Question 1

Telescoping Product with Odd Fractions Evaluate the product (2 − 1/3) (2 − 3/5) (2 − 5/7) … (2 − 997/999).

Accepted Answer

Introduction / Context: This product has a telescoping structure. Each factor is of the form 2 − n/(n + 2) with n running over odd numbers 1, 3, 5, …, 997. Recognizing and converting each factor to a simple ratio produces massive cancellation. Given Data / Assumptions: General term: 2 − n/(n + 2), where n takes odd values from 1 to 997. Product spans consecutive odd n, so adjacent terms share canceling numerators and denominators after simplification. Concept / Approach: Rewrite each factor: 2 − n/(n + 2) = (2(n + 2) − n)/(n + 2) = (n + 4)/(n + 2). The product becomes Π (n + 4)/(n + 2). Because n increases by 2 each step, most factors cancel. Step-by-Step Solution: Transform factors: (2 − n/(n + 2)) → (n + 4)/(n + 2).Sequence of numerators: 5, 7, 9, …, 1001.Sequence of denominators: 3, 5, 7, …, 999.Telescoping cancellation leaves only 1001/3. Verification / Alternative check: Check initial and final fringes: first denominator 3 remains in the bottom; last numerator 1001 remains on top; every intermediate 5, 7, …, 999 cancels. Why Other Options Are Wrong: 5/999 and 1001/999 keep extraneous terms; “None of these” is incorrect since a clean closed form 1001/3 exists; 1000/3 is off by 1 from the final numerator. Common Pitfalls: Failing to recognize the (n + 4)/(n + 2) representation; truncating the product incorrectly at endpoints; arithmetic slips with large boundary values. Final Answer: 1001/3

Question 2

Telescoping Product of the Form (1 − 1/k) Compute the product (1 − 1/3) (1 − 1/4) (1 − 1/5) … (1 − 1/n) for integer n ≥ 3.

Accepted Answer

Introduction / Context: Products of the form (1 − 1/k) are best handled by rewriting as simple fractions that telescope. This saves time and avoids expanding long products manually. Given Data / Assumptions: Product index runs from k = 3 to k = n, where n ≥ 3. Each term simplifies to (k − 1)/k. Concept / Approach: Rewrite each factor and cancel successive numerator–denominator pairs. Only the earliest numerator and the last denominator survive the cancellations, revealing a compact closed-form expression. Step-by-Step Solution: (1 − 1/k) = (k − 1)/k.Product from k = 3 to n: Π (k − 1)/k = (2/3) * (3/4) * (4/5) * … * ((n − 1)/n).Telescoping leaves 2/n (all intermediate factors cancel). Verification / Alternative check: Test n = 5: (1 − 1/3)(1 − 1/4)(1 − 1/5) = (2/3)(3/4)(4/5) = 2/5 = 2/n, confirming the formula. Why Other Options Are Wrong: 1/n misses the factor 2; 2(n − 1)/n and 2(n + 1)/n do not arise from telescoping; 1/(n − 1) is a common mis-cancellation error. Common Pitfalls: Dropping the initial 2/3 factor; misaligning the cancellation ends; attempting to sum instead of multiplying. Final Answer: 2/n

Question 3

Divisibility test using 9 and 11 (for 99) Which of the following integers is exactly divisible by 99 (i.e., divisible by both 9 and 11)?

Accepted Answer

Introduction / Context: To be divisible by 99, a number must be divisible by both 9 and 11. This question tests your fluency with quick divisibility checks without doing long division. Given Data / Assumptions: Four candidate numbers are provided. Rule for 9: sum of digits must be a multiple of 9. Rule for 11: (sum of digits in alternating positions) must differ by a multiple of 11 (including 0). Concept / Approach: Apply the tests sequentially. First screen with the easy 9-test (digit sum). For those passing, check the 11-test using alternating sum starting from the leftmost digit with +, −, +, −, … signs. Step-by-Step Solution: 114345: digit sum = 1+1+4+3+4+5 = 18 → multiple of 9.11-test: 1−1+4−3+4−5 = 0 → multiple of 11 → passes.Other options fail at least one test: e.g., 135792 has digit sum 27 (ok) but 11-test gives 3 (not multiple of 11); 913464: 27 (ok) but 11-test gives 9; 3572404 digit sum 25 (fails 9-test). Verification / Alternative check: Since 114345 passes both 9 and 11 tests, it is divisible by 99 (because 9 and 11 are coprime). Why Other Options Are Wrong: 3572404: fails divisibility by 9.135792 and 913464: pass 9 but fail the 11-test. Common Pitfalls: Mixing the 11-test sign pattern or summing all digits for the 11-test instead of alternating signs. Also, stopping after the 9-test and forgetting to check 11. Final Answer: 114345

Question 4

Divisibility by 8 (last three digits rule) What least digit should replace * in 91876*2 so that the number is divisible by 8?

Accepted Answer

Introduction / Context: A number is divisible by 8 if and only if its last three digits form a number divisible by 8. This problem asks you to choose the least digit replacing * to satisfy that condition. Given Data / Assumptions: Number pattern: 91876*2. Only the last three digits affect divisibility by 8. We must find the smallest digit that works. Concept / Approach: Consider the last three-digit number formed: 6*2. Let * = d. Then the value is 600 + 10d + 2. Reduce this expression modulo 8 to enforce divisibility. Step-by-Step Solution: Compute (600 + 10d + 2) mod 8.Since 600 mod 8 = 0 and 10 mod 8 = 2, expression ≡ (0 + 2d + 2) mod 8.We need 2d + 2 ≡ 0 (mod 8) ⇒ 2d ≡ 6 (mod 8) ⇒ d ≡ 3 (mod 4).Digits satisfying this are 3 and 7; the least such digit is 3. Verification / Alternative check: Test d = 3: last three digits 632; 632 ÷ 8 = 79 exactly. Hence the full number is divisible by 8. Why Other Options Are Wrong: 1, 2, 4 lead to last-three-digit numbers 612, 622, 642, which are not divisible by 8. Common Pitfalls: Checking more than the last three digits; forgetting to reduce 10 to 2 modulo 8; picking 7 (also valid) when the question asks for the least such digit. Final Answer: 3

Question 5

Divisibility by 9 (digit-sum rule) What least digit should replace * in 6135*2 so that the number is exactly divisible by 9?

Accepted Answer

Introduction / Context: Divisibility by 9 is determined by the sum of digits: the digit sum must be a multiple of 9. Here we choose the smallest digit * making 6135*2 divisible by 9. Given Data / Assumptions: Digits fixed except * in 6135*2. Digit sum with * included must be 9k. We want the least digit that works. Concept / Approach: Compute the current digit sum and then find * so that the total is the next multiple of 9. This is faster than trial division. Step-by-Step Solution: Digit sum without *: 6+1+3+5+2 = 17.We need 17 + * to be a multiple of 9.The next multiple of 9 after 17 is 18, so * = 1.Therefore, the least suitable digit is 1. Verification / Alternative check: With * = 1, full sum is 18; numbers with digit sum 18 are divisible by 9. Quick test by dividing confirms exact divisibility. Why Other Options Are Wrong: 0 gives sum 17; 2 gives 19; 3 gives 20. None are multiples of 9. Common Pitfalls: Choosing 10 − (17 mod 9) = 1 by mistake for 10-based rounding rather than 9; forgetting that the target must be a multiple of 9. Final Answer: 1

Question 6

Divisibility by 11 (alternating-sum test) Find the least digit * so that 97215*6 is divisible by 11.

Accepted Answer

Introduction / Context: A number is divisible by 11 if the alternating-sum of its digits is a multiple of 11 (including 0). This problem asks for the smallest digit * that makes 97215*6 divisible by 11. Given Data / Assumptions: Number pattern: 9 7 2 1 5 * 6. Alternating sum taken left to right with +, −, +, −, … Result must be 0, ±11, ±22, … Concept / Approach: Compute S = 9 − 7 + 2 − 1 + 5 − * + 6. Set S equal to a multiple of 11 and solve for * among digits 0–9. Choose the minimal valid digit. Step-by-Step Solution: Compute without *: 9−7=2; +2=4; −1=3; +5=8; +6=14.Expression becomes 14 − *.Set 14 − * ≡ 0 (mod 11) ⇒ 14 − * = 11 (since 14 and * are digits-sized apart).Hence * = 3 (least valid digit). Verification / Alternative check: With * = 3, S = 14 − 3 = 11 → divisible by 11, so the number is divisible by 11. Why Other Options Are Wrong: 1, 2, 5 give S = 13, 12, 9 respectively, none of which are multiples of 11. Common Pitfalls: Reversing the ± pattern; forcing S = 0 only, when any multiple of 11 suffices; choosing the largest valid digit instead of the least. Final Answer: 3

Question 7

Make 1056 divisible by 23 What is the least number that must be added to 1056 to obtain a number exactly divisible by 23?

Accepted Answer

Introduction / Context: Finding the least addition to reach a multiple of a divisor is a standard modulus exercise. Compute the remainder and add the complement to the next multiple. Given Data / Assumptions: N = 1056; divisor d = 23. We need k such that 1056 + k is a multiple of 23. k should be the smallest nonnegative integer that works. Concept / Approach: Let r = 1056 mod 23. If r = 0, k = 0. Otherwise, k = 23 − r. This ensures 1056 + k ≡ 0 (mod 23). Step-by-Step Solution: Compute 23 × 46 = 1058 ⇒ 1056 is 2 less than 1058.Therefore, 1056 mod 23 = 23 − 2 = 21, or simply note 1056 = 23×45 + 21.To reach the next multiple, add k = 23 − 21 = 2.Thus, the least number to add is 2. Verification / Alternative check: 1056 + 2 = 1058 = 23 × 46, confirming exact divisibility. Why Other Options Are Wrong: 3 or 21 overshoot the next multiple or correspond to different steps; 25 is unnecessary since 2 already works. Common Pitfalls: Confusing the remainder 21 with the needed addition; adding 21 instead of 2; not checking the very next multiple. Final Answer: 2

Question 8

Divisibility of 10^n − 1 by 11 For which values of n is (10^n − 1) divisible by 11?

Accepted Answer

Introduction / Context: Understanding periodic behavior modulo 11 is essential in number theory. This question explores when numbers of the form 10^n − 1 are divisible by 11. Given Data / Assumptions: We analyze 10^n modulo 11. Use modular exponent properties. n is a positive integer (including n = 0 is trivial but not required). Concept / Approach: Since 10 ≡ −1 (mod 11), we have 10^n ≡ (−1)^n (mod 11). Therefore 10^n − 1 ≡ (−1)^n − 1. This equals 0 exactly when n is even (because (−1)^{even} = 1). Step-by-Step Solution: Compute 10 ≡ −1 (mod 11).Then 10^n ≡ (−1)^n (mod 11).So 10^n − 1 ≡ (−1)^n − 1.If n is even: (−1)^n = 1 ⇒ expression ≡ 0 ⇒ divisible by 11.If n is odd: (−1)^n = −1 ⇒ expression ≡ −2 ⇒ not divisible by 11. Verification / Alternative check: Try n = 2: 10^2 − 1 = 99 divisible by 11. Try n = 1: 9 not divisible by 11. Pattern holds. Why Other Options Are Wrong: All values / odd / multiples of 11 contradict the modular result based on (−1)^n. Common Pitfalls: Assuming divisibility for all n because 99 is divisible by 11; forgetting 10 ≡ −1 (mod 11) and its parity effect. Final Answer: Even values of n

Question 9

Prime identification by quick factoring Which of the following numbers is a prime number?

Accepted Answer

Introduction / Context: Determining primality quickly often involves testing divisibility by small primes. Here, each candidate can be checked by trying primes up to its square root and spotting easy factorizations. Given Data / Assumptions: Candidates: 119, 187, 247. Test divisibility by 7, 11, 13, 17, 19, etc., up to √n. Composite if any nontrivial factor is found. Concept / Approach: Use small-prime trials: check 7, 11, 13, 17, 19. Products of familiar pairs help (e.g., 7×17, 11×17, 13×19). Step-by-Step Solution: 119 = 7 × 17 ⇒ composite.187 = 11 × 17 ⇒ composite.247 = 13 × 19 ⇒ composite.Therefore, none of the listed numbers is prime. Verification / Alternative check: Approximate square roots: √119 ≈ 10.9, √187 ≈ 13.7, √247 ≈ 15.7. All found factors are ≤ these, confirming valid checks. Why Other Options Are Wrong: Each specific number factors nontrivially; selecting any would incorrectly assert primality. Common Pitfalls: Stopping after testing 2, 3, 5 only; overlooking composite forms from recognizable products like 7×17 or 13×19. Final Answer: None of these

Question 10

Reconstruct the divisor from long-division remainders In a long division, the dividend is 380606 and the successive remainders (first to last) are 434, 125, and 413. What is the divisor?

Accepted Answer

Introduction / Context: In multi-digit long division, each partial remainder is strictly less than the divisor, and the final remainder relates the dividend, quotient, and divisor. Sometimes, recognizing compatible quotient–divisor pairs solves the puzzle rapidly. Given Data / Assumptions: Dividend N = 380606. Successive remainders: 434 → 125 → 413 (final). Remainders must be less than the divisor. Concept / Approach: Try the plausible three-digit divisor exceeding the observed remainders. Check whether N can be expressed as divisor * some integer + final remainder. If an option also matches a natural quotient seen in long division, it is very likely correct. Step-by-Step Solution: Test divisor d = 843 (greater than 434, 125, 413).Compute a trial quotient: 380606 − 413 = 380193.Check 380193 ÷ 843 = 451 exactly (since 843 × 451 = 380193).Hence N = 843 × 451 + 413, consistent with the given final remainder and the long-division structure. Verification / Alternative check: The presence of 451 among options hints it is the quotient for d = 843. Moreover, all intermediate remainders are less than 843, which is necessary for validity. Why Other Options Are Wrong: 451 is a quotient, not a divisor; 4215, 3372 are unreasonably large and do not align cleanly with the final remainder structure. Common Pitfalls: Forgetting that remainders must be less than the divisor; trying to compute every long-division step instead of verifying N = d*q + r_final. Final Answer: 843

Question 11

Two-digit number with digit sum 10 A two-digit number has digits that sum to 10. Reversing the digits decreases the number by 36. What is the product of the two digits?

Accepted Answer

Introduction / Context: Digit problems translate neatly into linear equations. With a two-digit number 10a + b (tens digit a, units digit b), reversing produces 10b + a. Conditions on sum and difference yield a and b directly. Given Data / Assumptions: a + b = 10. (10a + b) − (10b + a) = 36 (number decreases by 36 on reversal). Digits a, b are integers 0–9; a ≥ 1 for a two-digit number. Concept / Approach: Subtract to eliminate place-value weights: (10a + b) − (10b + a) = 9a − 9b = 36 ⇒ a − b = 4. Then solve the 2×2 system with the sum condition to get a and b. Step-by-Step Solution: From a + b = 10 and a − b = 4, add the equations.2a = 14 ⇒ a = 7.Then b = 10 − 7 = 3.Product = a * b = 7 * 3 = 21. Verification / Alternative check: Original number 73; reversed 37; difference 73 − 37 = 36; digit sum 7 + 3 = 10 — fully consistent. Why Other Options Are Wrong: 24, 36, 42 correspond to other pairs that do not satisfy both the sum and difference conditions simultaneously. Common Pitfalls: Reversing the difference sign (using 36 the wrong way around); forgetting that a is the tens digit (must be ≥ 1). Final Answer: 21

Question 12

Recovery fix (even numbers): sum equals 190 Five consecutive even numbers sum to 190. What is the sum of the largest and the smallest numbers?

Accepted Answer

Introduction / Context: The original stem mentioning “odd” with sum 190 is inconsistent (sum of five odd integers is odd). Applying the Recovery-First Policy, we minimally repair the item to “five consecutive even numbers,” which aligns with 190. Now we can solve using averages. Given Data / Assumptions: Five consecutive even integers. Total sum S = 190. We need the sum of the smallest and largest. Concept / Approach: For any five consecutive even numbers, the middle number equals the average (S/5). The sequence can be written as m−4, m−2, m, m+2, m+4 (with m even). The extremes sum to (m−4) + (m+4) = 2m, which is also 2 × (average). Step-by-Step Solution: Compute average: m = S / 5 = 190 / 5 = 38.Numbers: 34, 36, 38, 40, 42.Smallest + largest = 34 + 42 = 76.Therefore, the required sum is 76. Verification / Alternative check: Check total: 34 + 36 + 38 + 40 + 42 = 190; extremes sum 76 as computed. Any five-term symmetric AP has extremes sum = 2 × average. Why Other Options Are Wrong: 73, 75, 77 arise from using consecutive odds or incorrect averaging; they do not match the repaired even-number scenario consistent with S = 190. Common Pitfalls: Not repairing the stem; using 5 consecutive integers instead of even integers; miscomputing average as 190/4. Final Answer: 76

Question 13

Evaluate the rational expression for given values:  If a = 16 and b = 5, compute the exact value of (a^2 + b^2 + a*b) / (a^3 − b^3). Show the simplified fraction.

Accepted Answer

Introduction / Context: This problem checks fluency with powers, factorization, and reducing rational expressions. By plugging in a = 16 and b = 5 carefully and then simplifying the fraction, we avoid large, unnecessary computations and reach the exact simplified answer. Given Data / Assumptions: a = 16, b = 5. Expression: (a^2 + b^2 + a*b) / (a^3 − b^3). We must produce the simplest exact fraction (no decimals). Concept / Approach: First evaluate the numerator and denominator separately. After substituting, look for common factors. If both numerator and denominator share a nontrivial factor, reduce to lowest terms. Basic identities like a^3 − b^3 = (a − b)(a^2 + a*b + b^2) can also guide simplification logic even if we compute directly from numbers. Step-by-Step Solution: Compute numerator: a^2 + b^2 + a*b = 16^2 + 5^2 + 16*5 = 256 + 25 + 80 = 361.Compute denominator: a^3 − b^3 = 16^3 − 5^3 = 4096 − 125 = 3971.Form the fraction: 361 / 3971.Note 361 = 19^2 and 3971 = 19 * 209 (since 19*200 = 3800 and 19*9 = 171; sum = 3971).Reduce: (19^2)/(19*209) = 19/209. Since 209 = 11*19, we get 19/209 = 1/11. Verification / Alternative check: Using the identity a^3 − b^3 = (a − b)(a^2 + a*b + b^2) shows immediately that (a^2 + a*b + b^2) cancels, giving 1/(a − b) = 1/(16 − 5) = 1/11. This matches our arithmetic reduction. Why Other Options Are Wrong: 1/19 and 121/3971 are unreduced or incorrect simplifications; 19/209 is an intermediate unreduced form; “None of these” is invalid because 1/11 is attainable. Common Pitfalls: Forgetting the cube-difference identity; stopping at 361/3971 and not reducing; arithmetic slips when computing powers (16^3 or 5^3). Final Answer: 1/11

Question 14

Form and solve the quadratic equation:  “Five times a positive integer equals 3 less than twice the square of that integer.” Find the integer.

Accepted Answer

Introduction / Context: This is a classic translate-to-equation problem. Converting the sentence into an algebraic equation and then solving the resulting quadratic will yield the required positive integer. Such questions assess equation formation and careful arithmetic. Given Data / Assumptions: The unknown is a positive integer n. Statement: 5*n equals 2*n^2 − 3. We seek the positive integer solution. Concept / Approach: Translate the sentence into an equation, collect like terms, and solve the quadratic using factoring or the quadratic formula. Check which root is positive and integral, as required by the problem statement. Step-by-Step Solution: Write the equation: 5n = 2n^2 − 3.Rearrange: 2n^2 − 5n − 3 = 0.Compute discriminant: D = (−5)^2 − 4*2*(−3) = 25 + 24 = 49.Roots: n = [5 ± 7] / (2*2) = (5 ± 7)/4 → n = 12/4 = 3 or n = −2/4 = −0.5.Only the positive integer solution is allowed → n = 3. Verification / Alternative check: Test n = 3 in the original sentence: LHS = 5*3 = 15; RHS = 2*(3^2) − 3 = 18 − 3 = 15. Equality holds. Why Other Options Are Wrong: 13, 23, 33 do not satisfy the equation; “None of these” is incorrect because n = 3 works exactly. Common Pitfalls: Sign mistakes when moving terms; arithmetic errors on the discriminant; forgetting to reject non-integer or negative roots. Final Answer: 3

Question 15

Error analysis with fractions:  A student was asked to multiply a number by 3/2 but instead divided the number by 3/2 (i.e., multiplied by 2/3). The result was 10 less than the correct result. Find the original number.

Accepted Answer

Introduction / Context: This problem focuses on comparing the intended operation with the erroneous one. By modeling both results in algebra and equating their difference to 10, we can solve directly for the original number without guesswork. Given Data / Assumptions: Correct operation: multiply by 3/2. Actual operation: divide by 3/2, equivalent to multiply by 2/3. Wrong result is 10 less than the correct result. Concept / Approach: Let the original number be x. Then correct result is (3/2)*x, and wrong result is (2/3)*x. Their difference is specified, so form an equation and solve for x. Keep fractions exact to avoid rounding errors. Step-by-Step Solution: Let correct − wrong = 10 → (3/2)x − (2/3)x = 10.Compute the coefficient: 3/2 − 2/3 = 9/6 − 4/6 = 5/6.Thus (5/6)x = 10.Solve: x = 10 * (6/5) = 12. Verification / Alternative check: Correct result for x = 12 is (3/2)*12 = 18; wrong result is (2/3)*12 = 8; difference = 10, consistent with the statement. Why Other Options Are Wrong: 10, 15, and 20 do not satisfy the setup; “None of these” is unnecessary since 12 is exact. Common Pitfalls: Adding rather than subtracting the expressions; flipping 3/2 incorrectly; arithmetic slips when working with fractions. Final Answer: 12

Question 16

Five consecutive odd numbers have a total sum of 175.  Identify the second-largest number, square the smallest number, and report the sum of these two results.

Accepted Answer

Introduction / Context: This question combines sequences with arithmetic operations. Consecutive odd numbers centered on a middle term add up conveniently, enabling fast recovery of the five terms. Then we perform the requested operations to obtain the final sum. Given Data / Assumptions: Five consecutive odd integers sum to 175. We need: (second-largest) + (smallest)^2. Consecutive odd numbers are equally spaced with common difference 2. Concept / Approach: For five consecutive odds written as n−4, n−2, n, n+2, n+4, the sum is 5n. This determines the middle term immediately. Then list the set, pick out the smallest and second-largest, compute the square and the addition. Step-by-Step Solution: Let the five numbers be n−4, n−2, n, n+2, n+4.Sum = 5n = 175 → n = 35.Thus the numbers are 31, 33, 35, 37, 39.Smallest = 31; second-largest = 37.Compute 31^2 = 961; add 37 to get 961 + 37 = 998. Verification / Alternative check: Recheck the list and positions: second-largest is indeed 37 (only 39 is larger). The arithmetic 961 + 37 gives 998 consistently. Why Other Options Are Wrong: 989, 997, and 979 are close distractors but do not match the exact computation. Since 998 is not listed among the main options, “none of the above” is correct. Common Pitfalls: Picking the largest instead of second-largest; squaring the wrong term; using even numbers or mis-centering the five odds. Final Answer: none of the above

Question 17

Children’s Day distribution:  Sweets were to be equally distributed among 300 children. On the day, 50 children were absent, so each attending child received exactly one extra sweet. How many sweets were prepared for distribution?

Accepted Answer

Introduction / Context: This is a ratio and difference-per-recipient problem. A fixed total of sweets is split among different headcounts, leading to a per-child change of exactly one sweet. Setting up a simple equation in terms of total sweets solves it cleanly. Given Data / Assumptions: Originally planned recipients = 300 children. Actual recipients = 250 children (since 50 were absent). Each of the 250 received 1 more sweet than originally planned. Concept / Approach: Let total sweets be S. Then planned share per child is S/300, and actual share per child is S/250. The given condition S/250 = S/300 + 1 yields a simple linear equation for S. Step-by-Step Solution: Set S/250 = S/300 + 1.Multiply through by 1500 (LCM of 250 and 300): 6S = 5S + 1500.Solve: S = 1500.Therefore, 1500 sweets were prepared. Verification / Alternative check: Original per-child share: 1500/300 = 5. Actual per-child share: 1500/250 = 6. The difference is exactly 1 sweet, as required. Why Other Options Are Wrong: 1450, 1650, 1700, and 1800 do not satisfy S/250 − S/300 = 1. Common Pitfalls: Using 300 − 50 = 200 by mistake; mixing up “one extra per child” as “one extra total.” Final Answer: 1500

Question 18

Two-number system with sum–difference relation:  For positive numbers P and Q, the sum equals 2.5 times their difference, and their product is 84. Find P + Q.

Accepted Answer

Introduction / Context: This problem links sum, difference, and product of two positive numbers. Expressing the difference in terms of the sum and using the identity for product in terms of sum and difference leads to a single solvable equation in one variable (the sum). Given Data / Assumptions: P and Q are positive. Sum S = P + Q, difference D = |P − Q|. Condition: S = 2.5 * D, and product P*Q = 84. Concept / Approach: Use the relation for two numbers: P*Q = (S^2 − D^2)/4. Replace D by S/2.5 = (2/5)S to get a single equation in S. Solve for S, then choose the matching option. Step-by-Step Solution: Given S = 2.5D ⇒ D = S/2.5 = (2/5)S.Use product identity: P*Q = (S^2 − D^2)/4.Substitute D^2 = (4/25)S^2 ⇒ P*Q = (S^2 − (4/25)S^2)/4 = ((21/25)S^2)/4 = (21/100)S^2.Given P*Q = 84 ⇒ (21/100)S^2 = 84 ⇒ S^2 = 84 * (100/21) = 400.Thus S = 20 (positive case). Verification / Alternative check: If S = 20 and D = (2/5)*20 = 8, then numbers are (S ± D)/2 = (20 ± 8)/2 = 14 and 6. Their product is 84, confirming consistency. Why Other Options Are Wrong: 26, 24, and 22 do not satisfy the derived relation (21/100)S^2 = 84; only S = 20 works. Common Pitfalls: Mixing S and D; using S = 2.5/D instead of S = 2.5*D; forgetting the product identity. Final Answer: 20

Question 19

Three-number relation with sum repair (clarified):  The first number is twice the second and also three times the third. The sum of the three numbers is 154. What is the difference between the first and the third number?

Accepted Answer

Introduction / Context: (Recovery-first clarification) The original stem was ambiguous. To make it solvable without altering the core intent, we clarify the standard setup: the first number relates to the second and third by simple multiples, and the total sum is 154. We are asked for the difference between the first and third numbers. Given Data / Assumptions: Let the three numbers be F (first), S (second), T (third). Relations: F = 2S and F = 3T. Sum: F + S + T = 154. Concept / Approach: Use the relations to express S and T in terms of F, then compute the sum to solve for F. Once F is known, recover T and compute F − T. This linear system approach is direct and avoids unnecessary complexity. Step-by-Step Solution: From F = 2S ⇒ S = F/2. From F = 3T ⇒ T = F/3.Sum: F + F/2 + F/3 = 154.Combine fractions: F*(1 + 1/2 + 1/3) = F*(11/6) = 154.Solve for F: F = 154 * 6 / 11 = 84.Then T = F/3 = 28. Difference: F − T = 84 − 28 = 56. Verification / Alternative check: Compute S = F/2 = 42. Check sum: 84 + 42 + 28 = 154. All conditions are satisfied. Why Other Options Are Wrong: 42, 62, 68, and 74 do not equal F − T under the validated relations and sum. Common Pitfalls: Reversing the multipliers; forgetting to include all three terms in the sum; arithmetic slips when handling fractions 1/2 and 1/3. Final Answer: 56

Question 20

Planting trees along a straight road:  Find the maximum number of trees that can be planted 20 m apart on both sides of a straight road 1760 m long, including trees at both ends.

Accepted Answer

Introduction / Context: Spacing problems rely on the spacing formula: when placing items at fixed intervals along a segment and including both ends, the count on one side is (length/spacing) + 1. For two sides, double that number. Careful inclusion of endpoints is crucial. Given Data / Assumptions: Road length L = 1760 m. Spacing = 20 m. Trees are placed on both sides; endpoints are included. Concept / Approach: For one side: count = L/spacing + 1. Then multiply by 2 for both sides. Do not double-count endpoints across sides because each side is independent. Step-by-Step Solution: One-side count: 1760 / 20 + 1 = 88 + 1 = 89.Both sides: 2 * 89 = 178.Therefore, the maximum number of trees is 178. Verification / Alternative check: Reverse reasoning: If there are 89 trees on one side, there are 88 intervals (88 * 20 = 1760 m), confirming the formula. Why Other Options Are Wrong: 174 and 176 undercount by missing endpoints; 180 overcounts by adding extra positions; 172 undercounts more severely. Common Pitfalls: Using L/spacing instead of L/spacing + 1; forgetting to multiply by 2 for both sides; assuming shared endpoints reduce the total (they do not across different sides). Final Answer: 178

Number System Questions