Question 1

The arithmetic mean of the numbers 7, 5, 13, x and 9 is 10. What is the value of x?

Accepted Answer

Introduction / Context: This question is a straightforward application of the concept of arithmetic mean (average). You are given four explicit numbers and one unknown number x, and told that their overall average is 10. The task is to determine the value of x that makes this true. Such problems are common in basic quantitative aptitude and help reinforce the formula for the mean and simple equation solving. Given Data / Assumptions: - The five numbers are 7, 5, 13, x and 9. - Their arithmetic mean is 10. - We must solve for the unknown number x. Concept / Approach: The arithmetic mean of a set of numbers is the sum of the numbers divided by the count of numbers. Here, there are five numbers, so the mean is (7 + 5 + 13 + x + 9) / 5. We are told that this expression equals 10. By equating the expression to 10 and simplifying, we get a linear equation in x. Solving this equation gives the required value of x. This is a simple one step algebra problem after computing the sum of the known numbers. Step-by-Step Solution: Step 1: Write the formula for the mean of the five numbers: (7 + 5 + 13 + x + 9) / 5 = 10. Step 2: Compute the sum of the known numbers: 7 + 5 + 13 + 9. Step 3: 7 + 5 = 12, 12 + 13 = 25, 25 + 9 = 34. So the total sum is 34 + x. Step 4: Substitute into the mean formula: (34 + x) / 5 = 10. Step 5: Multiply both sides by 5 to clear the denominator: 34 + x = 50. Step 6: Subtract 34 from both sides to isolate x: x = 50 - 34. Step 7: Compute 50 - 34 = 16. Step 8: Therefore, x = 16. Verification / Alternative check: Substitute x = 16 back into the list and check the average. The five numbers become 7, 5, 13, 16 and 9. Their sum is 7 + 5 + 13 + 16 + 9 = 50. The mean is 50 / 5 = 10, exactly matching the given condition. This confirms that x = 16 is correct and there is no mistake in the computation. Why Other Options Are Wrong: If x were 10, 12 or 14, the total sum of the five numbers would be different from 50, and the resulting average would not be 10. For example, with x = 10, the sum would be 44 and the average would be 44 / 5 = 8.8, which is too small. Only x = 16 produces the required sum of 50 and therefore the correct mean of 10. Common Pitfalls: Some learners may forget to multiply the mean by the number of terms to find the total sum or may make a mistake when adding the known numbers. Others might divide by 4 instead of 5, forgetting that there are five numbers in total. Always count the terms carefully and double check the arithmetic when summing the numbers to avoid such simple but costly mistakes. Final Answer: The value of x for which the mean is 10 is 16, which corresponds to option D.

Question 2

In an arithmetic progression, the 3rd term is 13 and the 6th term is -5. What is the 11th term of this progression?

Accepted Answer

Introduction / Context: This question involves an arithmetic progression (A.P.), which is a sequence of numbers where the difference between consecutive terms is constant. You are given the 3rd and 6th terms of an A.P. and are asked to find the 11th term. The problem assesses your ability to apply the general n-th term formula of an A.P. and solve for both the first term and the common difference using the information provided. Given Data / Assumptions: - Let the first term be a. - Let the common difference be d. - The 3rd term T_3 is 13. - The 6th term T_6 is -5. - We need to determine the 11th term T_11. Concept / Approach: The n-th term of an arithmetic progression is given by T_n = a + (n - 1)d. Using this formula, T_3 can be written as a + 2d, and T_6 as a + 5d. Since their values are known, we get two linear equations in a and d. Solving these equations gives us specific values of a and d. Then we substitute these into the formula for T_11, which is a + 10d. This straightforward use of the general term formula is the standard way to approach such problems. Step-by-Step Solution: Step 1: Write T_3 in terms of a and d: T_3 = a + (3 - 1)d = a + 2d. Step 2: We are told T_3 = 13, so a + 2d = 13. Call this Equation (1). Step 3: Write T_6 in terms of a and d: T_6 = a + (6 - 1)d = a + 5d. Step 4: We are told T_6 = -5, so a + 5d = -5. Call this Equation (2). Step 5: Subtract Equation (1) from Equation (2): (a + 5d) - (a + 2d) = -5 - 13. Step 6: This simplifies to 3d = -18, so d = -6. Step 7: Substitute d = -6 back into Equation (1): a + 2 * (-6) = 13. Step 8: So a - 12 = 13, hence a = 25. Step 9: Now find T_11 using T_11 = a + (11 - 1)d = a + 10d. Step 10: Substitute a = 25 and d = -6: T_11 = 25 + 10 * (-6) = 25 - 60. Step 11: Compute 25 - 60 = -35, so T_11 = -35. Verification / Alternative check: To verify, write the first few terms using a = 25 and d = -6: 25, 19, 13, 7, 1, -5, -11, and so on. The 3rd term is 13 and the 6th term is -5, as required. From the 1st term to the 11th term, we move 10 steps, each subtracting 6, so T_11 = 25 - 10 * 6 = -35. This confirms the earlier calculation and shows the progression is consistent with the given data. Why Other Options Are Wrong: Values such as -29, -41 and -47 result from incorrect differences or miscounting the number of steps between terms. For example, using the wrong formula a + nd or misapplying the subtraction of equations can produce such incorrect results. Only -35 satisfies the progression defined by T_3 = 13 and T_6 = -5 with a constant difference of -6. Common Pitfalls: Students sometimes mistakenly write the general term as a + nd instead of a + (n - 1)d, which shifts all term indices and leads to incorrect answers. Others may mis-handle the signs when subtracting the equations for T_3 and T_6. Carefully writing each equation with full detail and double checking the subtraction step helps avoid these common issues. Final Answer: The 11th term of the arithmetic progression is -35, which corresponds to option D.

Question 3

What will be the quotient when 2143 is divided by 38?

Accepted Answer

Introduction / Context: This question tests your ability to perform integer division with multi digit numbers. You are asked to find the quotient when 2143 is divided by 38. The focus is on accurate multiplication and subtraction rather than any advanced concept. Such questions are common in aptitude tests to check mental arithmetic and basic numerical handling skills. Given Data / Assumptions: - Dividend = 2143. - Divisor = 38. - We are interested only in the quotient, not the remainder. Concept / Approach: We want to know how many whole times the number 38 fits into 2143. A practical way is to estimate around convenient multiples of the divisor. We first try a rough multiple such as 50 times 38 and then adjust up or down, checking nearby multiples until we find the largest multiple of 38 that does not exceed 2143. The corresponding factor is the quotient, and the leftover part is the remainder. Although the remainder is not required, it can help verify that we chose the correct multiple. Step-by-Step Solution: Step 1: Start by estimating 38 * 50, because 50 is a convenient round number. Step 2: Compute 38 * 50 = 1900. Step 3: Compare 1900 with 2143. The difference is 2143 - 1900 = 243, so we can still multiply 38 by a few more units. Step 4: Try adding 6 more multiples of 38. Compute 38 * 6 = 228. Step 5: Add this to 1900: 1900 + 228 = 2128. Step 6: Now compare 2128 with 2143. The difference is 2143 - 2128 = 15. We cannot add another full 38 because 2128 + 38 = 2166, which would exceed 2143. Step 7: Therefore, the largest multiple of 38 that is less than or equal to 2143 is 38 * 56 = 2128. Step 8: The quotient is 56 and the remainder is 15, though only the quotient is required. Verification / Alternative check: We can verify the quotient by performing the division 2143 ÷ 38 in the standard long division format. Repeated subtraction of 38 or stepwise division will also show that after 56 groups of 38, there is a leftover of 15, and that any attempt to take a 57th group of 38 would exceed 2143. Another quick sanity check is to approximate: 2143 ÷ 40 is a little over 53, so a quotient around 54 to 56 is reasonable. The more precise calculation confirms that 56 is correct. Why Other Options Are Wrong: If the quotient were 55, the product 38 * 55 would be 2090, leaving a remainder of 53; while this is less than 2143, it is not the largest possible quotient because 56 still fits. A quotient of 57 would give 38 * 57 = 2166, which is greater than 2143 and therefore impossible. Similarly, 54 is too small because we already know 56 is achievable. Thus, only 56 is both valid and maximal. Common Pitfalls: Students sometimes stop at 50 because 1900 feels close enough, or they miscalculate products like 38 * 6 or 38 * 7, leading to wrong conclusions. Others may confuse the remainder with the quotient. Carefully calculating nearby multiples and explicitly comparing them with the dividend helps avoid these mistakes. Final Answer: The quotient when 2143 is divided by 38 is 56, which corresponds to option A.

Question 4

The sum of the first 17 terms of an arithmetic progression is required. If the first term is -20 and the 17th (last) term is 28, what is this sum?

Accepted Answer

Introduction / Context: This question is about summing the terms of an arithmetic progression (A.P.). You are told the first term and the 17th term of an A.P. and asked to find the sum of the first 17 terms. This is a standard application of the formula for the sum of the first n terms of an A.P. and tests your understanding of series formulas and basic arithmetic. Given Data / Assumptions: - The number of terms n is 17. - The first term a is -20. - The 17th term, which is also the last term in the sum, is 28. - We need the sum S_17 of the first 17 terms. Concept / Approach: For an arithmetic progression, the sum S_n of the first n terms can be computed using the formula S_n = (n / 2) * (first term + last term). This works because pairing the first and last terms, second and second last terms and so on produces n / 2 pairs, each with the same sum. In this question, the first and 17th terms are given directly, so we can apply the formula without needing to find the common difference explicitly. Step-by-Step Solution: Step 1: Note the values: n = 17, a_1 = -20 and a_17 = 28. Step 2: Use the sum formula for an A.P.: S_n = (n / 2) * (first term + last term). Step 3: Substitute n = 17, first term = -20 and last term = 28 into the formula: S_17 = (17 / 2) * (-20 + 28). Step 4: Compute the sum inside the brackets: -20 + 28 = 8. Step 5: So S_17 = (17 / 2) * 8. Step 6: Simplify (17 / 2) * 8. First compute 8 / 2 = 4, then 17 * 4 = 68. Step 7: Therefore the sum of the first 17 terms is 68. Verification / Alternative check: We can also verify by finding the common difference and checking a few terms. The common difference d can be found from the formula for the n-th term: a_17 = a_1 + (17 - 1)d. Substitute 28 = -20 + 16d, which gives 16d = 48 and hence d = 3. The sequence starts as -20, -17, -14, ..., and increases by 3 each time. Using the sum formula with n = 17, a_1 = -20 and a_17 = 28 still gives S_17 = 17 * ( -20 + 28 ) / 2 = 68, confirming the earlier result. Why Other Options Are Wrong: Values 156, 142 and 242 result from miscalculations such as adding the first and last terms incorrectly, using the wrong number of terms or forgetting to divide by 2 in the sum formula. None of these values match the correctly computed sum of 68. Only 68 is consistent with both the formula and the specific terms of the progression. Common Pitfalls: A frequent mistake is to forget the division by 2 in the formula S_n = n * (first + last) / 2 or to use the wrong last term when n is given. Some learners also incorrectly assume the last term is n multiplied by the common difference without checking the given value. Always confirm the correct first and last terms and carefully follow the formula steps to avoid such errors. Final Answer: The sum of the first 17 terms of the progression is 68, which corresponds to option A.

Question 5

The point P(a, b) is first reflected in the origin to get point P1, and then P1 is reflected in the Y-axis to obtain the point (4, -3). What are the coordinates of the original point P?

Accepted Answer

Introduction / Context: This coordinate geometry question involves reflections of a point in the origin and in the Y-axis. You are told that a point P(a, b) is reflected in the origin to give P1, and then P1 is reflected in the Y-axis to produce the final point with coordinates (4, -3). You must determine the original coordinates (a, b). This tests your understanding of transformation rules for reflections in the coordinate plane. Given Data / Assumptions: - Original point is P(a, b). - Reflection of P in the origin gives P1. - Reflection of P1 in the Y-axis gives the final point with coordinates (4, -3). - We must find the coordinates (a, b). Concept / Approach: Reflections in the coordinate plane follow simple algebraic rules. Reflecting a point (x, y) in the origin changes it to (-x, -y). Reflecting a point (x, y) in the Y-axis changes it to (-x, y). Starting from P(a, b), we apply these transformations step by step: first reflect in the origin, then reflect in the Y-axis. This will express the final coordinates in terms of a and b. Setting these expressions equal to the given final coordinates allows us to solve for a and b directly. Step-by-Step Solution: Step 1: Start with the original point P(a, b). Step 2: Reflect P in the origin. The rule for reflection in the origin is (x, y) becomes (-x, -y). Step 3: Therefore, P1, the reflection of P in the origin, has coordinates (-a, -b). Step 4: Next, reflect P1 in the Y-axis. The rule for reflection in the Y-axis is (x, y) becomes (-x, y). Step 5: Apply this rule to P1 = (-a, -b). Its reflection in the Y-axis is (a, -b). Call this final point P2. Step 6: We are told that this final point P2 has coordinates (4, -3). Step 7: Therefore, (a, -b) = (4, -3). Step 8: Equate coordinates: a = 4 and -b = -3. Step 9: From -b = -3, we get b = 3. Step 10: Thus, the original point P has coordinates (4, 3). Verification / Alternative check: Verify by performing the transformations explicitly. Start from P(4, 3). Reflect in the origin to get P1(-4, -3). Then reflect P1(-4, -3) in the Y-axis: (x, y) becomes (-x, y), so (-4, -3) becomes (4, -3). This matches the given final point. The transformation chain is therefore correct, confirming that P is indeed (4, 3). Why Other Options Are Wrong: Option B (-4, 3) would reflect in the origin to (4, -3) and then in the Y-axis to (-4, -3), not (4, -3). Options C (3, 4) and D (-3, 4) produce entirely different final coordinates after the two reflections. Only (4, 3) results in (4, -3) after reflecting first in the origin and then in the Y-axis. Common Pitfalls: Some learners confuse reflection in the origin with reflection in one of the axes, or they apply the transformations in the wrong order. Another common mistake is to think that reflecting in the origin followed by the Y-axis is equivalent to reflecting in the X-axis, which it is not. The safest approach is to apply each transformation one at a time using the correct coordinate rules and then equate the result to the given point. Final Answer: The original coordinates of point P are (4, 3), which corresponds to option A.

Question 6

In an arithmetic progression, the 1st term is -10 and the 3rd term is -4. What is the 12th term of this progression?

Accepted Answer

Introduction / Context: This question deals with an arithmetic progression (A.P.), where each term increases by a constant difference. You are given the 1st and 3rd terms of the progression and asked to find the 12th term. The problem tests your understanding of how to use the general term formula of an arithmetic progression and how to compute the common difference from two known terms. Given Data / Assumptions: - First term a_1 is -10. - Third term a_3 is -4. - The progression has a constant common difference d. - We need to determine the 12th term a_12. Concept / Approach: The n-th term T_n of an arithmetic progression is given by T_n = a_1 + (n - 1)d. Knowing the first term and the third term allows us to compute the common difference. Once we have the common difference, we can directly calculate any term, including the 12th, by substituting n = 12 into the general term formula. This is a straightforward application of the definition of an arithmetic progression. Step-by-Step Solution: Step 1: Use the general term formula for T_3: T_3 = a_1 + (3 - 1)d = a_1 + 2d. Step 2: Substitute the known values: a_1 = -10 and T_3 = -4, so -10 + 2d = -4. Step 3: Solve for d: 2d = -4 + 10 = 6, so d = 3. Step 4: Now use the formula for the 12th term: T_12 = a_1 + (12 - 1)d = a_1 + 11d. Step 5: Substitute a_1 = -10 and d = 3: T_12 = -10 + 11 * 3. Step 6: Compute 11 * 3 = 33. Step 7: So T_12 = -10 + 33 = 23. Step 8: Therefore, the 12th term of the progression is 23. Verification / Alternative check: We can list some terms of the sequence to verify. Starting with -10 and adding 3 each time, we get: -10, -7, -4, -1, 2, 5, 8, 11, 14, 17, 20, 23, ... The 1st term is -10, the 3rd term is -4, as given. Counting to the 12th term, we indeed reach 23. This confirms that our calculated common difference and resulting 12th term are correct. Why Other Options Are Wrong: Options 20, 17 and 26 correspond to incorrect calculations, such as using the wrong number of steps (for example, 10d instead of 11d) or miscomputing the common difference. Only 23 is consistent with both the given terms and the constant step size of 3 between consecutive terms. Common Pitfalls: Some learners may incorrectly use d = (T_3 - a_1) / 3 instead of dividing by (3 - 1), leading to the wrong common difference. Others miscount when finding the 12th term and use 10 or 12 multipliers of d incorrectly. Remember that T_n uses (n - 1)d, and always double check the subtraction when computing the common difference. Final Answer: The 12th term of the arithmetic progression is 23, which corresponds to option D.

Question 7

In an arithmetic progression, the 3rd term is -14 and the 8th term is 1. What is the 11th term of this progression?

Accepted Answer

Introduction / Context: This arithmetic progression (A.P.) problem gives you the 3rd and 8th terms and asks for the 11th term. Since every term in an A.P. differs from the previous one by a fixed common difference, two known terms allow you to determine both the first term and the common difference. Once these are known, you can compute any term, including the 11th. This question tests the systematic use of the general term formula for an A.P. Given Data / Assumptions: - Let the first term be a. - Let the common difference be d. - The 3rd term T_3 is -14. - The 8th term T_8 is 1. - We need to find the 11th term T_11. Concept / Approach: The general formula for the n-th term of an A.P. is T_n = a + (n - 1)d. Using this, we can write expressions for T_3 and T_8 in terms of a and d. These produce two linear equations that we solve to find a and d. Then we substitute these values into T_11 = a + 10d to obtain the 11th term. This method is consistent and avoids guesswork, even when terms are negative. Step-by-Step Solution: Step 1: Express T_3 using the general formula: T_3 = a + (3 - 1)d = a + 2d. Step 2: Given T_3 = -14, we have a + 2d = -14. Call this Equation (1). Step 3: Express T_8 using the general formula: T_8 = a + (8 - 1)d = a + 7d. Step 4: Given T_8 = 1, we have a + 7d = 1. Call this Equation (2). Step 5: Subtract Equation (1) from Equation (2) to eliminate a: (a + 7d) - (a + 2d) = 1 - (-14). Step 6: This leads to 5d = 15. Step 7: Solve for d: d = 15 / 5 = 3. Step 8: Substitute d = 3 back into Equation (1): a + 2 * 3 = -14, so a + 6 = -14. Step 9: Thus a = -14 - 6 = -20. Step 10: Now compute T_11 using T_11 = a + (11 - 1)d = a + 10d. Step 11: Substitute a = -20 and d = 3: T_11 = -20 + 10 * 3. Step 12: Compute 10 * 3 = 30, giving T_11 = -20 + 30 = 10. Verification / Alternative check: List some terms of the progression using a = -20 and d = 3. The sequence begins -20, -17, -14, -11, -8, -5, -2, 1, 4, 7, 10, ... The 3rd term is -14 and the 8th term is 1, matching the problem statement. Continuing, the 11th term in this list is 10, which agrees with our formula calculation. This confirms the correctness of the computed 11th term. Why Other Options Are Wrong: Values 14, 16 and 20 arise from incorrect computations of the common difference or miscounting the number of steps from a known term. Only 10 is consistent with the progression defined by T_3 = -14 and T_8 = 1 and the constant increase of 3 between consecutive terms. Common Pitfalls: Learners sometimes subtract equations in the wrong order, introducing sign errors in the value of d. Others may use the wrong n when computing T_11, using 11d instead of 10d. Remember that T_n = a + (n - 1)d, and always recheck the arithmetic when handling negative numbers to avoid accidental sign mistakes. Final Answer: The 11th term of the arithmetic progression is 10, which corresponds to option D.

Question 8

In an arithmetic progression, the 3rd term is -9 and the 7th term is 11. What is the 15th term of this progression?

Accepted Answer

Introduction / Context: This arithmetic progression question gives two non consecutive terms and asks for another term further along the sequence. Specifically, you know the 3rd and 7th terms and must find the 15th term. Since an arithmetic progression has a constant common difference, the two known terms are enough to determine both the first term and the common difference, after which finding any term is straightforward. The question reinforces the use of the general term formula and solving simple simultaneous equations. Given Data / Assumptions: - Let the first term be a. - Let the common difference be d. - The 3rd term T_3 is -9. - The 7th term T_7 is 11. - We must find the 15th term T_15. Concept / Approach: The formula for the n-th term of an arithmetic progression is T_n = a + (n - 1)d. Using T_3 and T_7, we can write two equations involving a and d. Subtracting these equations removes a and lets us solve for d, the common difference. Once we know d, we can substitute back to find a. With both a and d known, computing T_15 using the same general formula becomes straightforward. This structured approach works for all similar A.P. problems. Step-by-Step Solution: Step 1: Write T_3 using the general formula: T_3 = a + (3 - 1)d = a + 2d. Step 2: Given T_3 = -9, we get a + 2d = -9. Call this Equation (1). Step 3: Write T_7 using the general formula: T_7 = a + (7 - 1)d = a + 6d. Step 4: Given T_7 = 11, we get a + 6d = 11. Call this Equation (2). Step 5: Subtract Equation (1) from Equation (2): (a + 6d) - (a + 2d) = 11 - (-9). Step 6: This simplifies to 4d = 20. Step 7: Solve for d: d = 20 / 4 = 5. Step 8: Substitute d = 5 into Equation (1): a + 2 * 5 = -9, so a + 10 = -9. Step 9: Thus a = -9 - 10 = -19. Step 10: Now find T_15 using T_15 = a + (15 - 1)d = a + 14d. Step 11: Substitute a = -19 and d = 5: T_15 = -19 + 14 * 5. Step 12: Compute 14 * 5 = 70. Therefore T_15 = -19 + 70 = 51. Verification / Alternative check: List several terms of the progression to verify. With a = -19 and d = 5, the sequence begins -19, -14, -9, -4, 1, 6, 11, 16, 21, 26, 31, 36, 41, 46, 51, ... The 3rd term is -9 and the 7th term is 11, matching the given information. The 15th term in this list is indeed 51, confirming that the computation is correct. Why Other Options Are Wrong: Values 28, 87 and 17 do not match the term obtained from a consistent arithmetic progression with T_3 = -9 and T_7 = 11. These incorrect answers typically come from an incorrect common difference or misusing the term index in the formula (for example, using 15d instead of 14d). Only 51 fits both the data given and the constant difference of 5 between consecutive terms. Common Pitfalls: Students often mis-handle the subtraction step when solving for d, particularly with negative numbers, leading to an incorrect common difference. Another frequent error is to confuse T_n = a + (n - 1)d with T_n = a + nd. Carefully writing down and checking each equation, especially the indices and signs, helps avoid such mistakes. Final Answer: The 15th term of the arithmetic progression is 51, which corresponds to option C.

Question 9

A certain real number is greater than 58 times its own reciprocal by 3/4. What is the value of this number?

Accepted Answer

Introduction / Context: This question tests your understanding of reciprocal numbers and how to convert a word statement involving a number and its reciprocal into a quadratic equation. Problems of this type are very common in aptitude exams because they combine algebraic manipulation, equation formation, and the ability to select the correct solution from the given options. Here we are told that a number is greater than 58 times its reciprocal by a certain amount, and we must carefully translate that language into a mathematical equation and then solve it. Given Data / Assumptions: Let the required real number be x. The reciprocal of the number is 1 / x, assuming x is not equal to 0. The number is greater than 58 times its reciprocal by 3/4. We assume standard real number arithmetic and look for values of x among the given options. Concept / Approach: The key idea is to translate the statement "a number is greater than 58 times its reciprocal by 3/4" into an algebraic equation. The phrase "greater by 3/4" means that if we subtract 58 times the reciprocal from the number, the result is 3/4. This leads to an equation of the form x - 58 * (1 / x) = 3/4. Once this equation is formed, we multiply through by a suitable factor to clear denominators and obtain a quadratic equation in x. Solving the quadratic gives two possible values, and we then match these with the given options to choose the correct answer. Step-by-Step Solution: Let the number be x and its reciprocal be 1 / x. According to the question, x is greater than 58 * (1 / x) by 3/4. So we write the equation: x - 58 / x = 3 / 4. Multiply both sides by 4x to remove denominators: 4x * x - 4x * (58 / x) = 3 * x. This simplifies to 4x^2 - 232 = 3x. Rearrange to standard quadratic form: 4x^2 - 3x - 232 = 0. Now solve using the quadratic formula: x = [3 ± sqrt(3^2 - 4 * 4 * (-232))] / (2 * 4). Compute the discriminant: 3^2 - 4 * 4 * (-232) = 9 + 3712 = 3721. The square root of 3721 is 61, so x = (3 ± 61) / 8. Thus x = (3 + 61) / 8 = 64 / 8 = 8, or x = (3 - 61) / 8 = -58 / 8 = -29 / 4. Among the options, only x = 8 is present, so the required number is 8. Verification / Alternative Check: We can verify the answer by substituting x = 8 back into the original relation. The reciprocal of 8 is 1 / 8. Then 58 times the reciprocal is 58 * (1 / 8) = 58 / 8 = 29 / 4. The difference x - 58 / x becomes 8 - 29 / 4. Write 8 as 32 / 4 to simplify the subtraction. Then 32 / 4 - 29 / 4 = 3 / 4, which matches exactly with the given condition. This confirms that x = 8 satisfies the statement. The other root, -29 / 4, is not listed among the options, so it cannot be the answer in this multiple choice setting. Why Other Options Are Wrong: Option -8: If x = -8, then the reciprocal is -1 / 8 and x - 58 / x becomes -8 - 58 / (-8) = -8 + 29 / 4, which does not simplify to 3 / 4. Option 12: For x = 12, the reciprocal is 1 / 12 and x - 58 / x becomes 12 - 58 / 12, which does not equal 3 / 4. Option -12: For x = -12, the difference between x and 58 times its reciprocal is again not equal to 3 / 4. Option 4: With x = 4, the reciprocal is 1 / 4 and x - 58 / x becomes 4 - 58 / 4, clearly not 3 / 4. Common Pitfalls: A common mistake is to interpret the phrase "greater than 58 times its reciprocal by 3/4" incorrectly and write x = 58 / x * (3 / 4) or something similar. Another frequent error is to forget to multiply every term by the common denominator when clearing fractions, which leads to incorrect coefficients in the quadratic. Students also sometimes compute the discriminant incorrectly or make sign errors when applying the quadratic formula. Finally, many forget to check which root actually appears in the options, and they may choose a mathematically valid root that is not available as an answer choice. Final Answer: Therefore, the required number that is greater than 58 times its reciprocal by 3/4 is 8.

Question 10

The first term of an arithmetic progression is -10 and the thirteenth term is 26. What is the sum of the first 13 terms of this arithmetic progression?

Accepted Answer

Introduction / Context: This problem checks your knowledge of arithmetic progressions and in particular the formula for the sum of the first n terms of an arithmetic progression. You are given the first term and the 13th term and asked to find the sum of the first 13 terms. This is a direct application of the standard sum formula and is a typical question in aptitude tests to see whether you remember formulas and can apply them correctly. Given Data / Assumptions: First term of the arithmetic progression, a1 = -10. Thirteenth term of the arithmetic progression, a13 = 26. Number of terms to be summed, n = 13. The sequence is a standard arithmetic progression with constant common difference. Concept / Approach: For any arithmetic progression, the sum of the first n terms is given by the formula S_n = n * (a1 + a_n) / 2, where a1 is the first term and a_n is the nth term. In this question, we are provided directly with a1 and a13, so we do not even need to calculate the common difference. The question reduces to substituting n = 13, a1 = -10, and a13 = 26 into the formula and simplifying carefully. Step-by-Step Solution: We have n = 13, a1 = -10, and a13 = 26. The sum of the first n terms of an arithmetic progression is S_n = n * (a1 + a_n) / 2. Here S_13 = 13 * (a1 + a13) / 2. Substitute the values: S_13 = 13 * (-10 + 26) / 2. Compute the sum in the numerator: -10 + 26 = 16. So S_13 = 13 * 16 / 2. Divide 16 by 2 first: 16 / 2 = 8. Now multiply: S_13 = 13 * 8 = 104. Therefore, the sum of the first 13 terms of the arithmetic progression is 104. Verification / Alternative Check: As a quick verification, we can find the common difference and a few terms. The common difference d is (a13 - a1) / 12 = (26 - (-10)) / 12 = 36 / 12 = 3. So the sequence is -10, -7, -4, -1, 2, 5, 8, 11, 14, 17, 20, 23, 26. We can sum a few terms in pairs from the ends: (-10 + 26) = 16, (-7 + 23) = 16, (-4 + 20) = 16, (-1 + 17) = 16, (2 + 14) = 16, (5 + 11) = 16 and the middle term is 8. There are 6 such pairs each summing to 16 and one middle term. So total sum is 6 * 16 + 8 = 96 + 8 = 104, which matches the formula calculation. Why Other Options Are Wrong: Option 140: This would require either a wrong value for the number of terms or a miscalculation in the formula and does not match the correct sum. Option 84: This is smaller than the correct value and usually appears if someone mistakenly uses 12 instead of 13 for the number of terms. Option 98: This could arise from an arithmetic slip when multiplying or dividing, but it does not satisfy the exact sum when checked. Option 156: This would correspond to a much larger average term and does not fit the actual sequence values. Common Pitfalls: Some students forget the correct sum formula and instead try to add many terms manually, which is time consuming and prone to error. Others may confuse the formula for the sum with the formula for the nth term, or mistakenly use n - 1 instead of n in the sum formula. Arithmetic errors, especially when dealing with negative first terms, are also common. It is important to first compute a1 + a_n correctly, then multiply by n, and only then divide by 2 to avoid mistakes. Final Answer: The sum of the first 13 terms of the arithmetic progression is 104.

Question 11

In an arithmetic progression, the third term is -8 and the ninth term is 10. What is the value of the sixteenth term of this progression?

Accepted Answer

Introduction / Context: This problem is a classic arithmetic progression question where two nonconsecutive terms are given and you are asked to find another term. Such questions test your understanding of the formula for the nth term of an arithmetic progression and your ability to form and solve simple linear equations using the given information. Given Data / Assumptions: Let the first term of the arithmetic progression be a. Let the common difference be d. The third term a3 is -8. The ninth term a9 is 10. We are required to find the sixteenth term a16. Concept / Approach: For an arithmetic progression, the nth term is given by a_n = a + (n - 1) * d. Using this formula, we can express the third term and the ninth term in terms of a and d. Since both of these terms are provided, we obtain two linear equations in a and d. Solving these equations gives us the values of a and d. Once we know the first term and the common difference, we can substitute into the term formula again to find the sixteenth term. Step-by-Step Solution: Write the third term using the formula: a3 = a + 2d = -8. Write the ninth term using the formula: a9 = a + 8d = 10. Subtract the first equation from the second to eliminate a: (a + 8d) - (a + 2d) = 10 - (-8). This simplifies to 6d = 18, so d = 18 / 6 = 3. Substitute d = 3 back into a + 2d = -8 to find a. Then a + 2 * 3 = -8, so a + 6 = -8. Therefore a = -8 - 6 = -14. Now find the sixteenth term using a16 = a + 15d. So a16 = -14 + 15 * 3 = -14 + 45 = 31. Hence, the sixteenth term of the arithmetic progression is 31. Verification / Alternative Check: We can verify our result by generating a few terms. With a = -14 and d = 3, the sequence starts as -14, -11, -8, -5, -2, 1, 4, 7, 10, ... We see that the third term is indeed -8 and the ninth term is 10, matching the data. Continuing the pattern, the tenth term is 13, the eleventh term is 16, the twelfth term is 19, the thirteenth term is 22, the fourteenth term is 25, the fifteenth term is 28, and the sixteenth term is 31. This confirms that our computed sixteenth term is correct. Why Other Options Are Wrong: Option 34: This would require either a different common difference or a different first term and does not fit with the given third and ninth terms. Option 28: This value actually appears as the fifteenth term in the correct progression, not the sixteenth term. Option 25: This value matches the fourteenth term, not the sixteenth, so it is incorrect for this question. Option 37: This would correspond to an even larger term and is inconsistent with the established pattern of d = 3. Common Pitfalls: A common mistake is to misapply the nth term formula and write a_n = a + nd instead of a + (n - 1) * d. Another frequent error is to subtract the equations in the wrong order or handle negative signs carelessly, leading to an incorrect value for the common difference. Some students also try to guess the sequence without forming equations, which can lead to confusion, especially if negative terms are involved. Working systematically with equations avoids these issues. Final Answer: The sixteenth term of the arithmetic progression is 31.

Question 12

What is the least number that must be added to 4131 so that the resulting sum is completely divisible by 19?

Accepted Answer

Introduction / Context: This question is based on basic divisibility and remainders. The aim is to find the smallest nonnegative integer that should be added to a given number so that the result becomes exactly divisible by a specified divisor. Such problems are very common in aptitude tests because they test understanding of modular arithmetic in a very simple and practical way. Given Data / Assumptions: Given number: 4131. Divisor: 19. We need to find the least integer k such that 4131 + k is divisible by 19. We assume standard properties of division and remainders. Concept / Approach: If a number N leaves a remainder r when divided by a divisor d, then N can be written as N = d * q + r. To make the number divisible by d, we need to add a number k so that the new remainder becomes zero. The smallest such k is simply d - r if r is not zero, or 0 if the number is already divisible. So the strategy is to first compute the remainder when 4131 is divided by 19 and then find k = 19 - remainder. Step-by-Step Solution: Let N = 4131 and d = 19. Compute the remainder when 4131 is divided by 19. Perform the division: 19 * 217 = 4123. Subtract this product from 4131 to get the remainder: 4131 - 4123 = 8. So 4131 leaves a remainder of 8 when divided by 19. We need to add a number k such that the new remainder becomes zero. The smallest positive k that does this is k = 19 - 8 = 11. Thus 4131 + 11 = 4142, and 4142 divided by 19 gives a whole number. So the least number that must be added is 11. Verification / Alternative Check: We can quickly verify our result. Compute 4142 / 19. Since 19 * 200 = 3800 and 19 * 218 = 4142, the division is exact. Hence 4142 is exactly divisible by 19, confirming that adding 11 to 4131 achieves divisibility. If we test a smaller candidate such as adding 10, we get 4141. Dividing 4141 by 19 gives a remainder of 18, so 10 is not enough. This confirms that 11 is indeed the smallest integer that works. Why Other Options Are Wrong: Option 10: 4131 + 10 = 4141, which is not divisible by 19 since it leaves a remainder of 18. Option 9: 4131 + 9 = 4140, which leaves a remainder of 17 when divided by 19. Option 12: 4131 + 12 = 4143, which is larger than needed and is not the least such number. Option 8: This is exactly the remainder of 4131 when divided by 19, but adding 8 gives 4139, which still leaves a remainder of 16. It does not make the number divisible. Common Pitfalls: A common error is to confuse the remainder itself with the required addition and choose 8 instead of 11. Another mistake is to perform the division incorrectly or to approximate instead of calculating the exact remainder. Some students may also test options one by one, which is slower and more error prone than using the simple formula k = d - remainder. Understanding the relationship between a number, its remainder, and the divisor is the key to solving such questions quickly. Final Answer: The least number that must be added to 4131 to make it divisible by 19 is 11.

Question 13

The first term of an arithmetic progression is -19 and the twelfth term is 36. What is the sum of the first 12 terms of this arithmetic progression?

Accepted Answer

Introduction / Context: This is another straightforward arithmetic progression sum question. We are given the first term and the twelfth term and asked to find the sum of the first twelve terms. It reinforces the basic formula for the sum of an arithmetic series and is typical of quick scoring questions in many competitive exams. Given Data / Assumptions: First term a1 = -19. Twelfth term a12 = 36. Number of terms n = 12. The sequence is an arithmetic progression with constant common difference. Concept / Approach: The sum S_n of the first n terms of an arithmetic progression is given by S_n = n * (a1 + a_n) / 2. The convenience in this question is that both the first term and the twelfth term are directly given, so we do not need to compute the common difference for the sum. By substituting n = 12, a1 = -19 and a12 = 36 into the formula, we can obtain the required sum quickly. Step-by-Step Solution: We have n = 12, a1 = -19, and a12 = 36. Use the sum formula S_n = n * (a1 + a_n) / 2. So S_12 = 12 * (a1 + a12) / 2. Substitute values: S_12 = 12 * (-19 + 36) / 2. Compute the sum inside the brackets: -19 + 36 = 17. So S_12 = 12 * 17 / 2. Simplify by dividing 12 by 2: 12 / 2 = 6. Now S_12 = 6 * 17 = 102. Therefore, the sum of the first 12 terms is 102. Verification / Alternative Check: As a quick check, we can find the common difference to see whether the numbers look reasonable. The common difference d is (a12 - a1) / 11 = (36 - (-19)) / 11 = 55 / 11 = 5. The sequence is then -19, -14, -9, -4, 1, 6, 11, 16, 21, 26, 31, 36. If we roughly average the first and last term, we get ( -19 + 36 ) / 2 = 17 / 2 = 8.5. Multiplying this average by 12 terms gives 8.5 * 12 = 102, which matches our formula result and confirms it is consistent. Why Other Options Are Wrong: Option 192: This would require a much larger average term and is not consistent with the actual terms of the sequence. Option 230: This is significantly higher than the correct sum and cannot be obtained from the given values. Option 214: This is also too large and would imply an incorrect application of the sum formula or wrong arithmetic. Option 120: Slightly higher than the correct sum but still incorrect and likely arises from a miscalculation when multiplying or dividing. Common Pitfalls: Some learners mistakenly use the formula S_n = n * a1 instead of including the last term, which is incorrect for sums of arithmetic progressions. Others may accidentally use 11 as the number of terms because they misread the given twelfth term as implying eleven intervals rather than twelve entries. The safest approach is to clearly identify n, a1, and a_n, substitute carefully into the formula, and simplify step by step, especially when negative numbers are involved. Final Answer: The sum of the first 12 terms of the arithmetic progression is 102.

Question 14

In an arithmetic progression, the fourth term is 11 and the seventh term is -4. What is the sum of the first 11 terms of this arithmetic progression?

Accepted Answer

Introduction / Context: This question tests your ability to use information about two nonconsecutive terms of an arithmetic progression to find both the first term and the common difference, and then compute the sum of a given number of terms. It is slightly more involved than a direct sum question, so it is a good test of conceptual understanding of arithmetic sequences. Given Data / Assumptions: Let the first term of the arithmetic progression be a. Let the common difference be d. The fourth term a4 is 11. The seventh term a7 is -4. We must find the sum of the first 11 terms, S_11. Concept / Approach: We use the nth term formula of an arithmetic progression: a_n = a + (n - 1) * d. From the given fourth term and seventh term, we build two equations in a and d. Solving these equations gives us a and d. Once we know a and d, we can either apply the sum formula S_n = n * (2a + (n - 1) * d) / 2 or S_n = n * (a1 + a_n) / 2 to find S_11. This approach combines equation solving and use of the sum formula. Step-by-Step Solution: Write the fourth term: a4 = a + 3d = 11. Write the seventh term: a7 = a + 6d = -4. Subtract the first equation from the second: (a + 6d) - (a + 3d) = -4 - 11. This simplifies to 3d = -15, so d = -15 / 3 = -5. Substitute d = -5 into a + 3d = 11. Then a + 3 * (-5) = 11, so a - 15 = 11. Therefore a = 11 + 15 = 26. Now use the sum formula S_n = n * (2a + (n - 1) * d) / 2 for n = 11. Compute S_11 = 11 * (2 * 26 + 10 * (-5)) / 2. Inside the brackets: 2 * 26 = 52 and 10 * (-5) = -50, so 52 - 50 = 2. Thus S_11 = 11 * 2 / 2 = 11. So the sum of the first 11 terms is 11. Verification / Alternative Check: To verify, let us list several terms. With a = 26 and d = -5, the sequence begins 26, 21, 16, 11, 6, 1, -4, -9, -14, -19, -24. We can check the given data: the fourth term is 11 and the seventh term is -4, which matches. Now add the terms in pairs from both ends: (26 + -24) = 2, (21 + -19) = 2, (16 + -14) = 2, (11 + -9) = 2, (6 + -4) = 2 and the middle term is 1. So we have five pairs each summing to 2 plus the middle 1. The total sum is 5 * 2 + 1 = 10 + 1 = 11, confirming our earlier calculation. Why Other Options Are Wrong: Option -75: This would correspond to an incorrect calculation of the sum or a wrong common difference, and it does not match the actual series sum. Option 55: This would arise if someone mistakenly treated the first term as 11 and assumed a simple pattern, but it does not satisfy the correct equations. Option 100: This is significantly larger than the correct value and is inconsistent with the decreasing nature of the sequence. Option -11: This would have the right magnitude but wrong sign and usually arises from carelessly pairing terms or misplacing negative signs. Common Pitfalls: Students often forget that the nth term formula uses (n - 1) * d, not n * d, which leads to incorrect equations for a4 and a7. Another common mistake is to mishandle negative differences, especially when substituting back to find the first term. When using the sum formula, it is easy to forget to multiply by the correct number of terms or to apply the negative sign incorrectly inside the brackets. Writing each step clearly and checking signs at each stage helps avoid these problems. Final Answer: The sum of the first 11 terms of the arithmetic progression is 11.

Question 15

Three numbers are in arithmetic progression with a total sum of 30 and their product equal to 910. What is the greatest of these three numbers?

Accepted Answer

Introduction / Context: This question combines properties of arithmetic progressions with algebraic manipulation involving sums and products. You are given that three numbers are in arithmetic progression, their sum is known, and their product is also known. You need to determine the largest of the three numbers. This type of problem checks your ability to model numbers in an arithmetic progression and to use algebra to simplify and solve the resulting equations. Given Data / Assumptions: Three numbers are in arithmetic progression. The sum of the three numbers is 30. The product of the three numbers is 910. We are asked to find the greatest of the three numbers. All three numbers are real numbers and the options suggest integer solutions. Concept / Approach: When three numbers are in arithmetic progression, they can be written as a - d, a, and a + d, where a is the middle term and d is the common difference. The sum of these three numbers is (a - d) + a + (a + d) = 3a, so we can quickly find the middle term from the sum. Then we use the product condition to determine d. Once we know a and d, we can write down all three numbers and identify the greatest. This method neatly uses the structure of an arithmetic progression. Step-by-Step Solution: Let the three numbers be a - d, a, and a + d. Their sum is (a - d) + a + (a + d) = 3a. Given that the sum is 30, we have 3a = 30, so a = 30 / 3 = 10. Therefore the three numbers are 10 - d, 10, and 10 + d. Their product is (10 - d) * 10 * (10 + d) = 910. Notice that (10 - d) * (10 + d) = 100 - d^2 by the difference of squares formula. So 10 * (100 - d^2) = 910. Divide both sides by 10 to simplify: 100 - d^2 = 91. Rearrange to find d^2: d^2 = 100 - 91 = 9. So d = 3 or d = -3, but in either case the three numbers are 7, 10, and 13. The greatest of these three numbers is 13. Verification / Alternative Check: We can verify by checking the sum and product directly. For the numbers 7, 10, and 13, the sum is 7 + 10 + 13 = 30, which matches the given sum. The product is 7 * 10 * 13 = 70 * 13 = 910, which matches the given product. The three numbers are equally spaced by 3, confirming they form an arithmetic progression. Since 13 is clearly larger than 7 and 10, our identification of the greatest number is correct. Why Other Options Are Wrong: Option 17: If 17 were the greatest, numbers like 13, 15, 17 might be considered, but their sum is 45, not 30, and their product is not 910. Option 15: For 9, 12, 15 the sum is 36, not 30, and the product is not 910 either. Option 10: This is the middle term, not the greatest term, in the correct arithmetic progression. Option 19: Any triplet with 19 as the largest term and in arithmetic progression would lead to a different sum and product, so it cannot satisfy the given conditions. Common Pitfalls: One common mistake is to assume the three numbers are a, a + d, and a + 2d instead of centering them as a - d, a, a + d. That approach can also work but leads to different algebra and sometimes more complex equations. Another error is to misapply the difference of squares formula or to mishandle the product condition. Students also sometimes forget that d can be positive or negative, but in this question both possibilities give the same set of three numbers, just in reversed order. Being systematic about the algebra avoids confusion. Final Answer: The greatest of the three numbers in arithmetic progression is 13.

Question 16

If A = 1 - 10 + 3 - 12 + 5 - 14 + 7 - 16 + ... up to 60 terms, what is the value of A?

Accepted Answer

Introduction / Context: This question involves an alternating series of positive and negative terms where the positive and negative terms each follow their own simple pattern. Such problems test your ability to recognize patterns, group terms intelligently, and sum a finite series without writing out every term individually. Handling signs carefully is crucial in series like this. Given Data / Assumptions: The series is A = 1 - 10 + 3 - 12 + 5 - 14 + 7 - 16 + ... The pattern continues in the same manner up to 60 terms in total. We are asked to find the final value of A after summing all 60 terms. The sequence of positive terms is 1, 3, 5, 7, ... (odd numbers). The sequence of negative terms is -10, -12, -14, -16, ... (negative even numbers starting from 10). Concept / Approach: The key idea is to group the series into pairs, each consisting of one positive odd number and the following negative even number. Each pair has the form (2k - 1) - (8 + 2k) for some integer k. We can simplify each pair algebraically and notice that the result is a constant, independent of k. Once we know how many such pairs appear in 60 terms, we can multiply the value of one pair by the number of pairs to get the total sum. Step-by-Step Solution: Write the first few terms with indices to see the pattern clearly. First pair: 1 - 10. Second pair: 3 - 12. Third pair: 5 - 14. Fourth pair: 7 - 16, and so on. Notice that the positive terms are odd numbers: 1, 3, 5, 7, ... which can be written as 2k - 1. The corresponding negative terms are even numbers starting from 10: 10, 12, 14, 16, ... which can be written as 8 + 2k. So a typical pair is (2k - 1) - (8 + 2k) = 2k - 1 - 8 - 2k. This simplifies to -9 for every pair. The series has 60 terms in total, and each pair uses 2 terms, so there are 60 / 2 = 30 pairs. Therefore, A = 30 * (-9) = -270. Verification / Alternative Check: We can verify the pattern by explicitly calculating the first few pair sums. For the first pair, 1 - 10 = -9. For the second pair, 3 - 12 = -9. For the third pair, 5 - 14 = -9. For the fourth pair, 7 - 16 = -9. This confirms that every pair contributes -9 to the sum. Since the pattern of odd positives and even negatives continues consistently, and we have a total of 30 such pairs, the total sum must be 30 * (-9) = -270. There is no leftover single term because 60 is an even number. Why Other Options Are Wrong: Option -360: This would correspond to 40 terms of -9 or some miscount of the number of pairs, so it does not match the correct pairing. Option -310: This is not a multiple of -9 and therefore cannot result from 30 identical pair sums of -9. Option -240: This equals -8 * 30, which would only occur if each pair summed to -8, which is not the case here. Option -300: This equals -10 * 30 and could result from a mistaken assumption that each pair sums to -10, but actual computations show that each pair sums to -9. Common Pitfalls: Students often try to handle such series term by term instead of recognizing the repeating pattern in pairs. This can lead to arithmetic errors or confusion. Another common issue is misidentifying the general form of the terms and thereby computing the pair sum incorrectly. Some may forget to divide the total number of terms by 2 to get the number of pairs, leading to an incorrect multiplier. Careful identification of the pair structure and verification of the pattern with the first few pairs helps avoid these mistakes. Final Answer: The value of A for the given 60 term series is -270.

Question 17

If A = 2^32, B = 2^31 + 2^30 + 2^29 + ... + 2^0 and C = 3^15 + 3^14 + 3^13 + ... + 3^0, which of the following is the correct order of A, B and C?

Accepted Answer

Introduction / Context: This question examines your understanding of powers of numbers and geometric series. You are asked to compare three large quantities defined in terms of powers of 2 and powers of 3. Even though the actual numbers are very large, you can use algebraic formulas and approximations to compare their sizes without computing exact decimal values. This type of comparison is common in aptitude tests to check conceptual reasoning with exponents. Given Data / Assumptions: A is defined as 2^32. B is defined as the sum 2^31 + 2^30 + 2^29 + ... + 2^0. C is defined as the sum 3^15 + 3^14 + 3^13 + ... + 3^0. All numbers are positive real numbers. We must determine which inequality among A, B, and C is correct. Concept / Approach: The sum of a finite geometric series with first term a and common ratio r is given by S_n = a * (r^n - 1) / (r - 1), when r is not equal to 1. For B, the series is a geometric sequence with first term 1 (that is 2^0) and ratio 2, from power 0 up to power 31. For C, the sequence has first term 1 (that is 3^0) and ratio 3, from power 0 up to power 15. Using the geometric series formula, we can express both B and C in compact form and then compare them to A. Step-by-Step Solution: First express B as a geometric sum with a = 1 and r = 2. B = 2^0 + 2^1 + 2^2 + ... + 2^31. For such a series, S_n = (r^n - 1) / (r - 1) with n = 32 and r = 2. Therefore B = (2^32 - 1) / (2 - 1) = 2^32 - 1. Thus A = 2^32 and B = 2^32 - 1, so A is larger than B by exactly 1. Now express C as a geometric sum with a = 1 and r = 3. C = 3^0 + 3^1 + 3^2 + ... + 3^15. Using S_n = (r^n - 1) / (r - 1) with n = 16 and r = 3, we get C = (3^16 - 1) / (3 - 1) = (3^16 - 1) / 2. Clearly 2^32 is enormously larger than 3^16, and B is just one less than A. Also, 3^16 is far smaller than 2^32, so (3^16 - 1) / 2 is much smaller than 2^32 - 1. Therefore A > B and B > C, giving A > B > C. Verification / Alternative Check: To gain more intuition, compare sizes roughly. Since 2^10 is 1024, which is about 10^3, we get 2^30 approximately equal to 10^9, and 2^32 is about 4 * 10^9. For 3^5, the value is 243, so 3^10 is about 243^2 which is roughly 59000, and 3^15 is about 59000 * 243 which is around 14 million. Hence 3^16 is about 3 * 14 million or roughly 40 million. Even if we double or triple this to account for the sum in C, we still stay in the order of tens of millions. By contrast, 2^32 is in the order of billions. So 2^32 and 2^32 - 1 are both far larger than any combination of the 3 powers in C, confirming the ordering A > B > C. Why Other Options Are Wrong: Option C > B > A: This would mean C is the largest, but C is based on 3 powers only up to 3^15, which yields much smaller values than powers of 2 up to 2^32. Option C > A > B: Again this claims C is largest and that A is larger than B, but although A larger than B is true, C is nowhere near their scale. Option A > C > B: This suggests C is between A and B, but B equals 2^32 - 1 while A is 2^32, so there is no room for another number between them when comparing exact values. Option B > A > C: This reverses A and B, but from B = 2^32 - 1 it is obvious that A = 2^32 is larger than B. Common Pitfalls: Some students may be tempted to approximate 3^n as larger than 2^n due to the bigger base and conclude that C could be comparable or larger, without noticing that the exponents are much smaller for powers of 3. Others might forget or misapply the geometric series formula, leading to incorrect closed forms. Also, failing to recognize that B is almost equal to A (just one less) can cause confusion when judging relative sizes. Careful use of formulas and approximate magnitude comparisons helps avoid such errors. Final Answer: The correct order of the three quantities is A > B > C.

Question 18

Let N = 1 + 11 + 111 + 1111 + ... + 111111111 (up to nine 1s). What is the sum of the digits of N?

Accepted Answer

Introduction / Context: This question involves a sum of numbers made entirely of the digit 1 repeated different numbers of times, and then asks for the sum of the digits of the result. Such numbers are often called repunits. The problem tests your ability to handle pattern based sums and to work efficiently with large looking numbers by focusing on their structure rather than raw magnitude. Given Data / Assumptions: The expression N is defined as N = 1 + 11 + 111 + 1111 + ... up to the term with nine 1s. The last term is 111111111 (nine 1s). There are nine terms in total in this sum. We must find the sum of the digits of the resulting number N. Concept / Approach: One direct approach is to recognize that adding 1, 11, 111, and so on in this pattern produces a very symmetric final number. In fact, if we add these terms carefully, we find that the result is 123456789. Once this is known or derived, the sum of the digits is easy to compute. Alternatively, we can simulate the process of adding these terms by looking at how many times each place value contributes to the final sum. However, for exam purposes, recognizing or verifying that N equals 123456789 is the fastest route. Step-by-Step Solution: Write out the nine terms explicitly: 1, 11, 111, 1111, 11111, 111111, 1111111, 11111111, and 111111111. Start adding them in a column format to observe the pattern in the sum. If you perform the addition carefully, you will see that carrying over digits creates a final number with digits increasing from 1 to 9. The final result of the sum is N = 123456789. Now find the sum of the digits of N: 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9. Compute this digit sum: 1 + 2 + 3 + 4 + 5 = 15 and 6 + 7 + 8 + 9 = 30. Add the two partial sums: 15 + 30 = 45. Therefore, the sum of the digits of N is 45. Verification / Alternative Check: To verify that N really is 123456789, you can add a few terms step by step. For example, 1 + 11 = 12. Then 12 + 111 = 123. Next 123 + 1111 = 1234. Continue this pattern and you will see the sum after four terms is 1234, after five terms is 12345, and so on, until after nine terms the sum is 123456789. This pattern arises because each new term ends with a 1 and creates a chain of carries that push the previous digits one place to the left while adding 1 to the new leading position. Why Other Options Are Wrong: Option 18: This is much smaller than the true digit sum and may come from adding only a subset of the digits or miscalculating the final number. Option 36: This is the sum of 1 to 8 but ignores the ninth digit or results from an arithmetic slip. Option 5: This is far too small and suggests that only one or two digits were considered. Option 27: This is the sum of the digits of 999999999, not of 123456789, so it does not match this specific series. Common Pitfalls: The main difficulty is that the numbers look large and many students are tempted to give up without attempting the addition. Others may add quickly and make carrying mistakes, leading to an incorrect final value for N. Some may also mistakenly interpret the question and think they only need the number of terms or the last digit. Writing the addition vertically and checking carries carefully makes the process straightforward. Recognizing the well known pattern 1 + 11 + 111 + ... up to nine 1s yielding 123456789 is a useful shortcut. Final Answer: The sum of the digits of N is 45.

Question 19

In the number sequence 1, 3, 4, 5, 7, 7, 10, 9, ... two interleaved arithmetic progressions are formed. What is the sum of the first 40 terms of this sequence?

Accepted Answer

Introduction / Context: This problem gives a sequence that at first glance looks irregular, but it actually consists of two interleaved arithmetic progressions, one on the odd positions and another on the even positions. You are asked to find the sum of the first 40 terms. This type of question tests pattern recognition, understanding of arithmetic progressions, and the ability to split a single sequence into multiple simpler subsequences. Given Data / Assumptions: The sequence starts as 1, 3, 4, 5, 7, 7, 10, 9, ... We need the sum of the first 40 terms of this sequence. We observe that odd position terms and even position terms follow different patterns. We assume the pattern continues in the same way beyond the shown terms. Concept / Approach: Separate the sequence into two subsequences: one containing terms at odd positions (1st, 3rd, 5th, ...) and the other containing terms at even positions (2nd, 4th, 6th, ...). By examining the first few terms of each subsequence, we can identify them as arithmetic progressions and find their explicit formulas. Since the first 40 terms include 20 odd-position terms and 20 even-position terms, we can calculate the sum of each subsequence separately and then add the results to obtain the total sum. Step-by-Step Solution: List the positions and terms: position 1: 1, 2: 3, 3: 4, 4: 5, 5: 7, 6: 7, 7: 10, 8: 9, and so on. Odd positions (1, 3, 5, 7, ...): 1, 4, 7, 10, ... Even positions (2, 4, 6, 8, ...): 3, 5, 7, 9, ... The odd-position subsequence 1, 4, 7, 10, ... is an arithmetic progression with first term a_odd = 1 and common difference d_odd = 3. The even-position subsequence 3, 5, 7, 9, ... is an arithmetic progression with first term a_even = 3 and common difference d_even = 2. Among the first 40 terms, there are 20 odd positions and 20 even positions. Sum of the first 20 odd-position terms: S_odd = 20 * (2 * a_odd + (20 - 1) * d_odd) / 2. So S_odd = 20 * (2 * 1 + 19 * 3) / 2 = 20 * (2 + 57) / 2 = 20 * 59 / 2 = 10 * 59 = 590. Sum of the first 20 even-position terms: S_even = 20 * (2 * a_even + (20 - 1) * d_even) / 2. So S_even = 20 * (2 * 3 + 19 * 2) / 2 = 20 * (6 + 38) / 2 = 20 * 44 / 2 = 10 * 44 = 440. Total sum of the first 40 terms is S_total = S_odd + S_even = 590 + 440 = 1030. Verification / Alternative Check: We can quickly check the first few terms generated by the formulas for both subsequences. For odd positions, the nth odd term is 1 + (n - 1) * 3. For n = 1, this gives 1; for n = 2, it gives 4; for n = 3, it gives 7; and for n = 4, it gives 10, matching the actual odd position terms. For even positions, the nth even term is 3 + (n - 1) * 2. For n = 1, this gives 3; for n = 2, it gives 5; for n = 3, it gives 7; and for n = 4, it gives 9, matching the observed even terms. Since both subsequences follow consistent arithmetic progressions, using their sum formulas is valid, and the calculated total of 1030 is reliable. Why Other Options Are Wrong: Option 1010: This is smaller than the correct sum and could result from incorrectly using 19 instead of 20 terms in one of the subsequences. Option 1115: This is larger than the correct value and suggests an overcount of one subsequence or incorrect common difference. Option 1031: This is only one more than the correct sum and might appear if a single term is miscounted twice or a minor arithmetic slip occurs. Option 990: This is noticeably smaller and indicates that the structure of the two interleaved progressions was not properly recognized. Common Pitfalls: Many students attempt to treat the given sequence as a single arithmetic progression, which it is not. Others may fail to notice that odd and even positions follow different patterns. Miscounting the number of terms in each subsequence is another frequent issue, especially when working with 40 total terms. To avoid these mistakes, always check whether a mixed sequence can be decomposed into simpler subsequences and carefully count how many terms of each type fall within the required range. Final Answer: The sum of the first 40 terms of the given sequence is 1030.

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