Question 1

Always divisible — For any positive integer n, the number 6^(2n) - 1 is always divisible by which of the following?

Accepted Answer

Introduction / Context:Expressions of the form a^(2n) - 1 often have predictable divisibility features because they factor as (a^n - 1)(a^n + 1). This question focuses on 6^(2n) - 1 and asks whether 5, 7, both, or neither always divide it. Modular arithmetic provides a fast, general proof. Given Data / Assumptions: n is any positive integer. Expression: 6^(2n) - 1. We test divisibility by 5 and by 7. Concept / Approach:Reduce the base 6 modulo 5 and modulo 7. For mod 5, note that 6 ≡ 1 (mod 5), so powers of 6 simplify drastically. For mod 7, note that 6 ≡ -1 (mod 7), and even exponents of -1 give 1. These two quick observations resolve the question without heavy computation. Step-by-Step Solution:Modulo 5: 6 ≡ 1 (mod 5) → 6^(2n) ≡ 1^(2n) ≡ 1 (mod 5) → 6^(2n) - 1 ≡ 0 (mod 5).Modulo 7: 6 ≡ -1 (mod 7) → 6^(2n) ≡ (-1)^(2n) ≡ 1 (mod 7) → 6^(2n) - 1 ≡ 0 (mod 7).Therefore, the expression is divisible by both 5 and 7 for every positive integer n. Verification / Alternative check:Use factorization: 6^(2n) - 1 = (6^n - 1)(6^n + 1). Since consecutive even-odd neighbors differ by 2, at least one factor is divisible by 3, but here the mod 5 and mod 7 arguments already confirm both divisibilities independently. Why Other Options Are Wrong: 11: Not always a divisor for general n. 5 or 7 alone: Both actually divide for all n, so singling out only one is incomplete. Neither: Contradicts the modular results. Common Pitfalls:Overcomplicating with large exponents; forgetting that 6 ≡ 1 (mod 5) and 6 ≡ -1 (mod 7); misapplying exponent rules to modular arithmetic. Final Answer:Both (B) and (C)

Question 2

Perfect-square spotting — Which of the following is not a perfect square of any natural number? Use quick tests and known squares.

Accepted Answer

Introduction / Context:Identifying non-squares quickly is an important test skill. While some options may be recognizable as well-known squares, others require quick checks: digit rules, last-two-digit patterns, and approximating square roots. This question asks you to spot the one number that is not a perfect square. Given Data / Assumptions: Candidates: 34692, 4096, 15129, 15376, 16641. We need to determine which is not a square of a natural number. Use recognition and short checks rather than long calculations. Concept / Approach:Recall known squares: 64^2 = 4096; 123^2 = 15129; 124^2 = 15376; 129^2 = 16641. For 34692, check nearby squares by estimating its square root and comparing to exact squares. Also, use quick filters like last-digit feasibility and digital patterns (optional). Step-by-Step Solution:Confirm known squares: 64^2 = 4096 → square.123^2 = 15129 → square.124^2 = 15376 → square.129^2 = 16641 → square.Evaluate 34692: sqrt(34692) is between 186 (186^2 = 34596) and 187 (187^2 = 34969). Since 34692 lies between these squares, it is not itself a perfect square. Verification / Alternative check:Compute nearest squares precisely: 186^2 = 34596 and 187^2 = 34969; 34692 is strictly between them, confirming it is not a square. Why Other Options Are Wrong: 4096 = 64^2; 15129 = 123^2; 15376 = 124^2; 16641 = 129^2 — all are exact squares, so they are not the correct choice. Common Pitfalls:Assuming an unfamiliar number is not a square without checking; rounding square roots too loosely; misremembering common squares like 64^2. Final Answer:34692

Question 3

Three numbers are related as follows: the first number is exactly twice the second number, and at the same time the first number is half of the third number. If the arithmetic average of all three numbers is 56, determine the smallest number among t…

Accepted Answer

Introduction / Context:This problem tests proportional reasoning with simultaneous relationships between three unknowns and the use of an average. By translating each sentence into equations, we can solve systematically for each number and then identify the smallest value. Given Data / Assumptions: The first number equals twice the second: let first = a, second = b, third = c, so a = 2b. The first is half of the third: a = c / 2, so c = 2a. The average is 56, so (a + b + c) / 3 = 56 → a + b + c = 168. All numbers are real; context suggests positive values. Concept / Approach:Use substitution to reduce the number of variables. Express a and c in terms of b, then substitute into the sum a + b + c = 168. Solve for b, recover a and c, and compare to find the minimum. This approach ensures no step is skipped and that relationships are consistent. Step-by-Step Solution: From a = 2b and c = 2a, we also have c = 4b.Sum: a + b + c = 2b + b + 4b = 7b.Set 7b = 168 → b = 168 / 7 = 24.Then a = 2b = 48 and c = 4b = 96.Among 48, 24, and 96, the smallest is 24. Verification / Alternative check:Average check: (48 + 24 + 96) / 3 = 168 / 3 = 56, matching the given average. Relationship checks: a = 2b → 48 = 2 * 24 (true) and a = c / 2 → 48 = 96 / 2 (true). Why Other Options Are Wrong:36, 40, and 48 do not arise from the solved system. 30 would yield sums and relationships that do not satisfy both conditions simultaneously. Common Pitfalls:Confusing which variable is double or half of which, or averaging incorrectly (e.g., dividing by 2 instead of 3). Another frequent error is forgetting that c depends on a, which itself depends on b. Final Answer:24

Question 4

Let x be exactly divisible by 5 (remainder 0) and let y leave a remainder of 1 when divided by 5. When (x + y) is divided by 5, what remainder is obtained?

Accepted Answer

Introduction / Context:Remainder or modular arithmetic problems often ask how remainders combine under addition. If one addend is a multiple of 5 and the other leaves remainder 1 modulo 5, the sum’s remainder follows directly from basic modular properties. Given Data / Assumptions: x ≡ 0 (mod 5), meaning x is a multiple of 5. y ≡ 1 (mod 5), meaning dividing y by 5 leaves remainder 1. We need (x + y) mod 5. Concept / Approach:Use the modular identity: (a + b) mod m = [(a mod m) + (b mod m)] mod m. This rule allows us to compute the remainder of a sum from the remainders of the terms. No explicit values for x or y are needed, only their equivalence classes modulo 5. Step-by-Step Solution: x mod 5 = 0, y mod 5 = 1.(x + y) mod 5 = (0 + 1) mod 5.Hence (x + y) mod 5 = 1. Verification / Alternative check:Pick examples: let x = 10 (clearly ≡ 0 mod 5) and y = 11 (≡ 1 mod 5). Then x + y = 21, and 21 divided by 5 leaves remainder 1. Any other valid pair will behave the same due to modular equivalence. Why Other Options Are Wrong:0, 2, 3, and 4 contradict the direct application of modular addition. They typically arise from guessing or ignoring the given remainder for y. Common Pitfalls:Assuming that any multiple of 5 changes the remainder of the sum (it does not). Another mistake is to add quotients rather than remainders. Final Answer:1

Question 5

In the product 75? × 49 × 867 × 943, the units digit of the result is 1. Determine the missing digit ? in 75?.

Accepted Answer

Introduction / Context:Determining a missing units digit often relies on modular arithmetic base 10. Only the units digits of the factors affect the final units digit of a product, which allows a compact calculation without full multiplication. Given Data / Assumptions: Number forms: 75? has units digit d (unknown). Units digits of the other factors: 49 → 9, 867 → 7, 943 → 3. The final product’s units digit equals 1. Concept / Approach:Compute the product of units digits modulo 10. Multiply progressively and keep only the last digit. Then solve for d such that d * (last-digit-product) ≡ 1 (mod 10). This uses the idea of multiplicative inverses modulo 10 for the final stage. Step-by-Step Solution: Let d be the units digit of 75?.Multiply known units: 9 * 7 = 63 → units 3.Then 3 * 3 = 9 → units 9.Overall units before including d: 9.We need d * 9 ≡ 1 (mod 10). Since 9 ≡ −1 (mod 10), this gives −d ≡ 1 → d ≡ 9 (mod 10).Hence d = 9. Verification / Alternative check:Replace ? with 9: units(759 * 49 * 867 * 943) = units(9 * 9 * 7 * 3) = units(9 * 9) = 1 after full modular multiplication, confirming the requirement. Why Other Options Are Wrong:Digits 1, 3, 5, or 7 fail d * 9 ≡ 1 (mod 10); their products have units 9, 7, 5, and 3 respectively, not 1. Common Pitfalls:Multiplying entire numbers unnecessarily or forgetting that only units digits matter for determining the final units digit. Final Answer:9

Question 6

The sum of three numbers is 132. The first number is twice the second, and the third number is one-third of the first. Find the value of the second number.

Accepted Answer

Introduction / Context:This is a classic system-of-equations problem stated in words. By expressing each relationship algebraically and using the total sum, we can solve for the second number directly and check consistency afterward. Given Data / Assumptions: Let the numbers be A (first), B (second), and C (third). A + B + C = 132. A = 2B. C = (1/3) * A. Concept / Approach:Reduce to a single variable. Replace A with 2B and C with A/3 = (2B)/3 in the sum. Solve for B. Then back-substitute to verify each relation and the overall sum. Step-by-Step Solution: Substitute: A + B + C = 2B + B + (2B)/3.Combine: 3B + (2/3)B = (11/3)B.Set (11/3)B = 132 → B = 132 * 3 / 11 = 36.Therefore, the second number is 36. Verification / Alternative check:Compute A = 2B = 72 and C = A/3 = 24. Check sum: 72 + 36 + 24 = 132; all conditions hold. Why Other Options Are Wrong:32, 48, and 60 contradict the equation (11/3)B = 132. 24 is actually the value of C, not B. Common Pitfalls:Misplacing the one-third relation (e.g., taking C = 3A) or adding coefficients incorrectly when combining fractional terms. Final Answer:36

Question 7

A two-digit number equals three times the sum of its digits, and adding 45 to the number reverses its digits. What is the number?

Accepted Answer

Introduction / Context:Puzzles about two-digit numbers typically translate into equations using the tens digit and units digit. Here we use two conditions: a value condition (three times the sum of digits) and a digit-reversal condition (adding 45 reverses the digits). Together they identify the number uniquely. Given Data / Assumptions: Let the number be 10a + b, where a is the tens digit and b is the units digit. Condition 1: 10a + b = 3(a + b). Condition 2: (10a + b) + 45 = 10b + a (reversed digits). Concept / Approach:Convert both conditions into linear equations in a and b. Solve one to relate a and b, then use the other to confirm specific integer digits. Because digits are 0–9 and tens digit is 1–9, we can also check divisibility or parity constraints quickly. Step-by-Step Solution: From Condition 1: 10a + b = 3a + 3b → 7a − 2b = 0 → 7a = 2b.This implies b = (7/2)a; therefore, a must be even. Try a = 2 → b = 7.Number candidate: 27. Check Condition 2.27 + 45 = 72, which is the exact reversal of digits (a and b swap positions).Therefore, the number is 27. Verification / Alternative check:Confirm Condition 1 numerically: 27 = 3 * (2 + 7) = 3 * 9 = 27; satisfied. The reversal check also holds precisely. Why Other Options Are Wrong:32, 72, 23, and 54 do not satisfy both conditions simultaneously. For example, 72 is the reversed result after adding 45, not the original number. Common Pitfalls:Forgetting digits must be integers, or misapplying the reversal equation as subtracting 45 instead of adding. Final Answer:27

Question 8

A fraction becomes 4 when 1 is added to both numerator and denominator, and it becomes 7 when 1 is subtracted from both numerator and denominator. What is the numerator of the original fraction?

Accepted Answer

Introduction / Context:Transformations that add or subtract the same amount to both the numerator and denominator create linear equations in the original numerator and denominator. Solving these simultaneously reveals the original fraction’s parts without guesswork. Given Data / Assumptions: Let the fraction be n / d with positive integers n and d. (n + 1) / (d + 1) = 4. (n − 1) / (d − 1) = 7. Concept / Approach:Convert each condition to an equation in n and d, then solve the pair. Because both outcomes are integers (4 and 7), n and d will be small whole numbers that satisfy both equations. Substitution or elimination are straightforward here. Step-by-Step Solution: From (n + 1) / (d + 1) = 4 → n + 1 = 4d + 4 → n = 4d + 3.From (n − 1) / (d − 1) = 7 → n − 1 = 7d − 7 → n = 7d − 6.Equate: 4d + 3 = 7d − 6 → 3d = 9 → d = 3.Then n = 4 * 3 + 3 = 15 (also n = 7 * 3 − 6 = 15).Hence the numerator is 15. Verification / Alternative check:Original fraction 15/3 = 5. Check: (15 + 1)/(3 + 1) = 16/4 = 4 and (15 − 1)/(3 − 1) = 14/2 = 7, both matching the statements. Why Other Options Are Wrong:2, 3, 7, and 9 do not satisfy both linear conditions simultaneously when paired with an integer denominator. Common Pitfalls:Adding 1 to only the numerator or denominator, or assuming the result must remain a proper fraction rather than allowing integer results. Final Answer:15

Question 9

The sum of squares of two numbers is 68, and the square of their difference is 36. Find the product of the two numbers.

Accepted Answer

Introduction / Context:This question leverages the identity for the square of a difference and its relationship to the product term. With aggregate information (sum of squares and squared difference), we can deduce the product directly without solving for individual numbers. Given Data / Assumptions: a^2 + b^2 = 68. (a − b)^2 = 36. We need ab. Concept / Approach:Use the identity (a − b)^2 = a^2 + b^2 − 2ab. Plug in the given values and solve for ab. This bypasses solving a quadratic for a and b and goes straight to the required product. Step-by-Step Solution: Start: (a − b)^2 = a^2 + b^2 − 2ab.Substitute: 36 = 68 − 2ab.Rearrange: 2ab = 68 − 36 = 32.Therefore, ab = 32 / 2 = 16. Verification / Alternative check:Optional: (a + b)^2 = a^2 + b^2 + 2ab = 68 + 32 = 100 → a + b = 10 is consistent with real solutions (e.g., numbers satisfying sum 10 and product 16 exist). Why Other Options Are Wrong:32 represents 2ab, not ab. 58 and 104 are inconsistent with the identity. 22 is arbitrary here. Common Pitfalls:Forgetting to divide by 2 after finding 2ab, or mixing identities for (a − b)^2 and (a + b)^2. Final Answer:16

Question 10

There are two numbers x and y such that 2x + 3y = 18 and 3x + 2y = 17. What is the larger of the two numbers?

Accepted Answer

Introduction / Context:This is a linear system in two variables. Solving simultaneously (by elimination or substitution) yields the unique pair. The final request is to select the larger of x and y from the solution pair. Given Data / Assumptions: 2x + 3y = 18. 3x + 2y = 17. We seek max{x, y}. Concept / Approach:Use elimination: align coefficients of one variable and subtract equations to isolate the other variable. After finding y, back-substitute to compute x. Compare the two values to determine the larger. Step-by-Step Solution: Multiply the first equation by 3: 6x + 9y = 54.Multiply the second equation by 2: 6x + 4y = 34.Subtract: (6x + 9y) − (6x + 4y) = 54 − 34 → 5y = 20 → y = 4.Substitute into 2x + 3y = 18: 2x + 12 = 18 → 2x = 6 → x = 3.Thus the larger number is 4. Verification / Alternative check:Check in the second equation: 3x + 2y = 9 + 8 = 17, correct. Consistency confirms the solution. Why Other Options Are Wrong:6, 8, and 12 are not values of x or y here. 5 is not obtained from the system. The correct larger value among (3, 4) is 4. Common Pitfalls:Arithmetic slips during elimination, or mixing which equation to multiply—both can create sign errors and wrong results. Final Answer:4

Question 11

Three numbers satisfy: the sum of the first and second is 45, the sum of the second and third is 55, and the sum of the third and thrice the first is 90. Find the third number.

Accepted Answer

Introduction / Context:Word problems with multiple sums over three unknowns can be untangled by expressing each variable in terms of another and substituting step by step. This yields a simple linear equation whose solution leads directly to the requested value. Given Data / Assumptions: Let the numbers be a (first), b (second), c (third). a + b = 45. b + c = 55. c + 3a = 90. Concept / Approach:Express b from the first equation and c from the second in terms of a, then substitute into the third equation. Solve for a, then compute c. This approach minimizes simultaneous juggling of all three variables at once. Step-by-Step Solution: From a + b = 45 → b = 45 − a.From b + c = 55 → c = 55 − b = 55 − (45 − a) = 10 + a.Plug into c + 3a = 90: (10 + a) + 3a = 90.Simplify: 4a + 10 = 90 → 4a = 80 → a = 20.Then c = 10 + a = 30. Verification / Alternative check:Check sums: a + b = 45 → b = 25; then b + c = 25 + 30 = 55; and c + 3a = 30 + 60 = 90. All conditions are satisfied. Why Other Options Are Wrong:20 and 25 are values of a and b respectively, not the third number. 35 and 40 do not satisfy the third equation when combined with the derived values. Common Pitfalls:Swapping which variable is expressed or sign mistakes when substituting b into the expression for c, leading to off-by-10 errors. Final Answer:30

Question 12

A number leaves remainder 3 when divided by 5. What remainder does the square of the same number leave upon division by 5?

Accepted Answer

Introduction / Context:Remainders of powers are efficiently handled using modular arithmetic. Knowing the remainder of a number modulo 5 allows quick determination of the remainder of its square modulo 5 by squaring within the modular system. Given Data / Assumptions: Let n ≡ 3 (mod 5), meaning n leaves remainder 3 upon division by 5. We seek n^2 mod 5. All operations are within integers modulo 5. Concept / Approach:Apply the rule: if n ≡ r (mod m), then n^2 ≡ r^2 (mod m). Therefore, it suffices to compute 3^2 modulo 5 and report the remainder. This avoids dealing with any specific value of n. Step-by-Step Solution: Given n ≡ 3 (mod 5).Compute r^2: 3^2 = 9.Reduce modulo 5: 9 mod 5 = 4.Hence n^2 ≡ 4 (mod 5). Verification / Alternative check:Example: n = 8 (since 8 ≡ 3 mod 5). Then n^2 = 64, and 64 mod 5 = 4, confirming the result. Why Other Options Are Wrong:9 is the unsimplified square; modulo 5 it equals 4. 3 and 1 correspond to squaring 3 incorrectly or confusing with other residues; 2 is unrelated here. Common Pitfalls:Forgetting to reduce the squared value modulo 5, or mistakenly thinking the remainder remains 3 after squaring. Final Answer:4

Question 13

Divisor–Dividend Relation with Given Remainder In a division problem, the divisor is two-thirds (2/3) of the dividend and also equals two times the remainder. If the remainder is 5, determine the value of the dividend.

Accepted Answer

Introduction / Context:This problem tests understanding of the standard division identity and how to interpret verbal relationships among dividend, divisor, quotient, and remainder. It is a classic number system application that rewards setting up clean equations before computing. Given Data / Assumptions: Remainder R = 5. Divisor d is two times the remainder, so d = 2 * R. Divisor d is also two-thirds of the dividend D, so d = (2/3) * D. Division identity: D = d * q + R for some integer quotient q. Concept / Approach:Use the two given equalities involving the divisor to solve for D directly. Then verify with the division identity that an integer quotient exists and the remainder matches 5. Step-by-Step Solution:From d = 2 * R and R = 5 → d = 10.Also d = (2/3) * D → 10 = (2/3) * D → D = 10 * 3 / 2 = 15.Check the division identity: 15 = 10 * q + 5 → q = 1 (an integer), valid. Verification / Alternative check:Confirm consistency: divisor equals two-thirds of the dividend (10 = 2/3 of 15) and also equals twice the remainder (10 = 2 * 5). Both conditions hold simultaneously. Why Other Options Are Wrong:25, 18, 24, and 30 fail at least one of the relationships d = (2/3)D or d = 2R with R = 5, or fail the identity D = d * q + R with integer q. Common Pitfalls:Confusing “two-thirds of the dividend” with “dividend is two-thirds of divisor,” or forgetting to check the division identity after computing D. Final Answer:15

Question 14

Counting Digits Used in Page Numbering How many individual digits are required to number all pages of a book from page 1 through page 200 (inclusive)?

Accepted Answer

Introduction / Context:Digit-counting problems are common in aptitude tests. The idea is to partition the page numbers by digit-length groups (1-digit, 2-digit, 3-digit) and sum their contributions. Given Data / Assumptions: Pages run consecutively from 1 to 200 inclusive. Digits needed for a k-digit page number equals k. We compute by ranges: 1–9, 10–99, 100–200. Concept / Approach:Use place-value counting: the number of pages in each range times the digits per page in that range. Then add all contributions. Step-by-Step Solution:Pages 1–9: 9 pages * 1 digit = 9.Pages 10–99: 90 pages * 2 digits = 180.Pages 100–200: 101 pages * 3 digits = 303.Total digits = 9 + 180 + 303 = 492. Verification / Alternative check:Sanity check: A large portion comes from 3-digit pages (just over 300 digits), which aligns with the final total near 500. Why Other Options Are Wrong:200 and 372 underestimate by ignoring digit-length growth; 600 overshoots; 489 is a near-miss from miscounting pages 100–200. Common Pitfalls:Excluding page 200 or miscounting 100–199 as an even 100 pages (it is 101 pages from 100 through 200 inclusive). Final Answer:492

Question 15

Two-Digit Number from Digit Constraints and Halving Rule A two-digit number has its units digit equal to the tens digit of half the number. Its tens digit is one less than the units digit of half the number. If the sum of the two digits is 7, determ…

Accepted Answer

Introduction / Context:This puzzle ties the digits of a two-digit number to the digits of half that number. It assesses digit reasoning, parity (to ensure halving stays integral), and translating statements into equations. Given Data / Assumptions: Let the number be 10a + b (tens digit a, units digit b). Half the number is (10a + b)/2, which must be an integer, so b is even. Units digit of the number equals the tens digit of half: b equals tens( (10a + b)/2 ). Tens digit of the number is one less than the units digit of half: a = units( (10a + b)/2 ) − 1. a + b = 7. Concept / Approach:Let half be 10x + y. Then b = x and a = y − 1. Also 10x + y = (10a + b)/2. Substitute x = b and y = a + 1 to relate a and b directly. Step-by-Step Solution:Set 10b + (a + 1) = (10a + b)/2.Multiply by 2: 20b + 2a + 2 = 10a + b.Rearrange: 19b − 8a + 2 = 0.Use a + b = 7 → b = 7 − a.Substitute: 19(7 − a) − 8a + 2 = 0 → 133 − 19a − 8a + 2 = 0 → 135 − 27a = 0 → a = 5.Then b = 2. Number = 52. Verification / Alternative check:Half of 52 is 26; tens(26) = 2 equals b, units(26) = 6 and a = 6 − 1 = 5; digit sum 5 + 2 = 7. All conditions satisfied. Why Other Options Are Wrong:16, 34, and 25 do not satisfy both digit constraints simultaneously; “Data in inadequate” is incorrect because the constraints determine a unique solution. Common Pitfalls:Forgetting half must be an integer; swapping the roles of tens and units; ignoring the digit-sum condition. Final Answer:52

Question 16

Two-Digit Number from Digit-Sum Conditions A two-digit number equals seven times the sum of its digits. If each digit is increased by 2, the resulting number is 4 more than six times the sum of its new digits. Find the original number.

Accepted Answer

Introduction / Context:This is a standard two-equation, two-unknowns problem using digit variables for a two-digit number. Such problems appear frequently in aptitude tests, requiring careful translation from words to equations. Given Data / Assumptions: Let tens digit be a and units digit be b; number = 10a + b. Condition 1: 10a + b = 7(a + b). Condition 2: New number after increasing each digit by 2 is 10(a + 2) + (b + 2), which equals 4 more than 6 times the sum of new digits. Concept / Approach:Convert each statement to an equation and solve simultaneously for integers a and b between 0 and 9, with a ≥ 1 for a valid two-digit number. Step-by-Step Solution:From Condition 1: 10a + b = 7a + 7b → 3a = 6b → a = 2b.From Condition 2: 10a + b + 22 = 6(a + b + 4) + 4 = 6a + 6b + 28.Rearrange: 4a − 5b = 6.Substitute a = 2b → 8b − 5b = 6 → 3b = 6 → b = 2 → a = 4.Number = 42. Verification / Alternative check:Check Condition 1: 42 = 7*(4 + 2) = 42. Check Condition 2: New number 64; sum of new digits = 6 + 4 = 10; 6*10 + 4 = 64. Both satisfied. Why Other Options Are Wrong:24, 48, and 36 fail one or both equations; “Data adequate” is not an answer value and does not identify the number. Common Pitfalls:Misplacing constants (22 vs 28) in the second equation; assuming a = b instead of using a = 2b from the first equation. Final Answer:42

Question 17

Sequential Transformations on a Number A number, when decreased by 4 and then divided by 6, gives 9. What is the result if instead the number is decreased by 3 and then divided by 5?

Accepted Answer

Introduction / Context:Translate stepwise verbal operations into algebraic equations. First, recover the original number from the given condition. Then apply the new transformation and compute the final value exactly. Given Data / Assumptions: (N − 4)/6 = 9. We must then compute (N − 3)/5. N is real (and will turn out to be integer). Concept / Approach:Solve the first linear equation for N. Substitute that value into the second expression to obtain the requested result. Mixed-number options indicate that exact division may produce an integer or a fractional result, so compute carefully. Step-by-Step Solution:(N − 4)/6 = 9 → N − 4 = 54 → N = 58.Compute (N − 3)/5 = (58 − 3)/5 = 55/5 = 11. Verification / Alternative check:Plug back: (58 − 4)/6 = 54/6 = 9, matching the initial condition. The second calculation is straightforward division yielding an integer. Why Other Options Are Wrong:9 2/5, 10 1/5, 11 2/5, and 10 4/5 correspond to incorrect substitutions or arithmetic slips. Only 11 matches the correct evaluation. Common Pitfalls:Subtracting 3 from 9 instead of from N; mixing order of operations; performing 55/5 incorrectly as a mixed number. Final Answer:11

Question 18

Three Variables with Close Ratios Numbers x, y, z satisfy xy = 96050 and xz = 95625, with y exceeding z by one (y = z + 1). Find the value of z.

Accepted Answer

Introduction / Context:This problem blends ratio reasoning with products of unknowns. The key is to divide the two product equations to relate y and z directly; then use the given difference condition to pin down exact values. Given Data / Assumptions: xy = 96050. xz = 95625. y = z + 1. Concept / Approach:Divide the equations: (xy)/(xz) = y/z = 96050/95625. Reduce the fraction to simplest terms to express y : z. Then insert y = z + 1 and solve for both as integer multiples of the simplified ratio. Step-by-Step Solution:y/z = 96050/95625.Reduce 96050/95625 by gcd 425 → 226/225.Thus y : z = 226 : 225.But y = z + 1 fits perfectly with 226 and 225 → let z = 225 and y = 226.Hence z = 225. Verification / Alternative check:With z = 225 and y = 226, the ratio y/z = 226/225 matches the reduced fraction. Existence of some x is guaranteed because xy and xz are given; x = (xz)/z = 95625/225 is consistent. Why Other Options Are Wrong:425, 220, 525, and 221 do not satisfy both the ratio 226:225 and the “y = z + 1” constraint simultaneously. Common Pitfalls:Failing to reduce the fraction fully; assuming y − z is greater than 1; mixing y and z when applying the ratio. Final Answer:225

Question 19

Numbers with Fixed Ratio and a Composite Condition Two numbers are in the ratio 3 : 2. If 10 and the sum of the two numbers are added to their product, the result is 16 squared (256). What is the smaller number?

Accepted Answer

Introduction / Context:Here, ratio information lets us parametrize the numbers in terms of a single variable. The combined condition involving product, sum, and a constant then determines that variable precisely. Given Data / Assumptions: Let the numbers be 3k and 2k. Product + sum + 10 = 256. We need the smaller number (2k). Concept / Approach:Compute product and sum in terms of k, form a quadratic equation, and solve for k. Select the positive integer solution that makes sense in the context and then report 2k. Step-by-Step Solution:Product = (3k)(2k) = 6k^2.Sum = 3k + 2k = 5k.Condition: 6k^2 + 5k + 10 = 256 → 6k^2 + 5k − 246 = 0.Discriminant = 5929 = 77^2; k = (−5 ± 77)/12 → k = 6 (positive), or negative root (discard).Smaller number = 2k = 12. Verification / Alternative check:Numbers are 18 and 12. Product 216; sum 30; product + sum + 10 = 256 = 16^2, satisfied. Why Other Options Are Wrong:14, 16, and 18 are not equal to 2k with k = 6; 10 breaks the ratio with the corresponding partner. Common Pitfalls:Mistaking which is smaller; arithmetic slips forming the quadratic; forgetting to add the 10 constant. Final Answer:12

Question 20

Two-Digit Number with Units Digit Tied to Tens Digit A two-digit number has its units digit equal to (2 × tens digit) + 1. After interchanging the digits, the difference between the new number and the original equals the original number minus 1. Wha…

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Introduction / Context:This digit puzzle encodes the number's structure into algebra. Recognizing the standard difference formula for reversed two-digit numbers accelerates the solution: (10b + a) − (10a + b) = 9(b − a). Given Data / Assumptions: Let number be 10a + b with digits a (tens) and b (units). Digit rule: b = 2a + 1. After reversing digits, difference (new − original) equals (original − 1). Concept / Approach:Use the reversal difference identity and substitute the linear relation between b and a. Solve for a and b in digit range constraints (a from 1 to 9, b from 0 to 9). Step-by-Step Solution:Difference after reversal: 9(b − a).Given: 9(b − a) = (10a + b) − 1.Substitute b = 2a + 1: LHS = 9((2a + 1) − a) = 9a + 9; RHS = 10a + (2a + 1) − 1 = 12a.Equation: 9a + 9 = 12a → 3a = 9 → a = 3.Then b = 2*3 + 1 = 7 → number = 37. Verification / Alternative check:Reverse 37 → 73; difference 73 − 37 = 36; original − 1 = 37 − 1 = 36. Both conditions hold. Why Other Options Are Wrong:52, 25, and 39 do not satisfy b = 2a + 1; 73 is the reversed number, not the original. Common Pitfalls:Misinterpreting the difference statement, or using b = 2a − 1 instead of b = 2a + 1. Final Answer:37

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