Q: Find the least positive integer that must be multiplied with 74088 so that the product becomes a perfect square.

Introduction:To make a number a perfect square, each prime’s exponent in its factorization must be even. We determine the smallest factor that \"balances\" odd exponents for 74088. Given Data / Assumptions: Number: 74088 Goal: smallest positive integer k such that 74088 * k is a perfect square Concept / Approach:Prime factorize 74088. If a prime has an odd exponent, multiply by that prime once to make it even. The product over all such primes is the least k. Step-by-Step Solution:Factorize: 74088 = 2^3 * 3^3 * 7^3All exponents are 3, which is odd.To get even exponents, multiply by 2^1 * 3^1 * 7^1 = 42Then 74088 * 42 = 2^4 * 3^4 * 7^4, which is a perfect square. Verification / Alternative check:Any smaller factor would leave at least one prime with an odd exponent; hence 42 is minimal. Why Other Options Are Wrong:44: = 2^2 * 11 adds an unnecessary 11 and over-fixes 2.46: = 2 * 23 introduces 23, not needed.48: = 2^4 * 3 adds excess powers. Common Pitfalls:Confusing \"dividing to make a square\" with \"multiplying\"; forgetting that each prime’s exponent must be even. Final Answer:42

Q: Find the number whose square equals the difference between 75.15^2 and 60.12^2 (i.e., (75.15)^2 − (60.12)^2).

Introduction:This problem uses the identity a^2 − b^2 = (a − b)(a + b). We then take the square root of the result to get the required number. Given Data / Assumptions: a = 75.15 b = 60.12 We seek N such that N^2 = a^2 − b^2 Concept / Approach:Compute a − b and a + b, multiply, and then take the positive square root (principal value) since the context refers to a magnitude. Step-by-Step Solution:a − b = 75.15 − 60.12 = 15.03a + b = 75.15 + 60.12 = 135.27a^2 − b^2 = (a − b)(a + b) = 15.03 * 135.27 = 2033.1081N = sqrt(2033.1081) = 45.09 Verification / Alternative check:Squaring 45.09 gives 2033.1081, matching a^2 − b^2 exactly. Why Other Options Are Wrong:225.9: This is (a + b) * (a − b) divided or scaled incorrectly.67.635: Likely half of 135.27; not the square root of the product.15.03: Equals a − b, not sqrt((a − b)(a + b)). Common Pitfalls:Mistaking a − b for the final answer; ignoring the square root step after using the identity. Final Answer:45.09

Q: A general wants to form the largest possible square array from 16160 soldiers but finds 31 soldiers left over. Find the number of soldiers in the front row (the side of the square).

Introduction:Arranging soldiers in a square means using the greatest perfect square not exceeding the total. Any leftovers indicate how far the count is from a perfect square. Given Data / Assumptions: Total soldiers = 16160 Left over = 31 Front row size equals the square root of the used perfect square Concept / Approach:If 31 are left, then the number used is 16160 − 31 = 16129. If 16129 is a perfect square, its square root gives the front row size. Step-by-Step Solution:Used soldiers = 16160 − 31 = 16129Check: 127^2 = 16129Therefore, side of the square (front row) = 127 Verification / Alternative check:Compute (126)^2 = 15876 and (128)^2 = 16384 to see that 16129 sits exactly at 127^2. Why Other Options Are Wrong:125, 126, 128: Their squares are not 16129. Common Pitfalls:Forgetting to subtract the leftover before taking the square root; confusing nearest higher vs lower perfect square. Final Answer:127

Q: A group decides that each subscriber will pay as many rupees as there are subscribers. The total subscription collected is Rs 2,582,449. Find the number of subscribers.

Introduction:This is a direct squares problem. If there are n subscribers and each pays n rupees, the total collected becomes n^2 rupees. Given Data / Assumptions: Total subscription = Rs 2,582,449 Each subscriber pays = n rupees Total = n * n = n^2 Concept / Approach:Set n^2 = 2,582,449 and take the (positive) square root to find n. Step-by-Step Solution:n^2 = 2,582,449n = sqrt(2,582,449) = 1607 (since 1607^2 = 2,582,449) Verification / Alternative check:Compute 1607 * 1607 to verify the exact total. Why Other Options Are Wrong:1603, 1503: Their squares do not equal 2,582,449.1291225: That is a large square-like number but not n; it resembles n^2 in format, not the count. Common Pitfalls:Confusing total paid per person with total collected; not recognizing the square relation. Final Answer:1607

Q: Each student in Class IX contributes an amount equal to the square of the total number of students. If the total collection is Rs 29,791, find the total number of students.

Introduction:The original stem is ambiguous. Applying the recovery-first policy, interpret it so the total becomes a perfect cube: if there are n students and each pays n^2 rupees, then the total T = n * n^2 = n^3. This matches the provided total of 29,791 (which is 31^3). Given Data / Assumptions: Total amount collected T = Rs 29,791 Recovered assumption: contribution per student = n^2 rupees Total students = n Concept / Approach:Use T = n^3. Take the cube root of 29,791 to find n. Step-by-Step Solution:T = n^3n = cube_root(29,791) = 31Check: 31^3 = 29,791 Verification / Alternative check:If one instead assumed each pays n rupees, then T = n^2, but 29,791 is not a perfect square; thus the recovered assumption (n^2 per student) is consistent with the data. Why Other Options Are Wrong:30, 33, 40 do not satisfy n^3 = 29,791. Common Pitfalls:Misreading the contribution rule and treating 29,791 as a square instead of a cube; failing to apply the recovery-first interpretation to align with given total. Final Answer:31

Question 1

Surd manipulation: Let x = √3 + √2. Evaluate (x + 1)/(x − 1) − (x − 1)/(x + 1).

Accepted Answer

Introduction / Context: This problem tests algebraic manipulation of surds and rationalizing expressions involving sums and differences with square roots. Given Data / Assumptions: x = √3 + √2; compute (x + 1)/(x − 1) − (x − 1)/(x + 1). Concept / Approach: Use the identity A/B − B/A = (A^2 − B^2)/(AB). Set A = x + 1 and B = x − 1 to simplify quickly. Step-by-Step Solution: Let A = x + 1, B = x − 1.Then A/B − B/A = (A^2 − B^2)/(AB).A^2 − B^2 = [(x + 1)^2 − (x − 1)^2] = [x^2 + 2x + 1] − [x^2 − 2x + 1] = 4x.AB = (x + 1)(x − 1) = x^2 − 1.So the expression = 4x/(x^2 − 1).Compute x^2: (√3 + √2)^2 = 3 + 2 + 2√6 = 5 + 2√6.Thus x^2 − 1 = 4 + 2√6 = 2(2 + √6) and 4x = 4(√3 + √2) = 2·2(√3 + √2).Hence expression = [2·2(√3 + √2)] / [2(2 + √6)] = [2(√3 + √2)] / (2 + √6).Rationalize: multiply top and bottom by (2 − √6). Simplification yields √2. Verification / Alternative check: Substitute approximate values √3 ≈ 1.732, √2 ≈ 1.414 to confirm numerically ≈ 1.414. Why Other Options Are Wrong: √3 and √6 do not match the simplified exact value; “None of above” is incorrect because a specific root fits. Common Pitfalls: Expanding incorrectly or forgetting to use A/B − B/A identity, and errors while rationalizing. Final Answer: √2

Question 2

Comparison of cube roots: Which is greater, ∛8 or ∛1000?

Accepted Answer

Introduction / Context: Estimating cube roots of perfect cubes is straightforward and useful in quick comparisons. Given Data / Assumptions: Compare ∛8 and ∛1000. Concept / Approach: Recognize that 8 = 2^3 and 1000 = 10^3. Then cube roots are exact integers. Step-by-Step Solution: ∛8 = 2.∛1000 = 10.Therefore, ∛1000 > ∛8. Verification / Alternative check: Obvious by comparing 2 and 10. Why Other Options Are Wrong: ∛8 is smaller; equality does not hold. Common Pitfalls: Confusing square roots with cube roots. Final Answer: ∛1000

Question 3

Comparison of square roots: Which is smaller, √81 or √1?

Accepted Answer

Introduction / Context: Basic comparison of square roots of perfect squares. Given Data / Assumptions: Compare √81 and √1. Concept / Approach: Compute both values directly. Step-by-Step Solution: √81 = 9.√1 = 1.Thus the smaller value is √1. Verification / Alternative check: None needed beyond direct evaluation. Why Other Options Are Wrong: √81 is larger; they are not equal. Common Pitfalls: None significant here. Final Answer: √1

Question 4

Sum of a cube and a square equals 4425: If (14)^3 is added to the square of a number, the result is 4425. Find the number.

Accepted Answer

Introduction / Context: This problem mixes perfect cubes and perfect squares, requiring inverse operations to isolate the unknown square. Given Data / Assumptions: 14^3 + n^2 = 4425. Concept / Approach: Compute 14^3 first, subtract from 4425 to get n^2, then take the square root to get n. Step-by-Step Solution: 14^3 = 2744.n^2 = 4425 − 2744 = 1681.n = √1681 = 41. Verification / Alternative check: 41^2 = 1681. Adding 2744 returns 4425. Why Other Options Are Wrong: 43 is not the number (though 1849 = 43^2 is a distractor). 1681 and 1849 are square values, not the number n itself. Common Pitfalls: Reporting n^2 instead of n, or miscomputing 14^3. Final Answer: 41

Question 5

Compare roots: Which is greater: the fourth root of 10 (∜10) or the cube root of 8 (∛8)?

Accepted Answer

Introduction / Context: Comparing different radicals often reduces to evaluating known perfect powers or estimating via basic properties. Given Data / Assumptions: Compare ∜10 and ∛8. Concept / Approach: 8 is a perfect cube, so ∛8 = 2 exactly. Estimate ∜10 ≈ 10^(1/4). Step-by-Step Solution: ∛8 = 2 (since 2^3 = 8).∜10 ≈ 10^0.25 ≈ 1.78 (since 1.78^4 ≈ 10).Therefore, 2 > 1.78, so ∛8 is greater. Verification / Alternative check: Quick reference: ∜16 = 2, and since 10 < 16, ∜10 < 2, confirming the comparison. Why Other Options Are Wrong: ∜10 is smaller; they are not equal; “None of these” is inapplicable. Common Pitfalls: Confusing indices (3rd vs 4th roots) or assuming higher index always yields a larger value. Final Answer: ∛8

Question 6

Find the least positive integer that must be multiplied with 74088 so that the product becomes a perfect square.

Accepted Answer

Introduction:To make a number a perfect square, each prime’s exponent in its factorization must be even. We determine the smallest factor that \"balances\" odd exponents for 74088. Given Data / Assumptions: Number: 74088 Goal: smallest positive integer k such that 74088 * k is a perfect square Concept / Approach:Prime factorize 74088. If a prime has an odd exponent, multiply by that prime once to make it even. The product over all such primes is the least k. Step-by-Step Solution:Factorize: 74088 = 2^3 * 3^3 * 7^3All exponents are 3, which is odd.To get even exponents, multiply by 2^1 * 3^1 * 7^1 = 42Then 74088 * 42 = 2^4 * 3^4 * 7^4, which is a perfect square. Verification / Alternative check:Any smaller factor would leave at least one prime with an odd exponent; hence 42 is minimal. Why Other Options Are Wrong:44: = 2^2 * 11 adds an unnecessary 11 and over-fixes 2.46: = 2 * 23 introduces 23, not needed.48: = 2^4 * 3 adds excess powers. Common Pitfalls:Confusing \"dividing to make a square\" with \"multiplying\"; forgetting that each prime’s exponent must be even. Final Answer:42

Question 7

Find the number whose square equals the difference between 75.15^2 and 60.12^2 (i.e., (75.15)^2 − (60.12)^2).

Accepted Answer

Introduction:This problem uses the identity a^2 − b^2 = (a − b)(a + b). We then take the square root of the result to get the required number. Given Data / Assumptions: a = 75.15 b = 60.12 We seek N such that N^2 = a^2 − b^2 Concept / Approach:Compute a − b and a + b, multiply, and then take the positive square root (principal value) since the context refers to a magnitude. Step-by-Step Solution:a − b = 75.15 − 60.12 = 15.03a + b = 75.15 + 60.12 = 135.27a^2 − b^2 = (a − b)(a + b) = 15.03 * 135.27 = 2033.1081N = sqrt(2033.1081) = 45.09 Verification / Alternative check:Squaring 45.09 gives 2033.1081, matching a^2 − b^2 exactly. Why Other Options Are Wrong:225.9: This is (a + b) * (a − b) divided or scaled incorrectly.67.635: Likely half of 135.27; not the square root of the product.15.03: Equals a − b, not sqrt((a − b)(a + b)). Common Pitfalls:Mistaking a − b for the final answer; ignoring the square root step after using the identity. Final Answer:45.09

Question 8

A general wants to form the largest possible square array from 16160 soldiers but finds 31 soldiers left over. Find the number of soldiers in the front row (the side of the square).

Accepted Answer

Introduction:Arranging soldiers in a square means using the greatest perfect square not exceeding the total. Any leftovers indicate how far the count is from a perfect square. Given Data / Assumptions: Total soldiers = 16160 Left over = 31 Front row size equals the square root of the used perfect square Concept / Approach:If 31 are left, then the number used is 16160 − 31 = 16129. If 16129 is a perfect square, its square root gives the front row size. Step-by-Step Solution:Used soldiers = 16160 − 31 = 16129Check: 127^2 = 16129Therefore, side of the square (front row) = 127 Verification / Alternative check:Compute (126)^2 = 15876 and (128)^2 = 16384 to see that 16129 sits exactly at 127^2. Why Other Options Are Wrong:125, 126, 128: Their squares are not 16129. Common Pitfalls:Forgetting to subtract the leftover before taking the square root; confusing nearest higher vs lower perfect square. Final Answer:127

Question 9

A group decides that each subscriber will pay as many rupees as there are subscribers. The total subscription collected is Rs 2,582,449. Find the number of subscribers.

Accepted Answer

Introduction:This is a direct squares problem. If there are n subscribers and each pays n rupees, the total collected becomes n^2 rupees. Given Data / Assumptions: Total subscription = Rs 2,582,449 Each subscriber pays = n rupees Total = n * n = n^2 Concept / Approach:Set n^2 = 2,582,449 and take the (positive) square root to find n. Step-by-Step Solution:n^2 = 2,582,449n = sqrt(2,582,449) = 1607 (since 1607^2 = 2,582,449) Verification / Alternative check:Compute 1607 * 1607 to verify the exact total. Why Other Options Are Wrong:1603, 1503: Their squares do not equal 2,582,449.1291225: That is a large square-like number but not n; it resembles n^2 in format, not the count. Common Pitfalls:Confusing total paid per person with total collected; not recognizing the square relation. Final Answer:1607

Question 10

Each student in Class IX contributes an amount equal to the square of the total number of students. If the total collection is Rs 29,791, find the total number of students.

Accepted Answer

Introduction:The original stem is ambiguous. Applying the recovery-first policy, interpret it so the total becomes a perfect cube: if there are n students and each pays n^2 rupees, then the total T = n * n^2 = n^3. This matches the provided total of 29,791 (which is 31^3). Given Data / Assumptions: Total amount collected T = Rs 29,791 Recovered assumption: contribution per student = n^2 rupees Total students = n Concept / Approach:Use T = n^3. Take the cube root of 29,791 to find n. Step-by-Step Solution:T = n^3n = cube_root(29,791) = 31Check: 31^3 = 29,791 Verification / Alternative check:If one instead assumed each pays n rupees, then T = n^2, but 29,791 is not a perfect square; thus the recovered assumption (n^2 per student) is consistent with the data. Why Other Options Are Wrong:30, 33, 40 do not satisfy n^3 = 29,791. Common Pitfalls:Misreading the contribution rule and treating 29,791 as a square instead of a cube; failing to apply the recovery-first interpretation to align with given total. Final Answer:31

Question 11

Let x = (√2 + 1)/(√2 − 1) and y = (√2 − 1)/(√2 + 1). Evaluate 3 (xy + 1) in terms of x and y.

Accepted Answer

Introduction:This problem leverages conjugate pairs and simplification. Recognize that x and y are reciprocals because they are formed by swapping numerator and denominator with conjugates of √2 ± 1. Given Data / Assumptions: x = (√2 + 1)/(√2 − 1) y = (√2 − 1)/(√2 + 1) Expression: 3(xy + 1) Concept / Approach:Compute xy. Since y is the reciprocal of x, xy = 1. Substitute into 3(xy + 1) and compare with expressions involving x and y to find identity matches. Step-by-Step Solution:xy = [(√2 + 1)/(√2 − 1)] * [(√2 − 1)/(√2 + 1)] = 13(xy + 1) = 3(1 + 1) = 6Compute x + y: it simplifies to 6 (by rationalization or numeric evaluation), hence 3(xy + 1) = x + y. Verification / Alternative check:Numerically, x ≈ 5.828 and y ≈ 0.172, so x + y ≈ 6 and 3(xy + 1) = 6. Why Other Options Are Wrong:x − y: ≈ 5.6568, not 6.2(x + y): ≈ 12, not 6.None of above: incorrect since x + y matches. Common Pitfalls:Missing that xy = 1 and overcomplicating the algebra; arithmetic slip when rationalizing surds. Final Answer:x + y

Question 12

Find the least perfect square that is divisible by each of 21, 36, and 66.

Accepted Answer

Introduction:The least perfect square divisible by given integers is found by taking the LCM, then adjusting prime exponents to be even (as required for a square). Given Data / Assumptions: Numbers: 21, 36, 66 Concept / Approach:LCM prime factorization method: 21 = 3 * 7, 36 = 2^2 * 3^2, 66 = 2 * 3 * 11. LCM collects the highest exponents across the set. Step-by-Step Solution:LCM = 2^2 * 3^2 * 7^1 * 11^1 = 2772To get a perfect square, make all exponents even: multiply by 7^1 * 11^1 = 77Least perfect square = 2772 * 77 = 213444 Verification / Alternative check:Prime factors of 213444 are 2^2 * 3^2 * 7^2 * 11^2, clearly a perfect square and divisible by each original number. Why Other Options Are Wrong:223444, 213442, 223445: Do not correspond to the necessary exact product or square factorization. Common Pitfalls:Using GCD instead of LCM; forgetting to square-complete the odd exponents (7 and 11). Final Answer:213444

Question 13

An organizer has 7777 chairs and wants a square arrangement (same number of rows and columns). What minimum number of chairs must be removed to form a perfect square layout?

Accepted Answer

Introduction:To form a square layout, the total must be a perfect square. We remove the smallest possible number to reach the nearest perfect square not exceeding 7777. Given Data / Assumptions: Total chairs = 7777 Concept / Approach:Find k = floor(sqrt(7777)). Then k^2 is the nearest lower perfect square. The minimum removal is 7777 − k^2. Step-by-Step Solution:k = floor(sqrt(7777)) = 88k^2 = 88^2 = 7744Minimum removal = 7777 − 7744 = 33 Verification / Alternative check:The next higher square is 89^2 = 7921, which would require adding chairs, not removing. Why Other Options Are Wrong:25, 44, 121 do not equal 7777 − 7744. Common Pitfalls:Rounding sqrt upward; attempting to remove to the nearest higher square rather than lower. Final Answer:33

Question 14

A shopkeeper displays bangles in a perfect square. He has 38 bangles left over. If he increases the side of the square by 1, he falls short by 25 bangles. Find the actual number of bangles he has.

Accepted Answer

Introduction:This is a classic perfect-square boundary problem involving leftovers and shortfalls when moving between consecutive squares k^2 and (k + 1)^2. Given Data / Assumptions: Total bangles = N N = k^2 + 38 (left over after forming k by k square) N = (k + 1)^2 − 25 (25 short of the next square) Concept / Approach:Equate the two expressions for N and solve for k; then substitute back to get N. Step-by-Step Solution:k^2 + 38 = (k + 1)^2 − 25k^2 + 38 = k^2 + 2k + 1 − 2538 = 2k − 24 ⇒ 2k = 62 ⇒ k = 31N = k^2 + 38 = 31^2 + 38 = 961 + 38 = 999 Verification / Alternative check:(k + 1)^2 − N = 32^2 − 999 = 1024 − 999 = 25, consistent with the shortfall. Why Other Options Are Wrong:1690, 538, 1024: Do not satisfy both square-boundary conditions.Cannot be determined: The equations determine N uniquely. Common Pitfalls:Arithmetic errors when expanding (k + 1)^2; mixing up leftover vs shortfall signs. Final Answer:999

Question 15

The sum of two numbers, when multiplied by each number separately, produces 2418 and 3666 respectively. Find the absolute difference between the two numbers.

Accepted Answer

Introduction:Let the numbers be a and b with sum s = a + b. We are given s * a and s * b; this permits solving for s and then |a − b| without finding a and b individually. Given Data / Assumptions: s * a = 2418 s * b = 3666 s = a + b Concept / Approach:Add the two equations and use s = a + b to solve for s, then use the difference to find |a − b|. Step-by-Step Solution:s * a + s * b = 2418 + 3666 = 6084s(a + b) = s * s = s^2 = 6084s = sqrt(6084) = 78Compute |a − b| = |(s * b − s * a)| / s = |3666 − 2418| / 78 = 1248 / 78 = 16 Verification / Alternative check:One can set a = 2418/78 and b = 3666/78, then verify the difference is 16. Why Other Options Are Wrong:21, 19, 23: Do not match the computed difference from the given products. Common Pitfalls:Forgetting that s^2 = 6084; making arithmetic slips with 1248/78 simplification. Final Answer:16

Question 16

One-fourth of a herd of camels is seen in a forest, twice the square root of the herd went to the mountains, and the remaining 15 camels were by the river. Find the total number of camels.

Accepted Answer

Introduction:This is a classic herd puzzle forming an equation with a square-root term. We translate the story into algebra and solve for the total. Given Data / Assumptions: Total camels = x Forest: x/4 Mountains: 2 * sqrt(x) River bank: 15 Concept / Approach:Set up x/4 + 2*sqrt(x) + 15 = x and solve for x ≥ 0. Prefer integer solutions relevant to counts. Step-by-Step Solution:x/4 + 2*sqrt(x) + 15 = xRearrange: 2*sqrt(x) = (3/4)x − 15Trial or algebraic manipulation shows x = 36 satisfies: 36/4 + 2*6 + 15 = 9 + 12 + 15 = 36 Verification / Alternative check:Check nearby integers; only 36 balances the equation exactly. Why Other Options Are Wrong:32, 34, 35 do not satisfy the equation when substituted. Common Pitfalls:Algebraic mistakes while isolating sqrt(x); forgetting the integer and nonnegative context for a count of camels. Final Answer:36

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