Q: In a percentage problem, 60% of 3/5 of a certain number is equal to 36. What is the value of that original number?

Introduction: This question checks your understanding of percentages and fractions applied together to an unknown number. You are told that 60% of three fifths of a number equals 36, and you must find the original number. Such problems are very common in quantitative aptitude exams. Given Data / Assumptions: Let the original number be N. Three fifths of the number is (3/5) * N. Sixty percent (60%) of this quantity is equal to 36. We assume N is a real positive number. Concept / Approach: Percentages can be written as fractions or decimals. Here, 60% = 60/100 = 3/5. The statement “60% of 3/5 of N is 36” becomes a simple linear equation in N. We multiply the fractions, set equal to 36, and then solve for N using basic algebra. Step-by-Step Solution: Step 1: Write the mathematical expression.60% of (3/5 of N) = 36Step 2: Convert 60% to fraction.(60/100) * (3/5 * N) = 36Step 3: Simplify the fractions.(3/5) * (3/5) * N = 36(9/25) * N = 36Step 4: Solve for N.N = 36 * (25/9)N = (36/9) * 25 = 4 * 25 = 100 Verification / Alternative check: Check by working forward from N = 100. Three fifths of 100 is (3/5) * 100 = 60. Then 60% of 60 is (60/100) * 60 = 36. This confirms that N = 100 satisfies the condition perfectly. Why Other Options Are Wrong: 94, 115, 86 and 120 do not satisfy the relationship when you compute 3/5 of the number and then take 60% of that result. Only 100 produces 36 at the end of the two-step percentage calculation. Common Pitfalls: A common mistake is to treat the statement as “60% of the number is 36” or as “3/5 of 36 is 60%”, which reverses the operations. Another mistake is to add percentages instead of multiplying them when they are applied successively to the same number. Final Answer: The original number is 100.

Q: Between 300 and 700 (inclusive), how many integers are divisible simultaneously by 2, 3 and 7, that is, by their common multiple?

Introduction: This divisibility and counting question asks you to determine how many numbers in a given interval are divisible by three given integers simultaneously: 2, 3 and 7. Such problems test your understanding of least common multiple (LCM) and counting multiples in a range. Given Data / Assumptions: We are considering integers from 300 to 700 inclusive. Each required integer must be divisible by 2, 3 and 7 at the same time. Divisibility by all three means divisibility by their least common multiple. Concept / Approach: The LCM of 2, 3 and 7 captures simultaneous divisibility. Once we find this LCM, we count how many multiples of this LCM lie between 300 and 700 inclusive using a simple floor and ceiling method. Step-by-Step Solution: Step 1: Compute the LCM of 2, 3 and 7.2, 3 and 7 are pairwise co-prime.LCM = 2 * 3 * 7 = 42.Step 2: Find the first multiple of 42 not less than 300.42 * 7 = 294 (too small).42 * 8 = 336 (within the range, this is the first valid multiple).Step 3: Find the last multiple of 42 not greater than 700.42 * 16 = 672 (within the range).42 * 17 = 714 (greater than 700, so stop at 16).Step 4: Count the multiples from 8 to 16 inclusive.Number of terms = 16 − 8 + 1 = 9. Verification / Alternative check: List the actual multiples: 336, 378, 420, 462, 504, 546, 588, 630, 672. There are exactly nine such numbers, all between 300 and 700 inclusive, each divisible by 42 and therefore by 2, 3 and 7. Why Other Options Are Wrong: The other counts (7, 11, 5, 8) do not match the actual number of valid multiples. Only 9 corresponds to the correct count obtained from both formula and direct listing. Common Pitfalls: Some candidates attempt to count multiples of 2, 3 and 7 separately and then try to intersect them manually, which is inefficient and error-prone. Using the LCM avoids double-counting and simplifies the problem greatly. Final Answer: There are 9 integers between 300 and 700 that are divisible by 2, 3 and 7 together.

Q: The sum of two numbers is 12 and their product is 35. What is the sum of the reciprocals of these two numbers?

Introduction: This question explores the relationship between the sum and product of two numbers and the sum of their reciprocals. It is a standard algebraic identity that is very useful in many aptitude problems involving symmetric expressions in two variables. Given Data / Assumptions: Two numbers, say a and b, have a sum a + b = 12. Their product is a * b = 35. We are asked to find 1/a + 1/b. Concept / Approach: Use the identity for the sum of reciprocals of two non-zero numbers: 1/a + 1/b = (a + b) / (a * b)Since we already know a + b and a * b, we can directly compute the sum of the reciprocals without solving for a and b individually. Step-by-Step Solution: Step 1: Recall the identity.1/a + 1/b = (a + b) / (a * b)Step 2: Substitute given values.a + b = 12a * b = 351/a + 1/b = 12 / 35Step 3: Simplify if possible.12/35 is already in simplest form (no common factor between 12 and 35). Verification / Alternative check: If you want, you can find the actual numbers. They satisfy t^2 − 12t + 35 = 0. The roots are t = 5 and t = 7 (since 5 + 7 = 12 and 5 * 7 = 35). Then 1/5 + 1/7 = (7 + 5) / 35 = 12/35, confirming the identity-based result. Why Other Options Are Wrong: 1/35 is far too small; 35/8 and 7/32 are unrelated to the ratio of sum to product here. 47/35 does not arise from any correct manipulation of the given data. Only 12/35 matches the derived expression (a + b)/(a * b). Common Pitfalls: Some students attempt to find the two numbers first every time, which is not necessary. Others mistakenly compute (a * b)/(a + b) instead of (a + b)/(a * b). Memorizing the simple identity saves time and prevents sign and inversion errors. Final Answer: The sum of the reciprocals of the two numbers is 12/35.

Q: A number when divided by 779 leaves a remainder of 47. If the same number is divided by 19, what remainder will it leave?

Introduction: This question tests your understanding of modular arithmetic and remainders. You are given the remainder when a number is divided by a large divisor and asked to find the remainder when the same number is divided by a smaller related divisor. Given Data / Assumptions: Let the unknown number be N. When N is divided by 779, the remainder is 47. We need the remainder when N is divided by 19. All divisions are in the standard integer sense. Concept / Approach: We express N in terms of the first division and then reduce modulo 19. Also, we notice that 779 is a multiple of 19, which greatly simplifies the calculation because any multiple of 779 is automatically a multiple of 19. Step-by-Step Solution: Step 1: Express N using the given remainder.N = 779k + 47 for some integer k.Step 2: Note that 779 = 19 * 41.Therefore, 779k = 19 * 41k is divisible by 19.Step 3: Reduce N modulo 19.N mod 19 = (779k + 47) mod 19.Since 779k is divisible by 19, it contributes remainder 0.So N mod 19 = 47 mod 19.Step 4: Compute 47 mod 19.19 * 2 = 38, 47 − 38 = 9.So the remainder is 9. Verification / Alternative check: You can pick a random k, say k = 1: N = 779 + 47 = 826. Now 826 ÷ 19 = 43 remainder 9. For any other integer k, the term 779k always adds a multiple of 19, so the remainder will stay 9. Why Other Options Are Wrong: Remainders 6, 5, 8 and 3 are not equal to 47 mod 19. Only 9 matches the correct reduction of 47 modulo 19. Common Pitfalls: Some students try to work directly with the large number 779, forgetting to factor it or see its relation to 19. Others mistakenly divide 779 by 19 and then mishandle the remainder instead of using the expression N = 779k + 47. Final Answer: When the number is divided by 19, the remainder is 9.

Q: The seven-digit number 517324 is completely divisible by 3. What is the smallest whole number that can replace the asterisk () to make the number divisible by 3?

Introduction: This is a basic divisibility test problem with a missing digit. You are given a large number with one unknown digit and asked to choose that digit so the entire number becomes divisible by 3. Such problems train you to apply divisibility rules quickly without long division. Given Data / Assumptions: The number has the form 517*324, where * represents a single digit from 0 to 9. The complete number must be divisible by 3. We are specifically asked for the smallest whole number that can replace *. Concept / Approach: The rule for divisibility by 3 states that a number is divisible by 3 if and only if the sum of its digits is divisible by 3. We therefore sum the known digits, include the unknown digit x in the sum, and choose x so that the total is a multiple of 3. Step-by-Step Solution: Step 1: Let the missing digit be x.Digits are 5, 1, 7, x, 3, 2, 4.Step 2: Sum the known digits.5 + 1 + 7 + 3 + 2 + 4 = 22.Step 3: Include x in the total sum.Total sum = 22 + x.Step 4: Apply the divisibility by 3 rule.We need 22 + x to be a multiple of 3.The next multiples of 3 after 22 are 24, 27, 30, etc.So 22 + x = 24, 27, 30, ...This gives x = 2, 5, 8, ...Step 5: Choose the smallest whole number.The smallest x in {2, 5, 8, ...} is 2. Verification / Alternative check: Substitute x = 2 to get the number 5172324. Sum of digits = 5 + 1 + 7 + 2 + 3 + 2 + 4 = 24, which is divisible by 3. Therefore, the full number is divisible by 3 as required. Why Other Options Are Wrong: 0 and 1 make the digit sum 22 and 23 respectively, which are not multiples of 3. 5 and 8 also give valid divisibility by 3, but they are larger than 2. The question explicitly asks for the smallest such digit, so 2 is the correct choice. Common Pitfalls: Some learners forget the “smallest” condition and might choose any digit that works, such as 5 or 8. Others try full long division rather than using the far quicker digit-sum rule. Final Answer: The smallest whole number that can replace * is 2.

Q: In the mixed number series 3, 4, 5, 5, 12, 13, 7, 24, 25, 8, 15, ..., what number should come next to correctly continue the pattern?

Introduction: This question presents a number series that is not a simple arithmetic or geometric progression. Instead, it is constructed from known Pythagorean triples. Recognizing this structure is essential for finding the next term in the sequence. Given Data / Assumptions: The series is: 3, 4, 5, 5, 12, 13, 7, 24, 25, 8, 15, ... We need to identify the pattern and determine the next number. Concept / Approach: Group the numbers to see if they form meaningful sets. Notice that they can be grouped as triples: (3, 4, 5), (5, 12, 13), (7, 24, 25), (8, 15, ...)Each of the first three groups is a well-known Pythagorean triple where a^2 + b^2 = c^2. Step-by-Step Solution: Step 1: Identify the triples.First triple: 3, 4, 5 → 3^2 + 4^2 = 9 + 16 = 25 = 5^2.Second triple: 5, 12, 13 → 5^2 + 12^2 = 25 + 144 = 169 = 13^2.Third triple: 7, 24, 25 → 7^2 + 24^2 = 49 + 576 = 625 = 25^2.Step 2: Look at the next group.Numbers so far: 8, 15, ...We recognise another Pythagorean triple: 8, 15, 17.8^2 + 15^2 = 64 + 225 = 289 = 17^2.Step 3: Conclude the next term.The missing number after 8, 15 should be 17. Verification / Alternative check: The pattern consistently follows primitive Pythagorean triples: (3, 4, 5), (5, 12, 13), (7, 24, 25), and now (8, 15, 17). There is no break or inconsistency in this structure, so the triple (8, 15, 17) naturally completes the next group. Why Other Options Are Wrong: 26, 16, 18 and 20 do not complete a standard Pythagorean triple with 8 and 15. Only 17 satisfies 8^2 + 15^2 = 17^2, so only 17 fits the observable pattern of the series. Common Pitfalls: Some learners try to look at differences between terms or use arithmetic progressions, which do not work here. Recognizing well-known triples is a major advantage in such pattern questions. Final Answer: The next number in the series is 17.

Q: The difference between two numbers is 11 and one fifth of their sum is 9. What are the two numbers?

Introduction: This is a classic linear equations problem involving two numbers. You are given their difference and a relation involving their sum. Solving such problems efficiently is crucial for many aptitude and reasoning exams. Given Data / Assumptions: Let the two numbers be x and y, with x > y. Their difference is x − y = 11. One fifth of their sum is 9, so (x + y)/5 = 9. We must find the values of x and y. Concept / Approach: We are given two linear equations in two unknowns. The standard method is to solve this system by addition or substitution. First, we determine the sum from the second condition, then use the sum and difference to solve for each number. Step-by-Step Solution: Step 1: Use the sum condition.(x + y)/5 = 9 ⇒ x + y = 45.Step 2: Use the difference condition.x − y = 11.Step 3: Add the two equations.(x + y) + (x − y) = 45 + 11.2x = 56 ⇒ x = 28.Step 4: Substitute x back into one equation to find y.x + y = 45 ⇒ 28 + y = 45 ⇒ y = 17. Verification / Alternative check: Check both conditions: Difference = 28 − 17 = 11 (correct). Sum = 28 + 17 = 45; one fifth of 45 is 45/5 = 9 (also correct). Thus both original statements are satisfied by the pair (28, 17). Why Other Options Are Wrong: Each other option pair either does not have a difference of 11, or if it does, its sum does not give (sum)/5 = 9. For example, 31 and 20 have sum 51, and one fifth of 51 is 10.2, not 9. Common Pitfalls: Students sometimes mis-handle the fraction and set the sum equal to 9 instead of 45. Others may mix up sum and difference while solving. Writing equations clearly and proceeding stepwise prevents these errors. Final Answer: The two numbers are 28 and 17.

Question 1

Consider all two digit natural numbers that are divisible by 5. What is the sum of all such two digit numbers?

Accepted Answer

Introduction: This problem asks you to find the sum of all two digit numbers that are divisible by 5. Recognizing that these numbers form an arithmetic progression allows you to apply the standard formula for the sum of an arithmetic series instead of adding each number individually. Given Data / Assumptions: We consider only two digit natural numbers (from 10 to 99 inclusive). We select only those that are divisible by 5. We must compute the sum of these selected numbers. Concept / Approach: Two digit numbers divisible by 5 must end in 0 or 5. The smallest such number is 10, and the largest such two digit number is 95. These numbers form an arithmetic progression with a common difference of 5. For an arithmetic progression with first term a, last term l, and n terms, the sum is S = n * (a + l) / 2. Step-by-Step Solution: First term a = 10. Common difference d = 5. Last term l = 95. Find the number of terms n using l = a + (n − 1)d. 95 = 10 + (n − 1) * 5. 95 − 10 = (n − 1) * 5 ⇒ 85 = 5(n − 1). Divide by 5: 17 = n − 1 ⇒ n = 18. Now compute the sum: S = n * (a + l) / 2. S = 18 * (10 + 95) / 2 = 18 * 105 / 2. 105 / 2 = 52.5, so S = 18 * 52.5. 18 × 50 = 900 and 18 × 2.5 = 45, so total S = 900 + 45 = 945. Therefore, the required sum is 945. Verification / Alternative check: You can quickly verify by grouping terms from both ends: (10 + 95) + (15 + 90) + (20 + 85) + ... . Each pair sums to 105. With 18 terms, there are 9 such pairs. So S = 9 * 105 = 945, which confirms our earlier result. Why Other Options Are Wrong: The values 678, 439, and 568 do not match the sum produced by the arithmetic progression formula and typically arise from miscounting the terms or omitting numbers ending in 0 or 5. Only 945 is consistent with all two digit multiples of 5 from 10 through 95 inclusive. Common Pitfalls: Common mistakes include forgetting that 95 is the largest two digit multiple of 5, using 100 incorrectly, or miscalculating n by not using the formula for the n-th term correctly. Careful use of the A.P. formulas avoids these errors. Final Answer: The sum of all two digit numbers divisible by 5 is 945.

Question 2

When each of the numbers 2272 and 875 is divided by the same three digit number N, the remainders obtained are equal. What is the sum of the digits of this divisor N?

Accepted Answer

Introduction: This question tests your understanding of remainders and the use of the highest common factor (HCF) when dealing with the same remainder condition. It is a classic number theory problem often seen in aptitude examinations. Given Data / Assumptions: When 2272 is divided by N, the remainder is some value R. When 875 is divided by the same N, the remainder is also R. N is a three digit number. We must find the sum of the digits of N. Concept / Approach: If two numbers leave the same remainder when divided by N, then their difference is exactly divisible by N. That is, N divides (2272 − 875). Therefore, N must be a factor of this difference. Because N is also specified to be a three digit divisor, we can factor the difference and choose the appropriate factor that has three digits, then sum its digits. Step-by-Step Solution: Let the common remainder be R. Then 2272 = N * a + R and 875 = N * b + R for some integers a and b. Subtract: 2272 − 875 = N(a − b). Compute the difference: 2272 − 875 = 1397. Thus, N divides 1397 exactly. Now factor 1397 to find its divisors. Check small primes: 1397 is not divisible by 2, 3, or 5. Test 7: 7 × 199 = 1393, so not divisible by 7. Test 11: 11 × 127 = 1397, so 1397 = 11 × 127. The positive divisors are 1, 11, 127, and 1397. Among these, the only three digit divisor is N = 127. So N = 127. Sum of digits of N = 1 + 2 + 7 = 10. Verification / Alternative check: You can verify that both 2272 and 875 leave the same remainder when divided by 127. Compute 2272 ÷ 127 and 875 ÷ 127, find their remainders and confirm they match. Since 127 is a three digit divisor of 1397, it correctly satisfies the given condition. Why Other Options Are Wrong: Digit sums 11, 12, and 13 would correspond to different three digit numbers, none of which divide 1397 exactly. Because 127 is the only three digit factor of 1397, any other choice would contradict the divisibility requirement implied by the equal remainders. Common Pitfalls: Learners sometimes try to find N by trial dividing 2272 and 875 directly, which is inefficient and error prone. Others forget to use the key idea that equal remainders imply that the difference of the numbers is exactly divisible by the divisor. Always reduce such problems to a difference and then factor that difference. Final Answer: The sum of the digits of the divisor N is 10.

Question 3

Let n and n + 2 be two consecutive odd integers. The difference between the squares of these two consecutive odd integers is always divisible by which of the following numbers?

Accepted Answer

Introduction: This conceptual number theory question asks about a general pattern involving consecutive odd integers and their squares. Rather than plugging in specific examples, we can use algebra to prove a property that holds for all such pairs of odd numbers. Given Data / Assumptions: We consider two consecutive odd integers. They can be written as k and k + 2 for some integer k. We look at their squares: k^2 and (k + 2)^2. We must determine a number that always divides the difference between these squares. Concept / Approach: The difference of squares formula is crucial: a^2 − b^2 = (a − b)(a + b). For consecutive odd integers, the difference between the two numbers is 2, and their sum is an even multiple of 2, which leads to a factor of 4. Careful simplification reveals that the difference between their squares is always a multiple of 8. Step-by-Step Solution: Let the two consecutive odd integers be (2n + 1) and (2n + 3), where n is any integer. Their squares are (2n + 1)^2 and (2n + 3)^2. Compute the difference D = (2n + 3)^2 − (2n + 1)^2. Use the difference of squares formula: D = [(2n + 3) − (2n + 1)] * [(2n + 3) + (2n + 1)]. Simplify the first factor: (2n + 3) − (2n + 1) = 2. Simplify the second factor: (2n + 3) + (2n + 1) = 4n + 4 = 4(n + 1). Thus D = 2 * 4(n + 1) = 8(n + 1). Therefore, the difference between the squares is always 8 times an integer. So D is always divisible by 8. Verification / Alternative check: Test with a few examples. Take 3 and 5: 5^2 − 3^2 = 25 − 9 = 16, which is divisible by 8. Take 7 and 9: 9^2 − 7^2 = 81 − 49 = 32, also divisible by 8. These examples support the general algebraic proof. Why Other Options Are Wrong: While the difference is also divisible by 2 and 4 (since 8 is a multiple of both), the question asks for a number that always divides the difference, and 8 is the strongest choice here. The option 6 does not always divide the difference; for instance, 16 and 32 are not divisible by 6. Therefore, 8 is the correct and most informative answer. Common Pitfalls: Some learners guess from small examples and choose 2 or 4, missing that 8 always works. Others may incorrectly define consecutive odd numbers or forget to use the difference of squares identity. Writing the numbers in the form (2n + 1) and (2n + 3) is a systematic way to avoid mistakes. Final Answer: The difference between the squares of two consecutive odd integers is always divisible by 8.

Question 4

When 4150 is divided by a certain positive integer, the quotient obtained is 55 and the remainder is 25. What is the value of the divisor?

Accepted Answer

Introduction: This question examines your understanding of the basic division algorithm and how quotient, remainder, dividend, and divisor are related. It is a direct application of a simple formula that is fundamental in arithmetic and number theory. Given Data / Assumptions: Dividend (the number being divided) = 4150. Quotient = 55. Remainder = 25. Divisor = unknown positive integer D. We must find D. Concept / Approach: The division algorithm states that: Dividend = Divisor × Quotient + Remainder. Using this formula and the given values, we can set up an equation in D and solve for the divisor directly. Because the numbers are small, this is straightforward and does not require trial and error. Step-by-Step Solution: Use the division algorithm: 4150 = D × 55 + 25. Subtract the remainder from both sides: 4150 − 25 = D × 55. Compute 4150 − 25: 4125 = D × 55. Now solve for D: D = 4125 / 55. Compute 55 × 75 to check: 55 × 70 = 3850 and 55 × 5 = 275, total 3850 + 275 = 4125. Therefore, 4125 / 55 = 75. So the divisor D is 75. Verification / Alternative check: Verify by performing the division directly: 4150 ÷ 75 = 55 remainder 25. 75 × 55 = 4125, and 4150 − 4125 = 25. This matches the given quotient and remainder exactly, confirming D = 75. Why Other Options Are Wrong: If D = 68, 65, or 70, then D × 55 + 25 would not equal 4150. For example, 70 × 55 + 25 = 3850 + 25 = 3875, which is less than 4150. Similarly, using 65 or 68 leads to incorrect dividends that do not match 4150, so they cannot be the correct divisor. Common Pitfalls: Some students attempt to divide 4150 by each option and then check the remainder manually, which is slower and more error prone than using the division formula directly. Others forget to subtract the remainder before dividing, or mistakenly divide by the quotient. Always start from the standard equation: Dividend = Divisor × Quotient + Remainder. Final Answer: The required divisor in this division problem is 75.

Question 5

Consider the numeric sequence: 125, 80, 45, 20, ... What is the next number that should appear in this series?

Accepted Answer

Introduction: This question tests your ability to recognize patterns in a number sequence. Such sequences often involve simple arithmetic operations like adding, subtracting, or using a constant difference that itself follows a pattern. Here, we must determine how the terms are changing to predict the next term correctly. Given Data / Assumptions: The sequence starts as: 125, 80, 45, 20, ... We must find the next term after 20. The pattern is assumed to be consistent throughout the sequence. Concept / Approach: First, look at the differences between consecutive terms: 80 − 125, 45 − 80, and 20 − 45. If these differences follow a simple pattern, such as an arithmetic progression, we can then predict the next difference and hence the next term. Recognizing that the differences themselves form a sequence is very common in such problems. Step-by-Step Solution: Compute the first difference: 80 − 125 = −45. Compute the second difference: 45 − 80 = −35. Compute the third difference: 20 − 45 = −25. So the sequence of differences is: −45, −35, −25, ... Now examine this difference sequence. The change from −45 to −35 is +10, and from −35 to −25 is also +10. Thus, the differences themselves form an arithmetic progression with common difference +10. The next difference should be: −25 + 10 = −15. To find the next term in the original series, add this next difference to the last term 20. Next term = 20 + (−15) = 5. Therefore, the next number in the sequence is 5. Verification / Alternative check: We can now write the extended sequence including the new term: 125, 80, 45, 20, 5. The differences are: −45, −35, −25, −15. The difference of differences remains +10 throughout, confirming the pattern is consistent. Why Other Options Are Wrong: A next term of 15, −5, or 10 would produce differences that break the arithmetic pattern of the difference sequence. For example, if the next term were 10, the last difference would be −10, and the step between −25 and −10 would be +15 instead of +10, which does not match the established pattern. Common Pitfalls: Some learners look only at the original terms and guess a random rule such as halving or subtracting a fixed number, which does not hold for all terms. Others may miss that the differences form a clear arithmetic series. Always compute and examine the difference sequence when the pattern is not immediately obvious. Final Answer: The next number that should appear in the series is 5.

Question 6

In number system aptitude, the difference between a positive proper fraction and its reciprocal is 9/20. What is the value of that fraction?

Accepted Answer

Introduction: This question tests your understanding of fractions and their reciprocals in basic number system aptitude. You are given the difference between a positive proper fraction and its reciprocal and asked to determine the original fraction. This involves setting up and solving a quadratic equation in fractional form. Given Data / Assumptions: The fraction is a positive proper fraction, so its value is between 0 and 1. The difference between the reciprocal of the fraction and the fraction itself is 9/20. Let the fraction be x, where 0 < x < 1. We need to find the exact value of x. Concept / Approach: A reciprocal of a non-zero number x is 1/x. Since the fraction is proper, 1/x will be greater than x. The statement “difference is 9/20” becomes an equation: 1/x − x = 9/20. We then clear denominators and solve the resulting quadratic equation to find the valid fractional solution between 0 and 1. Step-by-Step Solution: Step 1: Let the fraction be x.Step 2: Translate the statement.1/x − x = 9/20Step 3: Multiply both sides by 20x to clear denominators.20 − 20x^2 = 9xStep 4: Rearrange into standard quadratic form.20x^2 + 9x − 20 = 0Step 5: Solve using the quadratic formula.x = [−9 ± √(9^2 + 42020)] / (40)x = [−9 ± √1681] / 40 = [−9 ± 41] / 40Step 6: Compute both roots.x₁ = (32) / 40 = 4/5, x₂ = (−50) / 40 = −5/4 Verification / Alternative check: Since the fraction must be a positive proper fraction, we discard −5/4. Now check x = 4/5: reciprocal is 5/4. Difference = 5/4 − 4/5 = (25 − 16)/20 = 9/20, which matches the given condition. Why Other Options Are Wrong: 3/10 and 3/5 do not give a difference of 9/20 when compared with their reciprocals. 4/3 is not a proper fraction (it is greater than 1). 5/8 is less than 1 but its reciprocal difference does not equal 9/20. Common Pitfalls: Common errors include writing x − 1/x instead of 1/x − x, or forgetting that a proper fraction must lie between 0 and 1. Others try trial and error with options without forming the equation, which is slower and more error prone. Final Answer: The required proper fraction is 4/5.

Question 7

In a percentage problem, 60% of 3/5 of a certain number is equal to 36. What is the value of that original number?

Accepted Answer

Introduction: This question checks your understanding of percentages and fractions applied together to an unknown number. You are told that 60% of three fifths of a number equals 36, and you must find the original number. Such problems are very common in quantitative aptitude exams. Given Data / Assumptions: Let the original number be N. Three fifths of the number is (3/5) * N. Sixty percent (60%) of this quantity is equal to 36. We assume N is a real positive number. Concept / Approach: Percentages can be written as fractions or decimals. Here, 60% = 60/100 = 3/5. The statement “60% of 3/5 of N is 36” becomes a simple linear equation in N. We multiply the fractions, set equal to 36, and then solve for N using basic algebra. Step-by-Step Solution: Step 1: Write the mathematical expression.60% of (3/5 of N) = 36Step 2: Convert 60% to fraction.(60/100) * (3/5 * N) = 36Step 3: Simplify the fractions.(3/5) * (3/5) * N = 36(9/25) * N = 36Step 4: Solve for N.N = 36 * (25/9)N = (36/9) * 25 = 4 * 25 = 100 Verification / Alternative check: Check by working forward from N = 100. Three fifths of 100 is (3/5) * 100 = 60. Then 60% of 60 is (60/100) * 60 = 36. This confirms that N = 100 satisfies the condition perfectly. Why Other Options Are Wrong: 94, 115, 86 and 120 do not satisfy the relationship when you compute 3/5 of the number and then take 60% of that result. Only 100 produces 36 at the end of the two-step percentage calculation. Common Pitfalls: A common mistake is to treat the statement as “60% of the number is 36” or as “3/5 of 36 is 60%”, which reverses the operations. Another mistake is to add percentages instead of multiplying them when they are applied successively to the same number. Final Answer: The original number is 100.

Question 8

Consider the number series where each term is formed by concatenating two consecutive perfect squares: 14, 916, 2536, .... What is the next term in this series?

Accepted Answer

Introduction: This is a pattern-recognition number series question. Instead of a simple arithmetic or geometric progression, each term is constructed from two familiar mathematical objects: perfect squares. You must identify the rule and then extend the pattern to find the next term. Given Data / Assumptions: Given series: 14, 916, 2536, ... Each term appears to be more than a simple single number; it looks like a combination of two smaller numbers. We need to determine the next term after 2536. Concept / Approach: Look carefully at each term and try to break it into two meaningful parts. The first term “14” can be seen as “1” and “4”. The second term “916” can be seen as “9” and “16”. The third term “2536” can be seen as “25” and “36”. These are all perfect squares of consecutive integers: 1^2, 2^2, 3^2, 4^2, 5^2, 6^2. Step-by-Step Solution: Step 1: Break each term into two parts.14 → 1 and 4916 → 9 and 162536 → 25 and 36Step 2: Recognize the squares.1 = 1^2, 4 = 2^29 = 3^2, 16 = 4^225 = 5^2, 36 = 6^2Step 3: Spot the pattern.Each term is formed by concatenating two consecutive perfect squares.Term 1: 1^2 and 2^2 → 1, 4 → 14Term 2: 3^2 and 4^2 → 9, 16 → 916Term 3: 5^2 and 6^2 → 25, 36 → 2536Step 4: Find the next pair of squares.Next consecutive squares are 7^2 = 49 and 8^2 = 64.Concatenate 49 and 64 → 4964 Verification / Alternative check: We have used consecutive squares: (1,2), (3,4), (5,6), (7,8). At each stage, the concatenation rule is consistent. This confirms that 4964 follows logically as the next term. Why Other Options Are Wrong: 4981, 6481 and 6449 do not represent concatenations of two consecutive perfect squares. 36100 is unrelated to the observed structure. Only 4964 matches 7^2 and 8^2 placed side by side. Common Pitfalls: Students often look for differences between terms (916 − 14, 2536 − 916, etc.) and miss the structural pattern inside each term. Whenever numbers look “glued together,” always consider concatenation of known sequences like squares, cubes or primes. Final Answer: The next number in the series is 4964.

Question 9

In the six-digit number 4760, the two asterisks represent unknown non-zero digits in the hundreds and tens places respectively. If the complete number 4760 is divisible by both 3 and 11, what are these two digits (hundreds digit and tens digit) resp…

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Introduction: This divisibility question combines two classic rules: divisibility by 3 and divisibility by 11. You are given a six-digit number with two missing digits and asked to determine those digits so that the resulting number is divisible by both 3 and 11, with the additional condition that the missing digits are non-zero. Given Data / Assumptions: The number has the form 4 7 6 a b 0, where a is the hundreds digit and b is the tens digit. Digits a and b are non-zero digits from 1 to 9. The complete number 476ab0 is divisible by 3. The same number is also divisible by 11. Concept / Approach: We apply the two divisibility rules: Divisibility by 3: Sum of digits must be divisible by 3.Divisibility by 11: (Sum of digits in odd positions) − (sum of digits in even positions) must be a multiple of 11 (including 0).Positions are counted from the rightmost digit (units) as position 1. Step-by-Step Solution: Step 1: Use the rule for 3.Digits: 4, 7, 6, a, b, 0 → sum = 17 + a + b.For divisibility by 3, 17 + a + b must be a multiple of 3.Step 2: Use the rule for 11.Label positions from right: 0 (1), b (2), a (3), 6 (4), 7 (5), 4 (6).Odd positions: 0 + a + 7 = a + 7.Even positions: b + 6 + 4 = b + 10.Difference: (a + 7) − (b + 10) = a − b − 3.This difference must be 0, 11, −11, etc.Step 3: Solve for non-zero digits from the options.Check option (8, 5): a = 8, b = 5.For 11: a − b − 3 = 8 − 5 − 3 = 0 → multiple of 11 satisfied.For 3: digit sum = 4 + 7 + 6 + 8 + 5 + 0 = 30 → divisible by 3. Verification / Alternative check: Construct the number 476850. Check 476850 ÷ 3 = 158950 (integer). For 11, apply the 11-rule difference: (0 + 8 + 7) − (5 + 6 + 4) = 15 − 15 = 0, which is a multiple of 11. Hence 476850 is divisible by both 3 and 11. Why Other Options Are Wrong: Other pairs (6, 2), (8, 2), (6, 5) and (5, 2) either fail the 11-rule or do not satisfy the 3-rule when substituted as (a, b). Only (8, 5) satisfies both divisibility conditions simultaneously. Common Pitfalls: Students may mislabel positions for the 11-rule (starting from the left instead of the right), or forget that the difference can be zero. Some try to test divisibility by long division for each option, which is time consuming compared to using the rules directly. Final Answer: The hundreds and tens digits are 8 and 5 respectively.

Question 10

Among the following numbers, which one cannot possibly be the perfect square of a natural number?

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Introduction: This question tests your ability to quickly recognize whether a given integer can be a perfect square of a natural number without actually finding its square root. Using quick checks like the last digit and digital root helps you eliminate impossible candidates very efficiently. Given Data / Assumptions: You are given five positive integers: 32761, 42437, 81225, 20164 and 28561. Exactly one of these numbers cannot be the square of any natural number. You must identify which one is impossible to be a perfect square. Concept / Approach: Some simple rules for perfect squares are: 1. The last digit of a perfect square in base 10 can only be 0, 1, 4, 5, 6 or 9.2. If the last digit is 2, 3, 7 or 8, the number cannot be a perfect square.We will first apply the last-digit rule, which is extremely fast. Step-by-Step Solution: Step 1: Check the unit (ones) digit of each number.32761 → last digit 1 (possible for a square).42437 → last digit 7 (not possible for a square).81225 → last digit 5 (possible).20164 → last digit 4 (possible).28561 → last digit 1 (possible, in fact 169^2).Step 2: Apply the perfect square last-digit rule.Any number ending in 7 cannot be a perfect square in base 10.Therefore, 42437 cannot be a perfect square. Verification / Alternative check: If needed, we can cross check one of the others. For instance, 28561 is 169^2 and 81225 is 285^2, confirming that those are indeed perfect squares. 20164 and 32761 also pass more advanced tests and can correspond to actual squares, whereas 42437 fails immediately due to the last digit rule. Why Other Options Are Wrong: The other numbers all have permissible unit digits and in fact correspond to real square numbers or at least are not disqualified by the quick checks. Only 42437 ends in 7, which is impossible for the square of a natural number. Common Pitfalls: Candidates sometimes attempt to compute or approximate square roots for each option, which wastes time. Remembering the simple last-digit rule for squares lets you detect impossibilities in seconds. Final Answer: The number that cannot be a perfect square is 42437.

Question 11

Between 300 and 700 (inclusive), how many integers are divisible simultaneously by 2, 3 and 7, that is, by their common multiple?

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Introduction: This divisibility and counting question asks you to determine how many numbers in a given interval are divisible by three given integers simultaneously: 2, 3 and 7. Such problems test your understanding of least common multiple (LCM) and counting multiples in a range. Given Data / Assumptions: We are considering integers from 300 to 700 inclusive. Each required integer must be divisible by 2, 3 and 7 at the same time. Divisibility by all three means divisibility by their least common multiple. Concept / Approach: The LCM of 2, 3 and 7 captures simultaneous divisibility. Once we find this LCM, we count how many multiples of this LCM lie between 300 and 700 inclusive using a simple floor and ceiling method. Step-by-Step Solution: Step 1: Compute the LCM of 2, 3 and 7.2, 3 and 7 are pairwise co-prime.LCM = 2 * 3 * 7 = 42.Step 2: Find the first multiple of 42 not less than 300.42 * 7 = 294 (too small).42 * 8 = 336 (within the range, this is the first valid multiple).Step 3: Find the last multiple of 42 not greater than 700.42 * 16 = 672 (within the range).42 * 17 = 714 (greater than 700, so stop at 16).Step 4: Count the multiples from 8 to 16 inclusive.Number of terms = 16 − 8 + 1 = 9. Verification / Alternative check: List the actual multiples: 336, 378, 420, 462, 504, 546, 588, 630, 672. There are exactly nine such numbers, all between 300 and 700 inclusive, each divisible by 42 and therefore by 2, 3 and 7. Why Other Options Are Wrong: The other counts (7, 11, 5, 8) do not match the actual number of valid multiples. Only 9 corresponds to the correct count obtained from both formula and direct listing. Common Pitfalls: Some candidates attempt to count multiples of 2, 3 and 7 separately and then try to intersect them manually, which is inefficient and error-prone. Using the LCM avoids double-counting and simplifies the problem greatly. Final Answer: There are 9 integers between 300 and 700 that are divisible by 2, 3 and 7 together.

Question 12

The sum of two numbers is 12 and their product is 35. What is the sum of the reciprocals of these two numbers?

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Introduction: This question explores the relationship between the sum and product of two numbers and the sum of their reciprocals. It is a standard algebraic identity that is very useful in many aptitude problems involving symmetric expressions in two variables. Given Data / Assumptions: Two numbers, say a and b, have a sum a + b = 12. Their product is a * b = 35. We are asked to find 1/a + 1/b. Concept / Approach: Use the identity for the sum of reciprocals of two non-zero numbers: 1/a + 1/b = (a + b) / (a * b)Since we already know a + b and a * b, we can directly compute the sum of the reciprocals without solving for a and b individually. Step-by-Step Solution: Step 1: Recall the identity.1/a + 1/b = (a + b) / (a * b)Step 2: Substitute given values.a + b = 12a * b = 351/a + 1/b = 12 / 35Step 3: Simplify if possible.12/35 is already in simplest form (no common factor between 12 and 35). Verification / Alternative check: If you want, you can find the actual numbers. They satisfy t^2 − 12t + 35 = 0. The roots are t = 5 and t = 7 (since 5 + 7 = 12 and 5 * 7 = 35). Then 1/5 + 1/7 = (7 + 5) / 35 = 12/35, confirming the identity-based result. Why Other Options Are Wrong: 1/35 is far too small; 35/8 and 7/32 are unrelated to the ratio of sum to product here. 47/35 does not arise from any correct manipulation of the given data. Only 12/35 matches the derived expression (a + b)/(a * b). Common Pitfalls: Some students attempt to find the two numbers first every time, which is not necessary. Others mistakenly compute (a * b)/(a + b) instead of (a + b)/(a * b). Memorizing the simple identity saves time and prevents sign and inversion errors. Final Answer: The sum of the reciprocals of the two numbers is 12/35.

Question 13

A number when divided by 779 leaves a remainder of 47. If the same number is divided by 19, what remainder will it leave?

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Introduction: This question tests your understanding of modular arithmetic and remainders. You are given the remainder when a number is divided by a large divisor and asked to find the remainder when the same number is divided by a smaller related divisor. Given Data / Assumptions: Let the unknown number be N. When N is divided by 779, the remainder is 47. We need the remainder when N is divided by 19. All divisions are in the standard integer sense. Concept / Approach: We express N in terms of the first division and then reduce modulo 19. Also, we notice that 779 is a multiple of 19, which greatly simplifies the calculation because any multiple of 779 is automatically a multiple of 19. Step-by-Step Solution: Step 1: Express N using the given remainder.N = 779k + 47 for some integer k.Step 2: Note that 779 = 19 * 41.Therefore, 779k = 19 * 41k is divisible by 19.Step 3: Reduce N modulo 19.N mod 19 = (779k + 47) mod 19.Since 779k is divisible by 19, it contributes remainder 0.So N mod 19 = 47 mod 19.Step 4: Compute 47 mod 19.19 * 2 = 38, 47 − 38 = 9.So the remainder is 9. Verification / Alternative check: You can pick a random k, say k = 1: N = 779 + 47 = 826. Now 826 ÷ 19 = 43 remainder 9. For any other integer k, the term 779k always adds a multiple of 19, so the remainder will stay 9. Why Other Options Are Wrong: Remainders 6, 5, 8 and 3 are not equal to 47 mod 19. Only 9 matches the correct reduction of 47 modulo 19. Common Pitfalls: Some students try to work directly with the large number 779, forgetting to factor it or see its relation to 19. Others mistakenly divide 779 by 19 and then mishandle the remainder instead of using the expression N = 779k + 47. Final Answer: When the number is divided by 19, the remainder is 9.

Question 14

The seven-digit number 517324 is completely divisible by 3. What is the smallest whole number that can replace the asterisk () to make the number divisible by 3?

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Introduction: This is a basic divisibility test problem with a missing digit. You are given a large number with one unknown digit and asked to choose that digit so the entire number becomes divisible by 3. Such problems train you to apply divisibility rules quickly without long division. Given Data / Assumptions: The number has the form 517*324, where * represents a single digit from 0 to 9. The complete number must be divisible by 3. We are specifically asked for the smallest whole number that can replace *. Concept / Approach: The rule for divisibility by 3 states that a number is divisible by 3 if and only if the sum of its digits is divisible by 3. We therefore sum the known digits, include the unknown digit x in the sum, and choose x so that the total is a multiple of 3. Step-by-Step Solution: Step 1: Let the missing digit be x.Digits are 5, 1, 7, x, 3, 2, 4.Step 2: Sum the known digits.5 + 1 + 7 + 3 + 2 + 4 = 22.Step 3: Include x in the total sum.Total sum = 22 + x.Step 4: Apply the divisibility by 3 rule.We need 22 + x to be a multiple of 3.The next multiples of 3 after 22 are 24, 27, 30, etc.So 22 + x = 24, 27, 30, ...This gives x = 2, 5, 8, ...Step 5: Choose the smallest whole number.The smallest x in {2, 5, 8, ...} is 2. Verification / Alternative check: Substitute x = 2 to get the number 5172324. Sum of digits = 5 + 1 + 7 + 2 + 3 + 2 + 4 = 24, which is divisible by 3. Therefore, the full number is divisible by 3 as required. Why Other Options Are Wrong: 0 and 1 make the digit sum 22 and 23 respectively, which are not multiples of 3. 5 and 8 also give valid divisibility by 3, but they are larger than 2. The question explicitly asks for the smallest such digit, so 2 is the correct choice. Common Pitfalls: Some learners forget the “smallest” condition and might choose any digit that works, such as 5 or 8. Others try full long division rather than using the far quicker digit-sum rule. Final Answer: The smallest whole number that can replace * is 2.

Question 15

At a dinner party, every two guests share one bowl of rice between them, every three guests share one bowl of dal between them, and every four guests share one bowl of meat between them. If the total number of dishes served (rice bowls + dal bowls +…

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Introduction: This is a word problem involving ratios and inverse relationships. Instead of one dish per guest, dishes are shared among groups of guests, and you must deduce the number of guests from the total number of shared dishes. Such problems are common in aptitude exams to test algebraic modeling skills. Given Data / Assumptions: Total number of guests = n (unknown). Every 2 guests share 1 bowl of rice → number of rice bowls = n/2. Every 3 guests share 1 bowl of dal → number of dal bowls = n/3. Every 4 guests share 1 bowl of meat → number of meat bowls = n/4. Total number of dishes (bowls) = 65. We assume n is a positive integer and that divisions like n/2, n/3, n/4 give integer counts of bowls, meaning n is a common multiple where needed. Concept / Approach: The total dishes are the sum of the rice, dal and meat bowls. We translate the sharing statements into an equation in n and then solve for n. This is effectively solving a linear equation with fractional coefficients. Step-by-Step Solution: Step 1: Express dish counts in terms of n.Rice bowls = n/2.Dal bowls = n/3.Meat bowls = n/4.Total dishes = n/2 + n/3 + n/4 = 65.Step 2: Use a common denominator (12).n/2 = 6n/12, n/3 = 4n/12, n/4 = 3n/12.So total dishes = (6n + 4n + 3n) / 12 = 13n / 12.Step 3: Set equal to 65 and solve.13n / 12 = 65.Multiply both sides by 12: 13n = 65 * 12 = 780.n = 780 / 13 = 60. Verification / Alternative check: With 60 guests: rice bowls = 60/2 = 30; dal bowls = 60/3 = 20; meat bowls = 60/4 = 15. Total dishes = 30 + 20 + 15 = 65, matches the given condition exactly. Why Other Options Are Wrong: 74, 82, 58 and 48 do not satisfy the equation n/2 + n/3 + n/4 = 65 when substituted. For example, with n = 48, dishes would be 24 + 16 + 12 = 52, not 65. Common Pitfalls: Some students add the denominators incorrectly or forget to find a common denominator before combining fractions. Others misinterpret the wording and think each pair gets its own dish without using fractions at all. Proper algebraic translation is the key step. Final Answer: There were 60 guests at the party.

Question 16

In the mixed number series 3, 4, 5, 5, 12, 13, 7, 24, 25, 8, 15, ..., what number should come next to correctly continue the pattern?

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Introduction: This question presents a number series that is not a simple arithmetic or geometric progression. Instead, it is constructed from known Pythagorean triples. Recognizing this structure is essential for finding the next term in the sequence. Given Data / Assumptions: The series is: 3, 4, 5, 5, 12, 13, 7, 24, 25, 8, 15, ... We need to identify the pattern and determine the next number. Concept / Approach: Group the numbers to see if they form meaningful sets. Notice that they can be grouped as triples: (3, 4, 5), (5, 12, 13), (7, 24, 25), (8, 15, ...)Each of the first three groups is a well-known Pythagorean triple where a^2 + b^2 = c^2. Step-by-Step Solution: Step 1: Identify the triples.First triple: 3, 4, 5 → 3^2 + 4^2 = 9 + 16 = 25 = 5^2.Second triple: 5, 12, 13 → 5^2 + 12^2 = 25 + 144 = 169 = 13^2.Third triple: 7, 24, 25 → 7^2 + 24^2 = 49 + 576 = 625 = 25^2.Step 2: Look at the next group.Numbers so far: 8, 15, ...We recognise another Pythagorean triple: 8, 15, 17.8^2 + 15^2 = 64 + 225 = 289 = 17^2.Step 3: Conclude the next term.The missing number after 8, 15 should be 17. Verification / Alternative check: The pattern consistently follows primitive Pythagorean triples: (3, 4, 5), (5, 12, 13), (7, 24, 25), and now (8, 15, 17). There is no break or inconsistency in this structure, so the triple (8, 15, 17) naturally completes the next group. Why Other Options Are Wrong: 26, 16, 18 and 20 do not complete a standard Pythagorean triple with 8 and 15. Only 17 satisfies 8^2 + 15^2 = 17^2, so only 17 fits the observable pattern of the series. Common Pitfalls: Some learners try to look at differences between terms or use arithmetic progressions, which do not work here. Recognizing well-known triples is a major advantage in such pattern questions. Final Answer: The next number in the series is 17.

Question 17

The difference between two numbers is 11 and one fifth of their sum is 9. What are the two numbers?

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Introduction: This is a classic linear equations problem involving two numbers. You are given their difference and a relation involving their sum. Solving such problems efficiently is crucial for many aptitude and reasoning exams. Given Data / Assumptions: Let the two numbers be x and y, with x > y. Their difference is x − y = 11. One fifth of their sum is 9, so (x + y)/5 = 9. We must find the values of x and y. Concept / Approach: We are given two linear equations in two unknowns. The standard method is to solve this system by addition or substitution. First, we determine the sum from the second condition, then use the sum and difference to solve for each number. Step-by-Step Solution: Step 1: Use the sum condition.(x + y)/5 = 9 ⇒ x + y = 45.Step 2: Use the difference condition.x − y = 11.Step 3: Add the two equations.(x + y) + (x − y) = 45 + 11.2x = 56 ⇒ x = 28.Step 4: Substitute x back into one equation to find y.x + y = 45 ⇒ 28 + y = 45 ⇒ y = 17. Verification / Alternative check: Check both conditions: Difference = 28 − 17 = 11 (correct). Sum = 28 + 17 = 45; one fifth of 45 is 45/5 = 9 (also correct). Thus both original statements are satisfied by the pair (28, 17). Why Other Options Are Wrong: Each other option pair either does not have a difference of 11, or if it does, its sum does not give (sum)/5 = 9. For example, 31 and 20 have sum 51, and one fifth of 51 is 10.2, not 9. Common Pitfalls: Students sometimes mis-handle the fraction and set the sum equal to 9 instead of 45. Others may mix up sum and difference while solving. Writing equations clearly and proceeding stepwise prevents these errors. Final Answer: The two numbers are 28 and 17.

Question 18

In a class, approximately 4 out of every 5 students are left-handed. If there are exactly 40 students in the class, how many of them would be left-handed according to this ratio?

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Introduction: This is a simple ratio and proportion question. You are told that 4 out of every 5 students are left-handed, and you must use this information to find the exact number of left-handed students in a class of 40. Given Data / Assumptions: Ratio of left-handed students to total students is 4 : 5. Total number of students in the class = 40. We assume the ratio applies exactly to this class. We need to find the number of left-handed students. Concept / Approach: The statement “4 out of 5” corresponds to the fraction 4/5 of the total. Thus, the number of left-handed students is (4/5) of 40. This is a direct application of ratios or percentages (since 4/5 is 80%). Step-by-Step Solution: Step 1: Write the fraction representing left-handed students.Left-handed fraction = 4/5.Step 2: Apply it to the total number of students.Number of left-handed students = (4/5) * 40.Step 3: Compute the product.(4/5) * 40 = 4 * 8 = 32.So, there are 32 left-handed students. Verification / Alternative check: You can also think in terms of groups of 5 students. In each group of 5, 4 are left-handed. 40 students can be divided into 8 groups of 5. Each group contributes 4 left-handed students, so total left-handed = 8 * 4 = 32. Why Other Options Are Wrong: 40 would imply everyone is left-handed, which does not match 4 out of 5. 45 exceeds the total number of students. 30 or 28 would correspond to fractions smaller than 4/5 of 40. Only 32 exactly reflects the 4/5 proportion of 40 students. Common Pitfalls: Some students misread “4 out of 5” as 4 + 5 instead of a ratio, or incorrectly compute 40 * (5/4). Always remember that “4 out of 5” means 4/5 of the total quantity. Final Answer: According to the given ratio, 32 students would be left-handed.

Question 19

What number should replace the question mark in the series 3240, 540, 108, 27, ?, 4.5 so that the pattern is maintained?

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Introduction: This question is a number series puzzle where each term is obtained from the previous term by applying a specific operation. Recognizing the pattern, particularly when it involves division by decreasing integers, is key to finding the missing term. Given Data / Assumptions: The given series is: 3240, 540, 108, 27, ?, 4.5. Exactly one term is missing, marked by a question mark. We must deduce the rule connecting consecutive terms and then apply it to find the missing number. Concept / Approach: We compare each term with the previous one to see whether multiplication, division or a combination is being used. Here, the pattern turns out to be division by successively smaller integers: 6, 5, 4, 3, and 2. Step-by-Step Solution: Step 1: Compare 3240 and 540.3240 ÷ 540 = 6, so 540 = 3240 ÷ 6.Step 2: Compare 540 and 108.540 ÷ 108 = 5, so 108 = 540 ÷ 5.Step 3: Compare 108 and 27.108 ÷ 27 = 4, so 27 = 108 ÷ 4.Step 4: Continue the same logic.To go from 27 to the next term, divide by 3.Next term = 27 ÷ 3 = 9.Step 5: Confirm the last step.Then 9 ÷ 4.5 = 2, so 4.5 = 9 ÷ 2, consistent with dividing by 2 at the final step. Verification / Alternative check: The divisor sequence is 6, 5, 4, 3, 2, which is a simple descending pattern. Each division yields the next term correctly, and the missing value 9 fits perfectly into this scheme, maintaining the structure of the series. Why Other Options Are Wrong: Values like 7, 8, 10 or 6 would break the clean pattern of division by 6, 5, 4, 3 and 2. Only 9 ensures that the ratio between consecutive terms remains consistent with this rule. Common Pitfalls: Some candidates may look for addition or subtraction patterns instead of multiplicative ones. Others stop after checking only one or two steps and fail to notice the full descending divisor sequence. Final Answer: The missing number in the series is 9.

Question 20

The sum of three consecutive odd integers is 93. What is the value of the middle odd number?

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Introduction: This question is about consecutive odd numbers and simple algebra. When you know the sum of three consecutive odd integers, you can find the middle one quickly using a straightforward equation. This is a very common type of question in basic aptitude tests. Given Data / Assumptions: We have three consecutive odd integers. Their sum is 93. We must find the middle number. Concept / Approach: Consecutive odd numbers can be represented using a single variable. If the middle odd number is n, then the three consecutive odd numbers are n − 2, n and n + 2. Their sum becomes a simple linear expression in n, which we can solve easily. Step-by-Step Solution: Step 1: Represent the three consecutive odd integers.Let the middle number be n.Then the three consecutive odd numbers are n − 2, n, n + 2.Step 2: Write the sum expression.(n − 2) + n + (n + 2) = 93.Step 3: Simplify the left-hand side.(n − 2) + n + (n + 2) = 3n.So 3n = 93.Step 4: Solve for n.n = 93 / 3 = 31. Verification / Alternative check: Check the three numbers: n − 2 = 29, n = 31, n + 2 = 33. Sum = 29 + 31 + 33 = 93, which agrees with the given information. Thus the middle number is indeed 31. Why Other Options Are Wrong: 33, 29 and 27 do not produce a sum of 93 when used as the middle term of three consecutive odd integers. For instance, if 33 were the middle number, the numbers would be 31, 33 and 35, whose sum is 99, not 93. Common Pitfalls: Some students mistakenly take the numbers as n, n + 1, n + 2 instead of odd-only spacing, or try guessed triples rather than setting up the simple equation. Remember that consecutive odd numbers differ by 2, not 1. Final Answer: The middle odd number is 31.

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