730 - 1.730*1.023 - 2.247*1.023].

Accepted Answer

Introduction / Context:This problem directly uses a well-known algebraic identity. When x + y + z ≠ 0, the expression (x^3 + y^3 + z^3 - 3xyz) / (x^2 + y^2 + z^2 - xy - yz - zx) simplifies elegantly to x + y + z. Given Data / Assumptions: x = 2.247, y = 1.730, z = 1.023 We assume standard algebraic identities apply and no rounding is needed before substitution. Concept / Approach:Use the identity: (x^3 + y^3 + z^3 - 3xyz) / (x^2 + y^2 + z^2 - xy - yz - zx) = x + y + z, provided x + y + z ≠ 0. Here, the numbers are chosen so their sum is exact and simple. Step-by-Step Solution: Compute sum: x + y + z = 2.247 + 1.730 + 1.023 = 5.000By identity, the entire fraction simplifies to x + y + z = 5 Verification / Alternative check: You can numerically evaluate numerator and denominator separately, but the identity guarantees the result as long as x + y + z ≠ 0. Why Other Options Are Wrong: 1.730, 4, 5.247: These are distractors derived from partial sums or misapplied identities. Only the exact sum 5 satisfies the identity here. Common Pitfalls: Attempting to expand everything numerically, increasing chances of arithmetic error. Forgetting the identity or misremembering coefficients. Final Answer: 5

Question 18

Roots of unity style evaluation (corrected): If a satisfies a^2 + a + 1 = 0 (i.e., a is a non-real cube root of unity), compute the value of a^12 + a^6 + 1.

Accepted Answer

Introduction / Context:This problem uses properties of complex cube roots of unity. If a is a non-real cube root of unity, then a^3 = 1, a ≠ 1, and a^2 + a + 1 = 0. Powers of a cycle with period 3, enabling quick evaluation of higher powers. Given Data / Assumptions: a^2 + a + 1 = 0 a^3 = 1 and a ≠ 1 Concept / Approach:Use the periodicity: a^k repeats every 3 steps. Specifically, a^0 = 1, a^1 = a, a^2 = a^2, a^3 = 1, a^4 = a, etc. Reduce exponents modulo 3 to simplify a^12 and a^6. Step-by-Step Solution: a^3 = 1 ⇒ a^6 = (a^3)^2 = 1a^12 = (a^3)^4 = 1Therefore a^12 + a^6 + 1 = 1 + 1 + 1 = 3 Verification / Alternative check: From a^2 + a + 1 = 0 ⇒ a^2 = -a - 1. Iteratively computing powers confirms the period-3 cycle. Why Other Options Are Wrong: 2/1: Would require one of a^6 or a^12 to be 0, which never occurs for cube roots of unity. -3: Sign error; all three terms equal 1, summing to +3. Common Pitfalls: Misinterpreting the condition or forgetting that a ≠ 1. Not reducing exponents modulo 3 before evaluating. Final Answer: 3

Question 19

Custom operation evaluation: Given a * b = a + b + a / b, find the value of 12 * 4.

Accepted Answer

Introduction / Context:This problem defines a custom binary operation and asks for a direct evaluation. The key is to substitute exactly as defined and respect standard arithmetic precedence (division before addition). Given Data / Assumptions: Definition: a * b = a + b + a / b Compute: 12 * 4 Concept / Approach:Plug in a = 12 and b = 4 directly into the given definition. Do not overthink; there is no associativity or commutativity assumption needed for a single operation. Step-by-Step Solution: 12 * 4 = 12 + 4 + 12/4= 16 + 3= 19 Verification / Alternative check: Compute 12/4 first (3), then add to 12 + 4 = 16 to confirm 19. Why Other Options Are Wrong: 48: Treats * as multiplication; incorrect by definition. 20/21: Often arise from adding in a different order or misreading the division. Common Pitfalls: Assuming standard multiplication instead of the defined operation. Forgetting to perform a/b with correct precedence. Final Answer: 19

Question 20

Exponent rules and base conversion: Evaluate (4^6)^5 × (4^3)^8 ÷ 4^3 and express the result as (64)^k. Find k.

Accepted Answer

Introduction / Context:This problem checks familiarity with exponent laws and rewriting a power of 4 in terms of base 64. The goal is to simplify the left-hand side into a single power of 4 and then equate it to a power of 64 to find k. Given Data / Assumptions: Expression: (4^6)^5 × (4^3)^8 ÷ 4^3 Target form: (64)^k Recall: (a^m)^n = a^(m*n), a^m × a^n = a^(m+n), a^m ÷ a^n = a^(m-n) Concept / Approach:Simplify the left-hand side to 4^power. Then use 64 = 4^3 to convert 4^power into (4^3)^k = 4^(3k) and solve 3k = power. Step-by-Step Solution: (4^6)^5 = 4^(6*5) = 4^30(4^3)^8 = 4^(3*8) = 4^24Multiply: 4^30 × 4^24 = 4^(30 + 24) = 4^54Divide by 4^3: 4^54 ÷ 4^3 = 4^(54 - 3) = 4^51Set 4^51 = (64)^k = (4^3)^k = 4^(3k) ⇒ 3k = 51 ⇒ k = 17 Verification / Alternative check: Reverse: (64)^17 = (4^3)^17 = 4^51, which matches the simplified left-hand side. Why Other Options Are Wrong: 10/11/16: Result from arithmetic slips (e.g., adding exponents incorrectly or forgetting the final division by 4^3). Common Pitfalls: Misapplying (a^m)^n as a^(m + n) instead of a^(m*n). Not converting 64 to 4^3 before equating exponents. Final Answer: 17

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