Q: What is the remainder when 6910 is divided by 81?

Introduction / Context: This is another remainder problem that reinforces the concept of modular arithmetic. It is important for learners to be comfortable computing remainders quickly and accurately in quantitative exams. Given Data / Assumptions: - Dividend: 6910. - Divisor: 81. - We must determine the remainder when 6910 is divided by 81. Concept / Approach: We want to express 6910 in the form 81 * q + r, where q is the integer quotient and r is the remainder with 0 ≤ r < 81. We can find q by approximate or exact multiplication and then compute r as the difference between the dividend and 81 * q. Step-by-Step Solution: Let N = 6910 and d = 81. Compute an approximate quotient: 81 × 80 = 6480. Increase the multiple: 81 × 85 = 6885. Try 81 × 86 = 6966, which exceeds 6910, so 86 is too large. Thus q = 85 and the remainder r = 6910 − 6885 = 25. Verification / Alternative Check: We see that 6910 ÷ 81 is slightly greater than 85, because 6885 is close to 6910. The difference 25 is less than 81, so it is a valid remainder. Any attempt to use q = 86 would lead to a product higher than the dividend and therefore is invalid. This confirms that the correct remainder is 25. Why Other Options Are Wrong: 23, 21, and 19 are all smaller than 25 and would imply different quotients that do not match the exact multiplication with 81. 27 is larger than 25 and does not correspond to the computed remainder 6910 − 6885. Common Pitfalls: Errors often come from miscalculating the intermediate products 81 × 80, 81 × 85, or from subtracting incorrectly when finding the remainder. Some students also confuse quotient with remainder. Clear separation of these concepts and systematic checking of the largest valid multiple helps avoid these mistakes. Final Answer: The remainder when 6910 is divided by 81 is 25.

Question 1

The sum of three consecutive natural numbers, each divisible by 5, is 225. What is the largest of these three numbers?

Accepted Answer

Introduction / Context:This question combines consecutive numbers with divisibility by a fixed integer. The three natural numbers are consecutive multiples of 5 and their sum is given. You must find the largest of them. Problems like this build comfort with arithmetic progressions and basic divisibility rules in a natural numbers context. Given Data / Assumptions:- There are three consecutive natural numbers, each divisible by 5.- Being consecutive and divisible by 5 means they are consecutive multiples of 5, such as 5k, 5(k + 1), and 5(k + 2).- Their sum is 225.- We need to find the largest of these three numbers. Concept / Approach:Multiples of 5 can be written as 5 times an integer. If we denote the smallest multiple by 5k, the next two consecutive multiples are 5(k + 1) and 5(k + 2). Their sum can be simplified to a single expression in k, leading to a linear equation. Solving this equation determines k, and from there we can find each of the three numbers and identify the largest. Step-by-Step Solution:Step 1: Let the three consecutive multiples of 5 be 5k, 5(k + 1), and 5(k + 2).Step 2: Write their sum: 5k + 5(k + 1) + 5(k + 2).Step 3: Factor out 5: 5[k + (k + 1) + (k + 2)] = 5[3k + 3].Step 4: The sum is given as 225, so 5[3k + 3] = 225.Step 5: Divide both sides by 5 to get 3k + 3 = 45.Step 6: Subtract 3 from both sides: 3k = 42.Step 7: Divide both sides by 3 to solve for k: k = 14.Step 8: The three numbers are 5k = 5 * 14 = 70, 5(k + 1) = 5 * 15 = 75, and 5(k + 2) = 5 * 16 = 80.Step 9: Among these, the largest number is 80. Verification / Alternative check:Check the sum of the numbers 70, 75, and 80. Their sum is 70 + 75 + 80 = 225, which matches the given total. Also, all three numbers are divisible by 5, and they are consecutive multiples of 5 because the differences between them are 5. This confirms that our algebraic approach is correct and that 80 is indeed the largest of the three numbers. Why Other Options Are Wrong:If the largest number were 85, then the three consecutive multiples of 5 would be 75, 80, and 85, whose sum is 240, not 225. If the largest number were 75, the three numbers would be 65, 70, and 75, whose sum is 210. Similarly, a largest number of 70 or 90 would produce sums that do not equal 225. Therefore, those options do not satisfy the condition given in the question. Common Pitfalls:A typical mistake is to assume that the three numbers are 5 apart overall instead of 5 apart between consecutive terms. Another common error is setting them as 5k - 5, 5k, and 5k + 5 but then miscomputing the sum. Using the expression 5k, 5(k + 1), and 5(k + 2) and simplifying systematically avoids confusion and keeps the algebra very clear. Final Answer:The largest of the three numbers is 80.

Question 2

What is the greatest 6 digit number that is a perfect square?

Accepted Answer

Introduction / Context:This question involves perfect squares and place value. You are asked to find the largest 6 digit number that is a perfect square. Understanding how squares grow and how to bound them by digit length is an important skill that connects number sense with algebra and estimation. Given Data / Assumptions:- We are interested in 6 digit numbers, which range from 100000 to 999999 inclusive.- We want the greatest such number that is a perfect square of some integer n. - This means n^2 should have exactly 6 digits and be as large as possible. Concept / Approach:To find the largest 6 digit perfect square, we look for the largest integer n such that n^2 is less than 1000000 (since 1000000 is a 7 digit number). The next integer after n must produce a square that has 7 digits. Therefore, n should be one less than the first integer whose square is at least 1000000. Since 1000^2 = 1000000, we consider n = 999 and compute 999^2. That square will be the largest 6 digit perfect square. Step-by-Step Solution:Step 1: Note that 1000^2 = 1000000, which is a 7 digit number.Step 2: Therefore, any integer n greater than or equal to 1000 will give n^2 with at least 7 digits.Step 3: We need the largest integer n whose square still has 6 digits, so n must be 999.Step 4: Compute 999^2. Use the identity (1000 - 1)^2 = 1000^2 - 2 * 1000 * 1 + 1.Step 5: Calculate (1000 - 1)^2 = 1000000 - 2000 + 1.Step 6: Simplify: 1000000 - 2000 = 998000, then 998000 + 1 = 998001.Step 7: So 999^2 = 998001, which is a 6 digit number.Step 8: Check that it is the greatest 6 digit square by confirming that the next integer 1000 has square 1000000, which is 7 digits and therefore too large. Verification / Alternative check:You can compare 998001 with other options. For example, 999001 is close but would require the square root to be slightly more than 999, and 1000^2 already jumps to 1000000, so 999001 cannot be a perfect square. Similarly, 998009 and 998101 are not equal to 999^2 and would not have integer square roots. The calculation based on the binomial square formula is precise and confirms that 998001 is indeed the square of 999. Why Other Options Are Wrong:Option 999001 is not equal to 999^2, and there is no integer whose square produces 999001 within the 6 digit range. Options 998009 and 998101 are also not perfect squares of any integer (they lie between consecutive square values). The value 997969 is lower than 998001 and may be a square of a slightly smaller integer, but it is not the greatest 6 digit square. Only 998001 corresponds to 999^2, which is the largest 6 digit perfect square. Common Pitfalls:Some learners guess among the choices without connecting them to specific square roots, or they attempt to take square roots by trial and error without estimation. Others may incorrectly think 999999 is a square, but the square root of 999999 is not an integer. Understanding that 1000^2 is the first 7 digit square and stepping down to 999 is the systematic way to reach the correct answer. Final Answer:The greatest 6 digit perfect square number is 998001.

Question 3

The numbers x, y, and z are proportional to 2, 3, and 5 respectively, and x + y + z = 80. If z is given by the relation z = a x - 8, what is the value of a?

Accepted Answer

Introduction / Context:This algebra question involves proportional relationships and a linear equation relating two variables. You are told that three numbers x, y, and z are in proportion 2:3:5 and that their sum is 80. Additionally, z can be expressed in terms of x using the linear expression z = a x - 8. The task is to determine the value of the coefficient a. This type of question is common in problems on ratios, proportions, and linear equations. Given Data / Assumptions:- x : y : z = 2 : 3 : 5.- x + y + z = 80.- z can be written as z = a x - 8 for some constant a.- We must find the value of a that satisfies all these conditions. Concept / Approach:First, we use the proportional relationship to express x, y, and z in terms of a single scaling factor k. This will allow us to use the given sum to find specific numeric values for x, y, and z. Once we know x and z explicitly, we substitute them into z = a x - 8 and solve for a. This two step process moves from general ratios to concrete numbers and then to the linear relation between x and z. Step-by-Step Solution:Step 1: Since x, y, and z are proportional to 2, 3, and 5, let x = 2k, y = 3k, and z = 5k for some positive real number k.Step 2: Use the sum condition: x + y + z = 80.Step 3: Substitute the expressions in terms of k: 2k + 3k + 5k = 80.Step 4: Simplify the left side: 2k + 3k + 5k = 10k.Step 5: So 10k = 80, and dividing both sides by 10 gives k = 8.Step 6: Now compute the actual values of x, y, and z. We have x = 2k = 2 * 8 = 16.Step 7: y = 3k = 3 * 8 = 24, and z = 5k = 5 * 8 = 40.Step 8: Check that these values sum to 80: 16 + 24 + 40 = 80, which is correct.Step 9: Use the relation z = a x - 8. Substitute x = 16 and z = 40 into this equation.Step 10: This gives 40 = a * 16 - 8.Step 11: Add 8 to both sides: 40 + 8 = 16a, so 48 = 16a.Step 12: Divide both sides by 16: a = 48 / 16 = 3. Verification / Alternative check:Once we have a = 3, we can check the relation z = a x - 8 directly. With x = 16, a x - 8 = 3 * 16 - 8 = 48 - 8 = 40, which matches z = 40 that we obtained from the proportional relationship. The sum x + y + z remains 16 + 24 + 40 = 80, and the ratio 16 : 24 : 40 simplifies by dividing by 8 to 2 : 3 : 5. This consistency check confirms that our value of a is correct. Why Other Options Are Wrong:If a were 6, then z would be 6 * 16 - 8 = 96 - 8 = 88, which does not match z = 40 and would violate the sum condition. If a were 3/2, then z would be (3/2) * 16 - 8 = 24 - 8 = 16, again not equal to 40. Similar contradictions arise for a = 5/2 or a = 4. These values fail to satisfy both the ratio and the linear equation for the same x and z. Common Pitfalls:A common mistake is to misapply the ratio and write x = 2, y = 3, and z = 5 directly without using a scaling factor, which conflicts with the sum of 80. Another error is to forget to apply the sum condition and attempt to solve for a using only the ratio relationship. Always introduce a common factor for ratios, use the sum to find the exact values, and then apply any additional relations among the variables. Final Answer:The value of a is 3.

Question 4

Which one of the following is the minimum possible value of the sum of two positive integers whose product is 24?

Accepted Answer

Introduction / Context: This aptitude question tests understanding of factor pairs of a positive integer and how the sum of two numbers behaves when their product is fixed. Such questions are common in problems on numbers and optimisation in basic algebra. Given Data / Assumptions: - Two positive integers have a product equal to 24. - We need the smallest possible value of their sum a + b. - Only positive integers are considered, because the question explicitly asks for positive integers in the context of basic factor pairs. Concept / Approach: When the product a * b is fixed and positive, the sum a + b is smallest when the two positive integers are as close to each other as possible. Therefore, to minimise the sum, we should look for factor pairs of 24 that are nearest to each other in value. Step-by-Step Solution: List factor pairs of 24 in positive integers: (1, 24), (2, 12), (3, 8), (4, 6). Compute sums: 1 + 24 = 25, 2 + 12 = 14, 3 + 8 = 11, 4 + 6 = 10. Compare all sums: 25, 14, 11, and 10. The smallest among these sums is 10, which comes from the pair (4, 6). Therefore, the minimum possible value of the sum of the two positive integers is 10. Verification / Alternative Check: We can reason that for a constant positive product, making the numbers closer reduces their sum. The pair (4, 6) is the closest pair of positive factors of 24, so its sum should be minimal. Quick recomputation confirms that no other positive factor pair of 24 gives a sum smaller than 10. Why Other Options Are Wrong: 25 corresponds to the pair (1, 24) and is much larger than 10. 11 corresponds to the pair (3, 8) and is still greater than 10. 8 is not the sum of any positive integer factor pair of 24, so it is invalid in this context. 14 corresponds to the pair (2, 12) and is also larger than 10. Common Pitfalls: A common mistake is to ignore the word positive and consider negative factor pairs such as (-4, -6), which also multiply to 24 but give negative sums. In many school level aptitude questions, unless otherwise stated, factor pairs are taken as positive integers. Another error is to stop after checking only one or two factor pairs without systematically listing all possibilities. Final Answer: The minimum possible value of the sum of two positive integers whose product is 24 is 10.

Question 5

What is the smallest digit that can replace the symbol * in the number 63576*2 so that the resulting number is divisible by 8?

Accepted Answer

Introduction / Context: This problem checks the rule of divisibility by 8, specifically how only the last three digits of a number determine whether the entire number is divisible by 8. The question embeds an unknown digit represented by * and asks for the smallest digit that makes the number divisible by 8. Given Data / Assumptions: - The number is 63576*2, where * represents a single digit from 0 to 9. - We want the resulting full number to be divisible by 8. - We are asked for the smallest possible digit that satisfies this condition. Concept / Approach: A number is divisible by 8 if and only if the integer formed by its last three digits is divisible by 8. Therefore, we only need to examine the last three digits of 63576*2, which are 6, *, and 2. We will treat these as a three digit number 6*2 and search for the smallest digit that makes this three digit number divisible by 8. Step-by-Step Solution: Let the digit * be x, so the last three digits form the number 6x2. This can be written as 600 + 10x + 2, which is 602 + 10x. We need 602 + 10x to be divisible by 8. Test x from 0 to 9. When x = 3, we get 602 + 30 = 632, and 632 / 8 = 79 exactly. Therefore, 6 3 2 is divisible by 8, so the whole number 6357632 is divisible by 8. Verification / Alternative Check: We can also test x = 7, giving 602 + 70 = 672, which is also divisible by 8, but 7 is larger than 3. Since the question asks for the smallest possible digit, 3 is preferred. A quick long division of 632 by 8 confirms the divisibility without remainder. Why Other Options Are Wrong: If x = 1, the last three digits form 612, and 612 / 8 leaves a remainder. If x = 2, the last three digits form 622, which is not divisible by 8. If x = 4, the last three digits form 642, which also is not divisible by 8. Digit 0 gives 602, which is not divisible by 8 either. Therefore, only x = 3 works and is the smallest such digit. Common Pitfalls: Students often try to test the entire number instead of focusing only on the last three digits, which wastes time. Another common mistake is to misapply the divisibility rule by using the last digit or last two digits instead of the last three for 8. Some learners also forget to check all smaller digits once one working digit is found, which can lead to missing the smallest valid option. Final Answer: The smallest digit that can replace * so that 63576*2 is divisible by 8 is 3.

Question 6

What is the smallest number that must be added to 13851 so that the resulting number is exactly divisible by 87?

Accepted Answer

Introduction / Context: This question tests divisibility and remainders, specifically how to adjust a given number by the smallest amount so that it becomes exactly divisible by a given divisor. It is a standard type of problem in the number system and modular arithmetic section of aptitude exams. Given Data / Assumptions: - Original number: 13851. - Required divisor: 87. - We must find the least non negative integer k such that 13851 + k is divisible by 87. Concept / Approach: If a number N leaves a remainder r when divided by a divisor d, then the smallest number to be added so that N becomes divisible by d is (d − r), provided r is not zero. We therefore need to compute the remainder when 13851 is divided by 87 and adjust accordingly. Step-by-Step Solution: Compute 13851 ÷ 87 and find the remainder r. Performing the division, 87 × 159 = 13833. Subtract: 13851 − 13833 = 18, so the remainder r is 18. The smallest number to be added is 87 − 18 = 69. Therefore, 13851 + 69 = 13920, and 13920 ÷ 87 is an exact division. Verification / Alternative Check: Check directly: 13920 ÷ 87 = 160 exactly, with no remainder. If we added any smaller positive number, such as 54 or 43, the result would still not be divisible by 87 because the remainder would not be driven to zero. Hence, 69 is indeed the minimum such addition. Why Other Options Are Wrong: Adding 18 leads to 13869, which divided by 87 still leaves a remainder, since 87 × 159 = 13833 and 13869 − 13833 = 36. Adding 43 gives 13851 + 43 = 13894, which is not divisible by 87. Adding 54 gives 13905, which again is not an exact multiple of 87. Adding 33 is arbitrary and does not result in a multiple of 87 either. Common Pitfalls: A frequent error is to miscalculate the remainder or to forget that we need the smallest non negative adjustment. Some students also confuse the operation and subtract instead of adding, or they directly test options without understanding the relationship N + k ≡ 0 (mod d). Practising careful long division or modular arithmetic helps avoid such mistakes. Final Answer: The smallest number that must be added to 13851 to make it divisible by 87 is 69.

Question 7

What is the smallest number that must be subtracted from 63520 so that the resulting number becomes a perfect square?

Accepted Answer

Introduction / Context: This problem focuses on perfect squares and asks how far a given integer is from the nearest perfect square below it. Such questions are important in number theory and help candidates gain intuition about square numbers and their spacing. Given Data / Assumptions: - We are given the number 63520. - We must subtract the smallest possible non negative integer k so that 63520 − k is a perfect square. - The resulting number must be exactly equal to n^2 for some integer n. Concept / Approach: To find the nearest perfect square less than or equal to a number N, we can compute the integer square root of N, that is the largest integer n such that n^2 ≤ N. Then we compare N with n^2. The required subtraction is N − n^2. If N is already a perfect square, this difference is zero. Step-by-Step Solution: Let N = 63520. Compute the integer square root of N. Find n such that n^2 ≤ 63520 and (n + 1)^2 > 63520. Compute 252^2 = 63504 and 253^2 = 64009. Since 63504 ≤ 63520 and 64009 > 63520, the closest perfect square below 63520 is 63504. The difference is 63520 − 63504 = 16, so we must subtract 16 to reach a perfect square. Verification / Alternative Check: Verify: 63520 − 16 = 63504, and we already know 252^2 = 63504. If we subtracted any smaller positive number, such as 14 or 24, the result would lie between 63504 and 63520 or above 63520 and would not coincide with a known perfect square. Therefore 16 is the minimal subtraction that works. Why Other Options Are Wrong: Subtracting 30 yields 63490, which is not a perfect square between 251^2 = 63001 and 252^2 = 63504. Subtracting 24 gives 63496, again not equal to 251^2 or 252^2. Subtracting 14 results in 63506, which lies between 252^2 and 253^2 but is not equal to any n^2. Only subtracting 16 exactly hits 63504 which is 252^2. Common Pitfalls: Students often approximate square roots and may misidentify the nearest perfect square. Another mistake is to search both above and below the number when the question specifically asks for subtraction, implying the target square must be less than or equal to the original number. Careful computation of nearby squares is essential to avoid errors. Final Answer: The smallest number that must be subtracted from 63520 to make the result a perfect square is 16.

Question 8

What is the greatest four digit number that is exactly divisible by 81?

Accepted Answer

Introduction / Context: This problem tests divisibility and the idea of finding the greatest multiple of a given number within a specified range. Here the range is all four digit numbers, from 1000 to 9999, and the divisor is 81. Given Data / Assumptions: - We are asked for the largest four digit integer that is divisible by 81. - The largest four digit number overall is 9999. - We need the greatest integer m such that m ≤ 9999 and 81 divides m exactly. Concept / Approach: To find the greatest multiple of a number d that does not exceed N, we divide N by d, take the integer part of the quotient, and multiply back by d. That is, if q = floor(N ÷ d), then the required multiple is q * d. Here d = 81 and N = 9999. Step-by-Step Solution: Let N = 9999 and d = 81. Compute the integer quotient q = floor(9999 ÷ 81). Perform the division: 81 × 123 = 9963 and 81 × 124 = 10044 which exceeds 9999. Therefore q = 123, since 124 gives a product above the four digit range. The greatest multiple is q * d = 123 × 81 = 9963. Verification / Alternative Check: Check that 9963 is four digit and divisible by 81. Dividing 9963 by 81 gives 123 exactly. Any larger four digit number than 9963 would either not be divisible by 81 or would be above 9999. For example, 9993 ÷ 81 does not yield an integer quotient, so 9993 is not acceptable as the required multiple. Why Other Options Are Wrong: 9993 is not divisible by 81, since it does not give an integer quotient. 9936 and 9918 are smaller than 9963 and therefore cannot be the greatest multiple if a larger valid multiple exists. 9900 is also divisible by several numbers but is smaller than 9963 and thus cannot be the largest four digit multiple of 81. Common Pitfalls: Common errors include checking only the options without using the floor division idea, which can lead to missed or misjudged calculations. Another pitfall is assuming that the largest number ending with certain digits would be divisible, without verifying by exact division. Using the systematic method of floor division is more reliable and efficient. Final Answer: The largest four digit number exactly divisible by 81 is 9963.

Question 9

A non zero number is greater than seven times its reciprocal by 9.3. What is that number?

Accepted Answer

Introduction / Context: This question involves an equation with a number and its reciprocal. It tests understanding of how to translate a verbal description into an algebraic equation and then solve a quadratic equation that arises from the relationship. Given Data / Assumptions: - Let the required non zero number be x. - The phrase "greater than seven times its reciprocal by 9.3" means x − 7 * (1 / x) = 9.3. - x is non zero so that its reciprocal 1 / x is well defined. Concept / Approach: We convert the English statement into an algebraic equation involving x and 1 / x. Multiplying both sides of the equation by x eliminates the denominator and gives a quadratic equation in x. Solving this quadratic equation will provide all possible values of x that satisfy the condition, and then we compare them with the options provided. Step-by-Step Solution: Let x be the unknown number. According to the statement, x − 7 * (1 / x) = 9.3. Multiply both sides by x to clear the denominator: x^2 − 7 = 9.3x. Rewrite as a standard quadratic: x^2 − 9.3x − 7 = 0. Solve this quadratic. The solutions are x = 10 and x = −0.7. Verification / Alternative Check: Check x = 10: 7 times the reciprocal is 7 * (1 / 10) = 0.7. The number minus this value is 10 − 0.7 = 9.3, which matches the condition. For x = −0.7, seven times the reciprocal is 7 * (1 / −0.7) = −10, and −0.7 minus (−10) equals 9.3 as well. Both values satisfy the equation, but only 10 appears among the answer options, so it is the required choice. Why Other Options Are Wrong: If x = 20, then 20 − 7 * (1 / 20) = 20 − 0.35 = 19.65, not 9.3. If x = 5, then 5 − 7 * (1 / 5) = 5 − 1.4 = 3.6. If x = 14, then 14 − 7 * (1 / 14) is approximately 13.5, again not 9.3. Only x = 10 correctly satisfies the original relationship among the provided options. Common Pitfalls: Learners sometimes misread the phrase "greater than" and set 7 * (1 / x) − x = 9.3 instead of x − 7 * (1 / x) = 9.3. Another common error is to forget to multiply each term by x when clearing the denominator, which leads to an incorrect quadratic. Finally, after solving the quadratic, it is important to check which solution is presented in the options instead of assuming only one solution exists. Final Answer: The non zero number that is greater than seven times its reciprocal by 9.3 is 10.

Question 10

What is the sum of the first 12 terms of an arithmetic progression if the first term is 3 and the twelfth (last) term is 47?

Accepted Answer

Introduction / Context: This question checks knowledge of arithmetic progressions and, in particular, the formula for the sum of the first n terms when the first term and the nth term are known. Arithmetic progressions appear frequently in quantitative aptitude and algebra topics. Given Data / Assumptions: - First term of the arithmetic progression, a = 3. - Twelfth term is the last term mentioned, so l = 47. - Number of terms n = 12. - We need to calculate the sum S of the first 12 terms. Concept / Approach: The standard formula for the sum of the first n terms of an arithmetic progression is S = n / 2 * (first term + last term). This works whenever we know the first term a and the nth term l. We apply this formula directly using the given values. Step-by-Step Solution: Identify n = 12, a = 3, and l = 47. Use the sum formula S = n / 2 * (a + l). Compute a + l = 3 + 47 = 50. Compute n / 2 = 12 / 2 = 6. Therefore S = 6 * 50 = 300. Verification / Alternative Check: Another way is to note that the average of the first and last terms is (3 + 47) / 2 = 25. Since there are 12 terms, the sum is average × number of terms = 25 × 12 = 300. This matches the result obtained from the formula, verifying the correctness. Why Other Options Are Wrong: 260 would correspond to a smaller average term or fewer terms; it does not match 25 × 12. 280 also does not equal 25 × 12 and indicates miscalculation, perhaps mixing up the number of terms. 220 is far too small and might come from incorrectly using half the number of terms or an incorrect last term. 240 is again inconsistent with both the correct formula and simple averaging. Common Pitfalls: Students sometimes confuse the formula for the nth term with the formula for the sum. Another frequent mistake is misidentifying n; for example, treating 47 as if it were the number of terms instead of the last term. Carefully reading the statement and clearly labelling a, l, and n prevents these errors. Final Answer: The sum of the first 12 terms of the arithmetic progression is 300.

Question 11

What is the smallest number that must be subtracted from 210 so that the resulting number is exactly divisible by 11?

Accepted Answer

Introduction / Context: This problem involves divisibility and remainders and asks how to adjust a number by subtraction so that the result becomes divisible by a given divisor. It reinforces modular arithmetic concepts used in many aptitude tests. Given Data / Assumptions: - Original number: 210. - Divisor: 11. - We must find the least non negative integer k such that 210 − k is divisible by 11. Concept / Approach: If a number N leaves a remainder r when divided by d, then subtracting r from N makes the result exactly divisible by d. Here we first find the remainder when 210 is divided by 11. That remainder will be the smallest number that needs to be subtracted to obtain a multiple of 11. Step-by-Step Solution: Let N = 210 and d = 11. Divide 210 by 11: 11 × 19 = 209. Compute the remainder: 210 − 209 = 1. Therefore N leaves a remainder r = 1 when divided by 11. Subtracting r gives 210 − 1 = 209, which is divisible by 11. Verification / Alternative Check: We quickly check: 209 ÷ 11 = 19 exactly, with no remainder, confirming that 209 is a multiple of 11. Any smaller non negative subtraction than 1 would be 0, and 210 itself is not divisible by 11. Any larger subtraction such as 2 or 3 would also give multiples of 11, but they would not be minimal because 1 already works. Why Other Options Are Wrong: Subtracting 2 gives 208, which is not divisible by 11 because 11 × 18 = 198 and 11 × 19 = 209. Subtracting 3 produces 207, which is also not an exact multiple of 11. Subtracting 4 yields 206, again leaving a remainder when divided by 11. Subtracting 5 gives 205, which is divisible by 5 but not by 11. Only subtracting 1 gives the nearest lower multiple of 11. Common Pitfalls: Students sometimes mistakenly add the remainder rather than subtract it when the question explicitly asks for subtraction. Others may perform incorrect long division and find a wrong remainder. Carefully computing the product of the divisor and the integer quotient, then subtracting from the original number, helps avoid such calculation errors. Final Answer: The smallest number that must be subtracted from 210 to make it divisible by 11 is 1.

Question 12

The sum of a non zero number and nine times its reciprocal is 10. What is the value of the number?

Accepted Answer

Introduction / Context: This question again deals with a number and its reciprocal, leading to a quadratic equation. It checks the ability to translate a word statement into algebra, manipulate equations involving reciprocals, and then correctly interpret the resulting solutions. Given Data / Assumptions: - Let the non zero number be x. - The relationship given is x + 9 * (1 / x) = 10. - x is not zero, so 1 / x is defined. Concept / Approach: We represent the sentence "the sum of a non zero number and nine times its reciprocal is 10" by an equation in x. Multiplying both sides by x removes the denominator and yields a quadratic equation. Solving this quadratic gives all possible values of x that satisfy the condition, and we check which of these values match the options. Step-by-Step Solution: Start from x + 9 * (1 / x) = 10. Multiply both sides by x: x^2 + 9 = 10x. Rearrange into standard quadratic form: x^2 − 10x + 9 = 0. Factorise: x^2 − 10x + 9 = (x − 1)(x − 9) = 0. Thus, possible solutions are x = 1 or x = 9. Verification / Alternative Check: Check x = 1: 1 + 9 * (1 / 1) = 1 + 9 = 10, so x = 1 satisfies the equation. Check x = 9: 9 + 9 * (1 / 9) = 9 + 1 = 10, so x = 9 also satisfies the equation. Both values are mathematically valid, but among the given options, only 9 appears, so 9 is the correct choice for this multiple choice question. Why Other Options Are Wrong: 10 would give 10 + 9 * (1 / 10) = 10.9, not 10. 11 would give 11 + 9 / 11 which is greater than 11 and not equal to 10. 90 leads to 90 + 9 * (1 / 90) which is approximately 90.1, again not equal to 10. 8 yields 8 + 9 / 8 = 8 + 1.125 = 9.125, so it fails to satisfy the condition. Common Pitfalls: Students sometimes forget that quadratic equations can have two solutions and may stop after finding only one root. Another common mistake is incorrect multiplication when clearing denominators, which leads to a wrong quadratic. Finally, learners may not check which of the valid mathematical solutions actually appears among the options, which is crucial in multiple choice settings. Final Answer: The non zero number whose sum with nine times its reciprocal equals 10 is 9.

Question 13

The average of four consecutive odd numbers is 40. What is the largest of these four odd numbers?

Accepted Answer

Introduction / Context: This question is about averages and consecutive odd numbers. It checks how well candidates can link the arithmetic mean with an ordered set of consecutive terms and translate that understanding into a simple algebraic form. Given Data / Assumptions: - There are four consecutive odd integers. - Their average is 40. - We must identify the largest of these four odd numbers. Concept / Approach: For any set of equally spaced numbers such as consecutive odd integers, the average is the same as the middle point of the set. With four consecutive odd numbers, we can represent them in terms of a central value and then use the average to determine that central value. From there, we can easily find the largest number in the group. Step-by-Step Solution: Let the four consecutive odd numbers be x − 3, x − 1, x + 1, and x + 3. These are spaced by 2 and centred around x. The average of these four numbers is x, since symmetric deviations cancel out. We are told that the average is 40, so x = 40. Therefore, the four numbers are 37, 39, 41, and 43, and the largest is 43. Verification / Alternative Check: We can directly compute the average as follows: (37 + 39 + 41 + 43) ÷ 4. The sum of these four numbers is 160, and 160 ÷ 4 = 40, confirming both the average and the set of numbers. This establishes that 43 is indeed the largest number in the sequence. Why Other Options Are Wrong: 42 is even and does not fit the pattern of consecutive odd integers around an average of 40. 45 would be the largest number if the average were higher; it does not arise from the correct sequence with mean 40. 44 is even and also unsuitable for a set of odd numbers. 41 is part of the sequence but is not the largest; it is the third term in ascending order. Common Pitfalls: Some students forget that consecutive odd numbers differ by 2, not 1, and may incorrectly assume numbers like 39, 40, 41, 42. Others try to guess without using algebra, which can lead to confusion. Remembering that the average equals the central value for symmetric sequences helps solve such questions quickly and accurately. Final Answer: The largest of the four consecutive odd numbers is 43.

Question 14

When 2468 is divided by 37, what remainder is obtained?

Accepted Answer

Introduction / Context: This is a straightforward remainder problem in basic arithmetic and number theory. It checks comfort with long division or modular arithmetic, which is an essential skill for many quantitative aptitude questions. Given Data / Assumptions: - Dividend: 2468. - Divisor: 37. - We must find the remainder when 2468 is divided by 37. Concept / Approach: The remainder r in dividing a number N by d is defined as the difference N − d * q, where q is the integer quotient floor(N ÷ d). We can either perform standard long division or use approximate multiplication to find q and then compute the remainder. The idea is to identify the largest multiple of 37 that does not exceed 2468. Step-by-Step Solution: Start with N = 2468 and d = 37. Compute 37 × 60 = 2220. Increase the multiple: 37 × 66 = 2442. Try 37 × 67 = 2479, which is greater than 2468, so 67 is too large. Thus the greatest integer quotient q is 66 and the remainder is 2468 − 2442 = 26. Verification / Alternative Check: We can confirm by direct division: 2468 ÷ 37 is approximately 66.7, so the integer part is 66. Recomputing 37 × 66 = 2442 and subtracting from 2468 gives 26. Since 0 ≤ remainder < 37, and 26 is in this range, the value is consistent. Why Other Options Are Wrong: 36 is too close to 37 and would imply the dividend is just one less than the next multiple of 37, which is not true here. 18 and 14 are smaller remainders that do not match the calculation 2468 − 2442 = 26. 10 is also not obtained in any correct computation of 2468 mod 37. Common Pitfalls: A common error is miscalculating intermediate products such as 37 × 66, which leads to an incorrect remainder. Some learners may also incorrectly round rather than floor the quotient and then adjust, which is wrong in modular arithmetic. Careful and systematic multiplication avoids these mistakes. Final Answer: The remainder when 2468 is divided by 37 is 26.

Question 15

Of three numbers whose average is 40, the first number is one third of the sum of the other two. What is the value of the first number?

Accepted Answer

Introduction / Context: This question combines averages with simple algebraic relationships between three numbers. It checks whether the student can translate a verbal condition into equations and then use the given average to solve for specific values. Given Data / Assumptions: - There are three numbers: call them A, B, and C. - The average of A, B, and C is 40, so (A + B + C) ÷ 3 = 40. - The first number A is one third of the sum of the other two, so A = (B + C) ÷ 3. - We must find the value of A. Concept / Approach: From the average, we obtain the total sum of the three numbers. From the relationship involving the first number, we obtain another equation connecting A, B, and C. Combining these two equations allows us to express everything in terms of A and solve for it. Step-by-Step Solution: From the average: (A + B + C) ÷ 3 = 40, so A + B + C = 120. From the given condition: A = (B + C) ÷ 3, so B + C = 3A. Substitute B + C = 3A into the sum: A + (B + C) = A + 3A = 4A. We know A + B + C = 120, so 4A = 120. Therefore A = 120 ÷ 4 = 30. Verification / Alternative Check: If A = 30, then B + C = 3A = 90. The total sum A + B + C is 30 + 90 = 120. Dividing by 3 gives an average of 40, which matches the given average. Also, A is indeed one third of B + C, since 30 = 90 ÷ 3. Both conditions are satisfied, confirming the solution. Why Other Options Are Wrong: If A were 20, then B + C would be 60, making the total 80, whose average is 80 ÷ 3, not 40. If A were 50, then B + C would be 150, total 200, whose average is greater than 40. If A were 25, then B + C would be 75 and the average would not be 40. If A were 40, then the total 4A would be 160, giving an average of 160 ÷ 3, not 40. Common Pitfalls: Learners sometimes mix up which number is one third of the sum of the other two and may incorrectly write B = (A + C) ÷ 3 or a similar expression. Others forget to use the average to find the total sum first, which makes solving the system harder than necessary. Writing the equations step by step keeps the algebra simple and clear. Final Answer: The first number is 30.

Question 16

Convert the binary number 1101111 to its decimal (base 10) equivalent.

Accepted Answer

Introduction / Context: This is a basic computer science and number systems question that asks for conversion of a binary number to its decimal form. Understanding positional notation and powers of 2 is crucial for this type of conversion. Given Data / Assumptions: - The binary number given is 1101111. - We must convert this binary representation into its decimal (base 10) value. - In binary, each digit represents a power of 2, starting from 2^0 at the rightmost position. Concept / Approach: To convert from binary to decimal, we expand the binary number as a sum of powers of 2, including only those positions where the binary digit is 1. The general form for a binary number b6 b5 b4 b3 b2 b1 b0 is b6 * 2^6 + b5 * 2^5 + ... + b0 * 2^0. Step-by-Step Solution: Write 1101111 as digits from left to right: 1 1 0 1 1 1 1. Label powers of 2 from right to left: 2^0, 2^1, 2^2, 2^3, 2^4, 2^5, 2^6. Match digits to powers: 1 * 2^6, 1 * 2^5, 0 * 2^4, 1 * 2^3, 1 * 2^2, 1 * 2^1, 1 * 2^0. Compute each non zero term: 1 * 2^6 = 64, 1 * 2^5 = 32, 1 * 2^3 = 8, 1 * 2^2 = 4, 1 * 2^1 = 2, 1 * 2^0 = 1. Add them: 64 + 32 + 8 + 4 + 2 + 1 = 111. Verification / Alternative Check: We can also group intermediate sums: 64 + 32 = 96, 96 + 8 = 104, 104 + 4 = 108, 108 + 2 = 110, and finally 110 + 1 = 111. Another quick check is to use a calculator that supports binary to decimal conversion, but the stepwise method helps reinforce understanding of positional notation. Why Other Options Are Wrong: 101 corresponds to a different binary representation and is too small compared to 111. 110 is also smaller and would come from a slightly different binary pattern. 100 is even smaller and incorrect for this specific binary number. 115 is larger than the computed sum and does not correspond to 1101111 in binary. Common Pitfalls: Students sometimes misalign the powers of 2, starting from 2^1 instead of 2^0 at the rightmost bit. Others add incorrect powers or miscalculate exponents such as 2^5 or 2^6. Remembering that binary is base 2 and carefully mapping each bit to its power of 2 helps avoid mistakes. Final Answer: The decimal equivalent of binary 1101111 is 111.

Question 17

The sum of a fraction and seven times its reciprocal is 11/2. What is the value of the fraction?

Accepted Answer

Introduction / Context: This question involves an unknown fraction and its reciprocal in a linear relation that leads to a quadratic equation. It tests algebraic manipulation and solving quadratic equations that arise from reciprocal relationships. Given Data / Assumptions: - Let the fraction be x, where x is non zero. - The condition is x + 7 * (1 / x) = 11 / 2. - We must find the value of x that satisfies this equation and matches the style of an ordinary fraction. Concept / Approach: We convert the verbal statement into an equation and clear the denominator by multiplying both sides by x. This creates a quadratic equation in x. Solving the quadratic gives possible values for x, and then we interpret which of those values matches the intended answer in the context of the question and the options. Step-by-Step Solution: Start with x + 7 * (1 / x) = 11 / 2. Multiply both sides by x: x^2 + 7 = (11 / 2) * x. Multiply both sides by 2 to eliminate the fraction: 2x^2 + 14 = 11x. Rearrange into standard quadratic form: 2x^2 − 11x + 14 = 0. Factorise: 2x^2 − 11x + 14 = (2x − 7)(x − 2) = 0, so x = 7 / 2 or x = 2. Verification / Alternative Check: Check x = 7 / 2: the reciprocal is 2 / 7. Then x + 7 * (1 / x) becomes 7 / 2 + 7 * (2 / 7) = 7 / 2 + 2 = 7 / 2 + 4 / 2 = 11 / 2, which matches the given value. Check x = 2: the reciprocal is 1 / 2. Then 2 + 7 * (1 / 2) = 2 + 7 / 2 = 4 / 2 + 7 / 2 = 11 / 2, so it also satisfies the equation. Both values are mathematically correct, but among the answer options, only 7 / 2 is offered and it is explicitly written as a proper fraction form used in the statement, so this is chosen as the correct option. Why Other Options Are Wrong: 2 / 7 would give 2 / 7 + 7 * (7 / 2), which is much larger than 11 / 2. 3 / 4 leads to a different sum when combined with seven times its reciprocal and does not produce 11 / 2. 4 / 3 and 5 / 2 similarly do not satisfy the original equation when substituted. Common Pitfalls: Students may forget to multiply across correctly when clearing denominators and thus derive an incorrect quadratic equation. Another issue is assuming there will be only one solution and ignoring other roots. In multiple choice settings it is critical to solve accurately and then see which of the valid roots is present among the options. Final Answer: The fraction whose sum with seven times its reciprocal equals 11 / 2 is 7/2.

Question 18

What is the value of 9991 × 10009 when expressed as an exact product?

Accepted Answer

Introduction / Context: This problem seems to be a straightforward multiplication exercise, but the numbers 9991 and 10009 are chosen to encourage recognition of algebraic patterns. It tests both arithmetic skill and understanding of products involving numbers close to 10,000. Given Data / Assumptions: - We must compute 9991 × 10009 exactly. - Both factors are close to 10,000, which suggests looking for a pattern using (a − b)(a + b) if possible. Concept / Approach: We can write 9991 and 10009 in terms of 10,000. Note that 9991 = 10,000 − 9 and 10009 = 10,000 + 9. This fits the algebraic identity (a − b)(a + b) = a^2 − b^2, with a = 10,000 and b = 9. Using this identity simplifies the multiplication significantly. Step-by-Step Solution: Express the numbers: 9991 = 10,000 − 9 and 10009 = 10,000 + 9. Use the identity (a − b)(a + b) = a^2 − b^2. Here a = 10,000 and b = 9, so the product is 10,000^2 − 9^2. Compute 10,000^2 = 100,000,000. Compute 9^2 = 81, so the result is 100,000,000 − 81 = 99,999,919. Verification / Alternative Check: We can quickly compare this result with the options. The number 99,999,919 matches option 99999919 when the commas are removed, so it is consistent. Performing long multiplication would give the same answer, but using the difference of squares identity is much faster and less error prone, especially in an exam setting. Why Other Options Are Wrong: 99999099 is smaller than 99,999,919 and does not follow from the identity a^2 − b^2 with a = 10,000 and b = 9. 99999819 and 99999019 are also incorrect, as subtracting 81 from 100,000,000 gives 99,999,919 exactly, not these numbers. 100000000 would be the product if one factor were exactly 10,000 and the other were 10,000, which is not the case here because of the adjustment by 9. Common Pitfalls: One common mistake is miscomputing 10,000^2 or 9^2, or subtracting incorrectly. Another is failing to see the pattern and attempting full long multiplication, which increases the chance of arithmetic errors. Remembering key algebraic identities like (a − b)(a + b) = a^2 − b^2 allows for quick and reliable calculation in problems involving numbers symmetric around a central value. Final Answer: The exact value of 9991 × 10009 is 99999919.

Question 19

What is the quotient obtained when 7251 is divided by 66?

Accepted Answer

Introduction / Context: This question is a direct test of division skills, focused on finding the integer quotient when one number is divided by another. It reinforces basic arithmetic operations that underpin many more complex quantitative problems. Given Data / Assumptions: - Dividend: 7251. - Divisor: 66. - We are asked for the quotient, which is the integer part of 7251 ÷ 66. Concept / Approach: To find the quotient q and remainder r when N is divided by d, we write N = d * q + r, where 0 ≤ r < d. Here we only need q. Performing long division or approximating the quotient and refining it will give the correct integer value. Step-by-Step Solution: Let N = 7251 and d = 66. Approximate: 66 × 100 = 6600, which is less than 7251. Increase the multiple: 66 × 110 = 7260, which is slightly greater than 7251. Therefore the quotient must be just below 110, so try 66 × 109. Compute 66 × 109 = 66 × (100 + 9) = 6600 + 594 = 7194, which is less than 7251. Verification / Alternative Check: After computing 66 × 109 = 7194, we can find the remainder as 7251 − 7194 = 57. Because 57 is less than 66, the division is consistent with quotient 109 and remainder 57. If we tried quotient 110, the product 66 × 110 = 7260 would exceed 7251, which is not allowed. Therefore, the correct integer quotient is 109. Why Other Options Are Wrong: 110 would give a product greater than the dividend, which is impossible for a valid quotient. 111 and 112 are even larger and would lead to products further above 7251. 108 gives 66 × 108 = 7128, which is less than 7194 and thus not the greatest integer quotient; the actual quotient is higher at 109. Common Pitfalls: A common error is to stop at an approximate estimate such as 110 without checking whether the product exceeds the dividend. Another mistake is miscalculating 66 × 109 or similar products. Careful multiplication and a quick check that the product does not exceed the dividend ensure the correct quotient is chosen. Final Answer: The quotient when 7251 is divided by 66 is 109.

Question 20

What is the remainder when 6910 is divided by 81?

Accepted Answer

Introduction / Context: This is another remainder problem that reinforces the concept of modular arithmetic. It is important for learners to be comfortable computing remainders quickly and accurately in quantitative exams. Given Data / Assumptions: - Dividend: 6910. - Divisor: 81. - We must determine the remainder when 6910 is divided by 81. Concept / Approach: We want to express 6910 in the form 81 * q + r, where q is the integer quotient and r is the remainder with 0 ≤ r < 81. We can find q by approximate or exact multiplication and then compute r as the difference between the dividend and 81 * q. Step-by-Step Solution: Let N = 6910 and d = 81. Compute an approximate quotient: 81 × 80 = 6480. Increase the multiple: 81 × 85 = 6885. Try 81 × 86 = 6966, which exceeds 6910, so 86 is too large. Thus q = 85 and the remainder r = 6910 − 6885 = 25. Verification / Alternative Check: We see that 6910 ÷ 81 is slightly greater than 85, because 6885 is close to 6910. The difference 25 is less than 81, so it is a valid remainder. Any attempt to use q = 86 would lead to a product higher than the dividend and therefore is invalid. This confirms that the correct remainder is 25. Why Other Options Are Wrong: 23, 21, and 19 are all smaller than 25 and would imply different quotients that do not match the exact multiplication with 81. 27 is larger than 25 and does not correspond to the computed remainder 6910 − 6885. Common Pitfalls: Errors often come from miscalculating the intermediate products 81 × 80, 81 × 85, or from subtracting incorrectly when finding the remainder. Some students also confuse quotient with remainder. Clear separation of these concepts and systematic checking of the largest valid multiple helps avoid these mistakes. Final Answer: The remainder when 6910 is divided by 81 is 25.

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