Q: What are the two natural numbers whose squares add up to 52?

Introduction / Context:This question tests the ability to work with sums of squares of natural numbers. It is a typical number theory style problem that can be solved either by trial of small squares or by systematic reasoning about possible square values. Given Data / Assumptions:We need two natural numbers, say m and n.The condition is m^2 + n^2 = 52.Both m and n are positive integers. Concept / Approach:We look for pairs of natural numbers whose squares sum to 52. Since 52 is not very large, we can try small squares: 1, 4, 9, 16, 25, 36, and so on, and see which pair of these sums to 52. We must also ensure that the pairs match one of the options. Step-by-Step Solution:Step 1: List small squares of natural numbers: 1^2 = 1, 2^2 = 4, 3^2 = 9, 4^2 = 16, 5^2 = 25, 6^2 = 36, 7^2 = 49.Step 2: Look for two squares that sum to 52.Step 3: Check 16 + 36 = 52, which corresponds to 4^2 + 6^2 = 52.Step 4: Confirm that there is no other pair of squares in this range that sums to 52: 25 + 25 = 50, 9 + 36 = 45, 1 + 49 = 50, and so on.Step 5: Therefore, the only natural numbers whose squares add to 52 are 4 and 6. Verification / Alternative check:We can verify by direct substitution: 4^2 + 6^2 = 16 + 36 = 52. For completeness, we may quickly test the options: 2 and 7 give 4 + 49 = 53, 3 and 5 give 9 + 25 = 34, and 5 and 6 give 25 + 36 = 61. Only 4 and 6 satisfy the condition. Why Other Options Are Wrong:2, 7: 2^2 + 7^2 = 4 + 49 = 53, not 52.3, 5: 3^2 + 5^2 = 9 + 25 = 34.5, 6: 5^2 + 6^2 = 25 + 36 = 61. Common Pitfalls:Some students may mis square numbers or add incorrectly. It is also possible to overlook a valid pair if the list of squares is not written systematically. Writing all small squares clearly and checking each combination methodically avoids such mistakes. Final Answer:The two natural numbers whose squares sum to 52 are 4 and 6.

Q: $\sqrt{53824} = x$

Concept & Strategy: The fastest way to find the square root of a large perfect square is by observing its unit digit and estimating its range using known base squares. Base approximation logic: (xy)^2 \approx x^2 × 100 Step-by-Step Solution: - Identify the unit digit of the given number: The number 53824 ends in 4. - Therefore, its square root must end in either 2 or 8 (since 2^2 = 4 and 8^2 = 64). - Look at the options: All options end in 2, so we must estimate the magnitude. - Find the nearest base squares: 200^2 = 40000 300^2 = 90000 - Since 53824 is closer to 40000, the root is in the lower half of the 200-300 range. - Let's check 230 and 240: 230^2 = 52900 240^2 = 57600 - The number 53824 is just slightly larger than 52900. - The only option that fits this tight range and ends in 2 is 232. Exam Strategy & Shortcut: Use the range bounding method. 20^2 = 400, so 200^2 = 40000. 25^2 = 625, so 250^2 = 62500. The number 53824 lies between 40000 and 62500, which eliminates 332. Because 53824 is closer to 230^2 (52900) than 240^2 (57600), the answer must be 232. Common Pitfall: Students often try to calculate the square root using the traditional long division method, which wastes precious exam time. Always use options and approximation first. Final Answer: Therefore, the correct answer is 232.

Q: The square root of 41209 is equal to

Concept & Strategy: Determine the square root by analyzing the unit digit and using algebraic expansion to verify the closest base number. The algebraic identity used for verification: (a + b)^2 = a^2 + 2ab + b^2 Step-by-Step Solution: - The given number is 41209. - The unit digit is 9. A perfect square ending in 9 must have a square root ending in either 3 or 7 (since 3^2 = 9 and 7^2 = 49). - Looking at the given choices, all options end in 3. - Next, establish the base range by squaring hundreds: 100^2 = 10000 200^2 = 40000 300^2 = 90000 - The number 41209 is very close to 40000, which means its square root must be slightly greater than 200. - Verify 203 using the algebraic expansion: 203^2 = (200 + 3)^2 203^2 = 200^2 + 2(200)(3) + 3^2 203^2 = 40000 + 1200 + 9 = 41209. Exam Strategy & Shortcut: Immediately recognize that 41209 is just slightly above 40000 (which is 200^2). The only option slightly above 200 is 203. You can select this option without even writing down a single calculation. Common Pitfall: Choosing a number just based on the last digit without checking the magnitude. A student might guess 303 if they incorrectly estimate 300^2 as 30000 instead of 90000. Final Answer: Therefore, the correct answer is 203.

Q: $\sqrt{17956} + \sqrt{24025} = x$

Concept & Strategy: This requires finding the square roots of two separate large numbers and summing them. Use the unit digit technique combined with base range estimation for quick extractions. Base approximation logic: Estimate a^2 and b^2 to bound the root. Step-by-Step Solution: - Part 1: Find \sqrt{24025} - The number ends in 25, which means its square root must end in 5. - Remove the last two digits to get 240. - Find two consecutive integers whose product is close to but less than or equal to 240. 15 × 16 = 240. - This perfectly matches the pattern for numbers ending in 5. The square root is 155. - Part 2: Find \sqrt{17956} - The number ends in 6, so its root ends in 4 or 6. - Approximate the base: 130^2 = 16900 and 140^2 = 19600. - 17956 is between them. Let's test 134 and 136. - 135^2 = 13 × 14 appended with 25 = 18225. - Since 17956 < 18225, the root must be less than 135. Thus, it is 134. - Part 3: Addition - Sum the two roots: 134 + 155 = 289. Exam Strategy & Shortcut: Use the unit digit of the final sum. The first root ends in 4. The second root ends in 5. The sum of their unit digits is 4 + 5 = 9. Looking at the options, only 19 and 289 end in 9. Since both roots are clearly greater than 100, the sum must be over 200, making 289 the only logical choice. Common Pitfall: Miscalculating the square root of the first term by choosing the upper boundary (136) instead of the lower boundary (134). Comparing against the midpoint square (135^2) is the most reliable way to prevent this error. Final Answer: Therefore, the correct answer is 289.

Question 1

Which of the following numbers is the exact square root of 15376?

Accepted Answer

Introduction / Context:This question again checks mental computation of square roots of perfect squares. Being able to quickly identify square roots near familiar benchmark values such as 100 or 150 is useful in competitive exams. Given Data / Assumptions:We are given the number 15376.Options: 122, 124, 134, 128.We need the positive integer whose square is exactly 15376. Concept / Approach:First, estimate the rough size. Since 120^2 = 14400 and 130^2 = 16900, the square root of 15376 must lie between 120 and 130. We can test the values close to the middle, such as 124 or 128, using algebraic identities to avoid full multiplication. Step-by-Step Solution:Step 1: Note that 120^2 = 14400 and 130^2 = 16900, so the square root is between 120 and 130.Step 2: Check 124, which is slightly closer to the lower end.Step 3: Use the identity (a + b)^2 = a^2 + 2ab + b^2 with a = 120 and b = 4.Step 4: Compute 120^2 = 14400.Step 5: Compute 2ab = 2*120*4 = 960.Step 6: Compute b^2 = 4^2 = 16.Step 7: Add them: 14400 + 960 + 16 = 15376.Step 8: Since 124^2 = 15376, the square root of 15376 is 124. Verification / Alternative check:We can quickly check another nearby option, such as 122. Using (120 + 2)^2 = 14400 + 480 + 4 = 14884, which is smaller than 15376. That confirms the root must be larger than 122, and our computed value 124 fits correctly. Why Other Options Are Wrong:122: 122^2 = 14884, which is less than 15376.128: 128^2 = (120 + 8)^2 = 14400 + 1920 + 64 = 16384, larger than 15376.134: 134^2 is even larger than 16900, far above 15376. Common Pitfalls:Some test takers try to directly multiply large numbers without using identities, which is slow and error prone. Others may choose values based only on rough estimation without checking the exact square. Using (a + b)^2 helps compute squares efficiently and accurately. Final Answer:The exact square root of 15376 is 124.

Question 2

Which of the following numbers will have an irrational square root?

Accepted Answer

Introduction / Context:This question tests knowledge of perfect squares and the distinction between rational and irrational square roots. Recognizing perfect squares helps in many quantitative aptitude problems, especially those dealing with surds and simplification. Given Data / Assumptions:We are given four integers: 1024, 2401, 4096, and 2048.We must decide which one does not have a perfect integer square root.All numbers are positive. Concept / Approach:A number has a rational integer square root if and only if it is a perfect square. Many powers of 2 and certain powers of primes like 7 give perfect squares. We will check whether each option is a perfect square by remembering standard squares or by expressing the number as a known power. Step-by-Step Solution:Step 1: Consider 1024. We know 32^2 = 1024, so 1024 is a perfect square.Step 2: Consider 2401. This is equal to 7^4 since 7^2 = 49 and 49^2 = 2401, so its square root is 49, an integer.Step 3: Consider 4096. This equals 2^12, which can be written as (2^6)^2. Hence 4096 is a perfect square with integer square root 64.Step 4: Finally consider 2048. This equals 2^11, which is not an even power of 2. There is no integer n such that n^2 = 2^11, so 2048 is not a perfect square.Step 5: Therefore the square root of 2048 cannot be an integer and will be an irrational surd. Verification / Alternative check:We can approximate the square root of 2048. Since 45^2 = 2025 and 46^2 = 2116, the square root of 2048 lies between 45 and 46, and is not an integer. Any non perfect square positive integer has an irrational square root, so the classification is correct. Why Other Options Are Wrong:1024: Equal to 32^2, its square root is 32, which is rational.2401: Equal to 49^2, with rational square root 49.4096: Equal to 64^2, so the square root is rational and integral. Common Pitfalls:Students sometimes confuse large powers of 2 and forget which are squares. Remember that 2^(2k) is always a perfect square, while 2^(2k+1) is not. Here 1024 = 2^10 and 4096 = 2^12, both even exponents, but 2048 = 2^11 has an odd exponent, so it is not a perfect square. Final Answer:The number whose square root is irrational is 2048.

Question 3

What is the simplified form of (4x + 3)^2 * (3x - 5) - (4x^3 - 12x^2 + 9x - 20)?

Accepted Answer

Introduction / Context:This question tests polynomial expansion and simplification skills. Candidates must correctly apply algebraic identities, expand products, and combine like terms. Such algebra manipulation problems are common in both school level mathematics and aptitude tests. Given Data / Assumptions:Expression: (4x + 3)^2 * (3x - 5) - (4x^3 - 12x^2 + 9x - 20).x is a real variable.We must simplify the entire expression into standard polynomial form ax^3 + bx^2 + cx + d. Concept / Approach:First, expand (4x + 3)^2 using (a + b)^2 = a^2 + 2ab + b^2. Then multiply the resulting quadratic by (3x - 5) using distributive law. After obtaining the cubic polynomial from this product, subtract the polynomial (4x^3 - 12x^2 + 9x - 20), carefully combining like terms. Step-by-Step Solution:Step 1: Expand (4x + 3)^2 = (4x)^2 + 2*4x*3 + 3^2 = 16x^2 + 24x + 9.Step 2: Multiply this quadratic by (3x - 5): (16x^2 + 24x + 9)(3x - 5).Step 3: Distribute term by term: 16x^2*3x = 48x^3, 16x^2*(-5) = -80x^2, 24x*3x = 72x^2, 24x*(-5) = -120x, 9*3x = 27x, 9*(-5) = -45.Step 4: Combine like terms from the product: 48x^3 + (-80x^2 + 72x^2) + (-120x + 27x) - 45 = 48x^3 - 8x^2 - 93x - 45.Step 5: Subtract the polynomial (4x^3 - 12x^2 + 9x - 20):(48x^3 - 8x^2 - 93x - 45) - (4x^3 - 12x^2 + 9x - 20).Step 6: Distribute the minus sign: 48x^3 - 8x^2 - 93x - 45 - 4x^3 + 12x^2 - 9x + 20.Step 7: Combine like terms: (48x^3 - 4x^3) = 44x^3, (-8x^2 + 12x^2) = 4x^2, (-93x - 9x) = -102x, (-45 + 20) = -25.Step 8: Final simplified form is 44x^3 + 4x^2 - 102x - 25. Verification / Alternative check:We can verify by plugging in a simple value, for example x = 1, into both the original expression and the candidate simplified result. For x = 1, the original expression and 44*1^3 + 4*1^2 - 102*1 - 25 should give the same numerical value. This numerical check confirms there are no arithmetic mistakes. Why Other Options Are Wrong:6x^2 - 48x - 50: Missing cubic term, so it cannot match the product of a quadratic and linear expression.24x^3 + 2x^2 - 32x - 15: Coefficients do not agree with the computed expansion.24x^3 + 8x^2 - 42x - 50: Again, coefficients differ from the correct combination of like terms. Common Pitfalls:Frequent errors include missing terms during expansion, incorrect signs when subtracting polynomials, and failing to combine like terms accurately. Working systematically and writing each intermediate step clearly helps avoid these mistakes. Final Answer:The simplified form of the expression is 44x^3 + 4x^2 - 102x - 25.

Question 4

If x + 1/x = -2, then what is the value of x^7 + 1/x^7?

Accepted Answer

Introduction / Context:This question involves a classical algebraic technique of using identities to find higher powers of x and 1/x, given the value of x + 1/x. Such problems are standard in algebra sections of competitive examinations. Given Data / Assumptions:x + 1/x = -2.x is nonzero so that 1/x is defined.We are required to find x^7 + 1/x^7. Concept / Approach:When x + 1/x is known, we can often find higher symmetric expressions like x^2 + 1/x^2, x^3 + 1/x^3, and so on using identities. However, in this specific case, the value x + 1/x = -2 is special because it corresponds to a perfect square identity, which directly reveals the value of x without needing long chains of identities. Step-by-Step Solution:Step 1: Let S = x + 1/x. We are told S = -2.Step 2: Multiply the equation by x to remove the denominator: x^2 + 1 = -2x.Step 3: Rearrange to standard quadratic form: x^2 + 2x + 1 = 0.Step 4: Factorize this expression: x^2 + 2x + 1 = (x + 1)^2.Step 5: Set (x + 1)^2 = 0, so x + 1 = 0, giving x = -1.Step 6: Now compute x^7 + 1/x^7 for x = -1.Step 7: Since x = -1, x^7 = (-1)^7 = -1.Step 8: Also 1/x = 1/(-1) = -1, so 1/x^7 is also -1.Step 9: Therefore x^7 + 1/x^7 = -1 + (-1) = -2. Verification / Alternative check:We can confirm that x = -1 satisfies the original condition x + 1/x = -2: indeed, -1 + (-1) = -2. Since the quadratic equation had a repeated root, there is no other distinct value of x, so the result is unique. Why Other Options Are Wrong:1: This would require x^7 and 1/x^7 to be opposites, which does not happen for x = -1.-1: This would require only one term contributing, which is not the case here.0: This would imply x^7 = -1/x^7, which does not hold for x = -1. Common Pitfalls:Some students try to compute x^2 + 1/x^2, x^3 + 1/x^3, and so on using recursive identities, which is unnecessary and time consuming when x + 1/x equals -2. Recognizing that x + 1/x = -2 immediately leads to x = -1 and greatly simplifies the problem. Final Answer:The value of x^7 + 1/x^7 is -2.

Question 5

From two points on level ground that are 1 km apart, the angles of elevation of an aeroplane are 45° and 60°. What is the height of the aeroplane above the ground (in km)?

Accepted Answer

Introduction / Context:This is a height and distance problem from trigonometry, often appearing in aptitude tests. It involves angles of elevation observed from two points on level ground and requires the use of tangent ratios to determine the height of an aeroplane. Given Data / Assumptions:Two observation points on level ground are 1 km apart.The angles of elevation of the aeroplane from these points are 45 degrees and 60 degrees.The aeroplane is vertically above some point on the line connecting the two observers.We assume both observers and the foot of the perpendicular from the aeroplane lie on a straight horizontal line. Concept / Approach:We use the definition of tangent in a right triangle: tan(theta) = opposite / adjacent, where opposite is the height and adjacent is horizontal distance. If we let the nearer point (with larger angle 60 degrees) be B and the farther point (with smaller angle 45 degrees) be A, and the foot of the perpendicular from the aeroplane to the ground be D, we can form two right triangles A D and B D with common height but different base lengths. Step-by-Step Solution:Step 1: Let the height of the aeroplane above ground be h km.Step 2: Let A be the farther point and B be the nearer point, and let AD = x km.Step 3: Since A and B are 1 km apart on the same line, BD = x - 1 km.Step 4: From point A, tan 45° = h / x. Since tan 45° = 1, we get h = x.Step 5: From point B, tan 60° = h / (x - 1). Since tan 60° = √3, we have √3 = h / (x - 1).Step 6: Substitute h = x into the second equation to get √3 = x / (x - 1).Step 7: Cross-multiply: √3 (x - 1) = x.Step 8: Expand: √3 x - √3 = x.Step 9: Bring x terms together: √3 x - x = √3.Step 10: Factor x: x(√3 - 1) = √3, so x = √3 / (√3 - 1).Step 11: Rationalise the denominator: √3 / (√3 - 1) * (√3 + 1) / (√3 + 1) = (3 + √3) / (3 - 1) = (3 + √3) / 2.Step 12: Recall h = x, so the height h = (3 + √3) / 2 km. Verification / Alternative check:We can numerically approximate the height. √3 is about 1.732, so (3 + √3) / 2 is roughly (3 + 1.732) / 2 = 4.732 / 2 ≈ 2.366 km. If we use this approximate height and recompute the tangents for the two triangles, the angles will be close to 45 degrees and 60 degrees, confirming the solution. Why Other Options Are Wrong:√3 km: This is about 1.732 km, which would not satisfy the geometry with a 1 km separation between observers and the given angles.(√3 + 1) km: Approximately 2.732 km, which leads to inconsistent tangent ratios.2 km: This is a rough guess and does not satisfy both tan 45° and tan 60° relationships simultaneously. Common Pitfalls:Students may misplace the 1 km distance, assuming the aeroplane is directly over one of the observers, or they may swap which angle belongs to which point, leading to incorrect equations. Another frequent error is algebraic manipulation when solving for x and rationalising the denominator. Final Answer:The height of the aeroplane above the ground is (3 + √3) / 2 km.

Question 6

For the quadratic expression f(x) = 5 + 20x - 4x^2 with real x, what is the maximum value of f(x)?

Accepted Answer

Introduction / Context:This question tests understanding of quadratic functions and finding their maximum or minimum values. In aptitude exams, recognising that a quadratic with negative leading coefficient has a maximum at its vertex is very important. Given Data / Assumptions:Quadratic expression: f(x) = 5 + 20x - 4x^2.Coefficient of x^2 is -4, which is negative.x can be any real number.We need the maximum possible value of f(x). Concept / Approach:A general quadratic function f(x) = ax^2 + bx + c, with a negative, opens downwards and therefore has a maximum value at its vertex. The x coordinate of the vertex is given by x = -b / (2a). The maximum value is f of that x coordinate. We can apply this directly to the given quadratic. Step-by-Step Solution:Step 1: Rewrite f(x) in standard form: f(x) = -4x^2 + 20x + 5.Step 2: Identify a = -4, b = 20, c = 5.Step 3: Compute the x coordinate of the vertex using x = -b / (2a).Step 4: x = -20 / (2 * -4) = -20 / -8 = 2.5.Step 5: Evaluate f(2.5) to find the maximum value.Step 6: Compute f(2.5) = -4*(2.5)^2 + 20*(2.5) + 5.Step 7: (2.5)^2 = 6.25, so -4*6.25 = -25.Step 8: 20*(2.5) = 50.Step 9: Add the terms: -25 + 50 + 5 = 30.Step 10: Therefore, the maximum value of f(x) is 30. Verification / Alternative check:We can complete the square: f(x) = -4x^2 + 20x + 5 = -4(x^2 - 5x) + 5. Then write x^2 - 5x as (x - 2.5)^2 - 6.25. So f(x) = -4[(x - 2.5)^2 - 6.25] + 5 = -4(x - 2.5)^2 + 25 + 5 = -4(x - 2.5)^2 + 30. The term -4(x - 2.5)^2 is always less than or equal to 0, with maximum 0 at x = 2.5, so the maximum value is 30. Why Other Options Are Wrong:1 and 5: These are too small and correspond to values of x away from the vertex.25: This is close but arises from an intermediate step in completing the square, not the final maximum value. Common Pitfalls:Some students misuse the formula for the vertex or forget that a negative leading coefficient means a maximum, not a minimum. Others might plug random values of x instead of using the systematic vertex method, which can miss the true maximum. Final Answer:The maximum value of the quadratic expression is 30.

Question 7

If 3a = 4b = 6c and a + b + c = 279, then what is the value of a^2 + b^2 + c^2?

Accepted Answer

Introduction / Context:This question involves proportional relationships between three variables and asks for a sum of squares. It is typical of algebra and aptitude tests where candidates must convert a relation like 3a = 4b = 6c into useful expressions for a, b, and c, and then compute required symmetric combinations. Given Data / Assumptions:3a = 4b = 6c = k for some constant k.a + b + c = 279.We need to find a^2 + b^2 + c^2.All variables are assumed to be real numbers. Concept / Approach:When an expression like 3a = 4b = 6c is given, we usually set each equal to a common constant k. Then we express a, b, and c in terms of k. Using the sum a + b + c, we find k. Once we know a, b, and c explicitly, we can compute their squares and sum them. This method avoids solving simultaneous equations separately. Step-by-Step Solution:Step 1: Let 3a = 4b = 6c = k.Step 2: Express the variables: a = k / 3, b = k / 4, c = k / 6.Step 3: Use the given sum a + b + c = 279.Step 4: Substitute in terms of k: k/3 + k/4 + k/6 = 279.Step 5: Find a common denominator, which is 12. Then k/3 = 4k/12, k/4 = 3k/12, k/6 = 2k/12.Step 6: Add: (4k + 3k + 2k)/12 = 9k/12 = 3k/4.Step 7: So 3k/4 = 279, giving k = 279 * 4 / 3.Step 8: Compute k = (279 / 3) * 4 = 93 * 4 = 372.Step 9: Now a = 372 / 3 = 124, b = 372 / 4 = 93, c = 372 / 6 = 62.Step 10: Compute squares: a^2 = 124^2 = 15376, b^2 = 93^2 = 8649, c^2 = 62^2 = 3844.Step 11: Sum them: a^2 + b^2 + c^2 = 15376 + 8649 + 3844.Step 12: 15376 + 8649 = 24025, and 24025 + 3844 = 27869. Verification / Alternative check:We can verify by checking that our values of a, b, and c satisfy the original relationship 3a = 4b = 6c. Compute 3*124 = 372, 4*93 = 372, and 6*62 = 372. All are equal to 372, so the relationship holds, confirming that the values are correct and the sum of squares 27869 is reliable. Why Other Options Are Wrong:27969, 27789, 27896: These numbers are close to 27869 but arise from slight arithmetic errors such as mis adding the squares or mis squaring one of the numbers. Only 27869 exactly matches 124^2 + 93^2 + 62^2. Common Pitfalls:A common source of error is misinterpreting 3a = 4b = 6c as three separate equations rather than introducing a single constant k. Another frequent mistake is arithmetic error in squaring large numbers or in adding them. Breaking the calculation into clear steps reduces such mistakes. Final Answer:The value of a^2 + b^2 + c^2 is 27869.

Question 8

Determine the value of x when x^3 = 27.

Accepted Answer

Introduction / Context:This is a direct cube equation that tests understanding of cube roots. Recognising basic cubes and their roots is important for many aptitude questions, especially those involving higher degree equations or simplifications. Given Data / Assumptions:Equation: x^3 = 27.We look for the real value of x.Options: 1, 3, 9, 12. Concept / Approach:To solve x^3 = 27, we find the cube root of 27. Since 3^3 = 27, 3 is the real cube root. This is a basic fact that should be memorised: 1^3 = 1, 2^3 = 8, 3^3 = 27, 4^3 = 64, and so on. Among the options, we pick the value whose cube equals 27. Step-by-Step Solution:Step 1: Start with the equation x^3 = 27.Step 2: Take the cube root of both sides: x = cube root of 27.Step 3: Recall that 3^3 = 3 * 3 * 3 = 9 * 3 = 27.Step 4: Therefore cube root of 27 is 3.Step 5: So the solution is x = 3. Verification / Alternative check:We can verify by substituting x = 3 back into the equation. Compute 3^3 = 27, which matches the right hand side. None of the other options yield 27 when cubed. Why Other Options Are Wrong:1: 1^3 = 1, not 27.9: 9^3 = 729, much larger than 27.12: 12^3 = 1728, again much larger than 27. Common Pitfalls:Some students confuse square roots and cube roots, and might mistakenly think of 27 as 3^2 or as 9^2. It is essential to remember the basic cube values for small integers to avoid such mistakes. Final Answer:The value of x satisfying x^3 = 27 is 3.

Question 9

What is the square root of 1485961?

Accepted Answer

Introduction / Context:This question checks the ability to handle large perfect squares. Instead of performing long multiplication from scratch, using algebraic identities around a nearby convenient base, such as 1200, helps in efficient computation. Given Data / Assumptions:We must find the square root of 1485961.Options: 1213, 1219, 1229, 1239.One of these is the exact square root. Concept / Approach:Since 1200^2 = 1440000, our target number 1485961 is slightly above this. So the square root will be somewhat greater than 1200. Among the options, 1219, 1229, and 1239 are all close to 1200, but we can test 1219 first, using (1200 + 19)^2 or (1220 - 1)^2 to compute efficiently. Step-by-Step Solution:Step 1: Consider 1219 as a candidate.Step 2: Write 1219^2 as (1200 + 19)^2 = 1200^2 + 2*1200*19 + 19^2.Step 3: Compute 1200^2 = 1440000.Step 4: Compute 2*1200*19 = 2400*19 = 45600.Step 5: Compute 19^2 = 361.Step 6: Add: 1440000 + 45600 = 1485600.Step 7: Add 361 to get 1485961.Step 8: Therefore 1219^2 = 1485961 exactly, so the square root is 1219. Verification / Alternative check:To confirm, we can quickly test another nearby option, such as 1213 or 1229, and see that their squares do not match 1485961. This provides additional confidence that 1219 is the unique correct choice among the options. Why Other Options Are Wrong:1213: Its square will differ from 1485961 because changing the base from 1219 to 1213 changes the value by a significant margin.1229: This is larger than 1219, and its square will exceed 1485961.1239: This is even larger, yielding a much greater square. Common Pitfalls:Students may attempt direct multiplication without using identities, which is slow and error prone for such large numbers. Another mistake is estimating based on too wide an interval instead of testing the most reasonable candidate precisely. Final Answer:The square root of 1485961 is 1219.

Question 10

Which of the following quadratic equations has 3 as a root?

Accepted Answer

Introduction / Context:This question checks whether you can verify if a given number is a root of a quadratic equation. It is a basic skill in algebra, often used when factoring quadratics or solving multiple choice questions quickly. Given Data / Assumptions:Candidate root: x = 3.Four quadratic equations are provided.We need to find which equation is satisfied by x = 3. Concept / Approach:To test whether a number is a root of an equation, we substitute that number into the left hand side of the equation and check whether the result is zero. If the value is zero, the number is a root. If not, it is not a root. We will apply this check to each option. Step-by-Step Solution:Step 1: Check option (a): x^2 - 7x + 12 = 0 at x = 3.Step 2: Compute 3^2 - 7*3 + 12 = 9 - 21 + 12.Step 3: 9 - 21 = -12 and -12 + 12 = 0, so x = 3 satisfies equation (a).Step 4: Check option (b): x^2 - 5x + 12 at x = 3 gives 9 - 15 + 12 = 6, not zero.Step 5: Check option (c): x^2 + 3x + 12 at x = 3 gives 9 + 9 + 12 = 30, not zero.Step 6: Check option (d): x^2 + 3x - 12 at x = 3 gives 9 + 9 - 12 = 6, not zero.Step 7: Only option (a) yields zero, so it is the correct equation. Verification / Alternative check:We can factor equation (a): x^2 - 7x + 12 = 0. This factors as (x - 3)(x - 4) = 0, so the roots are x = 3 and x = 4. This confirms that x = 3 is indeed a root of this equation. Why Other Options Are Wrong:x^2 - 5x + 12 = 0: Substituting x = 3 gives 6, not 0.x^2 + 3x + 12 = 0: Substituting x = 3 gives 30, not 0.x^2 + 3x - 12 = 0: Substituting x = 3 gives 6, not 0. Common Pitfalls:Some candidates may try to factor each quadratic blindly, which takes more time than simply substituting the given value. Others may make simple arithmetic mistakes while evaluating the expressions at x = 3, so careful calculation is essential. Final Answer:The equation that has 3 as a root is x^2 - 7x + 12 = 0.

Question 11

A number has 27 decimal places after the decimal point. How many decimal places will its cube root have?

Accepted Answer

Introduction / Context:This question focuses on the relationship between decimal places of a number and its cube root. It is a concept from number system and powers of ten, frequently used when dealing with very small or very large decimal numbers. Given Data / Assumptions:A decimal number has exactly 27 decimal places after the decimal point.We consider the cube root of this number.We assume the number is positive and written in standard decimal form. Concept / Approach:If a number is expressed as 10^(-n), then its cube root is 10^(-n/3). The number of decimal places corresponds to the magnitude of the negative exponent on 10. If a number has 27 decimal places, it behaves like 10^(-27) multiplied by some nonzero digit pattern. Taking the cube root divides the exponent by 3, so the cube root will behave like 10^(-9) times some digits, giving 9 decimal places. Step-by-Step Solution:Step 1: Suppose a number has 27 decimal places; for example, it can be thought of in the form k * 10^(-27), where k is not divisible by 10.Step 2: The cube root of this number is (k * 10^(-27))^(1/3).Step 3: This equals k^(1/3) * (10^(-27))^(1/3).Step 4: For powers of ten, (10^(-27))^(1/3) = 10^(-27/3) = 10^(-9).Step 5: So the cube root is some nonzero factor multiplied by 10^(-9).Step 6: A factor of 10^(-9) corresponds to 9 digits after the decimal point in standard decimal representation.Step 7: Therefore, the cube root of a number with 27 decimal places will have exactly 9 decimal places. Verification / Alternative check:As a simple illustration, consider the number 10^(-27) itself, which is 0.000000000000000000000000001 with 27 zeros after the decimal point before the 1. Its cube root is 10^(-9), which is 0.000000001, with 9 decimal places. This matches our reasoning. Why Other Options Are Wrong:3 and 6: These come from dividing 27 by 9 or by a wrong factor, not by 3.27: This would mean the cube root has the same number of decimal places as the original number, which is not consistent with exponent rules. Common Pitfalls:Students sometimes confuse square roots and cube roots and may divide 27 by 2 instead of by 3. Others may think that the number of decimal places stays unchanged under roots. Understanding exponents of 10 and how roots affect them is crucial. Final Answer:The cube root will have 9 decimal places.

Question 12

What is the value of 105^3 (the cube of 105)?

Accepted Answer

Introduction / Context:This problem asks for the cube of 105. Direct multiplication is possible, but using algebraic identities can simplify the calculation. Recognising patterns like (100 + 5)^3 helps in quick mental math and is very useful for competitive exams. Given Data / Assumptions:We need to compute 105^3.Options: 1157625, 1175625, 1185625, 1158625.We assume standard rules of arithmetic. Concept / Approach:We can express 105 as 100 + 5 and use the binomial expansion for (a + b)^3. The identity is (a + b)^3 = a^3 + 3a^2b + 3ab^2 + b^3. This allows us to use simpler cubes and squares like 100^3 and 5^3. Step-by-Step Solution:Step 1: Write 105^3 as (100 + 5)^3.Step 2: Apply the identity (a + b)^3 = a^3 + 3a^2b + 3ab^2 + b^3 with a = 100 and b = 5.Step 3: Compute a^3 = 100^3 = 1000000.Step 4: Compute 3a^2b = 3 * 100^2 * 5 = 3 * 10000 * 5 = 150000.Step 5: Compute 3ab^2 = 3 * 100 * 5^2 = 3 * 100 * 25 = 7500.Step 6: Compute b^3 = 5^3 = 125.Step 7: Add all contributions: 1000000 + 150000 = 1150000.Step 8: Add 7500 to get 1157500.Step 9: Finally add 125 to get 1157625.Step 10: Therefore, 105^3 = 1157625. Verification / Alternative check:We can quickly check the last digit. Since 5^3 ends in 125, any cube of a number ending in 5 will end in 125. All options end with 625, consistent with this rule, so we rely on the detailed binomial computation above to distinguish the correct value. Why Other Options Are Wrong:1175625, 1185625, 1158625: These differ by thousands from the correct value 1157625 and result from incorrect middle term calculations or arithmetic errors in adding the contributions from the binomial expansion. Common Pitfalls:Errors often occur in computing 3a^2b and 3ab^2, or in adding the large numbers at the end. Writing each step carefully and checking intermediate sums is essential to avoid misplacing digits. Final Answer:The value of 105^3 is 1157625.

Question 13

If the numbers 93, 204, and 256 are arranged in ascending order, which of the following is correct?

Accepted Answer

Introduction / Context:This question tests basic understanding of ordering numbers. While it looks simple, such questions are important in early parts of aptitude tests and can also appear in data interpretation where quick recognition of ascending and descending order matters. Given Data / Assumptions:Numbers: 93, 204, 256.We need to list them in ascending order.All numbers are positive integers and are compared on the usual number line. Concept / Approach:Ascending order means arranging numbers from smallest to largest. We compare their magnitudes. Since all three numbers have different sizes, we identify the smallest, the middle, and the largest. Step-by-Step Solution:Step 1: Compare 93 and 204. Since 93 is a two digit number and 204 is a three digit number, 93 is smaller than 204.Step 2: Compare 204 and 256. Both are three digit numbers, but 204 has smaller hundreds and tens portions than 256, so 204 is smaller than 256.Step 3: Combine these observations: the numbers in increasing order are 93, then 204, then 256.Step 4: Write in inequality form: 93 < 204 < 256. Verification / Alternative check:On a mental number line, numbers grow as we move to the right. 93 is less than 100, 204 is a little above 200, and 256 is above 250, so clearly the order 93, 204, 256 is correct. Why Other Options Are Wrong:256 < 204 < 93: This reverses the true magnitude relationship, placing the largest number first.204 < 256 < 93: This incorrectly places 93, the smallest number, at the end.256 < 93 < 204: This puts the largest number first and does not represent ascending order. Common Pitfalls:Although this problem is straightforward, in timed tests candidates sometimes misread the symbols or quickly tick a choice without carefully checking the order of all three numbers. It is wise to read inequalities slowly and confirm that each symbol points from smaller to larger in ascending order questions. Final Answer:The correct ascending order is 93 < 204 < 256.

Question 14

What is the value of 997^3 (the cube of 997)?

Accepted Answer

Introduction / Context:This question asks for the cube of 997, a number close to 1000. Instead of multiplying 997 three times directly, we can use the algebraic identity for the cube of a difference to simplify the calculation. This technique is very useful for handling numbers close to powers of 10. Given Data / Assumptions:We must compute 997^3.Options: 991026973, 991029673, 991029773, 991097273.All calculations use standard arithmetic rules. Concept / Approach:Notice that 997 = 1000 - 3. We can apply the identity (a - b)^3 = a^3 - 3a^2b + 3ab^2 - b^3 with a = 1000 and b = 3. This leverages simple powers like 1000^3 and small cubes like 3^3 for an easier computation. Step-by-Step Solution:Step 1: Write 997^3 as (1000 - 3)^3.Step 2: Use (a - b)^3 = a^3 - 3a^2b + 3ab^2 - b^3 with a = 1000, b = 3.Step 3: Compute a^3 = 1000^3 = 1000000000.Step 4: Compute 3a^2b = 3 * 1000^2 * 3 = 3 * 1000000 * 3 = 9000000.Step 5: Compute 3ab^2 = 3 * 1000 * 3^2 = 3 * 1000 * 9 = 27000.Step 6: Compute b^3 = 3^3 = 27.Step 7: Substitute into the identity: 997^3 = 1000000000 - 9000000 + 27000 - 27.Step 8: First subtract 9000000: 1000000000 - 9000000 = 991000000.Step 9: Add 27000: 991000000 + 27000 = 991027000.Step 10: Subtract 27: 991027000 - 27 = 991026973.Step 11: Therefore, 997^3 = 991026973. Verification / Alternative check:We can check the last digits. Since 7^3 ends in 343, any cube of a number ending in 7 will end with 343 when considering only the last three digits. Here, 991026973 ends with 973, which matches the pattern from our binomial expansion and aligns with the other terms. The detailed identity based computation is the most reliable method in this context. Why Other Options Are Wrong:991029673 and 991029773: These are close to the correct answer but involve incorrect middle digits, usually from mis adding 27000 or mis subtracting 27.991097273: This has a different pattern in the hundred thousands and millions place, indicating an error in the 3a^2b term. Common Pitfalls:Typical mistakes include sign errors in the formula for (a - b)^3, mis computing 3a^2b or 3ab^2, and mis handling the subtraction of b^3. Writing each contribution separately and using place value carefully helps prevent such errors. Final Answer:The value of 997^3 is 991026973.

Question 15

What are the two natural numbers whose squares add up to 52?

Accepted Answer

Introduction / Context:This question tests the ability to work with sums of squares of natural numbers. It is a typical number theory style problem that can be solved either by trial of small squares or by systematic reasoning about possible square values. Given Data / Assumptions:We need two natural numbers, say m and n.The condition is m^2 + n^2 = 52.Both m and n are positive integers. Concept / Approach:We look for pairs of natural numbers whose squares sum to 52. Since 52 is not very large, we can try small squares: 1, 4, 9, 16, 25, 36, and so on, and see which pair of these sums to 52. We must also ensure that the pairs match one of the options. Step-by-Step Solution:Step 1: List small squares of natural numbers: 1^2 = 1, 2^2 = 4, 3^2 = 9, 4^2 = 16, 5^2 = 25, 6^2 = 36, 7^2 = 49.Step 2: Look for two squares that sum to 52.Step 3: Check 16 + 36 = 52, which corresponds to 4^2 + 6^2 = 52.Step 4: Confirm that there is no other pair of squares in this range that sums to 52: 25 + 25 = 50, 9 + 36 = 45, 1 + 49 = 50, and so on.Step 5: Therefore, the only natural numbers whose squares add to 52 are 4 and 6. Verification / Alternative check:We can verify by direct substitution: 4^2 + 6^2 = 16 + 36 = 52. For completeness, we may quickly test the options: 2 and 7 give 4 + 49 = 53, 3 and 5 give 9 + 25 = 34, and 5 and 6 give 25 + 36 = 61. Only 4 and 6 satisfy the condition. Why Other Options Are Wrong:2, 7: 2^2 + 7^2 = 4 + 49 = 53, not 52.3, 5: 3^2 + 5^2 = 9 + 25 = 34.5, 6: 5^2 + 6^2 = 25 + 36 = 61. Common Pitfalls:Some students may mis square numbers or add incorrectly. It is also possible to overlook a valid pair if the list of squares is not written systematically. Writing all small squares clearly and checking each combination methodically avoids such mistakes. Final Answer:The two natural numbers whose squares sum to 52 are 4 and 6.

Question 16

$\sqrt{53824} = x$

Accepted Answer

Concept & Strategy: The fastest way to find the square root of a large perfect square is by observing its unit digit and estimating its range using known base squares. Base approximation logic: (xy)^2 \approx x^2 × 100 Step-by-Step Solution: - Identify the unit digit of the given number: The number 53824 ends in 4. - Therefore, its square root must end in either 2 or 8 (since 2^2 = 4 and 8^2 = 64). - Look at the options: All options end in 2, so we must estimate the magnitude. - Find the nearest base squares: 200^2 = 40000 300^2 = 90000 - Since 53824 is closer to 40000, the root is in the lower half of the 200-300 range. - Let's check 230 and 240: 230^2 = 52900 240^2 = 57600 - The number 53824 is just slightly larger than 52900. - The only option that fits this tight range and ends in 2 is 232. Exam Strategy & Shortcut: Use the range bounding method. 20^2 = 400, so 200^2 = 40000. 25^2 = 625, so 250^2 = 62500. The number 53824 lies between 40000 and 62500, which eliminates 332. Because 53824 is closer to 230^2 (52900) than 240^2 (57600), the answer must be 232. Common Pitfall: Students often try to calculate the square root using the traditional long division method, which wastes precious exam time. Always use options and approximation first. Final Answer: Therefore, the correct answer is 232.

Question 17

The square root of 41209 is equal to

Accepted Answer

Concept & Strategy: Determine the square root by analyzing the unit digit and using algebraic expansion to verify the closest base number. The algebraic identity used for verification: (a + b)^2 = a^2 + 2ab + b^2 Step-by-Step Solution: - The given number is 41209. - The unit digit is 9. A perfect square ending in 9 must have a square root ending in either 3 or 7 (since 3^2 = 9 and 7^2 = 49). - Looking at the given choices, all options end in 3. - Next, establish the base range by squaring hundreds: 100^2 = 10000 200^2 = 40000 300^2 = 90000 - The number 41209 is very close to 40000, which means its square root must be slightly greater than 200. - Verify 203 using the algebraic expansion: 203^2 = (200 + 3)^2 203^2 = 200^2 + 2(200)(3) + 3^2 203^2 = 40000 + 1200 + 9 = 41209. Exam Strategy & Shortcut: Immediately recognize that 41209 is just slightly above 40000 (which is 200^2). The only option slightly above 200 is 203. You can select this option without even writing down a single calculation. Common Pitfall: Choosing a number just based on the last digit without checking the magnitude. A student might guess 303 if they incorrectly estimate 300^2 as 30000 instead of 90000. Final Answer: Therefore, the correct answer is 203.

Question 18

$\sqrt{17956} + \sqrt{24025} = x$

Accepted Answer

Concept & Strategy: This requires finding the square roots of two separate large numbers and summing them. Use the unit digit technique combined with base range estimation for quick extractions. Base approximation logic: Estimate a^2 and b^2 to bound the root. Step-by-Step Solution: - Part 1: Find \sqrt{24025} - The number ends in 25, which means its square root must end in 5. - Remove the last two digits to get 240. - Find two consecutive integers whose product is close to but less than or equal to 240. 15 × 16 = 240. - This perfectly matches the pattern for numbers ending in 5. The square root is 155. - Part 2: Find \sqrt{17956} - The number ends in 6, so its root ends in 4 or 6. - Approximate the base: 130^2 = 16900 and 140^2 = 19600. - 17956 is between them. Let's test 134 and 136. - 135^2 = 13 × 14 appended with 25 = 18225. - Since 17956 < 18225, the root must be less than 135. Thus, it is 134. - Part 3: Addition - Sum the two roots: 134 + 155 = 289. Exam Strategy & Shortcut: Use the unit digit of the final sum. The first root ends in 4. The second root ends in 5. The sum of their unit digits is 4 + 5 = 9. Looking at the options, only 19 and 289 end in 9. Since both roots are clearly greater than 100, the sum must be over 200, making 289 the only logical choice. Common Pitfall: Miscalculating the square root of the first term by choosing the upper boundary (136) instead of the lower boundary (134). Comparing against the midpoint square (135^2) is the most reliable way to prevent this error. Final Answer: Therefore, the correct answer is 289.

Question 19

$\sqrt{44944} + \sqrt{52441} = x$

Accepted Answer

Concept & Strategy: Solve this by approximating the individual square roots using their unit digits and bounding them between known multiples of 10. Base approximation logic: Find bounds (10x)^2 < N < (10(x+1))^2 Step-by-Step Solution: - Part 1: Find \sqrt{44944} - Unit digit is 4, so the root ends in 2 or 8. - Find base boundaries: 200^2 = 40000 and 220^2 = 48400. 210^2 = 44100. - 44944 is slightly larger than 44100. - This points to the root being 212. - Part 2: Find \sqrt{52441} - Unit digit is 1, so the root ends in 1 or 9. - Find base boundaries: 220^2 = 48400 and 230^2 = 52900. - 52441 is very close to 52900. - This means it is at the higher end of the range, pointing to 229. - Part 3: Addition - Sum the two calculated roots: 212 + 229 = 441. Exam Strategy & Shortcut: Use unit digit analysis for the final sum. The first root ends in 2. The second root ends in 9. The sum of the unit digits is 2 + 9 = 11, which means the final answer must end in 1. Among the given options, only 441 ends in 1. Common Pitfall: Getting bogged down in long division for square roots. Even if you aren't entirely sure of the exact roots, estimating them as \sim 212 and \sim 230 gives a sum around 442, making 441 the obvious answer by approximation alone. Final Answer: Therefore, the correct answer is 441.

Question 20

One-fourth of the sum of prime numbers, greater than 4 but less than 16, is the square of

Accepted Answer

Concept & Strategy: This is a multi-step arithmetic reasoning problem. It requires identifying prime numbers within a specific range, performing basic operations, and then finding the number whose square matches the result. Prime Number Definition: A number divisible only by 1 and itself. Step-by-Step Solution: - Step 1: Identify the prime numbers. - We need prime numbers greater than 4 and less than 16. - These prime numbers are: 5, 7, 11, and 13. - Step 2: Find their sum. - Sum = 5 + 7 + 11 + 13 = 36. - Step 3: Calculate one-fourth of the sum. - One-fourth of 36 = (36/4) = 9. - Step 4: Relate back to the question. - The result is 9. The question asks: "9 is the square of what number?" - Since 3^2 = 9, the required number is 3. Exam Strategy & Shortcut: Write down the primes quickly: 5, 7, 11, 13. Add them in pairs to be faster: (13+7) + (11+5) = 20 + 16 = 36. Divide by 4 to get 9. The square root of 9 is 3. This entire process should take less than 15 seconds. Common Pitfall: Including numbers that are not prime, such as 9 or 15, in the initial list. Remember that 9 is divisible by 3, and 15 is divisible by 3 and 5. Also, failing to read the final part of the prompt "is the square of" and looking for 9 in the options. Final Answer: Therefore, the correct answer is 3.

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