Question 1

The difference of the squares of two numbers is 180, and the square of the smaller number is 8 times the larger number. What are the two numbers?

Accepted Answer

Introduction / Context: This is a number theory and algebra question involving two conditions: a difference of squares and a relationship between one square and the other number. Such problems train you to set up equations based on words and then solve them systematically. It also shows that solutions can involve negative numbers even when squares are positive. Given Data / Assumptions: Let the two numbers be x and y. The difference of their squares is 180, that is y^2 - x^2 = 180 (assuming y is the larger number in magnitude of square). The square of the smaller number is 8 times the larger number, so x^2 = 8y if x is the smaller number. The numbers can be positive or negative. We must find a pair (x, y) that satisfies both conditions. Concept / Approach: We convert the statements into algebraic equations and solve them. Because squares remove signs, the relationship between x^2 and y can allow negative solutions as well. The first equation involves difference of squares, and the second links x^2 directly to y. Substituting y from one equation into the other leads to a single equation in x. We then factor this equation to find possible values of x, and for each x we compute y and check both conditions. Step-by-Step Solution: Step 1: Let the two numbers be x and y, with x taken as the smaller number.Step 2: From the statement the difference of their squares is 180, we write y^2 - x^2 = 180.Step 3: From the statement the square of the smaller number is 8 times the larger number, we write x^2 = 8y.Step 4: From x^2 = 8y, express y in terms of x: y = x^2 / 8.Step 5: Substitute y = x^2 / 8 into the first equation: (x^2 / 8)^2 - x^2 = 180.Step 6: Simplify: x^4 / 64 - x^2 = 180.Step 7: Multiply through by 64 to clear the denominator: x^4 - 64x^2 - 11520 = 0.Step 8: Treat this as a quadratic in x^2. Let t = x^2, then t^2 - 64t - 11520 = 0.Step 9: Factorise or solve this quadratic. The factorisation gives (t - 144)(t + 80) = 0.Step 10: So t = 144 or t = -80. Since t = x^2 cannot be negative, we discard t = -80 and keep x^2 = 144.Step 11: Thus x = 12 or x = -12.Step 12: For each x, compute y = x^2 / 8 = 144 / 8 = 18.Step 13: So possible pairs are (12, 18) or (-12, 18). The smaller number, by value, can be -12 if we allow negatives. Verification / Alternative check: Check the pair (-12, 18). First, y^2 - x^2 = 18^2 - (-12)^2 = 324 - 144 = 180, satisfying the difference of squares condition. Second, the square of the smaller number is (-12)^2 = 144, and 8 times the larger number is 8 * 18 = 144, so the second condition is also met. Therefore, (-12, 18) is a correct pair. The pair (12, 18) does not match the wording if we insist that the smaller number is x, because 12 is smaller than 18 and then x^2 = 144 is not 8 times 18 unless we reassign roles. With the given options, the pair -12 and 18 is the correct choice. Why Other Options Are Wrong: Option 10 and 12: Their squares are 100 and 144. The difference is 44, not 180, and 10^2 is not 8 times 12. Option 10 and 18: Squares are 100 and 324. The difference is 224, not 180. Option 12 and -18: Squares are 144 and 324. The difference is still 180, but the statement about which number is smaller and the relation x^2 = 8y do not line up when matched with the wording. Option 12 and 18: Not listed as correct in the given options and does not align with the specific condition when assigning smaller and larger as described in the options. Common Pitfalls: One common mistake is to treat only positive numbers as possible answers and therefore ignore negative solutions, even when they satisfy the conditions. Another error is to misinterpret the phrase the square of the smaller number is 8 times the larger number and write y^2 = 8x instead of x^2 = 8y. Careful reading and consistent assignment of variables to larger and smaller numbers is essential. Final Answer: The two numbers that satisfy both conditions are -12 and 18.

Question 2

In the number series 9, 11, 16, 26, ?, 69, what number should replace the question mark to continue the pattern correctly?

Accepted Answer

Introduction / Context: This problem is a typical missing-term number series question, often asked in aptitude tests to assess pattern recognition and logical reasoning. The series is 9, 11, 16, 26, ?, 69, and we must identify the rule that governs how each term is formed from previous terms. Such questions encourage careful analysis of differences and second-level patterns in those differences. Given Data / Assumptions: The given series is: 9, 11, 16, 26, ?, 69. We are to find the single missing number in place of the question mark. All terms are integers. The pattern is assumed to be consistent throughout the series. Concept / Approach: The usual method for such series is to compute first differences (subtract each term from the next) and look for a pattern. If the first differences themselves follow some rule, we use that rule to find the missing term. Sometimes second differences or other operations are required. Here, we start with first differences and look for a simple progression or pattern in them. Step-by-Step Solution: Step 1: Write down the series: 9, 11, 16, 26, ?, 69.Step 2: Compute the differences between consecutive terms where possible.Step 3: 11 - 9 = 2, 16 - 11 = 5, 26 - 16 = 10.Step 4: Let the missing term be x. Then the remaining difference is x - 26 and 69 - x.Step 5: Looking at 2, 5, and 10, we try to detect a pattern. Notice that 2, 5, 10 could be followed by 17 and 26, because the jumps between them are 3, 5, 7, 9, which form an increasing sequence of odd numbers.Step 6: Specifically, start with 2, then add 3 to get 5, add 5 to get 10, add 7 to get 17, and add 9 to get 26.Step 7: So the sequence of differences should be: 2, 5, 10, 17, 26.Step 8: We already have 2, 5, and 10. Now we assign x - 26 = 17 and 69 - x = 26.Step 9: From x - 26 = 17, we get x = 43.Step 10: Check the last difference: 69 - 43 = 26, which matches the expected pattern. Verification / Alternative check: Now that we have x = 43, write the full series: 9, 11, 16, 26, 43, 69. The differences are 2, 5, 10, 17, 26. Next, calculate the differences between these differences: 5 - 2 = 3, 10 - 5 = 5, 17 - 10 = 7, 26 - 17 = 9. These second differences form the sequence 3, 5, 7, 9, which are consecutive odd numbers. This confirms that the pattern is consistent and that x = 43 is the correct missing term. Why Other Options Are Wrong: Option 31: Produces differences 2, 5, 10, 5, 38 which show no consistent pattern. Option 38: Gives differences 2, 5, 10, 12, 31; again, no regular structure in first or second differences. Option 45: Leads to differences 2, 5, 10, 19, 24; the second differences do not form a simple sequence. Option 41: Yields differences 2, 5, 10, 15, 28, which still breaks the neat pattern of second differences being consecutive odd numbers. Common Pitfalls: Students may stop after noticing a rough increase in differences and guess a value without checking higher-level patterns. Another common mistake is to try unrelated operations (such as multiplying or dividing terms) when a simpler difference-based pattern exists. It is usually best to start with first and second differences and only switch approaches if no reasonable pattern emerges. Final Answer: The missing number in the series is 43.

Question 3

In the number series 305, 338, 404, 503, 635, ( ? ), what number should come in place of the question mark to continue the pattern correctly?

Accepted Answer

Introduction / Context: This question presents a number series and asks for the next term. Recognising the pattern behind the series is a standard requirement in many aptitude tests. The given series shows steadily increasing numbers, so examining differences between consecutive terms is a natural first step. Often, these differences themselves form a simple pattern, such as an arithmetic progression. Given Data / Assumptions: The given sequence is: 305, 338, 404, 503, 635, ?. We are required to find the term that should replace the question mark. All numbers are positive integers. The rule governing the sequence is assumed to be consistent throughout. Concept / Approach: The standard method is to compute the first differences between successive terms and then look for a pattern such as a constant increment, a second-level arithmetic progression, or a multiplicative pattern. If the first differences themselves form an arithmetic progression, we can continue that progression to find the missing term. This approach keeps the reasoning systematic and avoids random guessing. Step-by-Step Solution: Step 1: Write the series: 305, 338, 404, 503, 635, ?.Step 2: Compute the differences between consecutive terms.Step 3: 338 - 305 = 33.Step 4: 404 - 338 = 66.Step 5: 503 - 404 = 99.Step 6: 635 - 503 = 132.Step 7: So the sequence of first differences is 33, 66, 99, 132.Step 8: Now, examine these differences. Notice that each difference increases by 33: 66 - 33 = 33, 99 - 66 = 33, 132 - 99 = 33.Step 9: This suggests the differences themselves form an arithmetic progression with first term 33 and common difference 33.Step 10: Therefore, the next difference should be 132 + 33 = 165.Step 11: The missing term is the previous term plus this new difference: ? = 635 + 165 = 800. Verification / Alternative check: Write out the extended series including the found term: 305, 338, 404, 503, 635, 800. The differences are now 33, 66, 99, 132, 165. The differences between these differences are 33, 33, 33, 33, showing a constant increase, which confirms that the pattern is consistent and our value 800 fits the rule correctly. Why Other Options Are Wrong: Option 820: Would lead to a last difference of 185, which breaks the constant increment of 33 between differences. Option 880: Gives a last difference of 245, which does not fit the established pattern. Option 890: Produces a difference of 255, again inconsistent with the step of adding 33 to each previous difference. Option 760: Would yield a difference of only 125, interrupting the clear progression of 33, 66, 99, 132, 165. Common Pitfalls: Some test-takers try to relate terms multiplicatively or look for more complicated patterns before checking simple differences. Others may miscalculate one of the differences, causing them to miss the simple arithmetic progression pattern. It is important to carefully compute and then inspect first differences and, if needed, differences of those differences. Final Answer: The number that should replace the question mark in the series is 800.

Question 4

Consider the pair series (2, 3), (3, 5), (5, 7), (7, 11), (11, 13), ___. What is the next ordered pair of numbers that should follow this pattern?

Accepted Answer

Introduction / Context: This question gives a sequence of ordered pairs and asks for the next pair. It is a pattern-recognition problem based on prime numbers. Such questions appear in reasoning and aptitude tests to check familiarity with basic number properties such as primes and the ability to notice how they are arranged in a given pattern. Given Data / Assumptions: The given pairs are (2, 3), (3, 5), (5, 7), (7, 11), (11, 13). We must find the next pair in the same format, (a, b). All entries are positive integers. The pattern is consistent and is related to prime numbers. Concept / Approach: We observe that each pair seems to consist of two prime numbers. Moreover, from one pair to the next, the second element of the previous pair often becomes the first element of the next pair. This creates a kind of overlapping sequence of consecutive primes. So we list consecutive prime numbers and see how they form the pairs. Step-by-Step Solution: Step 1: List the first few prime numbers in order: 2, 3, 5, 7, 11, 13, 17, 19, and so on.Step 2: Note the given pairs: (2, 3), (3, 5), (5, 7), (7, 11), (11, 13).Step 3: In each pair, both numbers are primes. Also, except for the first pair, the first number of each pair is the second number of the previous pair.Step 4: Observe the sequence of primes in the first and second positions: 2, 3, 5, 7, 11, 13.Step 5: Continuing with the prime sequence after 13, the next prime is 17.Step 6: Following the pattern, the next pair should start with 13 (the second element of the last given pair) and then use the next prime, which is 17.Step 7: Therefore, the next pair is (13, 17). Verification / Alternative check: Check that the pattern is consistent: pair 1 uses primes 2 and 3, pair 2 uses primes 3 and 5, pair 3 uses primes 5 and 7, pair 4 uses primes 7 and 11, and pair 5 uses primes 11 and 13. The first element of each new pair is the second element of the previous pair, and the second element is the next prime in the sequence. Following this rule naturally produces (13, 17) as the next pair. Why Other Options Are Wrong: Option 13, 15: The number 15 is not a prime number, so it breaks the pattern of using only primes. Option 15, 17: Again, 15 is not prime, so this pair does not follow the established structure. Option 13, 19: This skips the prime 17, so it does not use the next prime after 13. Option 17, 19: This does not continue from the second element of the previous pair, which should be 13, so it breaks the link between consecutive pairs. Common Pitfalls: Some learners may focus only on differences between the numbers, such as 1, 2, 2, 4, 2, and not recognise the prime number pattern. Others might think the second number always increases by 2, which is only true among some primes but not when the gap becomes larger, as between 7 and 11. Recognising primes and their natural order is crucial here. Final Answer: The next ordered pair in the series is 13, 17.

Question 5

All rational numbers belong to which broader set of numbers?

Accepted Answer

Introduction / Context: This is a conceptual question from basic number systems. It asks which larger set all rational numbers belong to. Understanding the hierarchy of number sets is fundamental in mathematics because it clarifies how different types of numbers relate to each other. Rational numbers, integers, whole numbers, irrational numbers, and real numbers are all part of a nested structure that appears frequently in school mathematics and aptitude exams. Given Data / Assumptions: We are considering rational numbers, that is numbers that can be written as p / q, where p and q are integers and q is not zero. Options include Integers, Whole numbers, Irrational numbers, Real numbers, and Prime numbers. We need to identify the most appropriate larger set that contains all rational numbers. No additional restrictions or special conditions are given. Concept / Approach: To answer this, we recall the standard classification of numbers. Whole numbers are non-negative integers (0, 1, 2, 3, ...). Integers include negative numbers, zero, and positive numbers (..., -2, -1, 0, 1, 2, ...). Rational numbers include all integers and fractions that can be expressed as p / q with q not equal to zero. Irrational numbers cannot be expressed as such fractions (for example, square root of 2, pi, and so on). The real numbers are all numbers that can be placed on the number line, including both rational and irrational numbers. Therefore, every rational number is also a real number, but not every real number is rational. Step-by-Step Solution: Step 1: Recall the definition of a rational number: any number that can be written as p / q where p and q are integers and q is not zero.Step 2: Consider whether all rational numbers are integers. This is not true, because fractions like 1 / 2 and 3 / 4 are rational but not integers.Step 3: Consider whether all rational numbers are whole numbers. This is also not true for the same reason and also because negative rationals exist, whereas whole numbers are non-negative.Step 4: Consider whether rational numbers are irrational numbers. This is false by definition. A number cannot be both rational and irrational; these sets are disjoint subsets of the real numbers.Step 5: Recall that real numbers consist of both rational and irrational numbers, essentially covering all possible decimal expansions that represent magnitudes on the number line.Step 6: Since every rational number can be plotted on the real number line, every rational number is a real number. Thus, the correct umbrella set that contains all rational numbers is the set of real numbers.Step 7: Prime numbers are only a subset of positive integers greater than 1 and do not cover general rational numbers such as fractions, so they cannot be the correct choice either. Verification / Alternative check: Take specific examples: 3, -5, 1 / 2, and -7 / 4 are all rational because they can be written as p / q with integer p and q not equal to zero. All of these can be located on the number line. Similarly, a decimal like 0.25 is rational, because it equals 1 / 4, and it also sits on the number line between 0 and 1. These examples show that rational numbers fit comfortably inside the set of real numbers. By contrast, irrational numbers like square root of 2 cannot be written as simple fractions, but they also lie on the number line. Together, rational and irrational numbers make up the set of real numbers. Why Other Options Are Wrong: Option Integers: While every integer is a rational number, not every rational number is an integer, because rationals also include fractions. Option Whole numbers: Whole numbers exclude negative numbers and fractions, so they do not include all rational numbers. Option Irrational numbers: These are specifically numbers that cannot be written as p / q, so they exclude all rational numbers. Option Prime numbers: These are a small subset of positive integers greater than 1 and certainly do not include general fractions. Common Pitfalls: Learners sometimes confuse the direction of inclusion and think that all real numbers are rational, which is wrong; only some real numbers are rational. Another misunderstanding is to assume that because integers are simpler or more familiar, rational numbers must all be integers, but this ignores fractions. Remembering the hierarchy Whole numbers ⊂ Integers ⊂ Rational numbers ⊂ Real numbers helps to avoid these mistakes, with irrational numbers forming the complementary part of real numbers that are not rational. Final Answer: All rational numbers are Real numbers.

Question 6

What is the least number that must be multiplied with 69120 in order to make the product a perfect cube?

Accepted Answer

Introduction / Context: This question is another example of working with perfect cubes using prime factorization. Instead of dividing, we now need to multiply the given number by the smallest possible integer so that the result becomes a perfect cube. This is a common type of problem in number systems that reinforces the idea that the exponents of prime factors in a perfect cube must all be multiples of 3. Given Data / Assumptions: The given number is 69120. We must find the smallest positive integer k such that 69120 * k is a perfect cube. All calculations are done with exact prime factorizations. A perfect cube has prime exponents that are multiples of 3. Concept / Approach: The strategy is to write 69120 as a product of prime powers. We then examine the exponents of the primes and determine how many more of each prime are needed to make all the exponents multiples of 3. The smallest multiplier k will be formed by the extra prime factors required to adjust these exponents. This method avoids trial and error and leads directly to the minimal solution. Step-by-Step Solution: Step 1: Factorize 69120 into prime factors.Step 2: The prime factorization of 69120 is 2^9 * 3^3 * 5^1.Step 3: For a number to be a perfect cube, the exponent of each prime in its factorization must be a multiple of 3.Step 4: Check the exponent of each prime in 69120:- For prime 2, the exponent is 9, which is already a multiple of 3 (9 = 3 * 3).- For prime 3, the exponent is 3, also a multiple of 3.- For prime 5, the exponent is 1, which is not a multiple of 3.Step 5: We need to increase the exponent of 5 from 1 to the next multiple of 3, which is 3.Step 6: To raise the exponent from 1 to 3, we need 2 more factors of 5, that is multiply by 5^2 = 25.Step 7: Thus, k must at least contain 25, and no other extra prime factors are needed because the exponents of 2 and 3 are already multiples of 3.Step 8: Therefore, the least number that must be multiplied to 69120 is 25. Verification / Alternative check: Compute the product 69120 * 25. In prime factors, this is (2^9 * 3^3 * 5^1) * 5^2 = 2^9 * 3^3 * 5^3. Now each exponent is a multiple of 3: 9, 3, and 3. So 69120 * 25 is a perfect cube. Specifically, we can write the result as (2^3)^3 * (3^1)^3 * (5^1)^3, which equals (8 * 3 * 5)^3 = 120^3, confirming that the product is indeed a perfect cube. Why Other Options Are Wrong: Option 10: This equals 2 * 5; multiplying by this gives 2^10 * 3^3 * 5^2, where the exponents 10 and 2 are not multiples of 3. Option 50: This is 2 * 5^2; multiplying gives 2^10 * 3^3 * 5^3, where the exponent of 2 is still not a multiple of 3. Option 5: This only increases the exponent of 5 from 1 to 2, which is still not a multiple of 3. Option 15: This is 3 * 5; multiplying gives 2^9 * 3^4 * 5^2, where both 4 and 2 are not multiples of 3. Common Pitfalls: A frequent mistake is to pick a number that makes one prime exponent a multiple of 3 but disrupts another exponent. Some students also forget to fully factor the given number, leading to incorrect exponent counts. Others try random multipliers instead of systematically adjusting exponents. The best practice is always to rely on prime factorization and exponent arithmetic when dealing with perfect powers. Final Answer: The least number that must be multiplied with 69120 to make a perfect cube is 25.

Question 7

What is the least positive number that, when divided by both 63 and 105, leaves a remainder of 4 in each case?

Accepted Answer

Introduction / Context: This problem is a standard application of greatest common divisors and least common multiples in modular arithmetic. It asks for the smallest number that leaves the same remainder when divided by two different divisors. Such questions are common in number system topics of aptitude exams and help learners understand congruences and how to use least common multiples to synchronise remainder conditions. Given Data / Assumptions: We are looking for a number N. When N is divided by 63, it leaves a remainder of 4. When N is divided by 105, it leaves the same remainder of 4. N must be the least positive integer satisfying both conditions. Concept / Approach: If a number N leaves the same remainder r when divided by two numbers, say m and n, then N - r is divisible by both m and n. In this problem, N leaves remainder 4 when divided by 63 and 105, so N - 4 must be divisible by both 63 and 105. Therefore, N - 4 is a common multiple of 63 and 105. The smallest positive N will correspond to N - 4 being the least common multiple of 63 and 105, and then we add the remainder back. Step-by-Step Solution: Step 1: Let the required number be N.Step 2: According to the problem, N ≡ 4 (mod 63) and N ≡ 4 (mod 105).Step 3: This means that N - 4 is divisible by both 63 and 105.Step 4: Therefore, N - 4 must be a common multiple of 63 and 105.Step 5: To get the smallest such N, find the least common multiple (LCM) of 63 and 105.Step 6: Factorise: 63 = 3^2 * 7 and 105 = 3 * 5 * 7.Step 7: The LCM takes the highest powers of all primes present: LCM = 3^2 * 5 * 7 = 9 * 5 * 7 = 315.Step 8: So N - 4 = 315 for the smallest positive solution.Step 9: Therefore, N = 315 + 4 = 319. Verification / Alternative check: Check divisibility conditions for N = 319. When 319 is divided by 63, we compute 63 * 5 = 315 and the remainder is 319 - 315 = 4, which matches the requirement. When 319 is divided by 105, 105 * 3 = 315 and again the remainder is 319 - 315 = 4. Since both conditions are satisfied and 319 is based on the least common multiple, there is no smaller positive N that works for both divisors simultaneously. Why Other Options Are Wrong: Option 634: 634 - 4 = 630, which is divisible by 63 and 105, but 630 is actually 2 * 315, so 634 is a larger valid solution, not the least one. Option 256: 256 - 4 = 252, which is divisible by 63 but not by 105. Option 214: 214 - 4 = 210, which is divisible by 105 but not by 63. Option 172: 172 - 4 = 168, which is divisible by 63 but not by 105. Common Pitfalls: A common mistake is to set up separate equations for N and try to solve them directly rather than shifting by the remainder. Some may also confuse greatest common divisor (GCD) with least common multiple (LCM) and attempt to use the GCD instead, which does not synchronise the remainder conditions properly. Remember that when the remainders are the same, subtracting the remainder reduces the problem to finding a common multiple. Final Answer: The least number that leaves a remainder of 4 when divided by both 63 and 105 is 319.

Question 8

Find two numbers whose sum is 29 and whose product is 100.

Accepted Answer

Introduction / Context: This is a classic quadratic-based word problem. We are told the sum and product of two numbers and asked to find the numbers themselves. Such questions map naturally to quadratic equations, since if the numbers are x and y, then x + y and xy correspond to the coefficients and constant term of a quadratic with roots x and y. Recognising this structure allows an efficient algebraic solution. Given Data / Assumptions: There are two numbers, call them x and y. Their sum is x + y = 29. Their product is x * y = 100. We must find the actual values of x and y. Numbers are assumed to be real, and the options suggest positive integer solutions. Concept / Approach: If x and y are the roots of a quadratic equation, then the equation can be written as t^2 - (sum of roots) * t + (product of roots) = 0. Here, the sum of roots is 29 and the product is 100, so we construct the quadratic t^2 - 29t + 100 = 0. Solving this quadratic will directly give the two numbers. This approach is systematic and avoids guesswork, though we can later cross-check with the options by simple substitution. Step-by-Step Solution: Step 1: Let the two numbers be x and y.Step 2: We know x + y = 29 and x * y = 100.Step 3: Form the quadratic whose roots are x and y: t^2 - (x + y)t + x * y = 0.Step 4: Substitute the known sum and product: t^2 - 29t + 100 = 0.Step 5: Solve this quadratic using the discriminant method. The discriminant D is 29^2 - 4 * 1 * 100 = 841 - 400 = 441.Step 6: The square root of D is sqrt(441) = 21.Step 7: The roots are t = [29 ± 21] / 2.Step 8: Compute the two roots: t1 = (29 + 21) / 2 = 50 / 2 = 25 and t2 = (29 - 21) / 2 = 8 / 2 = 4.Step 9: Therefore, the two numbers are 25 and 4. Verification / Alternative check: Check the sum: 25 + 4 = 29, which matches the given sum. Check the product: 25 * 4 = 100, which matches the given product. Both conditions are satisfied, so the pair (25, 4) is correct. Also, the order does not matter since (4, 25) is the same pair of numbers in a different order. Why Other Options Are Wrong: Option 20 and 5: Sum is 25 and product is 100, so the sum condition fails. Option 20 and 9: Sum is 29 but product is 180, not 100. Option 10 and 10: Sum is 20 and product is 100, failing the sum condition. Option 19 and 10: Sum is 29, but product is 190, not 100. Common Pitfalls: Some learners rely entirely on trial and error with the options and may not systematically check both sum and product. Others might form the quadratic equation incorrectly (for example, writing t^2 + 29t + 100). Another common mistake is miscomputing the discriminant or its square root. Using clear algebraic steps helps to avoid these errors. Final Answer: The required two numbers are 25 and 4.

Question 9

The product of the digits of a two-digit number is 72. If we add 9 to the number, the resulting number has its digits interchanged. What is this original two-digit number?

Accepted Answer

Introduction / Context: This question is a classic digit-based puzzle often found in quantitative aptitude tests. It involves a two-digit number with a constraint on the product of its digits and an additional condition about what happens when a fixed value is added to the number. The task is to translate these verbal conditions into equations involving the digits and then solve them systematically. Given Data / Assumptions: Let the two-digit number have tens digit a and units digit b. The number itself can be written as 10a + b. The product of its digits is a * b = 72. When 9 is added to the number, we get a new number whose digits are interchanged, that is 10b + a. We must find the original two-digit number. Concept / Approach: We use place value to express the original and the reversed numbers. The original is 10a + b, while the reversed number is 10b + a. Using the condition that adding 9 to the original number gives the reversed number, we can write an equation involving a and b. Together with the product condition a * b = 72, this forms a solvable system. Solving these equations will give the digits a and b and thus the original number. Step-by-Step Solution: Step 1: Let the tens digit be a and the units digit be b. Then the number is 10a + b.Step 2: The product of the digits is a * b = 72.Step 3: When 9 is added to the number, we obtain the reversed-digit number: 10a + b + 9 = 10b + a.Step 4: Rearrange the equation: 10a + b + 9 = 10b + a.Step 5: Bring like terms together: 10a - a + b - 10b + 9 = 0, which simplifies to 9a - 9b + 9 = 0.Step 6: Factor out 9: 9(a - b + 1) = 0, so a - b + 1 = 0.Step 7: Therefore, a = b - 1.Step 8: Use the product condition a * b = 72. Substitute a = b - 1 to get (b - 1) * b = 72.Step 9: Expand: b^2 - b - 72 = 0.Step 10: Solve the quadratic equation b^2 - b - 72 = 0. The discriminant is 1 + 288 = 289, whose square root is 17.Step 11: The roots are b = (1 ± 17) / 2. This gives b = 9 or b = -8.Step 12: A digit must be non-negative and at most 9, so b = 9 is valid and b = -8 is discarded.Step 13: Then a = b - 1 = 9 - 1 = 8.Step 14: Hence the original two-digit number is 10a + b = 10 * 8 + 9 = 89. Verification / Alternative check: Check the product of digits: 8 * 9 = 72, which matches the first condition. Now add 9 to the number: 89 + 9 = 98. The new number 98 has digits 9 and 8, which are exactly the reverse of the original digits 8 and 9. Thus, both conditions are satisfied, confirming that 89 is correct. Why Other Options Are Wrong: Option 98: Product of digits is 9 * 8 = 72, but if we add 9 to 98, we get 107, which does not have digits reversed. Option 78: Product of digits is 7 * 8 = 56, not 72. Option 87: Product of digits is 8 * 7 = 56, again not 72. Option 69: Product is 6 * 9 = 54, and adding 9 gives 78, whose digits are not the reverse of 69. Common Pitfalls: Some students may attempt to test each option without constructing equations, which can be slower. Others might misinterpret the reversal condition and write 10a + b - 9 = 10b + a or confuse the roles of a and b. Another common error is making algebraic mistakes when simplifying the equations. Setting up the equations carefully and solving step by step avoids these issues. Final Answer: The original two-digit number is 89.

Question 10

Find a two-digit number which is exactly three times the product of its digits.

Accepted Answer

Introduction / Context: This is another digit-based puzzle involving a two-digit number and a relationship between the number itself and the product of its digits. The given condition is that the two-digit number equals three times the product of its digits. Such problems help learners practice forming algebraic equations in terms of the digits and solving them logically rather than by random guessing. Given Data / Assumptions: Let the tens digit of the two-digit number be a and the units digit be b. The number itself is 10a + b. The product of its digits is a * b. The number is exactly three times the product of its digits: 10a + b = 3 * a * b. Digits a and b are integers with a from 1 to 9 and b from 0 to 9. Concept / Approach: We use place value to convert the verbal statement into an equation in a and b. Once we have 10a + b = 3ab, we can rearrange it into a standard algebraic form and either solve systematically or check possible digit combinations. Recognising that both digits are small allows us to reason quickly, but writing the equation helps maintain clarity and avoid missing solutions. Step-by-Step Solution: Step 1: Let the tens digit be a and the units digit be b, so the number is 10a + b.Step 2: The product of the digits is a * b.Step 3: According to the problem, 10a + b = 3ab.Step 4: Rearrange this equation: 10a + b - 3ab = 0.Step 5: Factor where possible. Group the terms involving b: 10a + b - 3ab = 10a + b(1 - 3a).Step 6: This shows that for integer digits a and b, we can test small values of a (1 to 9) and solve for b.Step 7: Another approach is to test each option given. Check option 24 first. Here, a = 2 and b = 4. The product of digits is 2 * 4 = 8. Three times the product is 3 * 8 = 24, which equals the original number.Step 8: Therefore, 24 satisfies the condition exactly: 10a + b = 3ab.Step 9: For completeness, note that if a = 1 and b = 5, we get 15 = 3 * (1 * 5) = 15, so 15 also satisfies the equation. However, among the options provided, only 24 appears as a valid answer. Verification / Alternative check: Verify 24 explicitly: The digits are 2 and 4. The product of the digits is 2 * 4 = 8. Three times this product is 3 * 8 = 24, which is indeed the original number. Therefore, 24 meets the condition perfectly. It is common in such questions that more than one mathematical solution exists, but the correct choice is determined by the available options. Why Other Options Are Wrong: Option 12: Digits are 1 and 2; product is 2, and 3 * 2 = 6, which is not 12. Option 48: Digits are 4 and 8; product is 32, and 3 * 32 = 96, not 48. Option 36: Digits are 3 and 6; product is 18, and 3 * 18 = 54, not 36. Option 15: As discussed, 15 actually satisfies the equation but is not provided as the correct choice in this particular question set. The key correct option given is 24. Common Pitfalls: Some learners may overlook that the problem is about a two-digit number and accidentally consider three-digit numbers. Others may try to solve the equation in general but make algebraic mistakes. Still others may not realise that there can be more than one mathematical solution and get confused if they find a second number like 15. In multiple choice settings, it is important to match the solution with the provided options and choose the one that fits. Final Answer: The required two-digit number which is three times the product of its digits is 24.

Question 11

Which of the following numbers is not a triangular number?

Accepted Answer

Introduction / Context: This question checks understanding of triangular numbers. A triangular number counts objects that can be arranged in an equilateral triangular pattern. Triangular numbers appear in many combinatorial and geometric contexts. Knowing how to recognise and test triangular numbers is useful in number theory and aptitude problems. Given Data / Assumptions: We are given the options 15, 10, 5, 3, and 21. We must determine which one is not a triangular number. A triangular number Tn is given by Tn = n * (n + 1) / 2 for some positive integer n. Numbers are positive integers. Concept / Approach: To test if a number N is triangular, we can either list triangular numbers or invert the formula Tn = n * (n + 1) / 2. If N is triangular, the equation n^2 + n - 2N = 0 will have a positive integer solution for n. Equivalently, 8N + 1 must be a perfect square, because the discriminant of the quadratic in n is 1 + 8N. We can use either the direct listing or the discriminant method to check each option. Step-by-Step Solution: Step 1: Recall the formula for triangular numbers: Tn = n * (n + 1) / 2.Step 2: List the first few triangular numbers: T1 = 1, T2 = 3, T3 = 6, T4 = 10, T5 = 15, T6 = 21, and so on.Step 3: Compare each option with this list.Step 4: 3 appears as T2, so 3 is triangular.Step 5: 10 appears as T4, so 10 is triangular.Step 6: 15 appears as T5, so 15 is triangular.Step 7: 21 appears as T6, so 21 is also triangular.Step 8: The number 5 does not appear among these triangular numbers, and it is not formed by n * (n + 1) / 2 for any small integer n.Step 9: Therefore, 5 is not a triangular number. Verification / Alternative check: Use the discriminant test. For a number N to be triangular, 8N + 1 should be a perfect square. For N = 5, compute 8 * 5 + 1 = 41, which is not a perfect square. For N = 3, 8 * 3 + 1 = 25 = 5^2, so 3 is triangular. For N = 10, 8 * 10 + 1 = 81 = 9^2, so 10 is triangular. For N = 15, 8 * 15 + 1 = 121 = 11^2, so 15 is triangular. For N = 21, 8 * 21 + 1 = 169 = 13^2, so 21 is triangular. This confirms that 5 alone fails the triangular test. Why Other Options Are Wrong: Option 15: It is T5, a known triangular number. Option 10: It is T4, clearly triangular. Option 3: It is T2, the second triangular number. Option 21: It is T6, also triangular. Common Pitfalls: Students sometimes mistakenly think that any number that can form some geometric pattern is triangular, without checking the formula. Others might confuse triangular numbers with square numbers or other figurate numbers. Using the straightforward formula Tn = n * (n + 1) / 2, or the discriminant test 8N + 1 being a perfect square, helps avoid such confusion and gives a quick, reliable check. Final Answer: The number that is not a triangular number is 5.

Question 12

The product of the digits of a two-digit number is 18. If we add 63 to this number, the new number formed has its digits interchanged. What is the original two-digit number?

Accepted Answer

Introduction / Context: This puzzle is similar in style to other digit problems: a two-digit number is described by conditions on the product of its digits and on how the number changes when a fixed value is added. Translating these conditions into equations involving the tens and units digits is a key skill in quantitative aptitude, especially in the topic of number puzzles and digit manipulations. Given Data / Assumptions: Let the two-digit number have tens digit a and units digit b. The number is 10a + b. The product of the digits is a * b = 18. When 63 is added to the number, the digits get interchanged, resulting in 10b + a. We must find the original number 10a + b. Concept / Approach: The number with digits interchanged is 10b + a. Using the condition that adding 63 to the original gives this reversed number leads to a linear equation in a and b. Combined with the product condition a * b = 18, this forms a system of two equations. Solving this system reveals the specific digits. This approach is systematic and avoids random guessing, though we can later confirm with the options easily. Step-by-Step Solution: Step 1: Let the original two-digit number be 10a + b, where a is the tens digit and b is the units digit.Step 2: We are given that a * b = 18.Step 3: When we add 63 to the original number, the digits are interchanged. So 10a + b + 63 = 10b + a.Step 4: Rearrange the equation: 10a + b + 63 = 10b + a.Step 5: Bring all terms to one side: 10a - a + b - 10b + 63 = 0.Step 6: Simplify: 9a - 9b + 63 = 0.Step 7: Factor out 9: 9(a - b + 7) = 0, so a - b + 7 = 0.Step 8: Therefore, a = b - 7.Step 9: Use the product condition a * b = 18. Substitute a = b - 7, giving (b - 7) * b = 18.Step 10: Expand: b^2 - 7b - 18 = 0.Step 11: Solve the quadratic b^2 - 7b - 18 = 0. The discriminant is 49 + 72 = 121, whose square root is 11.Step 12: The roots are b = (7 ± 11) / 2, giving b = 9 or b = -2.Step 13: A digit must be between 0 and 9, so b = 9 is valid and b = -2 is invalid.Step 14: Then a = b - 7 = 9 - 7 = 2.Step 15: Thus, the original number is 10a + b = 10 * 2 + 9 = 29. Verification / Alternative check: Check the product of the digits: 2 * 9 = 18, matching the first condition. Now add 63 to the number: 29 + 63 = 92. The number 92 has digits 9 and 2, which are the reverse of the original digits 2 and 9. Therefore, both the product condition and the digit reversal condition hold, confirming that 29 is the correct original number. Why Other Options Are Wrong: Option 92: This is actually the number obtained after adding 63 to the original, not the original itself. Option 36: Digits are 3 and 6, product is 18, but 36 + 63 = 99, which does not have digits 6 and 3. Option 63: Digits are 6 and 3, product is 18, but 63 + 63 = 126, which is not the reverse 36. Option 39: Digits are 3 and 9, product is 27, not 18. Common Pitfalls: Some students may directly test each option but forget to verify both conditions (product and reversal after addition). Others might mix up the digits in the equation and write 10b + a + 63 = 10a + b instead of the correct relation. Also, errors often occur in solving the quadratic equation if the discriminant or roots are computed incorrectly. Careful algebra and a quick double-check with the given conditions help avoid these mistakes. Final Answer: The original two-digit number is 29.

Question 13

Which of the following five digit numbers is exactly divisible by 99 without leaving any remainder?

Accepted Answer

Introduction / Context: This question tests basic number theory and divisibility rules, specifically for the composite number 99. Since 99 is equal to 9 * 11, any number that is exactly divisible by 99 must satisfy the divisibility rules for both 9 and 11 at the same time. Understanding and applying these rules quickly is very useful in aptitude exams where calculations need to be done without a calculator. Given Data / Assumptions: - We have four candidate five digit numbers: 51579, 51557, 55036 and 49984. - We need to find which number is exactly divisible by 99. - A number divisible by 99 must be divisible by 9 and by 11. Concept / Approach: - 99 = 9 * 11, so divisibility by 99 implies divisibility by 9 and by 11. - Divisibility by 9: a number is divisible by 9 if the sum of its digits is divisible by 9. - Divisibility by 11: for a number, take the alternating sum of its digits (sum of digits in odd positions minus sum of digits in even positions in the same order). If the result is 0 or a multiple of 11, the number is divisible by 11. Step-by-Step Solution: Step 1: Check divisibility of 51579 by 9. Sum of digits = 5 + 1 + 5 + 7 + 9 = 27. Since 27 is divisible by 9, the number passes the rule for 9. Step 2: Check divisibility of 51579 by 11. Alternating sum = (5 + 5 + 9) - (1 + 7) = 19 - 8 = 11. Since 11 is a multiple of 11, 51579 is divisible by 11. Step 3: Since 51579 is divisible by both 9 and 11, it is divisible by 99. Step 4: For completeness, quickly see that the other numbers fail at least one rule. For example, 51557 has sum of digits 23 which is not divisible by 9. So it cannot be divisible by 99. Verification / Alternative check: We can also verify by direct division: 51579 / 99 gives an integer 521 with no remainder if we perform long division or use modular checks for 99 directly. Since both divisibility tests are satisfied, and quick checks reject the other options, further heavy computation is not needed. Why Other Options Are Wrong: Option B: 51557 has sum of digits 23 which is not divisible by 9, so it cannot be divisible by 99. Option C: 55036 has sum of digits 19 which is not divisible by 9. It fails the rule for 9, so it is not divisible by 99. Option D: 49984 has sum of digits 34 which is not divisible by 9, so it is not divisible by 99. Option E: 51599 has sum of digits 29 which is not divisible by 9, so this candidate also fails. Common Pitfalls: - Some learners check only one of the factors, usually 9, and forget to verify divisibility by 11. - Another common mistake is to compute the alternating sum for the 11 test incorrectly by mixing up positions or signs. - Trying full long division for every option wastes time in an exam. Using divisibility rules is much faster and less error prone. Final Answer: The only number that is divisible by both 9 and 11, and hence by 99, is 51579.

Question 14

The sum of a non zero real number and its reciprocal is 2. What is the value of that number?

Accepted Answer

Introduction / Context: This problem checks understanding of simple algebraic equations involving a number and its reciprocal. Questions of the form x + 1 / x equal to some constant appear frequently in aptitude and competitive exams. Being comfortable with converting such verbal statements into equations and solving quadratic equations quickly is very important for speed and accuracy. Given Data / Assumptions: - Let the unknown non zero real number be x. - The sum of the number and its reciprocal is 2, so x + 1 / x = 2. - x is non zero, so division by x is valid. Concept / Approach: - Translate the statement into an equation: x + 1 / x = 2. - Clear the denominator by multiplying both sides by x, which is allowed because x is non zero. - This will give a quadratic equation in x. - Solve the quadratic and choose the solution that fits all conditions. Step-by-Step Solution: Step 1: Start from the equation x + 1 / x = 2. Step 2: Multiply both sides by x to remove the denominator: x * x + 1 = 2x. Step 3: This simplifies to x^2 + 1 = 2x. Step 4: Rearrange all terms to one side to get a standard quadratic equation: x^2 - 2x + 1 = 0. Step 5: Factor the quadratic: x^2 - 2x + 1 = (x - 1)^2 = 0. Step 6: Solve (x - 1)^2 = 0 to get the single solution x = 1. Verification / Alternative check: Substitute x = 1 back into the original relation. Then x + 1 / x becomes 1 + 1 / 1 which is 1 + 1 = 2. This matches the given condition exactly, so x = 1 is correct. There is no other solution because the quadratic has a repeated root, meaning there is only one distinct solution. Why Other Options Are Wrong: Option B: If x = 2, then x + 1 / x = 2 + 1 / 2 = 2.5, which is not equal to 2. Option C: If x = -1, then x + 1 / x = -1 + (-1) = -2, which does not satisfy the condition. Option D: If x = 4, then x + 1 / x = 4 + 1 / 4 = 4.25, which is incorrect. Option E: If x = 0.5, then x + 1 / x = 0.5 + 2 = 2.5, so this is also not valid. Common Pitfalls: - Forgetting that x must be non zero and multiplying by x without noting that condition. - Making sign errors when moving terms to one side of the equation. - Not recognizing that a perfect square quadratic like (x - 1)^2 = 0 has only one distinct solution. Final Answer: The required non zero number whose sum with its reciprocal equals 2 is 1.

Question 15

In the following number series 2, 3, 6, 15, 45, what number should replace the question mark as the next term in the sequence?

Accepted Answer

Introduction / Context: Number series questions are common in aptitude tests and are used to check pattern recognition skills. The given series 2, 3, 6, 15, 45, ? is a classic example where the pattern involves changing multiplication factors. The goal is to identify how each term is generated from the previous term and then extend the pattern to find the missing value. Given Data / Assumptions: - The series is: 2, 3, 6, 15, 45, ?. - Each term is obtained from the previous term using a consistent rule. - We assume there is a single underlying pattern, not a random jump. Concept / Approach: - First, look at the ratios or multipliers between consecutive terms. - Compute 3 / 2, 6 / 3, 15 / 6 and 45 / 15 to detect any trend in the multiplying factors. - If the multipliers follow a systematic pattern, extend that pattern to predict the next multiplier and hence the next term. Step-by-Step Solution: Step 1: Find the multiplier from 2 to 3. Here 2 * 1.5 = 3. Step 2: From 3 to 6, the multiplier is 2, since 3 * 2 = 6. Step 3: From 6 to 15, the multiplier is 2.5, since 6 * 2.5 = 15. Step 4: From 15 to 45, the multiplier is 3, since 15 * 3 = 45. Step 5: The sequence of multipliers is 1.5, 2, 2.5, 3. This clearly increases by 0.5 each time. Step 6: Following this pattern, the next multiplier should be 3.5. Step 7: Therefore, the next term is 45 * 3.5 = 45 * (7 / 2) = (45 * 7) / 2 = 315 / 2 = 157.5. Verification / Alternative check: We can verify by writing the general idea: each term is obtained from the previous one by multiplying with a factor that increases by 0.5 successively. So term n has multiplier 1.5 + 0.5 * (n - 2) when n starts from 2. Checking all current steps confirms the pattern, so using 3.5 for the next step is fully justified. Why Other Options Are Wrong: Option A (90): would require a multiplier of 2 from 45 to 90, which breaks the 0.5 increment pattern. Option B (135): would require a multiplier of 3 from 45 to 135, but we have already used 3 from 15 to 45 and the pattern demands 3.5 next. Option D (200): implies a multiplier of 4.44 which does not follow any simple regular increase. Option E (210): would correspond to a multiplier of 4.666 which again does not fit the simple arithmetic progression of multipliers. Common Pitfalls: - Many learners look only at differences of terms rather than ratios, which does not reveal the pattern in this case. - Some may assume a fixed multiplier, which fails once they see 2 to 3 and 3 to 6 are different. - Another mistake is to stop after noticing an approximate pattern rather than checking all transitions to confirm consistency. Final Answer: The next term in the series that follows the multiplier pattern is 157.5.

Question 16

In the number series 3, 4, 10, 33, 136, what number should replace the question mark as the next term in the sequence?

Accepted Answer

Introduction / Context: This question is a non standard number series problem designed to test the ability to identify a pattern where each term depends on the previous term multiplied by an increasing integer and then adjusted by the same integer. Recognizing such mixed operations is a core skill in number series questions used in many competitive exams. Given Data / Assumptions: - The series given is 3, 4, 10, 33, 136, ?. - We assume there is a unique consistent rule linking each term to the previous term. - We want the next term that maintains that rule. Concept / Approach: - Look for relationships of the form next term = current term * k + k, where k changes in a regular way. - Compare how the coefficient and constant term might depend on the position in the series. - Verify the pattern across all known transitions before extending it. Step-by-Step Solution: Step 1: From 3 to 4, we can write 3 * 1 + 1 = 4. Step 2: From 4 to 10, we can write 4 * 2 + 2 = 10. Step 3: From 10 to 33, we can write 10 * 3 + 3 = 33. Step 4: From 33 to 136, we can write 33 * 4 + 4 = 136. Step 5: The pattern is now clear: next term = current term * n + n, where n is 1, 2, 3, 4, and so on. Step 6: For the next transition, the multiplier and the addend should both be 5, so we compute 136 * 5 + 5. Step 7: 136 * 5 = 680, and 680 + 5 = 685. Verification / Alternative check: We can summarise the recurrence as a_(k+1) = a_k * (k) + (k) where k starts at 1 for the first step. Checking each step: - 3 to 4 uses k = 1. - 4 to 10 uses k = 2. - 10 to 33 uses k = 3. - 33 to 136 uses k = 4. The pattern holds for all known steps, so using k = 5 for the next term is logically consistent and yields 685. Why Other Options Are Wrong: Option A (150): This would require 136 * k + k = 150 for some single integer k, which is not an integer solution consistent with the earlier pattern. Option B (298): Does not fit the rule 136 * 5 + 5 and would break the linear increase of the coefficient. Option C (463): Cannot be expressed as 136 * 5 + 5 or 136 * 5 + any simple k consistent with earlier steps. Option D (572): Looks like 136 * 4 + 28, which does not match the clean rule of multiplying and then adding the same integer. Common Pitfalls: - Looking only at differences between terms instead of considering multiplicative patterns combined with additions. - Assuming the pattern is based on fixed multiplication or fixed addition, which fails in this series. - Not verifying the guessed pattern on every step, which can lead to selecting a number that coincidentally fits one gap but not the entire series. Final Answer: The correct next term that follows the pattern a_(k+1) = a_k * k + k is 685.

Question 17

In the number series 7, 10, 16, 28, ?, 100, what number should replace the question mark so that the pattern remains consistent?

Accepted Answer

Introduction / Context: This series question checks whether you can detect a pattern that uses both multiplication and subtraction in a consistent way. Such series require you to test hypotheses about how each term is generated from the previous one. The sequence 7, 10, 16, 28, ?, 100 hides a simple linear relation once you look at the operations between consecutive terms carefully. Given Data / Assumptions: - The given terms are 7, 10, 16, 28, ?, 100. - There is a consistent rule connecting each term to the next. - The goal is to find the missing term that preserves this rule. Concept / Approach: - Try to express each term in relation to the previous one using a simple operation such as multiplication plus or minus a constant. - Check if a single relation like next = 2 * current - constant works across several steps. - Once the pattern is found and verified, apply it to compute the missing term and the final term to ensure consistency. Step-by-Step Solution: Step 1: Look at 7 and 10. One way to get 10 from 7 is 7 * 2 - 4 = 14 - 4 = 10. Step 2: Check if the same relation holds from 10 to 16. Compute 10 * 2 - 4 = 20 - 4 = 16. This matches the series. Step 3: From 16 to 28, apply the same rule. 16 * 2 - 4 = 32 - 4 = 28. Again, it matches. Step 4: Continue the same pattern to find the missing term. From 28, next term should be 28 * 2 - 4 = 56 - 4 = 52. Step 5: Confirm that applying the same rule from 52 gives the final term 100. 52 * 2 - 4 = 104 - 4 = 100, which matches the given last term. Verification / Alternative check: The discovered pattern is a_n+1 = 2 * a_n - 4 for all n. We tested it across all available transitions. Since it works from the first to the second term, from the second to the third, from the third to the fourth, and from the fifth to the sixth once we insert 52, the rule is fully consistent. Therefore 52 is the only value that keeps the series coherent. Why Other Options Are Wrong: Option A (34): Using 34 would give 28 * 2 - 4 = 52, not 34, so the rule would be broken at that point. Option B (40): 28 * 2 - 4 is not 40, so 40 does not satisfy the established relation. Option D (60): If we use 60, then 60 * 2 - 4 = 116, which does not match 100, so the pattern fails later. Option E (46): Similarly, 28 * 2 - 4 is not 46, and the relation is inconsistent if we choose this value. Common Pitfalls: - Many students try difference patterns (3, 6, 12, ...) and get stuck, because the sequence is not based on differences but on a direct relation. - Some may assume changing multipliers rather than noticing the simple repeated operation of doubling and then subtracting 4. - Forgetting to verify the rule with the last term leads to incorrect choices that happen to fit only one gap. Final Answer: The value that must replace the question mark to maintain the pattern is 52.

Question 18

In the number series 24, 25, ?, 41, -8, 73, what value should replace the question mark so that the pattern of the series is preserved?

Accepted Answer

Introduction / Context: This number series problem uses an alternating pattern in the differences between consecutive terms. The key here is to recognize that the increments are not simple arithmetic or geometric progressions but follow a pattern involving alternating positive and negative squares. Mastering such series improves a learner capability to spot non obvious but systematic patterns. Given Data / Assumptions: - The series is 24, 25, ?, 41, -8, 73. - One middle term is missing. - The goal is to determine the missing term so that the overall pattern of the series stays consistent. Concept / Approach: - Compute the differences between consecutive known terms where possible. - Look for a pattern in these differences, such as squares of odd numbers with alternating signs. - Use this pattern to fill in the missing term and confirm the pattern with later terms in the series. Step-by-Step Solution: Step 1: The difference from 24 to 25 is 25 - 24 = 1, which is 1^2. Step 2: Let the missing term be x. Then the difference from 25 to x is x - 25, and from x to 41 is 41 - x. Step 3: Consider the later known differences. From 41 to -8, the difference is -8 - 41 = -49, which is -7^2. From -8 to 73, the difference is 73 - (-8) = 81, which is 9^2. Step 4: The pattern in the differences suggests alternating squares of odd numbers with alternating signs: +1^2, -3^2, +5^2, -7^2, +9^2. Step 5: Using this, from 24 to 25 we already have +1^2. Then from 25 to x we should have -3^2 = -9. So x - 25 = -9, giving x = 16. Step 6: Next difference from 16 to 41 should be +5^2 = +25. Indeed 16 + 25 = 41, matching the given term. Step 7: The next differences -7^2 and +9^2 produce -8 and 73 as already verified, so the pattern is consistent. Verification / Alternative check: Write the full list of differences including the missing ones using x = 16: 24 to 25 is +1, 25 to 16 is -9, 16 to 41 is +25, 41 to -8 is -49, -8 to 73 is +81. These are 1, -9, 25, -49, 81 which correspond to 1^2, -3^2, 5^2, -7^2, 9^2. This neat pattern confirms that the missing term 16 is correct. Why Other Options Are Wrong: Option A (13): With 13, the differences would not yield perfect odd squares and the later terms would not fit a clean pattern. Option C (25): Repeating 25 would give a zero difference at that step, which does not fit the odd square sequence. Option D (43): This would give 24 to 25 difference 1, 25 to 43 difference 18, which is not a perfect square, and the pattern collapses. Option E (19): Differences again become irregular and do not align with the alternating square structure. Common Pitfalls: - Looking only at the values, not at the differences between terms. - Checking linear or geometric patterns instead of considering squares or higher powers. - Not validating the pattern with all terms, especially the later ones where the pattern becomes clearer. Final Answer: The missing term in the series that preserves the alternating odd square difference pattern is 16.

Question 19

A three digit number has 3 in the tens place. When the digits in the hundreds and units places are interchanged symmetrically about the tens position, the new number becomes 396 more than the original number. The sum of the digits in the units and h…

Accepted Answer

Introduction / Context: This problem involves a three digit number and uses place value concepts along with basic algebra. The number is described in terms of its hundreds, tens and units digits, and a condition is given after interchanging two of the digits. Such questions test a learner understanding of how to represent numbers algebraically based on their digits and how to form and solve equations from word conditions. Given Data / Assumptions: - Let the original three digit number have hundreds digit a, tens digit 3 and units digit b. - The tens place is fixed as 3. - When the hundreds and units digits are interchanged, the new number is 396 more than the original number. - The sum of the hundreds and units digits is given as a + b = 14. Concept / Approach: - Use place value to express the original and new numbers in terms of a and b. - Write the algebraic equation that represents the condition that the new number is 396 more than the original number. - Combine this equation with the sum condition a + b = 14 to solve for the digits a and b. - Form the original number 100a + 30 + b and match with the options. Step-by-Step Solution: Step 1: Represent the original number as 100a + 30 + b. Step 2: After swapping the hundreds and units digits, the new number becomes 100b + 30 + a. Step 3: The new number is 396 more than the original, so write the equation: 100b + 30 + a = 100a + 30 + b + 396. Step 4: Simplify this equation. Subtract 30 from both sides and rearrange terms to get 100b + a = 100a + b + 396. Step 5: Bring like terms together: 100b - b = 100a - a + 396, which gives 99b = 99a + 396. Step 6: Divide every term by 99 to obtain b = a + 4. Step 7: Use the second condition a + b = 14. Substitute b = a + 4 to get a + (a + 4) = 14. Step 8: Simplify: 2a + 4 = 14 so 2a = 10 and a = 5. Then b = 5 + 4 = 9. Step 9: The original number is 100a + 30 + b = 100 * 5 + 30 + 9 = 539. Verification / Alternative check: Check the conditions with 539. The new number after swapping hundreds and units digits is 935. The difference 935 - 539 = 396, which matches the problem statement. The sum of the hundreds and units digits is 5 + 9 = 14, also matching the given condition. So the solution is fully consistent. Why Other Options Are Wrong: Option A (513): Here a + b = 5 + 3 = 8, not 14, so it fails the sum condition. Option C (439): The tens digit is 3, but a + b = 4 + 9 = 13 and the difference after swapping would not be 396. Option D (613): The tens digit is 1, which already violates the fact that the tens digit must be 3. Option E (731): Again, the tens digit is 3, but a + b = 7 + 1 = 8 and the required difference condition does not hold. Common Pitfalls: - Confusing the tens place with the hundreds or units place. - Forgetting that the tens digit is fixed and only the hundreds and units digits are swapped. - Making algebraic mistakes when simplifying the equation or solving the system for a and b. Final Answer: The original three digit number that satisfies all the conditions is 539.

Question 20

Four consecutive even integers are such that the sum of the reciprocals of the first two is 11 divided by 60. What is the reciprocal of the third largest number among these four even integers?

Accepted Answer

Introduction / Context: This question examines understanding of consecutive even integers, reciprocals and the ability to set up and solve a rational equation. It also checks whether the learner can interpret the phrase third largest correctly when the numbers are ordered. Such problems are common in number system and algebra sections of competitive exams. Given Data / Assumptions: - We have four consecutive even integers. - The sum of the reciprocals of the first two integers is 11 / 60. - We are asked to find the reciprocal of the third largest integer among the four. Concept / Approach: - Let the four consecutive even integers be n, n + 2, n + 4 and n + 6, where n is an even integer. - The first two integers are n and n + 2, so their reciprocals are 1 / n and 1 / (n + 2). - Use the given sum of reciprocals to form an equation and solve for n. - Once n is known, list the four integers in increasing order and identify the third largest integer, then compute its reciprocal. Step-by-Step Solution: Step 1: Write the condition on reciprocals: 1 / n + 1 / (n + 2) = 11 / 60. Step 2: Combine the fractions on the left side: ( (n + 2) + n ) / ( n * (n + 2) ) = (2n + 2) / ( n * (n + 2) ). Step 3: Factor the numerator as 2(n + 1), so the expression becomes 2(n + 1) / ( n * (n + 2) ). This equals 11 / 60. Step 4: Cross multiply: 2(n + 1) * 60 = 11 * n * (n + 2). Step 5: Simplify the left side to 120(n + 1). Expand the right side as 11(n^2 + 2n) = 11n^2 + 22n. Step 6: Set up the equation 120n + 120 = 11n^2 + 22n. Step 7: Rearrange to standard quadratic form: 11n^2 + 22n - 120n - 120 = 0 which simplifies to 11n^2 - 98n - 120 = 0. Step 8: Factor the quadratic. It can be written as (n - 10)(11n + 12) = 0. Step 9: The solutions are n = 10 or n = -12 / 11. Since n must be an integer and we are dealing with positive even integers in a typical aptitude context, we take n = 10. Step 10: The four consecutive even integers are therefore 10, 12, 14 and 16. Step 11: In increasing order, the smallest is 10 and the largest is 16. The third largest is 12 (the numbers from smallest to largest are 10, 12, 14, 16, so largest is 16, second largest is 14, third largest is 12). Step 12: The reciprocal of 12 is 1 / 12. Verification / Alternative check: Check the original condition with n = 10. Then 1 / 10 + 1 / 12 = (6 + 5) / 60 = 11 / 60, which matches the given value. The arithmetic is consistent, confirming that the chosen n is correct and that 1 / 12 is the required reciprocal. Why Other Options Are Wrong: Option B (2 / 14): This simplifies to 1 / 7, corresponding to a number 7 which is not part of the even sequence 10, 12, 14, 16. Option C (1 / 14): This is the reciprocal of 14, which is the second largest integer, not the third largest. Option D (2 / 13): This is not even a simple reciprocal of an integer and does not correspond to any of the four integers directly. Option E (1 / 16): This is the reciprocal of the largest number, whereas the question explicitly asks for the third largest. Common Pitfalls: - Misinterpreting third largest as third number instead of counting from the largest end. - Making algebraic mistakes when simplifying the rational equation. - Forgetting that the integers are even and consecutive, which restricts valid values of n. Final Answer: The reciprocal of the third largest integer in the sequence is 1/12.

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