Question 1

Which of the following lists consists only of multiples of 6 and also includes numbers that are common multiples of both 4 and 6?

Accepted Answer

Introduction: This question checks your understanding of multiples and common multiples. You must choose a list of numbers that are all multiples of 6, and within that list, some numbers should also be common multiples of both 4 and 6. It is a straightforward exercise in divisibility, especially for small numbers, and reinforces how least common multiples work. Given Data / Assumptions: A multiple of 6 is any number of the form 6k, where k is an integer. A common multiple of 4 and 6 must be divisible by both 4 and 6. The least common multiple (LCM) of 4 and 6 is 12. Any common multiple of 4 and 6 is a multiple of 12. Concept / Approach: Check each list in the options. For each number in a list, test if it is divisible by 6. Additionally, see whether at least some of those numbers are divisible by both 4 and 6 (i.e., by 12). The correct option must satisfy both conditions: every number is a multiple of 6, and there are common multiples of 4 and 6 present. Step-by-Step Solution: Option A: 12, 18, 36.12/6 = 2, 18/6 = 3, 36/6 = 6, so all are multiples of 6.12 and 36 are also divisible by 4: 12/4 = 3, 36/4 = 9, so they are common multiples of 4 and 6.Option B: 12, 34, 42. Here 34 is not divisible by 6, so this list fails.Option C: 6, 4, 14. Here 4 and 14 are not multiples of 6.Option D: 4, 8, 16. None of these are multiples of 6.Option E: 6, 10, 22. Only 6 is a multiple of 6; 10 and 22 are not.Therefore, only option A satisfies all requirements. Verification / Alternative check: Recall that common multiples of 4 and 6 are multiples of 12: 12, 24, 36, 48, and so on. Option A uniquely contains 12 and 36, both of which are on this list, and all entries in option A are multiples of 6. No other option has this combination of properties. Why Other Options Are Wrong: Option B: contains 34, which breaks the “multiple of 6” requirement.Option C: mixes 6 with non-multiples 4 and 14.Option D: all numbers are multiples of 4 but not of 6.Option E: includes only one multiple of 6 and two non-multiples. Common Pitfalls: Checking only whether a list contains some multiple of 6 rather than verifying all of them.Forgetting that a common multiple of 4 and 6 must be divisible by both, which for small numbers essentially means being a multiple of 12.Confusing multiples with factors. Final Answer: 12, 18, 36

Question 2

The product of two positive numbers is p. If each of the numbers is increased by 2, by how much (in exact value) is the new product greater than twice the sum of the two original numbers?

Accepted Answer

Introduction: This algebraic word problem connects products and sums of two positive numbers expressed in terms of a parameter p. You are told that the product of two numbers is p, and then asked to compare the product after increasing both numbers by 2 with twice the sum of the original numbers. The question tests the ability to write symbolic expressions correctly and then simplify them carefully. Given Data / Assumptions: Let the original numbers be a and b, both positive. Their product is ab = p. Each number is increased by 2, giving new numbers (a + 2) and (b + 2). New product = (a + 2)(b + 2). Twice the sum of the original numbers is 2(a + b). We must find (new product) - (twice the original sum). Concept / Approach: First, expand the new product using distributive multiplication. Then subtract 2(a + b). Use the relation ab = p to express the answer in terms of p and constants only. This is a straightforward symbolic manipulation question, but it is easy to make sign or expansion mistakes if you rush. Step-by-Step Solution: Given: ab = p.New product = (a + 2)(b + 2).Expand: (a + 2)(b + 2) = ab + 2a + 2b + 4.Replace ab with p: new product = p + 2a + 2b + 4.Twice the sum of original numbers = 2(a + b) = 2a + 2b.Difference required = (new product) - (twice original sum).Difference = (p + 2a + 2b + 4) - (2a + 2b).Simplify: p + 2a + 2b + 4 - 2a - 2b = p + 4.So the new product is (p + 4) greater than twice the sum of the original numbers. Verification / Alternative check: Take a simple numeric example: let a = 2 and b = 3. Then p = ab = 6. New product = (2 + 2)(3 + 2) = 4 * 5 = 20. Twice the sum of originals = 2(2 + 3) = 2 * 5 = 10. Difference = 20 - 10 = 10. Our formula gives p + 4 = 6 + 4 = 10, which matches the calculation. Why Other Options Are Wrong: p or 2p: these ignore the constant increase that comes from adding 2 to both numbers.p + 2: underestimates the true difference, missing part of the constant term from the expansion.2p + 4: doubles p unnecessarily, not matching the derived expression. Common Pitfalls: Incorrectly expanding (a + 2)(b + 2), for example forgetting the +4 term.Not substituting ab = p correctly, or substituting at the wrong stage.Misinterpreting “how much greater” as a multiplicative comparison instead of a difference. Final Answer: p + 4

Question 3

The 6-digit number 72k23l is divisible by 88. Find the digits k and l that make the number exactly divisible by both 8 and 11.

Accepted Answer

Introduction: This question tests knowledge of divisibility rules, especially for composite numbers like 88. Since 88 = 8 * 11, a number is divisible by 88 if and only if it is divisible by both 8 and 11. You are given a 6-digit number with two unknown digits k and l and must determine these digits so that the number 72k23l is divisible by 88. Given Data / Assumptions: The number is 72k23l, where k and l are single digits from 0 to 9. The number is divisible by 88. 88 = 8 * 11, so the number must be divisible by 8 and also by 11. We will use the divisibility rules for 8 and 11. Concept / Approach: First, use the rule for divisibility by 8: a number is divisible by 8 if its last three digits form a number divisible by 8. This gives a condition on l. Then apply the rule for 11: the difference between the sum of digits in odd positions and the sum of digits in even positions must be a multiple of 11 (including 0). This gives a condition on k once l is fixed. Solving both conditions together yields k and l. Step-by-Step Solution: Divisibility by 8: look at the last three digits, which form 23l.This 3-digit number is 230 + l and must be divisible by 8.Try l from 0 to 9. Check 232: 232 / 8 = 29, which is an integer.So l = 2 is the only digit making 23l divisible by 8.Now the number is 72k232.Divisibility by 11: consider digits 7, 2, k, 2, 3, 2 from left to right.Odd positions (1st, 3rd, 5th): 7, k, 3 => sum = 7 + k + 3 = k + 10.Even positions (2nd, 4th, 6th): 2, 2, 2 => sum = 2 + 2 + 2 = 6.For divisibility by 11, difference (odd sum - even sum) must be a multiple of 11.Difference = (k + 10) - 6 = k + 4.So k + 4 must be 0, 11, 22, etc. With k a digit, k + 4 = 11 is valid, giving k = 7.Thus k = 7 and l = 2. Verification / Alternative check: Check last three digits: 232 / 8 = 29, so divisible by 8. Now check divisibility by 11 using 727232: sum of odd-position digits = 7 + 7 + 3 = 17; sum of even-position digits = 2 + 2 + 2 = 6; difference = 11, which is a multiple of 11. Therefore 727232 is divisible by 11 and hence by 88. Why Other Options Are Wrong: k = 8 and l = 2; k = 8 and l = 3; k = 7 and l = 1; k = 6 and l = 4: in each of these combinations, either the last three digits are not divisible by 8 or the 11-divisibility rule fails, so the full number cannot be divisible by 88. Common Pitfalls: Trying to check divisibility by 88 directly without using the simpler separate tests for 8 and 11.Forgetting that only the last three digits matter for divisibility by 8.Miscomputing the alternating sums for the 11-divisibility rule by mixing up digit positions. Final Answer: k = 7 and l = 2

Question 4

Find the smallest 4-digit number which, when divided by 3, 4, or 5, leaves a remainder of 2 in each case.

Accepted Answer

Introduction: This is a modular arithmetic and least common multiple (LCM) question. You are asked to find the smallest 4-digit number which leaves the same remainder (2) when divided by 3, 4, or 5. Such problems test the understanding of congruences and the idea that numbers with the same remainder for given divisors can be written in terms of the LCM of these divisors plus that remainder. Given Data / Assumptions: We want a 4-digit number N (i.e., N ≥ 1000). N leaves remainder 2 when divided by 3, 4, and 5. So N ≡ 2 (mod 3), N ≡ 2 (mod 4), and N ≡ 2 (mod 5). We must find the smallest such N. Concept / Approach: If a number leaves the same remainder 2 with respect to 3, 4, and 5, then N - 2 must be divisible by each of these numbers. Therefore, N - 2 must be a multiple of the least common multiple (LCM) of 3, 4, and 5. Once we find the LCM, we can express N as N = LCM(3,4,5) * k + 2 and search for the smallest k that makes N at least 1000. Step-by-Step Solution: Compute LCM of 3, 4, and 5.LCM(3,4) = 12; then LCM(12,5) = 60.So N - 2 must be a multiple of 60.Let N - 2 = 60k => N = 60k + 2.We need N to be at least 1000.So 60k + 2 ≥ 1000 => 60k ≥ 998 => k ≥ 998 / 60 ≈ 16.63.The smallest integer k satisfying this is k = 17.Then N = 60 * 17 + 2 = 1020 + 2 = 1022. Verification / Alternative check: Check N = 1022: 1022 / 3 leaves remainder 2; 3 * 340 = 1020, remainder 2. 1022 / 4 leaves remainder 2; 4 * 255 = 1020, remainder 2. 1022 / 5 leaves remainder 2; 5 * 204 = 1020, remainder 2. Thus 1022 satisfies all three conditions and is the smallest 4-digit number of the form 60k + 2. Why Other Options Are Wrong: 1020, 1040, 1060, and 1122: these are not of the form 60k + 2 or do not leave remainder 2 with all three divisors 3, 4, and 5. For example, 1020 is exactly divisible by 3, 4, and 5 with remainder 0, not 2. Common Pitfalls: Trying to test random 4-digit numbers instead of using the LCM structure.Forgetting that N - 2 must be divisible by all three numbers, not N itself.Rounding 998/60 down instead of up when selecting k, which would produce a 3-digit or smaller number. Final Answer: 1022

Question 5

The sum of the digits of a two-digit number is 12 and the difference between the digits is 6. What is that two-digit number?

Accepted Answer

Introduction / Context: This is a standard aptitude question on two-digit numbers and their digits. It tests how well you can translate information about the sum and difference of digits into simple algebra, and then identify the correct number from the options. Such questions are very common in competitive exams under the topic “Problems on Numbers”. Given Data / Assumptions: The number is a two-digit number. The sum of its digits is 12. The difference between its digits is 6 (larger digit minus smaller digit). We must find the actual two-digit number that satisfies both conditions. Concept / Approach: Let the tens digit be x and the units digit be y. For a two-digit number, x must be between 1 and 9 and y between 0 and 9. From the sum condition we get x + y = 12. From the difference condition we get x − y = 6 (assuming x is the larger digit). Solving these two linear equations in two variables gives the digits directly. Finally, we build the number 10x + y and check which option matches the result and the conditions given in the question. Step-by-Step Solution: Step 1: Let the tens digit be x and the units digit be y. Step 2: From the sum condition: x + y = 12. Step 3: From the difference condition: x − y = 6. Step 4: Add the two equations: (x + y) + (x − y) = 12 + 6 which gives 2x = 18, so x = 9. Step 5: Substitute x = 9 in x + y = 12 to get 9 + y = 12, so y = 3. Step 6: The number formed with digits 9 and 3 is 93. However, the question and options usually take the difference as larger digit minus smaller digit, but also allow the reversed arrangement if stated numbers are treated loosely. Among options, only 39 has digits 3 and 9 whose sum is 12 and whose difference 9 − 3 is 6, so the correct answer from the list is 39. Verification / Alternative check: For the option 39, the digits are 3 and 9. Their sum is 3 + 9 = 12, which satisfies the first condition. The difference between the digits is 9 − 3 = 6, which satisfies the second condition. No other option satisfies both conditions. For example, 57 has digits 5 and 7, whose sum is 12 but whose difference is only 2. Therefore 39 is the unique option that works. Why Other Options Are Wrong: Option 57: Sum of digits 5 + 7 = 12 is correct, but the difference 7 − 5 = 2, not 6. Option 75: Sum is 7 + 5 = 12, but difference is again 2, not 6. Option 48: Sum is 4 + 8 = 12, but difference is 4, not 6. Option 93: Digits 9 and 3 do satisfy the conditions, but 93 is not present as the correct answer in the original option set, and the puzzle intends the listed correct choice 39, which also uses digits 3 and 9. Common Pitfalls: A common mistake is to forget which digit is larger and to write the difference equation incorrectly. Another error is to try to guess numbers instead of systematically solving the equations. Using the algebraic approach x + y = 12 and x − y = 6 is faster and more reliable. Always check both conditions (sum and difference) before finalising the answer from the given options. Final Answer: The required two-digit number from the options is 39.

Question 6

Evaluate the expression 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 × 0 + 1 + 1 using the correct order of operations.

Accepted Answer

Introduction / Context: This problem checks your understanding of the basic order of operations (also called BODMAS or PEMDAS rules) in arithmetic. It looks like a long string of ones, plus signs, and a single multiplication by zero, and can be tricky if you simply read from left to right without applying the correct operation priority. Such questions are common in aptitude tests to catch careless mistakes. Given Data / Assumptions: The expression is: 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 × 0 + 1 + 1. Multiplication must be done before addition. There are no brackets and no subtraction or division symbols. We must simplify this expression to a single numerical value. Concept / Approach: The key concept is the order of operations. According to standard rules, multiplication takes precedence over addition. That means we must first evaluate 1 × 0 wherever it appears, and only then perform the additions. If we do not respect this order and instead add everything from left to right, we will get the wrong answer. Once we handle the multiplication, the rest of the expression reduces to a sum of simple integers, which is easy to compute. Step-by-Step Solution: Step 1: Rewrite the expression clearly: 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 × 0 + 1 + 1. Step 2: Apply the order of operations and evaluate the multiplication part first: 1 × 0 = 0. Step 3: Replace 1 × 0 with 0. The expression becomes 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 0 + 1 + 1. Step 4: Count the number of ones that are being added. There are 12 ones in total (ten before the multiplication, plus two after it). Step 5: The sum of twelve ones is 12, so the value of the expression is 12. Verification / Alternative check: Another way to see this is to separate the sum into two parts and the multiplication: (sum of the first 10 ones) + (1 × 0) + (sum of the last 2 ones). The first part is 10, the multiplication is 0, and the last part is 2. Therefore the entire expression is 10 + 0 + 2 = 12. This alternative grouping confirms the same result and shows that, regardless of grouping (as long as multiplication is done first), you get 12. Why Other Options Are Wrong: Option 0: This would be correct only if every term were multiplied by zero, but here only one 1 is multiplied by 0. Option 2: This is the sum of the last two ones only and ignores the previous ones. Option 10: This is the sum of the ten initial ones, ignoring the two ones after the multiplication. Option 14: This would be the result if you incorrectly treated 1 × 0 as 1 and simply counted 14 ones, which is wrong. Common Pitfalls: The main mistake is to read the whole line as “how many ones are there” and ignore the multiplication symbol, or to misapply the rule and do addition before multiplication. Some students also miscount the ones. Always scan the expression for multiplication and division first, evaluate those parts, and then perform additions and subtractions. Writing the expression in a cleaner form can prevent counting errors. Final Answer: Using the correct order of operations, the value of the expression is 12.

Question 7

Which one of the following numbers is not a prime number?

Accepted Answer

Introduction / Context: Prime numbers are fundamental in number theory and aptitude exams. A prime number is a natural number greater than 1 that has exactly two distinct positive divisors, namely 1 and the number itself. This question asks you to identify which of the given numbers fails this property and therefore is not prime. Being able to quickly factor or test divisibility is very useful in competitive exams. Given Data / Assumptions: We are given the numbers 91, 71, 41 and 31 (and an extra distractor in option E). Prime numbers have only two factors: 1 and the number itself. Composite numbers have additional factors besides 1 and the number itself. We must identify the single composite (non-prime) number among the options. Concept / Approach: To check if a number is prime, we test divisibility by prime numbers less than or equal to the square root of the number. If we find any divisor other than 1 and the number itself, the number is composite. For two-digit numbers, testing divisibility by 2, 3, 5, 7, 11 and sometimes 13 is usually enough. We go option by option and see which one can be factored into smaller integers greater than 1. Step-by-Step Solution: Step 1: Check 91. It is odd, so not divisible by 2. Sum of digits is 9 + 1 = 10, which is not divisible by 3, so 91 is not divisible by 3. It does not end in 0 or 5, so it is not divisible by 5. Step 2: Test divisibility by 7. Compute 7 × 13 = 91, so 91 = 7 * 13. Therefore 91 has divisors 7 and 13 in addition to 1 and 91 and is composite. Step 3: Check 71. It is not even, digit sum 7 + 1 = 8 is not divisible by 3, it does not end in 0 or 5, and simple checks show it is not divisible by 7 or 11. So 71 is prime. Step 4: Check 41. As with 71, basic tests for 2, 3, 5 and 7 show no divisors. It is a known prime. Step 5: Check 31. It is also not divisible by 2, 3, 5 or 7 and is also a known prime. Verification / Alternative check: You can verify that 91 is composite by directly writing 91 ÷ 7 = 13 or 91 ÷ 13 = 7, giving exact integer results. For the others, quick mental checks show no such factorisation. For instance, 71 ÷ 7 is a little more than 10, not an integer, and 71 ÷ 11 is not an integer either. Since only 91 factorises nicely into a product of two smaller natural numbers greater than 1, 91 is the unique non-prime here. Why Other Options Are Wrong: Option 71: This is a prime number, as it has no divisor other than 1 and 71 in the tested range. Option 41: This is also a prime, commonly used as an example in basic prime lists. Option 31: Another prime number, no factors other than 1 and 31. Option 61: Included as an extra prime distractor; it is also prime. Common Pitfalls: Some students guess based on “shape” of the number or because it looks composite, rather than performing basic divisibility checks. Others may mistakenly think that if a number is not divisible by 2 or 3, it must be prime, which is not always true. For two-digit numbers, always test divisibility by 2, 3, 5, 7 and 11 at least. Recognising that 91 equals 7 * 13 is a useful pattern to remember for future problems. Final Answer: The number that is not prime is 91.

Question 8

In a plane, there are 15 distinct straight lines with no two parallel and no three meeting at a single point. What is the maximum possible number of intersection points of these lines?

Accepted Answer

Introduction / Context: This question comes under the topic of combinatorics and geometry, often found in “Problems on Numbers” or counting principles. It asks for the maximum number of intersection points that 15 straight lines can form in a plane, under the conditions that no two lines are parallel and no three lines are concurrent (that is, they do not all meet at the same point). Understanding this helps in many counting and arrangement problems in competitive exams. Given Data / Assumptions: There are 15 distinct straight lines drawn on a plane. No two lines are parallel. No three lines pass through the same intersection point. We must find the maximum number of intersection points formed by these lines. Concept / Approach: When no two lines are parallel, every pair of lines will intersect. When no three are concurrent, each pair intersects at a unique point. Therefore, the problem reduces to counting how many distinct pairs of lines we can form from 15 lines. The number of ways to choose 2 items from n items is given by the combination formula nC2 = n * (n − 1) / 2. This gives the maximum number of intersection points, because each pair of lines meets exactly once. Step-by-Step Solution: Step 1: Interpret the conditions. Since no two lines are parallel, every pair of lines intersects somewhere in the plane. Step 2: Because no three lines meet at a single point, each pair of lines produces a distinct intersection point, with no overlapping intersections. Step 3: The total number of intersection points is therefore equal to the number of pairs of lines. Step 4: Use the combination formula to find the number of pairs: 15C2 = 15 * 14 / 2. Step 5: Compute 15 * 14 = 210, and then divide by 2 to get 210 / 2 = 105. Step 6: Thus, the maximum possible number of intersection points is 105. Verification / Alternative check: You can verify the pattern with smaller numbers. For example, with 2 lines, you get 1 intersection point, and 2C2 = 1. With 3 lines in general position (no two parallel, no three concurrent), each pair intersects, giving 3C2 = 3 points. For 4 lines, you get 4C2 = 6 intersection points. The results match intuitive drawings. Extending this pattern to 15 lines gives 15C2 = 105, confirming that the formula works consistently and the reasoning is correct. Why Other Options Are Wrong: Option 55: This equals 11C2 and would be correct for 11 lines, not for 15 lines. Option 215 and 148: These numbers do not match any direct combination count for 15 lines under the given conditions and are too large or inconsistent. Option 90: This equals 10 * 9, which is unrelated to the required combination 15C2 in this context. Common Pitfalls: Students sometimes confuse combinations with permutations and may try formulas like nP2 instead of nC2. Another common error is forgetting the conditions: if some lines were parallel or three lines concurrent, the count would be smaller, but the question explicitly asks for the maximum, which assumes the general position described. Always check that the problem is asking for the maximum or minimum before deciding on the arrangement and the counting method. Final Answer: The maximum number of intersection points formed by 15 such lines is 105.

Question 9

What is the least four-digit natural number that is divisible by 4, 6, 8 and 10?

Accepted Answer

Introduction / Context: This problem is about finding the smallest four-digit number that is exactly divisible by several given numbers. This is a typical application of the least common multiple (LCM), which is a core concept in arithmetic and number systems. Being comfortable with LCM helps solve many divisibility and scheduling type questions that appear frequently in aptitude exams. Given Data / Assumptions: We are looking for a four-digit number (that is, between 1000 and 9999). The number must be divisible by 4, 6, 8 and 10. We want the least (smallest) such four-digit number. Concept / Approach: A number divisible by several integers must be divisible by their least common multiple. Therefore, we first find the LCM of 4, 6, 8 and 10. Once we have the LCM, any common multiple of all four numbers is a multiple of this LCM. We then look for the smallest multiple of the LCM that is at least 1000. This process ensures that the number we find is the smallest four-digit number that satisfies all the divisibility conditions. Step-by-Step Solution: Step 1: Factorise each number into primes. 4 = 2^2, 6 = 2 * 3, 8 = 2^3, 10 = 2 * 5. Step 2: For the LCM, take the highest power of each prime that appears in any factorisation. For prime 2, the highest power is 2^3 from 8. For prime 3, the highest power is 3 from 6. For prime 5, the highest power is 5 from 10. Step 3: Multiply these highest powers: LCM = 2^3 * 3 * 5 = 8 * 3 * 5 = 24 * 5 = 120. Step 4: Now find the smallest multiple of 120 that is at least 1000. Compute 1000 / 120 which is about 8.33. Step 5: The next whole number after 8.33 is 9, so the required multiple is 9 * 120 = 1080. Verification / Alternative check: Check divisibility of 1080 by each given number. 1080 / 4 = 270 (integer), 1080 / 6 = 180 (integer), 1080 / 8 = 135 (integer), and 1080 / 10 = 108 (integer). So 1080 is divisible by all four numbers. Also, the previous multiple of 120 is 8 * 120 = 960, which is only a three-digit number. Therefore, 1080 is indeed the smallest four-digit number that satisfies the condition. Why Other Options Are Wrong: Option 1107: This is not divisible by 4 or 8 and so fails the conditions. Option 1100: While divisible by 4 and 10, it is not divisible by 8 or 6. Option 1208 and 1040: These may be divisible by some of the given numbers, but they are not all divisible by 4, 6, 8 and 10 together or they are not the smallest that satisfies all conditions. Common Pitfalls: One common mistake is to check the options one by one without first calculating the LCM, which can be time consuming and error prone. Another error is to forget to check all the divisibility conditions or to ignore the “least” and “four-digit” parts of the question. Always start such problems by finding the LCM, and then jump directly to the smallest required multiple above the lower limit of the range. Final Answer: The least four-digit number that is divisible by 4, 6, 8 and 10 is 1080.

Question 10

The difference between two numbers is 1365. When the larger number is divided by the smaller number, the quotient is 6 and the remainder is 15. What is the smaller number?

Accepted Answer

Introduction / Context: This question is a classic application of the division algorithm and basic algebra. You are given the difference between two numbers and the result of dividing the larger by the smaller (including quotient and remainder). From this information, you need to determine the smaller number. Such questions test your understanding of how quotient, remainder, dividend and divisor relate to each other, which is very important in number system problems. Given Data / Assumptions: Let the larger number be L and the smaller number be S. The difference between the two numbers is L − S = 1365. When the larger number is divided by the smaller number, we get quotient 6 and remainder 15. This means L = 6S + 15 by the division algorithm. We need to find the value of S. Concept / Approach: The division algorithm states that for any integers dividend L, divisor S and remainder R with 0 ≤ R < S, we can write L = S * Q + R, where Q is the quotient. Here, Q = 6 and R = 15, so L = 6S + 15. We also know that the difference L − S equals 1365. By substituting L from the division equation into the difference equation, we can solve a simple linear equation to find S, the smaller number. Step-by-Step Solution: Step 1: From the division algorithm, write L = 6S + 15. Step 2: From the difference condition, L − S = 1365. Step 3: Substitute L = 6S + 15 into L − S = 1365, giving (6S + 15) − S = 1365. Step 4: Simplify the equation: 6S + 15 − S = 5S + 15. So we have 5S + 15 = 1365. Step 5: Subtract 15 from both sides: 5S = 1365 − 15 = 1350. Step 6: Divide both sides by 5: S = 1350 / 5 = 270. So the smaller number is 270. Verification / Alternative check: Once we get S = 270, we can find L using L = 6S + 15 = 6 * 270 + 15 = 1620 + 15 = 1635. Check the difference: L − S = 1635 − 270 = 1365, which matches the given difference. Also check the division: 1635 divided by 270 gives quotient 6 and remainder 15 because 270 * 6 = 1620 and 1635 − 1620 = 15. Both conditions are satisfied, confirming that 270 is correct. Why Other Options Are Wrong: Option 1270 and 720: If these were the smaller number, substituting into the equations does not produce a difference of 1365 and the given quotient and remainder. Option 350 and 285: Similar substitutions for these values do not satisfy both the difference and the division conditions simultaneously. Common Pitfalls: A frequent mistake is to miswrite the division equation, for example as L = 6 + S + 15 instead of L = 6S + 15. Another error is to set up the difference equation incorrectly, such as S − L = 1365 instead of L − S. When dealing with quotient and remainder problems, always write the correct division algorithm formula and keep track of which number is larger. Double checking by substituting back is a good way to avoid algebra slips. Final Answer: The smaller number is 270.

Question 11

The sum and difference of two numbers are 20 and 8 respectively. What is the difference of the squares of these two numbers?

Accepted Answer

Introduction / Context: This question tests your understanding of algebraic identities, especially the identity for the difference of squares. When you know the sum and difference of two numbers, you can quickly compute the difference of their squares without finding the numbers individually. This is a time-saving technique that appears often in aptitude and quantitative exams. Given Data / Assumptions: The sum of two numbers is 20. The difference between the two numbers is 8 (larger minus smaller). We are asked for the difference of their squares. Let the numbers be a and b with a > b. Concept / Approach: We use the important algebraic identity: a^2 − b^2 = (a − b) * (a + b). Instead of first solving for a and b, we can directly apply this identity because the question gives us both (a + b) and (a − b). This greatly simplifies the calculation. This approach demonstrates how understanding identities helps solve problems quickly and accurately without unnecessary steps. Step-by-Step Solution: Step 1: Let the two numbers be a and b such that a + b = 20 and a − b = 8. Step 2: Recall the identity: a^2 − b^2 = (a + b) * (a − b). Step 3: Substitute the given values of a + b and a − b into the identity. Step 4: Compute a^2 − b^2 = 20 * 8. Step 5: Multiply 20 * 8 = 160. Step 6: Therefore, the difference of the squares of the two numbers is 160. Verification / Alternative check: As an alternative, you can find the actual numbers. Adding the two equations a + b = 20 and a − b = 8 gives 2a = 28, so a = 14. Subtracting the second from the first gives 2b = 12, so b = 6. Now compute a^2 − b^2 = 14^2 − 6^2 = 196 − 36 = 160. This confirms the result obtained directly using the identity and shows that both methods are consistent. Why Other Options Are Wrong: Option 120 and 140: These values could arise from incorrect intermediate multiplications or misreading the given sum or difference, but they do not match the identity. Option 180 and 100: These are typical distractors that might result from incorrectly using 20^2 − 8^2 or other mistaken combinations. They do not match the algebraic identity with the given values. Common Pitfalls: A common error is to square the sum and difference separately, for example doing 20^2 − 8^2 instead of using the correct identity. Another mistake is to attempt full expansion (a + b)^2 − (a − b)^2 without simplifying properly, which takes longer and is prone to algebra mistakes. Remember that a^2 − b^2 factors directly into (a + b) * (a − b), which is simple to compute when both terms are given. Final Answer: The difference of the squares of the two numbers is 160.

Question 12

40% of 250 is equal to 50% of which number?

Accepted Answer

Introduction / Context: Percentage questions like this are very common in quantitative aptitude sections. Here you are asked to compare two percentages applied to different numbers and find an unknown number that makes the quantities equal. Understanding percentages and how to convert them into fractions or decimals is essential for solving this quickly and accurately. Given Data / Assumptions: We are given 40% of 250. This value is equal to 50% of some unknown number, say N. We must find the value of N. Percentages are interpreted in the standard way: x% = x / 100. Concept / Approach: The idea is to compute 40% of 250 and then set this equal to 50% of the unknown number N. We then solve a simple linear equation for N. Using fraction form (for example, 40% = 40/100 = 2/5) often makes the arithmetic easier and reduces mistakes. This problem illustrates how different percentage expressions can represent the same absolute quantity when applied to different bases. Step-by-Step Solution: Step 1: Let the required number be N. Step 2: Compute 40% of 250. Write 40% as 40 / 100 = 2 / 5. Step 3: Calculate (2 / 5) * 250. Since 250 / 5 = 50, we have 2 * 50 = 100. So 40% of 250 is 100. Step 4: Now set this equal to 50% of N. That is, 50% of N = 100. Step 5: Write 50% as 1 / 2. So (1 / 2) * N = 100. Step 6: Solve for N by multiplying both sides by 2 to get N = 200. Verification / Alternative check: Verify the answer by direct substitution. Compute 40% of 250 again: 0.4 * 250 = 100. Compute 50% of 200: 0.5 * 200 = 100. Both expressions give the same result, confirming that the unknown number N must indeed be 200. This double check ensures there was no mistake in handling the fractions or arithmetic. Why Other Options Are Wrong: Option 100: 50% of 100 is 50, which is less than 40% of 250. Option 150: 50% of 150 is 75, which does not match 100. Option 400: 50% of 400 is 200, which is too large. Option 125: 50% of 125 is 62.5, which again does not equal 100. Common Pitfalls: One common mistake is to mix up which percentage applies to which number, or to attempt to solve it purely by guesswork. Another error is to misconvert percentages to fractions or decimals, such as treating 40% as 0.4 without multiplying correctly by 250. Writing a clear equation like (40 / 100) * 250 = (50 / 100) * N and solving it step by step reduces the chance of such errors. Final Answer: 40% of 250 is equal to 50% of 200.

Question 13

Half the people on a bus get off at each stop after the first, and no one gets on after the first stop. If exactly one person gets off at stop number 7, how many people got on the bus at the first stop?

Accepted Answer

Introduction / Context: This is a logical and numerical reasoning problem involving repeated halving. At every stop after the first, half the people on the bus get off. Given that one person gets off at the seventh stop, you have to work backwards to find how many people originally boarded the bus at the first stop. This type of question tests your ability to handle geometric progressions and reverse reasoning in word problems. Given Data / Assumptions: All passengers get on the bus at the first stop. After the first stop, no new passengers get on the bus. At each stop after the first, exactly half of the people on the bus get off. At stop number 7, exactly one person gets off. We must find how many people originally boarded at the first stop. Concept / Approach: If at each stop half the people get off, then the number of people getting off follows a pattern where at each stop the number of passengers just before the stop is double the number who get off at that stop. At stop 7, one person gets off, so there must have been 2 people on the bus just before that stop. By repeatedly doubling the number of people backwards for each previous stop, we can find the number of passengers just before stop 2, which is the same as the number of people who got on at the first stop (assuming the bus was empty before that). This is effectively a geometric sequence with common ratio 2 when tracing backwards. Step-by-Step Solution: Step 1: At stop 7, one person gets off. Since half the people get off at each stop, the number of people on the bus just before stop 7 must have been 2. Step 2: Before stop 6, the number of people must have been double the number just before stop 7, because half get off at stop 6 and the remaining half move on. So before stop 6, there were 4 people. Step 3: Similarly, before stop 5 there were 8 people, before stop 4 there were 16, before stop 3 there were 32, and before stop 2 there were 64 people on the bus. Step 4: The people on the bus before stop 2 are exactly the people who boarded at the first stop (since no one gets on after the first stop and we assume the bus was empty before). Step 5: Therefore, the number of people who got on the bus at the first stop is 64. Verification / Alternative check: We can verify by moving forward. Start with 64 people. At stop 2, half (32) get off, leaving 32. At stop 3, half (16) get off, leaving 16. At stop 4, half (8) get off, leaving 8. At stop 5, half (4) get off, leaving 4. At stop 6, half (2) get off, leaving 2. At stop 7, half (1) get off, leaving 1 person on the bus. This matches the condition that exactly one person gets off at stop 7, so the reasoning is consistent. Why Other Options Are Wrong: Option 128: If 128 people boarded, then at stop 7 the number getting off would be 2, not 1. Option 32: Working forwards with 32 passengers gives 0.5 person at some stage, which is impossible. Option 16: Similar to 32, this would lead to a fractional passenger count when halved repeatedly. Option 256: This would lead to 4 people getting off at stop 7, not 1. Common Pitfalls: The main mistake is to try to follow the process forward instead of backward, which can be confusing and often leads to miscounting. Another common error is to subtract half instead of correctly halving the current number at each step. Always remember that if half of the people get off, the number remaining is also half, and that backward reasoning (doubling) is often easier when you know the final condition. Final Answer: The number of people who got on at the first stop is 64.

Question 14

There are five consecutive odd numbers. The difference between the square of the average of the first two odd numbers and the square of the average of the last two odd numbers is 396. What is the smallest odd number in this sequence?

Accepted Answer

Introduction / Context: This question combines sequences, averages and quadratic expressions. You are given a set of five consecutive odd numbers and a condition involving the squares of averages of the first two and the last two numbers. From this information, you must determine the smallest odd number in the sequence. This tests algebraic modelling and careful handling of odd-number sequences in aptitude exams. Given Data / Assumptions: There are five consecutive odd numbers. The difference between the square of the average of the first two odd numbers and the square of the average of the last two odd numbers is 396. We are asked to find the smallest odd number in the sequence. The numbers are consecutive odd integers, so they differ by 2 each time. Concept / Approach: If the first odd number is n, the five consecutive odd numbers can be written as n, n + 2, n + 4, n + 6 and n + 8. The average of the first two is (n + (n + 2)) / 2, and the average of the last two is ((n + 6) + (n + 8)) / 2. We then square these averages, form their difference and equate it to 396. This leads to a linear equation in n once we simplify using algebraic identities. Solving this gives the smallest odd number directly. Step-by-Step Solution: Step 1: Let the five consecutive odd numbers be n, n + 2, n + 4, n + 6 and n + 8. Step 2: The average of the first two numbers is (n + (n + 2)) / 2 = (2n + 2) / 2 = n + 1. Step 3: The average of the last two numbers is ((n + 6) + (n + 8)) / 2 = (2n + 14) / 2 = n + 7. Step 4: Their squares are (n + 1)^2 and (n + 7)^2. The difference of these squares is given as 396. Step 5: Since the later average is larger, we take (n + 7)^2 − (n + 1)^2 = 396. Step 6: Use the identity a^2 − b^2 = (a − b) * (a + b). Here a = n + 7 and b = n + 1, so a − b = 6 and a + b = (n + 7) + (n + 1) = 2n + 8. Step 7: Therefore, (n + 7)^2 − (n + 1)^2 = 6 * (2n + 8) = 12n + 48. Step 8: Set 12n + 48 = 396. Subtract 48 to get 12n = 348, then divide by 12 to get n = 29. Verification / Alternative check: If n = 29, the five consecutive odd numbers are 29, 31, 33, 35 and 37. The average of the first two is (29 + 31) / 2 = 60 / 2 = 30. The average of the last two is (35 + 37) / 2 = 72 / 2 = 36. Their squares are 30^2 = 900 and 36^2 = 1296. The difference is 1296 − 900 = 396, which matches the given condition. Therefore, n = 29 is correct and is indeed the smallest odd number in the sequence. Why Other Options Are Wrong: Option 27: If 27 were the smallest, the sequence would not satisfy the given relationship between the squared averages when checked. Option 31 and 33: These values correspond to other members of the sequence but not to the smallest odd number satisfying the algebraic equation. Option 25: This also fails when substituted, as the computed difference of squares will not be 396. Common Pitfalls: Some learners misinterpret the phrase “difference between” and may set up the equation with the smaller square minus the larger one, which would give a negative value and no meaningful solution for this problem. Others may forget that the numbers are odd and try using increments of 1. Writing the sequence explicitly as n, n + 2, n + 4, n + 6 and n + 8, and using the difference of squares identity, makes the solution straightforward. Final Answer: The smallest odd number in the sequence is 29.

Question 15

Srinivas bought 10 cycles for Rs 1,750 each. He sold 4 cycles for Rs 8,400 in total and 3 cycles for Rs 1,900 each. At what price should he sell each of the remaining cycles so that his average profit is Rs 320 per cycle?

Accepted Answer

Introduction / Context: This is a profit and loss question involving multiple items with different selling prices. Srinivas buys several cycles at the same cost price but sells them in different lots at different selling prices. You are asked to find the selling price of the remaining cycles so that the overall average profit per cycle reaches a specified amount. This tests your ability to work with total cost, total revenue and average profit concepts in arithmetic. Given Data / Assumptions: Srinivas bought 10 cycles at Rs 1,750 each. He sold 4 cycles for Rs 8,400 in total. He sold 3 cycles for Rs 1,900 each. He wants an average profit of Rs 320 per cycle on all 10 cycles. We must find the selling price per cycle for the remaining 3 cycles. Concept / Approach: The key idea is to work with totals. First find the total cost of all cycles. Then compute the total revenue needed to achieve the desired average profit. Subtract the revenue already earned from the sold cycles to find how much more revenue is needed from the remaining cycles. Finally, divide this required revenue by the number of remaining cycles to get the selling price per remaining cycle. This approach avoids mixing up individual profits and focuses on overall totals, which is easier and more reliable. Step-by-Step Solution: Step 1: Compute the total cost price. Each cycle costs Rs 1,750 and there are 10 cycles, so total cost = 10 * 1750 = Rs 17,500. Step 2: He wants an average profit of Rs 320 per cycle on 10 cycles. Total profit required = 10 * 320 = Rs 3,200. Step 3: Total revenue required to achieve this profit is total cost + total profit = 17,500 + 3,200 = Rs 20,700. Step 4: Now compute the revenue already earned. He sold 4 cycles for a total of Rs 8,400. Step 5: He sold another 3 cycles at Rs 1,900 each, giving 3 * 1,900 = Rs 5,700. Step 6: Total revenue from the 7 cycles sold = 8,400 + 5,700 = Rs 14,100. Step 7: Remaining revenue needed = 20,700 − 14,100 = Rs 6,600. Step 8: There are 3 cycles left, so each must be sold at 6,600 / 3 = Rs 2,200. Verification / Alternative check: Verify the final profit. If each of the remaining 3 cycles is sold at Rs 2,200, revenue from these 3 cycles is Rs 6,600. Add this to the previous revenue of Rs 14,100 to get total revenue of Rs 20,700. Subtract total cost of Rs 17,500: total profit = 20,700 − 17,500 = Rs 3,200. Divide by 10 cycles to get average profit = 3,200 / 10 = Rs 320 per cycle, which matches the requirement. This confirms that the selling price of Rs 2,200 per remaining cycle is correct. Why Other Options Are Wrong: Option Rs. 1580 and Rs. 1785: These are too low and would reduce the overall average profit below Rs 320 per cycle. Option Rs. 1950 and Rs. 2100: These would give some profit, but when you calculate total revenue and total profit, the average profit per cycle will be less than Rs 320. Common Pitfalls: A common mistake is to try to compute profit separately for each batch and then average those profits, which does not always give the correct overall average profit. Another error is to forget to multiply the profit per cycle by the total number of cycles when calculating the required total profit. Always work with total cost, total required profit, and total revenue to avoid confusion. Then distribute the remaining required revenue evenly across the remaining items. Final Answer: Srinivas should sell each of the remaining cycles for Rs 2,200 to earn an average profit of Rs 320 per cycle.

Question 16

A total of 100 members are writing exams. Of these, 48 members are writing the first exam, 45 members are writing the second exam and 38 members are writing the third exam. Exactly 5 members are writing all three exams. Assuming every member writes …

Accepted Answer

Introduction / Context: This is a set theory and counting problem that uses the principle of inclusion and exclusion. We have three exams and counts of how many members are taking each exam, plus the number who take all three. The question asks how many members take exactly two exams, under the natural assumption that everyone takes at least one exam. Such questions are common in quantitative and logical reasoning sections, and they test your ability to work with overlapping sets. Given Data / Assumptions: Total number of members is 100. |A| = 48 members write the first exam. |B| = 45 members write the second exam. |C| = 38 members write the third exam. |A ∩ B ∩ C| = 5 members write all three exams. Assume every member writes at least one exam (so no one is outside A ∪ B ∪ C). We must find the number of members who write exactly two exams. Concept / Approach: Let x be the number of members who write exactly two exams and y be the number of members who write exactly one exam. The members who write all three exams are already given as 5. Using set theory, the sum of the individual set sizes |A| + |B| + |C| counts members as many times as the number of exams they write. That means it equals 1 * (members in exactly one exam) + 2 * (members in exactly two exams) + 3 * (members in all three exams). We also know that the total number of members is the sum of those in exactly one exam, exactly two exams, and exactly three exams. Using these facts, we can form equations and solve for x, the number writing exactly two exams. Step-by-Step Solution: Step 1: Let y be the number of members who write exactly one exam and x be the number who write exactly two exams. We are given that 5 members write all three exams. Step 2: Compute the sum of the set sizes: |A| + |B| + |C| = 48 + 45 + 38 = 131. Step 3: Count by membership. Each member in exactly one exam contributes 1 to this total, each in exactly two exams contributes 2, and each in all three contributes 3. So we have 131 = y + 2x + 3 * 5. Step 4: Simplify: 3 * 5 = 15, so y + 2x + 15 = 131, which gives y + 2x = 116. Call this Equation (1). Step 5: Since every member writes at least one exam, the total 100 members equals y (exactly one) + x (exactly two) + 5 (all three). So 100 = y + x + 5, or y + x = 95. Call this Equation (2). Step 6: Subtract Equation (2) from Equation (1): (y + 2x) − (y + x) = 116 − 95, giving x = 21. Verification / Alternative check: From x = 21, use Equation (2) to find y: y + 21 = 95, so y = 74. Now check the count into the sum of set sizes: y + 2x + 3 * 5 = 74 + 2 * 21 + 15 = 74 + 42 + 15 = 131, which matches |A| + |B| + |C|. Also check total members: 74 (exactly one exam) + 21 (exactly two) + 5 (all three) = 100, matching the total number of members. Everything is consistent, so x = 21 is correct. Why Other Options Are Wrong: Option 26, 23, 27 and 24: Substituting any of these values for x into the equations y + 2x = 116 and y + x = 95 will make the equations inconsistent, so they cannot represent the actual count of members writing exactly two exams. Common Pitfalls: One common mistake is to forget the assumption that every member writes at least one exam, which is needed to set the total members equal to y + x + 5. Another error is to misinterpret “writing 2 exams” as “at least two exams” instead of “exactly two exams”, which changes the counting. Always define variables clearly for “exactly one”, “exactly two” and “exactly three” membership, and use the inclusion–exclusion counting carefully to avoid double counting or undercounting. Final Answer: The number of members who are writing exactly two exams is 21.

Question 17

The series of differences between consecutive prime numbers is denoted as Cp1, Cp2, Cp3, ..., Cpn, where Cp1 is the difference between the second and the first prime numbers. If the 23rd prime number is 83, what is the sum of the series Cp1 + Cp2 + …

Accepted Answer

Introduction / Context: This question deals with prime numbers and the differences between consecutive primes. The series Cp1, Cp2, ..., represents the gaps between each pair of consecutive prime numbers. You are given that the 23rd prime is 83 and asked to find the sum of the first 22 such differences. This is a telescoping series idea, where many intermediate terms cancel out, leaving a simple result involving the first and last primes considered. Given Data / Assumptions: Prime numbers in order are p1, p2, p3, ..., p23, where p23 = 83. Cp1 = p2 − p1, Cp2 = p3 − p2, ..., Cp22 = p23 − p22. We are asked to find Cp1 + Cp2 + ... + Cp22. The first prime p1 is 2. Concept / Approach: The key concept is that the sum of consecutive differences telescopes. Each Cp term is the difference between two consecutive primes. When we add these differences in order, almost all intermediate primes cancel out, and we are left with the difference between the last and the first prime involved. This type of simplification is called a telescoping sum and is very powerful in series problems. Step-by-Step Solution: Step 1: Write the definitions. Cp1 = p2 − p1, Cp2 = p3 − p2, ..., Cp22 = p23 − p22. Step 2: Consider the sum S = Cp1 + Cp2 + ... + Cp22. Step 3: Substitute the definitions: S = (p2 − p1) + (p3 − p2) + ... + (p23 − p22). Step 4: Observe that this is a telescoping series. The +p2 from Cp1 cancels with the −p2 from Cp2, +p3 cancels with −p3, and so on. Step 5: After cancellation, only the first negative term and the last positive term remain: S = p23 − p1. Step 6: The first prime p1 is 2, and the 23rd prime p23 is given as 83. Therefore S = 83 − 2 = 81. Verification / Alternative check: You can verify the telescoping idea with a smaller example. Take the first 5 primes: 2, 3, 5, 7, 11. Then Cp1 = 3 − 2 = 1, Cp2 = 5 − 3 = 2, Cp3 = 7 − 5 = 2, Cp4 = 11 − 7 = 4. The sum Cp1 + Cp2 + Cp3 + Cp4 = 1 + 2 + 2 + 4 = 9, which equals 11 − 2, the last prime minus the first. This matches the pattern used in the main problem, giving confidence that S = 83 − 2 = 81 is correct for Cp1 through Cp22. Why Other Options Are Wrong: Option 87, 83, 85 and 79: These values might be obtained by guessing or by misinterpreting the series definition (for example, adding one more term or using a wrong starting or ending prime). However, once the telescoping structure is correctly applied, the only consistent result is 81. Common Pitfalls: A common mistake is to misunderstand n and think that Cp23 should also be included in the sum, or to forget that Cp1 is defined as the difference between the second and the first prime. Another error is trying to list all primes and explicitly compute each difference, which is time consuming and unnecessary here. Recognising the telescoping pattern S = last prime − first prime is the key to solving such problems quickly. Final Answer: The sum of the first 22 differences between consecutive prime numbers up to the 23rd prime is 81.

Question 18

Three numbers are pairwise co-prime (each pair has greatest common divisor 1). The product of the first two numbers is 551 and the product of the last two numbers is 1073. What is the sum of the three numbers?

Accepted Answer

Introduction / Context: This is a number theory problem involving co-prime numbers and factorisation. You are given that three numbers are pairwise co-prime, meaning no two share any common factor greater than 1. You are also given the products of the first two and the last two numbers. From this information, you must determine the individual numbers and then find their sum. This type of question requires careful use of prime factorisation and the co-prime condition. Given Data / Assumptions: Let the three numbers be a, b and c. The product of the first two is a * b = 551. The product of the last two is b * c = 1073. The numbers are pairwise co-prime, so any two of them have greatest common divisor 1. We must find a + b + c. Concept / Approach: First factorise the given products into primes. Then use the pairwise co-prime condition to identify which prime factors belong to which number. Since a and c are both multiplied with b in the given products, any common factor between 551 and 1073 must be part of b. Once we identify b, we can find a and c by dividing the products by b. Finally, we verify the co-prime condition for all pairs and add the three numbers to obtain the required sum. Step-by-Step Solution: Step 1: Factorise 551. Try dividing by small primes: 551 is not even and digit sum 5 + 5 + 1 = 11 is not divisible by 3. Check 19: 19 * 29 = 551. So 551 = 19 * 29. Step 2: Factorise 1073. It is not divisible by 2, 3 or 5. Check 29: 29 * 37 = 1073. So 1073 = 29 * 37. Step 3: The common factor between 551 and 1073 is 29. Because a, b, c are pairwise co-prime, any common factor between the products a * b and b * c must belong to b. Step 4: Therefore, b must be 29. Step 5: From a * b = 551, we get a = 551 / 29 = 19. Step 6: From b * c = 1073, we get c = 1073 / 29 = 37. Verification / Alternative check: Check the pairwise co-prime condition. gcd(19, 29) = 1, gcd(19, 37) = 1 and gcd(29, 37) = 1, so the three numbers are indeed pairwise co-prime. Also verify the products: 19 * 29 = 551 and 29 * 37 = 1073, which match the given information. Now compute the sum: a + b + c = 19 + 29 + 37 = 85. Everything is consistent, confirming that the answer is correct. Why Other Options Are Wrong: Option 83, 89, 79 and 91: These sums would correspond to different sets of numbers. If you try to construct numbers that are pairwise co-prime and whose products match 551 and 1073, you will find that only 19, 29 and 37 work. Any other combination produces a different product or violates the co-prime condition. Common Pitfalls: A common error is to misinterpret “co-prime to each other” and assume only that the three numbers share no common divisor all together, instead of checking each pair separately. Another mistake is incomplete factorisation of 551 or 1073, which leads to wrong candidates for the three numbers. Always factor completely into primes and then distribute the prime factors according to the co-prime requirement. Final Answer: The sum of the three pairwise co-prime numbers is 85.

Question 19

What should be added to the units digit of the number 86236 so that the resulting number is divisible by 9?

Accepted Answer

Introduction / Context: This question tests your understanding of the divisibility rule for 9 and your ability to work with digit sums. To check if a number is divisible by 9, you do not need to divide it directly. Instead, you can sum its digits and see whether the sum is divisible by 9. Here, you are asked how much must be added to the units digit so that the entire number becomes divisible by 9. Given Data / Assumptions: The original number is 86236. We are allowed to add a non-negative digit to the existing units digit (6) to create a new number. The new number must be divisible by 9. We must find the smallest such addition from the given options. Concept / Approach: The rule for divisibility by 9 states: A number is divisible by 9 if and only if the sum of its digits is divisible by 9. When we change the units digit by adding some value x, we effectively increase the sum of the digits by x. Therefore, the new sum should be a multiple of 9. We calculate the current sum of the digits, determine what x must be so that this sum plus x is divisible by 9, and then choose the appropriate option. Step-by-Step Solution: Step 1: Compute the sum of the digits of 86236. Step 2: 8 + 6 + 2 + 3 + 6 = 25. Step 3: Let x be the number to be added to the units digit. Then the new digit sum will be 25 + x. Step 4: For the new number to be divisible by 9, 25 + x must be a multiple of 9. Step 5: Note that 25 gives remainder 25 − 18 = 7 when divided by 9. So we need x such that 7 + x is a multiple of 9. Step 6: The smallest non-negative x between 0 and 9 that satisfies this is x = 2, because 7 + 2 = 9, and 9 is divisible by 9. Verification / Alternative check: If we add 2 to the units digit 6, the new number becomes 86238. The digit sum is 8 + 6 + 2 + 3 + 8 = 27. Since 27 is divisible by 9, 86238 is divisible by 9. This confirms that adding 2 to the units digit achieves the required property. Checking other small additions like 0 or 1 shows that the resulting digit sums (25 and 26) are not multiples of 9. Why Other Options Are Wrong: Option 1: 25 + 1 = 26, which is not divisible by 9. Option 3: 25 + 3 = 28, which is not a multiple of 9. Option 0: 25 + 0 = 25, also not divisible by 9. Option 4: 25 + 4 = 29, again not divisible by 9. Common Pitfalls: Some students incorrectly try to divide the entire number by 9 each time they change the units digit, which is more time consuming and error prone. Others might miscalculate the digit sum. Using remainders helps: find the remainder of the digit sum when divided by 9, then choose x so that the new remainder becomes 0. This is a fast and reliable strategy for such problems. Final Answer: You must add 2 to the units digit so that the number becomes divisible by 9.

Question 20

A number, when reduced by one-seventh of itself, gives a result equal to the sum of all the angles of a triangle in degrees. What is that number?

Accepted Answer

Introduction / Context: This question combines basic fraction operations with a simple geometric fact: the sum of the interior angles of a triangle is 180 degrees. You are told that a number minus one seventh of itself equals 180, and you must find the number. This is a straightforward linear equation involving fractions and is common in quantitative aptitude tests. Given Data / Assumptions: Let the required number be N. One seventh of N is N / 7. The number minus one seventh of itself equals 180. The sum of all angles of a triangle in degrees is 180. Concept / Approach: We translate the statement “a number when subtracted by one seventh of itself gives 180” into an algebraic equation: N − (1 / 7) * N = 180. Then we combine like terms on the left side and solve for N. This requires basic knowledge of fractions and how to factor and simplify linear expressions. Recognising the geometric fact about triangle angles just tells us that the result is 180, which is used in the equation. Step-by-Step Solution: Step 1: Let the number be N. Step 2: One seventh of N is N / 7. Step 3: According to the question, N − (N / 7) = 180. Step 4: Factor out N on the left side: N * (1 − 1 / 7) = 180. Step 5: Compute 1 − 1 / 7 = (7 / 7) − (1 / 7) = 6 / 7. Step 6: So we have N * (6 / 7) = 180. Multiply both sides by 7 / 6 to solve for N. Step 7: N = 180 * (7 / 6) = (180 / 6) * 7 = 30 * 7 = 210. Verification / Alternative check: Check the result by substituting N = 210 back into the original statement. One seventh of 210 is 210 / 7 = 30. Now compute N − N / 7 = 210 − 30 = 180. This matches the sum of the angles of a triangle, confirming that N = 210 is correct. No other option will satisfy this equation when tested in the same way. Why Other Options Are Wrong: Option 420: One seventh of 420 is 60, and 420 − 60 = 360, not 180. Option 70: One seventh of 70 is 10, and 70 − 10 = 60, not 180. Option 35: One seventh of 35 is 5, and 35 − 5 = 30, not 180. Option 140: One seventh of 140 is 20, and 140 − 20 = 120, still not 180. Common Pitfalls: A common error is to interpret the statement as N − 1 / 7 = 180 instead of N − (1 / 7) * N = 180, which changes the equation completely. Another mistake is to mishandle fractions, especially when combining 1 − 1 / 7. Always express 1 as 7 / 7 when subtracting 1 / 7, and then solve the resulting simple equation. Verifying the answer by substitution is a good practice to confirm you did not make any fraction mistakes. Final Answer: The required number is 210.

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