Question 1

In the number series 3240, 540, 108, 27, ?, 4.5, what value should replace the question mark so that the series follows a consistent pattern?

Accepted Answer

Introduction / Context: This number series problem focuses on division by successively changing divisors. To solve such questions, it is important to examine how each term transforms into the next, especially when the numbers decrease significantly. The series 3240, 540, 108, 27, ?, 4.5 can be cracked by comparing ratios of consecutive terms. Given Data / Assumptions: - The series is 3240, 540, 108, 27, ?, 4.5. - There is one missing term between 27 and 4.5. - We assume a simple multiplicative or divisive pattern. Concept / Approach: - Compute the ratio of each term to the previous term. - If the ratios form a simple pattern, such as division by consecutive integers, extend that pattern to find the missing term. - Confirm that the same pattern works consistently through the entire series. Step-by-Step Solution: Step 1: Find the ratio from 3240 to 540. 3240 / 540 = 6, so 3240 has been divided by 6 to get 540. Step 2: Find the ratio from 540 to 108. 540 / 108 = 5, so 540 is divided by 5 to give 108. Step 3: Find the ratio from 108 to 27. 108 / 27 = 4, so 108 is divided by 4 to get 27. Step 4: We can see a pattern: the series divides successively by 6, 5, and 4. Step 5: Following this pattern, the next divisor should be 3, and after that 2. Step 6: Compute the unknown term after 27 by dividing 27 by 3. This gives 27 / 3 = 9. Step 7: Now check the final step: 9 divided by 2 should be 4.5, which matches the given last term. Verification / Alternative check: We can summarize the pattern as: 3240 / 6 = 540, 540 / 5 = 108, 108 / 4 = 27, 27 / 3 = 9, and 9 / 2 = 4.5. This shows a clear and consistent sequence of divisors: 6, 5, 4, 3, 2. Since the pattern works perfectly across all transitions, the missing term must be 9. Why Other Options Are Wrong: Option A (7): 27 / 3 is not 7, and 7 / 2 does not give 4.5, so it destroys the clean divisor pattern. Option B (10): Similarly, 27 / 3 is not 10 and 10 / 2 is 5, which does not match 4.5. Option D (8): 27 / 3 is 9, not 8, and 8 / 2 is 4, not 4.5. Option E (6): This would lead to 6 / 2 = 3 as the last term, which is inconsistent with the given 4.5. Common Pitfalls: - Looking at differences instead of ratios in a sequence where numbers shrink quickly. - Assuming a random or non systematic pattern without testing a simple descending divisor pattern. - Forgetting to verify that the pattern holds both before and after the missing term. Final Answer: The missing term that keeps the descending divisor pattern intact is 9.

Question 2

In the number series 2, 17, 89, 359, 1079, what number should come next in place of the question mark so that the pattern continues?

Accepted Answer

Introduction / Context: This number series question involves a non standard pattern where each term is obtained from the previous term using a changing multiplier and a small additive adjustment. The series 2, 17, 89, 359, 1079, ? is deliberately designed to be challenging and requires careful observation of how the terms grow and how the operations applied to each term change from step to step. Given Data / Assumptions: - The series of terms is 2, 17, 89, 359, 1079, ?. - The pattern relates each term to its immediate predecessor using multiplication and addition. - We seek a rule consistent across all transitions that also leads to one of the given options. Concept / Approach: - Look at how much each term increases compared to the previous term. This suggests trying expressions of the type next = previous * k + c where k and c may vary. - In some series, multipliers may decrease steadily and the added constants may follow a small pattern. - Once a plausible rule is identified for all known steps, apply it to obtain the next term and check with the options. Step-by-Step Solution: Step 1: Observe the jump from 2 to 17. One convenient way to write this is 2 * 8 + 1 = 17. Step 2: From 17 to 89 we can write 17 * 5 + 4 = 89. Step 3: From 89 to 359 we can write 89 * 4 + 3 = 359. Step 4: From 359 to 1079 we can write 359 * 3 + 2 = 1079. Step 5: Notice that the multipliers are 8, 5, 4, 3 and the added constants are 1, 4, 3, 2. After the first step, the multipliers are descending as 5, 4, 3 and the added constants are small integers that also change slowly. Step 6: The natural extension of this pattern is to use the next smaller multiplier, 2, together with the next small constant, 1. Step 7: Therefore, the next term should be 1079 * 2 + 1. Step 8: Compute this value: 1079 * 2 = 2158; adding 1 gives 2159. Verification / Alternative check: Even though this pattern is not as simple as a pure arithmetic or geometric progression, the sequence 2 * 8 + 1, 17 * 5 + 4, 89 * 4 + 3, 359 * 3 + 2, 1079 * 2 + 1 is internally consistent. The multipliers step down as 8, 5, 4, 3, 2 while the added constants 1, 4, 3, 2, 1 remain small and fit a natural decreasing pattern at the end. This yields a unique continuation 2159, which matches one of the options. Why Other Options Are Wrong: Option A (2137): This value cannot be obtained from 1079 by a simple integer multiplier and a small integer adjustment consistent with the observed pattern. Option B (2121): Similarly, 2121 is not equal to 1079 * 2 + c or 1079 * 3 + c for any small constant c that would fit earlier steps. Option C (2377): This is too large compared to 1079 and would require breaking the descending multiplier structure. Option E (2197): This number is notable as 13^3 but it does not follow from the relation used in the earlier steps of this specific series. Common Pitfalls: - Trying only simple difference or ratio based approaches without considering combined multiply and add relations. - Expecting the same multiplier throughout, which does not fit the data here. - Giving up on a pattern as soon as one simple idea fails rather than exploring more flexible forms like a_n+1 = a_n * k + c where k and c may change in a controlled way. Final Answer: The number that continues the observed pattern in the series is 2159.

Question 3

How many integers between 300 and 650 inclusive are divisible by both 5 and 7 (that is, divisible by 35)?

Accepted Answer

Introduction / Context: This problem checks understanding of divisibility and counting of multiples within a closed interval. Specifically, we are asked to find numbers that are divisible by both 5 and 7, which can be reduced to counting multiples of their least common multiple. Such questions occur frequently in quantitative aptitude tests and also reinforce the concept of least common multiple. Given Data / Assumptions: - We are considering all integers from 300 to 650 inclusive. - We want numbers that are divisible by 5 and by 7. - There is no other restriction such as positivity, but the interval is entirely positive anyway. Concept / Approach: - A number divisible by both 5 and 7 is divisible by the least common multiple of 5 and 7. - Since 5 and 7 are prime, their least common multiple is 5 * 7 = 35. - The task reduces to counting multiples of 35 between 300 and 650 inclusive. - Use the formula for counting multiples in an interval using integer division. Step-by-Step Solution: Step 1: Compute the least common multiple of 5 and 7. Since they are prime and share no common factors, lcm(5, 7) = 5 * 7 = 35. Step 2: Find the smallest multiple of 35 that is greater than or equal to 300. Divide 300 by 35 to get 300 / 35 which is approximately 8.57. Step 3: The smallest integer k such that 35k is at least 300 is k = 9. Then 35 * 9 = 315, which is the first valid multiple. Step 4: Find the largest multiple of 35 that is less than or equal to 650. Compute 650 / 35 which is approximately 18.57. Step 5: The largest integer k with 35k at most 650 is k = 18. Then 35 * 18 = 630, which is the last valid multiple. Step 6: Now count how many integers k from 9 to 18 inclusive there are. This is 18 - 9 + 1 = 10. Step 7: Therefore, there are 10 numbers between 300 and 650 inclusive that are divisible by both 5 and 7. Verification / Alternative check: List the multiples of 35 from 315 to 630 to verify: 315, 350, 385, 420, 455, 490, 525, 560, 595, 630. That is exactly 10 numbers. Each is clearly divisible by 5, as they end in 0 or 5, and also divisible by 7, since they are multiples of 35. This confirms our count. Why Other Options Are Wrong: Option A (8): This undercounts the multiples. There are more than 8 multiples of 35 in the given range. Option B (9): This still misses one valid multiple, so it is too small. Option D (12): This overcounts and would require extending beyond 630 or including numbers below 315. Option E (11): This is also too large; the explicit list confirms there are only 10. Common Pitfalls: - Forgetting to use the least common multiple and instead trying to filter by 5 and 7 separately without a systematic method. - Off by one errors when counting endpoints, especially when using floor and ceiling division. - Including numbers like 300 or 650 without checking divisibility carefully. Final Answer: The number of integers between 300 and 650 that are divisible by both 5 and 7 is 10.

Question 4

Find two natural numbers whose sum is 85 and whose least common multiple (LCM) is 102.

Accepted Answer

Introduction / Context: This problem tests understanding of the relationship between the greatest common divisor, least common multiple and the product of two numbers. It also requires checking candidate pairs against conditions on both sum and least common multiple. Such questions are standard in number system topics of aptitude exams. Given Data / Assumptions: - We are looking for two natural numbers. - Their sum must be 85. - Their least common multiple is 102. - Several candidate pairs are suggested in the options. Concept / Approach: - For two positive integers a and b, the relation a * b = gcd(a, b) * lcm(a, b) always holds. - However, here it may be faster to test the given options directly. - For each candidate pair, check if the sum is 85 and then compute its least common multiple. - The pair that simultaneously satisfies both conditions is the correct answer. Step-by-Step Solution: Step 1: Check option A, 30 and 55. The sum is 30 + 55 = 85, so the sum condition is satisfied. Step 2: Compute the greatest common divisor of 30 and 55. Their common factors are 1 and 5, so gcd(30, 55) = 5. Step 3: Use the relation lcm(a, b) = (a * b) / gcd(a, b) = (30 * 55) / 5 = 330. This does not match the required 102. Step 4: Check option B, 17 and 68. Sum is 17 + 68 = 85, sum condition holds. Step 5: Compute gcd(17, 68). Since 17 is a prime factor of 68, gcd(17, 68) = 17. Then lcm is (17 * 68) / 17 = 68, not 102. Step 6: Check option C, 35 and 55. Sum is 90, which already fails the sum condition 85, so this pair can be rejected immediately. Step 7: Check option D, 51 and 34. Sum is 51 + 34 = 85, so the sum condition is satisfied. Step 8: Compute gcd(51, 34). Both are divisible by 17, so gcd(51, 34) = 17. Step 9: Compute lcm(51, 34) using the product relation: lcm = (51 * 34) / 17 = (1734) / 17 = 102. Step 10: Since this pair satisfies both the sum and LCM conditions, 51 and 34 are the required numbers. Verification / Alternative check: We can also factor 102 as 2 * 3 * 17. The pair 51 and 34 are 3 * 17 and 2 * 17 respectively. Their product is 51 * 34 = 1734. Dividing by their gcd 17 gives 102, and their sum is 85, confirming that they perfectly meet both conditions. Why Other Options Are Wrong: Option A (30 and 55): Although the sum is 85, their LCM is 330, not 102. Option B (17 and 68): Sum equals 85 but the LCM is only 68. Option C (35 and 55): The sum is 90, which breaks the sum requirement. Option E (49 and 36): The sum is 85 but gcd and LCM do not yield 102. The LCM would be larger than 102 in this case. Common Pitfalls: - Assuming that matching the sum alone is enough and ignoring the LCM condition. - Miscalculating the greatest common divisor and hence getting the wrong LCM. - Forgetting the useful identity a * b = gcd(a, b) * lcm(a, b) which simplifies such problems. Final Answer: The required natural numbers are 51 and 34.

Question 5

The sum of two numbers is equal to three times their difference. If the smaller of the two numbers is 10, what is the value of the other number?

Accepted Answer

Introduction / Context: This algebra question focuses on forming and solving a simple linear equation involving the sum and difference of two numbers. It requires translating a verbal relationship into an algebraic equation and then solving for the unknown. The presence of a None of these option means we must check the result carefully instead of simply matching an option without working it out. Given Data / Assumptions: - There are two numbers, one smaller and one larger. - The smaller number is given as 10. - The sum of the two numbers is equal to three times their difference. - We are asked to find the value of the other (larger) number. Concept / Approach: - Let the larger number be x. Since the smaller is 10, the two numbers are 10 and x. - Their sum is 10 + x. - Their difference, taking larger minus smaller, is x - 10. - According to the statement, sum = 3 * difference, so set up the equation 10 + x = 3(x - 10) and solve for x. Step-by-Step Solution: Step 1: Let the larger number be x. Then the sum is x + 10. Step 2: The difference (larger minus smaller) is x - 10. Step 3: Use the given condition: sum equals three times their difference, so write x + 10 = 3(x - 10). Step 4: Expand the right side: 3(x - 10) = 3x - 30. Step 5: The equation becomes x + 10 = 3x - 30. Step 6: Rearrange to bring all x terms to one side: 10 + 30 = 3x - x, giving 40 = 2x. Step 7: Solve for x: 2x = 40 so x = 20. Step 8: Thus, the other number is 20. Verification / Alternative check: Check the original condition using the numbers 10 and 20. Their sum is 10 + 20 = 30. Their difference (larger minus smaller) is 20 - 10 = 10. Three times the difference is 3 * 10 = 30, which matches the sum. This confirms that the computed value 20 is correct mathematically. Why Other Options Are Wrong: Option A (15): If the numbers are 10 and 15, then sum is 25 and difference is 5. Three times the difference is 15, which does not equal the sum 25. Option B (30): With 10 and 30, the sum is 40 and the difference is 20, three times the difference is 60, not 40. Option C (40): With 10 and 40, the sum is 50 and the difference is 30, three times the difference is 90, not 50. Option E (20): This is the correct numeric value of the other number, but the option list labels only None of these as a text choice. Since 20 is not directly listed as a separate option among A, B or C, the correct marked option must be None of these. Common Pitfalls: - Mixing up the direction of the difference and writing 10 - x instead of x - 10; this changes the equation. - Trying to guess the answer from options without forming and solving the equation. - Ignoring the None of these choice and picking the closest looking number instead of checking the algebra carefully. Final Answer: The other number is 20, which is not explicitly listed, so the correct option is None of these.

Question 6

Which of the following fractions is the largest in value: 5 divided by 113, 7 divided by 120, 13 divided by 145 or 17 divided by 160?

Accepted Answer

Introduction / Context: Comparing fractions with different numerators and denominators is a classic aptitude skill. Instead of converting every fraction to decimal form with a calculator, we generally look for efficient comparison techniques such as cross multiplication or approximation. This question asks for the largest fraction among four given values. Given Data / Assumptions: - The fractions to be compared are 5 / 113, 7 / 120, 13 / 145 and 17 / 160. - All denominators are positive, so larger numerator to denominator ratios mean larger fractions. - We want the single fraction with the greatest value. Concept / Approach: - Direct decimal conversion is possible but time consuming. Instead, we can compare fractions pairwise using cross multiplication. - To compare a / b and c / d, compute a * d and c * b. The fraction with the larger product has the larger value. - Apply this method systematically to find the largest fraction among the four. Step-by-Step Solution: Step 1: First compare 5 / 113 and 7 / 120. Cross multiply: 5 * 120 = 600, and 7 * 113 = 791. Since 791 is greater than 600, 7 / 120 is greater than 5 / 113. Step 2: Now compare 7 / 120 and 13 / 145. Compute 7 * 145 = 1015 and 13 * 120 = 1560. Since 1560 is greater than 1015, 13 / 145 is greater than 7 / 120. Step 3: So far, 13 / 145 is larger than both 7 / 120 and 5 / 113. Step 4: Now compare 13 / 145 and 17 / 160. Compute 13 * 160 = 2080 and 17 * 145 = 2465. Since 2465 is greater than 2080, 17 / 160 is greater than 13 / 145. Step 5: Therefore, among the fractions 5 / 113, 7 / 120, 13 / 145 and 17 / 160, the largest value is 17 / 160. Verification / Alternative check: For additional confirmation, we can estimate decimal values. 5 / 113 is slightly larger than 5 / 115 which is about 0.043, so it is near 0.044. 7 / 120 is about 0.058. 13 / 145 is near 0.09 and 17 / 160 is about 0.10625. These approximations confirm that 17 / 160 is the largest fraction. Why Other Options Are Wrong: Option A (5 / 113): This is the smallest fraction because the numerator is small and the denominator is relatively large. Option B (7 / 120): This is larger than 5 / 113 but still significantly smaller than 13 / 145 and 17 / 160. Option C (13 / 145): This is fairly large but cross multiplication shows that 17 / 160 is larger. Option E (None of these): One of the listed fractions, namely 17 / 160, already satisfies the condition of being the largest, so None of these is not correct. Common Pitfalls: - Trying to work with approximate decimals by rough mental division without a systematic method, which can lead to mistakes. - Assuming that the fraction with the largest numerator or smallest denominator must automatically be the largest without checking the ratio. - Mixing up cross multiplication and comparing the wrong products. Final Answer: The largest fraction among the given options is 17/160.

Question 7

What is the least value that should be added to 1812 in order to make the resulting number exactly divisible by 7, 11 and 14?

Accepted Answer

Introduction / Context: This problem involves divisibility and least common multiple. We are asked to adjust a given number by adding the smallest possible value so that the new number becomes divisible by three given divisors. Understanding how to use least common multiple to combine divisibility conditions is very important in aptitude exams. Given Data / Assumptions: - We start with the number 1812. - We want 1812 plus some least non negative integer k to be divisible by 7, 11 and 14 simultaneously. - The required number must obey the divisibility conditions for all three divisors at once. Concept / Approach: - A number divisible by 14 and by 7 and by 11 must be divisible by the least common multiple of these three numbers. - Since 14 already contains the prime factor 7, the combined least common multiple is lcm(7, 11, 14) = 2 * 7 * 11 = 154. - Therefore, we need to find the smallest value k such that 1812 + k is a multiple of 154. - That is, we find the next multiple of 154 greater than or equal to 1812 and compute the difference. Step-by-Step Solution: Step 1: Factor the numbers. 14 = 2 * 7, while 7 and 11 are primes. Step 2: The least common multiple is 2 * 7 * 11 = 154. Step 3: We need a number of the form 154 * n that is just greater than or equal to 1812. Step 4: Compute 1812 divided by 154. 154 * 10 = 1540 and 154 * 11 = 1694, 154 * 12 = 1848. Step 5: 1848 is the first multiple of 154 that is greater than 1812. Step 6: The required addition k is 1848 - 1812 = 36. Step 7: Therefore, 1812 + 36 = 1848 is divisible by 7, 11 and 14. Verification / Alternative check: Check divisibility of 1848: 1848 / 7 = 264 exactly, so it is divisible by 7. Next, 1848 / 11 = 168 exactly, so it is divisible by 11. Since 1848 is divisible by 14 = 2 * 7 and we already know it is divisible by 7, checking evenness: 1848 is even, so it is divisible by 2 as well. Hence it satisfies all conditions. Why Other Options Are Wrong: Option A (12): 1812 + 12 = 1824; 1824 is not a multiple of 154 and fails divisibility by at least one of the three numbers. Option C (72): 1812 + 72 = 1884; 1884 divided by 154 is not an integer, so this is not a common multiple. Option D (154): 1812 + 154 = 1966; this is 154 more than 1848 and is not guaranteed to be the smallest such adjustment, and in fact 1966 is not a multiple of 154. Option E (18): 1812 + 18 = 1830 which is not divisible by 154, so it fails as well. Common Pitfalls: - Forgetting that 14 already includes a factor of 7 and incorrectly multiplying 7, 11 and 14 directly without reducing. - Trying to test each divisor separately on 1812 + k without using the least common multiple to simplify the problem. - Failing to choose the smallest positive difference to the next multiple, which can lead to a larger than necessary answer. Final Answer: The least value that must be added to 1812 to make it divisible by 7, 11 and 14 is 36.

Question 8

Which of the following values is the largest: 53, 64, 126 or 27612?

Accepted Answer

Introduction / Context: This question is a simple comparison of four positive integers. It ensures that learners can quickly identify the largest among a set of values, which is a basic but essential skill in quantitative reasoning. Although this is far easier than many other number problems, it still requires careful reading to avoid mistakes. Given Data / Assumptions: - The given numbers are 53, 64, 126 and 27612. - All are positive integers. - We must find the single largest value among them. Concept / Approach: - When comparing natural numbers, the number with more digits is automatically larger than any number with fewer digits. - If numbers have the same number of digits, compare digit by digit from left to right. - Here, only one number has five digits while the others have two or three digits. Step-by-Step Solution: Step 1: Note the number of digits in each candidate. 53 and 64 are two digit numbers, 126 is a three digit number and 27612 is a five digit number. Step 2: Any five digit number is larger than any three digit number, which in turn is larger than any two digit number because the place value of the leading digit is much higher. Step 3: Since 27612 is the only five digit number in the list, it must be greater than 126, 64 and 53. Step 4: Therefore, 27612 is the largest of the given values. Verification / Alternative check: We can also do rough comparisons using magnitude. 53 and 64 are less than 100, 126 is a little more than 100, while 27612 is greater than 27000. Clearly, a number of order tens of thousands is far larger than numbers around 100 or below. Why Other Options Are Wrong: Option A (53): This is only a two digit number and is smaller than 64, 126 and 27612. Option B (64): Larger than 53 but still only two digits, so it is far smaller than 126 and especially 27612. Option D (126): This is a three digit number and is larger than 53 and 64, but it is much smaller than 27612. Option E (None of these): Not correct because one of the listed options, namely 27612, is clearly the largest. Common Pitfalls: - Overthinking a simple comparison problem and trying to find hidden structure instead of straightforwardly comparing. - Misreading one of the numbers, especially 27612, and assuming it has fewer digits. - Forgetting that digit count is the quickest way to spot the largest among positive integers. Final Answer: The largest value among the given numbers is 27612.

Question 9

How many five digit numbers that are divisible by 125 can be formed using all the digits 2, 3, 8, 7 and 5 exactly once each?

Accepted Answer

Introduction / Context: This question combines ideas from permutations and divisibility rules. We must count how many five digit numbers divisible by 125 can be formed when each digit from a given set is used exactly once. Understanding the divisibility rule for 125 and constructing valid endings is the key step, followed by counting permutations of remaining digits. Given Data / Assumptions: - The digits available are 2, 3, 8, 7 and 5. - Each five digit number must use all five digits exactly once. - The number must be divisible by 125. Concept / Approach: - A number is divisible by 125 if and only if its last three digits form a number that is divisible by 125. - The multiples of 125 between 100 and 999 that use only the digits from the given set are limited. - First identify valid three digit endings that are multiples of 125 and use only the available digits. - For each valid ending, arrange the remaining two digits in the first two places in all possible orders. Step-by-Step Solution: Step 1: Recall common three digit multiples of 125: 125, 250, 375, 500, 625, 750, 875. Step 2: From these, pick those that use only digits from the set {2, 3, 8, 7, 5}. The candidates are 375 and 875 because 3, 7, 5 and 8, 7, 5 are all available digits. Step 3: The ending 125 uses digit 1 which is not in the set. Endings 250, 500, 625 and 750 involve digit 0 or 6 which are not allowed by the digit set. Step 4: Therefore, the only possible last three digit blocks for our five digit numbers are 375 and 875. Step 5: Consider the ending 375. The digits used at the end are 3, 7 and 5, leaving digits 2 and 8 for the first two positions. Step 6: The first two positions can be arranged as 28 or 82. So with ending 375, the valid numbers are 28375 and 82375. Step 7: Now consider the ending 875. The digits used are 8, 7 and 5, leaving digits 2 and 3 for the first two positions. Step 8: The first two positions can be arranged as 23 or 32. So with ending 875, the valid numbers are 23875 and 32875. Step 9: In total, we have 2 numbers from ending 375 and 2 numbers from ending 875, giving 2 + 2 = 4 valid five digit numbers. Verification / Alternative check: Check one example numerically. 28375 / 125 = 227 exactly, so it is divisible by 125. Similarly, 82375 / 125 = 659, 23875 / 125 = 191 and 32875 / 125 = 263, each an integer. All four constructed numbers are divisible by 125, and no other multiples of 125 using only these digits exist. Why Other Options Are Wrong: Option B (5): This would require one more valid arrangement from some other three digit ending, but no other eligible ending uses only the digits from the given set. Option C (6): Overcounts by assuming more patterns than actually exist for the three digit endings. Option D (7): Significantly overestimates the count and ignores the strict divisibility condition. Option E (3): Undercounts, since we have explicitly constructed four distinct valid numbers. Common Pitfalls: - Forgetting that divisibility by 125 is determined solely by the last three digits, not the whole number at once. - Trying to permute all five digits freely and then test divisibility one by one instead of narrowing down valid endings first. - Overlooking that digits cannot be repeated and must be used exactly once. Final Answer: Exactly 4 five digit numbers can be formed using the digits 2, 3, 8, 7 and 5 once each that are divisible by 125.

Question 10

Between 1 and 200 inclusive, how many integers are divisible by 3 but not divisible by 7?

Accepted Answer

Introduction / Context: This counting problem tests inclusion exclusion logic for divisibility. We need to count numbers in an interval that satisfy one divisibility condition (divisible by 3) but explicitly fail another (not divisible by 7). Such problems help build the habit of using least common multiple and simple counting formulas systematically. Given Data / Assumptions: - We consider all integers from 1 to 200 inclusive. - We want numbers that are divisible by 3. - Among those, we must exclude numbers that are also divisible by 7. - The answer should be a simple count of such integers. Concept / Approach: - First count how many numbers between 1 and 200 are divisible by 3. - Then count how many numbers between 1 and 200 are divisible by both 3 and 7, that is, by lcm(3, 7) = 21. - Subtract the count of numbers divisible by 21 from the count of numbers divisible by 3 to get the final answer. Step-by-Step Solution: Step 1: Count multiples of 3 between 1 and 200. The smallest multiple is 3 and the largest multiple less than or equal to 200 is 198. Step 2: The number of multiples of 3 is 200 divided by 3, taking the integer part. 200 / 3 is approximately 66.66, so there are 66 multiples of 3. Step 3: Now find numbers that are multiples of both 3 and 7. The least common multiple of 3 and 7 is 21. Step 4: Count multiples of 21 between 1 and 200. The smallest multiple is 21 and the largest is 189. Step 5: The number of multiples of 21 is 200 divided by 21, taking the integer part. 200 / 21 is approximately 9.52, so there are 9 such multiples. Step 6: All numbers divisible by both 3 and 7 are also divisible by 21, so we must exclude these 9 from the 66 multiples of 3. Step 7: Therefore, the count of numbers divisible by 3 but not by 7 is 66 - 9 = 57. Verification / Alternative check: As a check, imagine listing multiples of 3 up to 200: 3, 6, 9, ..., 198. Among these, every seventh multiple of 3 starting from 21 (that is, 21, 42, 63, 84, 105, 126, 147, 168, 189) is also divisible by 7. This list clearly has 9 terms, confirming that 9 multiples must be excluded, and thus 57 remain. Why Other Options Are Wrong: Option A (38): This is too small and would correspond to miscounting the multiples or using the wrong interval. Option B (47): This also undercounts, ignoring some valid multiples of 3. Option D (66): This counts all multiples of 3 and fails to remove those that are also divisible by 7. Option E (60): This does not match the precise inclusion exclusion calculation and is not backed by a consistent method. Common Pitfalls: - Forgetting to subtract the numbers that are divisible by both 3 and 7, leading to an overcount. - Miscalculating the floor of 200 / 3 or 200 / 21. - Confusing the interval boundaries (for example, accidentally using up to 201 instead of 200). Final Answer: There are 57 integers between 1 and 200 that are divisible by 3 but not by 7.

Question 11

By which smallest positive integer should 5000 be divided so that the resulting quotient is a perfect square number?

Accepted Answer

Introduction / Context: In this question, we are working with perfect squares and factorisation. The task is to find the smallest positive integer by which 5000 should be divided so that the quotient is a perfect square number. This tests understanding of prime factorisation and the condition for a number to be a perfect square in terms of its prime powers. Given Data / Assumptions: - Original number = 5000. - We divide 5000 by a positive integer N. - The quotient 5000 / N must be a perfect square. - We want the least possible value of N. Concept / Approach: A positive integer is a perfect square if, in its prime factorisation, all prime exponents are even. So, we factorise 5000 into primes, then decide what should be divided out so that the remaining exponents are all even. The smaller the divisor N, the better, provided the quotient still becomes a perfect square. Step-by-Step Solution: Step 1: Factorise 5000. 5000 = 5 * 1000 = 5 * (10^3) = 5 * (2^3 * 5^3) = 2^3 * 5^4. Step 2: Check exponents of prime factors. We have 2^3 and 5^4. The exponent of 5 is 4 (even), but the exponent of 2 is 3 (odd). Step 3: For 5000 / N to be a perfect square, the exponents in the quotient must all be even. Let 5000 / N = 2^(3 - e2) * 5^(4 - e5), where N = 2^e2 * 5^e5 and e2, e5 are non negative integers. Step 4: We want 3 - e2 to be even and non negative. The smallest such choice is e2 = 1, giving 3 - 1 = 2, which is even. Step 5: Since 4 is already even, we can take e5 = 0, so the exponent of 5 remains 4. Step 6: Thus, the smallest divisor that works is N = 2^1 * 5^0 = 2. Step 7: Check: 5000 / 2 = 2500 = 50^2, which is a perfect square. Verification / Alternative check: Try the other options quickly: 5000 / 5 = 1000 (not a perfect square). 5000 / 10 = 500 (not a perfect square). 5000 / 15 ≈ 333.33 (not an integer, so cannot be a perfect square). This confirms that 2 is the only correct and smallest divisor among the given options. Why Other Options Are Wrong: - 5: Leaves 1000, which is not a perfect square. - 10: Leaves 500, which is not a perfect square. - 15: Does not even give an integer quotient, so it is invalid for forming a perfect square. Common Pitfalls: Students sometimes mistakenly try to make 5000 itself a perfect square by multiplying it instead of considering division as asked. Others may guess values without fully factorising 5000 and checking exponents of primes, which can easily lead to wrong answers. Another error is ignoring that the quotient must be an integer perfect square, not just a square of a fraction. Final Answer: Thus, the smallest integer by which 5000 should be divided to obtain a perfect square is 2.

Question 12

A fraction becomes 6/5 when 5 is added to its numerator and becomes 1/2 when 4 is added to its denominator. What is the value of the original fraction?

Accepted Answer

Introduction / Context: This question tests algebraic thinking using fractions. We are given how the value of a fraction changes when we add specific numbers to its numerator and denominator. From these conditions, we must find the original fraction. Such problems are common in aptitude tests because they combine fractions, equations, and basic reasoning. Given Data / Assumptions: - Let the original fraction be x / y, where x and y are positive integers and y is not zero. - When 5 is added to the numerator, the fraction becomes 6/5. - When 4 is added to the denominator, the fraction becomes 1/2. - We must find the original fraction x / y. Concept / Approach: We translate the verbal conditions into algebraic equations. Then we solve the simultaneous equations to find x and y. Once we know x and y, we compute the fraction and compare it with the given options. This uses the basic idea that if two fractions are equal, their cross products are equal. Step-by-Step Solution: Step 1: Let the fraction be x / y. Step 2: From the first condition, (x + 5) / y = 6 / 5. Cross multiply: 5(x + 5) = 6y. This gives 5x + 25 = 6y. (Equation 1) Step 3: From the second condition, x / (y + 4) = 1 / 2. Cross multiply: 2x = y + 4. So y = 2x - 4. (Equation 2) Step 4: Substitute y from Equation 2 into Equation 1. 5x + 25 = 6(2x - 4). 5x + 25 = 12x - 24. Rearrange: 25 + 24 = 12x - 5x. 49 = 7x, so x = 7. Step 5: From y = 2x - 4, substitute x = 7. y = 2 * 7 - 4 = 14 - 4 = 10. Step 6: Therefore, the original fraction is 7 / 10. Verification / Alternative check: Check with the conditions: (7 + 5) / 10 = 12 / 10 = 6 / 5, which matches the first condition. 7 / (10 + 4) = 7 / 14 = 1 / 2, which matches the second condition. Hence the fraction 7 / 10 fully satisfies both conditions. Why Other Options Are Wrong: - 8/9: Adding 5 to numerator or 4 to denominator does not give 6/5 or 1/2. - 7/8: Fails both transformation conditions when tested. - 6/11: Also does not produce 6/5 and 1/2 under the given operations. Common Pitfalls: A frequent mistake is to treat the changes to numerator and denominator incorrectly, for example adding 5 and 4 to both numerator and denominator together. Another common error is forgetting to cross multiply correctly when equating fractions. Also, some students guess from options without checking both conditions, which leads to incorrect answers. Final Answer: The original fraction that satisfies both conditions is 7/10.

Question 13

How many positive integers less than 200 are multiples of both 5 and 6?

Accepted Answer

Introduction / Context: This question checks understanding of multiples and the least common multiple (LCM). We are asked to count how many numbers below 200 are multiples of both 5 and 6. A number that is a multiple of both 5 and 6 must be a multiple of their LCM. This is a standard idea in number systems and divisibility. Given Data / Assumptions: - We consider positive integers less than 200. - Each required number must be divisible by both 5 and 6. - We will use the least common multiple of 5 and 6 to find such numbers. Concept / Approach: To be a multiple of both 5 and 6, a number must be a multiple of LCM(5, 6). Once we find the LCM, we can list or count its multiples that are strictly less than 200. This avoids checking divisibility by 5 and 6 separately for every number. Step-by-Step Solution: Step 1: Find LCM of 5 and 6. Prime factorisation: 5 = 5, 6 = 2 * 3. So LCM(5, 6) = 2 * 3 * 5 = 30. Step 2: List multiples of 30 that are less than 200. They are 30, 60, 90, 120, 150, 180. Step 3: Count these multiples. There are 6 such numbers. Verification / Alternative check: Check one example, say 90: 90 / 5 = 18 and 90 / 6 = 15, so it is divisible by both. Similarly, 180 / 5 = 36 and 180 / 6 = 30. All listed numbers are clearly multiples of 30 and hence of 5 and 6. No other multiples of 30 less than 200 are missing. Why Other Options Are Wrong: - 21: Overestimates the count; there are not so many multiples of 30 below 200. - 7: Would imply 7 multiples, but 30 * 7 = 210, which is not less than 200. - 9: Also too large, since 9 multiples would reach 270 which exceeds 200. Common Pitfalls: Some learners mistakenly add or average 5 and 6 instead of using LCM. Others include 0 or include 200 itself despite the wording “less than 200”. Misinterpreting “both 5 and 6” as “either 5 or 6” is another typical error. Final Answer: The number of integers less than 200 that are multiples of both 5 and 6 is 6.

Question 14

What is the numerical value of 2 raised to the power (3^2), that is 2^(3^2)?

Accepted Answer

Introduction / Context: This question checks understanding of exponents and the correct interpretation of powers written in a compact format. The expression is 2^(3^2), meaning 2 raised to the power 3 squared, not (2^3)^2 unless explicitly bracketed that way. Recognising the order of evaluation in exponent towers is crucial in such problems. Given Data / Assumptions: - The base is 2. - The exponent is 3^2. - 3^2 means 3 raised to the power 2. - We must compute 2^(3^2). Concept / Approach: First evaluate the inner exponent 3^2. Then use that result as the power of 2. Exponentiation is performed from the top down in an exponent tower: 2^(3^2) is 2^9, not (2^3)^2. Once we have 2^9, we can compute it step by step using repeated doubling. Step-by-Step Solution: Step 1: Compute 3^2. 3^2 = 3 * 3 = 9. Step 2: Rewrite the expression using this value. 2^(3^2) = 2^9. Step 3: Compute 2^9. 2^1 = 2. 2^2 = 4. 2^3 = 8. 2^4 = 16. 2^5 = 32. 2^6 = 64. 2^7 = 128. 2^8 = 256. 2^9 = 512. Verification / Alternative check: You can group powers to check: 2^9 = (2^3)^3 = 8^3 = 8 * 8 * 8 = 64 * 8 = 512. This matches our earlier calculation, so the value 512 is consistent and confirmed. Why Other Options Are Wrong: - 64: This is 2^6, which would come from misreading the expression. - 128: This equals 2^7, another incorrect power. - 256: This equals 2^8, which is still below 2^9. None of these match 2^9. Common Pitfalls: A common mistake is to misinterpret the notation and compute (2^3)^2 = 2^6 = 64 instead of 2^(3^2). Another error is to stop early when computing powers of 2 by hand and miscount the steps. Careful reading and systematic computation prevent these mistakes. Final Answer: The value of 2^(3^2) is 512.

Question 15

What is the digit in the units place of the product 51 × 52 × 53 × ... × 59?

Accepted Answer

Introduction / Context: This question tests your ability to find the units digit of a large product without actually multiplying all the numbers. The key idea is to focus only on the units digits and to use the fact that a factor of 10 guarantees a units digit of 0 in the product. Given Data / Assumptions: - The product is 51 × 52 × 53 × ... × 59. - These are 9 consecutive integers from 51 up to 59. - We are only interested in the units digit of this product. Concept / Approach: If the product of several integers contains at least one factor 2 and one factor 5, then the product will be divisible by 10 and the units digit will be 0. In a longer product of consecutive integers, it is almost certain that both factors 2 and 5 appear, and often an entire factor of 10 appears directly. Step-by-Step Solution: Step 1: Look for numbers ending in 0 or 5 in the product. Among 51, 52, 53, 54, 55, 56, 57, 58, 59, the only number ending in 5 is 55. Step 2: Check for even numbers. We have many even numbers: 52, 54, 56, 58 are all even. Step 3: Note that 55 provides a factor 5 and any even number provides a factor 2. So, the product contains at least one factor 2 * 5 = 10. Step 4: Once there is a factor 10 in the product, the units digit of the entire product must be 0. Verification / Alternative check: You can also notice that 55 × 52 already ends with 0 because 55 × 52 = (5 × 11) × (4 × 13) and clearly includes 5 × 2 = 10. Multiplying further by any other integers will keep the units digit 0 because 10 times any integer ends with 0. Why Other Options Are Wrong: - 4, 3, 6: These digits arise in units digits of some smaller products, but once a full factor of 10 is present, the units digit cannot be anything other than 0. Common Pitfalls: Some students try to compute the entire product or track every multiplication, which is time consuming and error prone. Others look only at the first and last numbers instead of checking for factors 2 and 5. Remember that the appearance of 2 and 5 together is enough to fix the units digit as 0. Final Answer: The units digit of the product 51 × 52 × 53 × ... × 59 is 0.

Question 16

A number exceeds one sixth of itself by 30. What is the value of this number?

Accepted Answer

Introduction / Context: This question is a simple algebra problem involving fractions of a number. It asks you to express the relationship between a number and one sixth of that number, and then solve for the original number given the difference. This is common in aptitude tests and reinforces the idea of translating words into equations. Given Data / Assumptions: - Let the unknown number be N. - The number N exceeds one sixth of itself by 30. - In symbols: N is greater than N / 6 by 30. Concept / Approach: We convert the sentence into an algebraic equation. The phrase “exceeds its 1/6 by 30” means N - (N / 6) = 30. Once we have this equation, we can solve for N by simple manipulation. The key idea is understanding that the difference between a number and a fraction of itself can be written as a single multiple of the number. Step-by-Step Solution: Step 1: Let the number be N. Step 2: Translate the description into an equation. N - N / 6 = 30. Step 3: Combine like terms on the left side. N - N / 6 = (6N / 6) - (N / 6) = 5N / 6. So we get 5N / 6 = 30. Step 4: Solve for N. Multiply both sides by 6: 5N = 30 * 6 = 180. Now divide by 5: N = 180 / 5 = 36. Verification / Alternative check: Compute one sixth of 36: 36 / 6 = 6. Now check the condition: 36 exceeds 6 by 30 because 36 - 6 = 30. So N = 36 satisfies the given statement perfectly. Why Other Options Are Wrong: - 120: One sixth is 20; 120 - 20 = 100, not 30. - 150: One sixth is 25; 150 - 25 = 125, not 30. - 24: One sixth is 4; 24 - 4 = 20, not 30. Common Pitfalls: A common mistake is to set up the equation as N / 6 - N = 30 or N + N / 6 = 30, which reverses the intended relationship. Another typical error is mishandling the fraction when combining terms or solving the resulting equation. Writing the fraction with a common denominator helps avoid these errors. Final Answer: The required number is 36.

Question 17

The sum of three consecutive even integers is 84. What is the smallest of these three even integers?

Accepted Answer

Introduction / Context: This question is a classic example of working with consecutive even integers and forming a simple linear equation. It tests understanding of how to represent consecutive even numbers algebraically and how to use a given sum to find each number. Given Data / Assumptions: - We have three consecutive even integers. - The sum of these three integers is 84. - We must find the least (smallest) of these three integers. Concept / Approach: Consecutive even integers can be represented as x, x + 2, and x + 4, where x is an even integer. Their sum is given to be 84. So we can set up the equation x + (x + 2) + (x + 4) = 84, solve for x, and then interpret x as the smallest of the three integers. Step-by-Step Solution: Step 1: Represent the three consecutive even integers. Let the smallest even integer be x. Then the next two even integers are x + 2 and x + 4. Step 2: Use the given sum. x + (x + 2) + (x + 4) = 84. Step 3: Simplify the equation. Combine like terms: x + x + x = 3x and 2 + 4 = 6. So 3x + 6 = 84. Step 4: Solve for x. Subtract 6 from both sides: 3x = 84 - 6 = 78. Divide by 3: x = 78 / 3 = 26. Step 5: Identify the three integers. They are 26, 28, and 30. So the smallest integer is 26. Verification / Alternative check: Check the sum: 26 + 28 + 30 = 84. This matches the given condition exactly, so our result is consistent. Why Other Options Are Wrong: - 28: If the smallest were 28, the others would be 30 and 32, and the sum would be 90, not 84. - 30: With 30, 32, and 34, the sum would be 96. - 32: With 32, 34, and 36, the sum would be 102. Common Pitfalls: Some students mistakenly use x, x + 1, x + 2 for consecutive even integers, which is incorrect because that represents consecutive integers, not even ones. Others misread the question and give the middle or largest integer instead of the smallest. Keeping track of what is asked and representing the pattern correctly avoids these mistakes. Final Answer: The smallest of the three consecutive even integers is 26.

Question 18

Which smallest number must be subtracted from 400 so that the resulting number is exactly divisible by 7?

Accepted Answer

Introduction / Context: This is a divisibility and remainder problem. The question asks by how much we need to decrease 400 so that the new number is divisible by 7. It checks understanding of remainders and how they relate to adjusting a number to become a multiple of a divisor. Given Data / Assumptions: - Original number = 400. - We subtract a smallest positive integer k from 400. - The resulting number 400 - k must be divisible by 7. Concept / Approach: If a number leaves a remainder r when divided by 7, then subtracting r from that number will make it divisible by 7. So we first find 400 modulo 7, that is the remainder when 400 is divided by 7. That remainder itself will be the smallest non negative number to subtract. Step-by-Step Solution: Step 1: Divide 400 by 7 to find the remainder. 7 × 57 = 399. Subtract: 400 - 399 = 1. So the remainder when 400 is divided by 7 is 1. Step 2: Use the remainder to find the required subtraction. If 400 leaves remainder 1, then 400 - 1 = 399. Step 3: Check divisibility of 399 by 7. 399 / 7 = 57 exactly, so 399 is divisible by 7. Therefore, the smallest number to be subtracted from 400 is 1. Verification / Alternative check: Try other small subtractions to confirm: 400 - 2 = 398. 398 / 7 = 56 remainder 6, not divisible. 400 - 4 = 396. 396 / 7 = 56 remainder 4, not divisible. So none of the larger subtractions gives a multiple of 7 before subtracting 1, confirming that 1 is indeed the least. Why Other Options Are Wrong: - 6: 400 - 6 = 394, which is not divisible by 7. - 2: 398 is not divisible by 7. - 4: 396 is still not divisible by 7. Common Pitfalls: Some students attempt to divide each candidate (400 - option) by 7 directly, which works but is slower. Others confuse “subtract” with “add” and mistakenly look for a number to add to 400. Remember that the remainder gives an immediate clue to how much you need to subtract to reach the nearest lower multiple of 7. Final Answer: The smallest number that must be subtracted from 400 is 1.

Question 19

By what least positive integer should 1200 be multiplied so that the product becomes a perfect square?

Accepted Answer

Introduction / Context: This problem focuses on perfect squares and prime factorisation. We are asked to find the smallest integer by which 1200 must be multiplied to obtain a perfect square. This requires understanding how prime exponents must behave in the factorisation of a perfect square. Given Data / Assumptions: - Original number = 1200. - We multiply 1200 by a positive integer k. - The product 1200 * k must be a perfect square. - We want the smallest such k. Concept / Approach: A positive integer is a perfect square if, in its prime factorisation, all the exponents are even. So we factorise 1200 into primes, look at the parity (even or odd) of each exponent, and then decide what smallest factor k must be added so that all exponents in the product 1200 * k are even. Step-by-Step Solution: Step 1: Factorise 1200. 1200 = 12 * 100. 12 = 2^2 * 3, and 100 = 2^2 * 5^2. So 1200 = (2^2 * 3) * (2^2 * 5^2) = 2^4 * 3^1 * 5^2. Step 2: Examine the exponents of primes. Exponent of 2 is 4 (even). Exponent of 3 is 1 (odd). Exponent of 5 is 2 (even). Step 3: For a perfect square, all exponents must be even. Only the exponent of 3 is odd, equal to 1. Step 4: Multiply 1200 by the smallest factor that makes exponent of 3 even. Multiplying by 3 changes the exponent of 3 from 1 to 2. So we take k = 3. Step 5: Check the new factorisation. 1200 * 3 = 3600. 3600 = 36 * 100 = 6^2 * 10^2 = (6 * 10)^2 = 60^2, which is a perfect square. Verification / Alternative check: We can quickly check other options: 1200 * 2 = 2400, not a perfect square. 1200 * 4 = 4800, not a perfect square. 1200 * 5 = 6000, not a perfect square. Therefore, only k = 3 leads to a perfect square, and it is the smallest such value. Why Other Options Are Wrong: - 2: Leaves the exponent of 3 unchanged at 1, so the product is not a perfect square. - 4: Changes only powers of 2, not 3, so exponent of 3 is still 1. - 5: Changes exponent of 5 but exponent of 3 remains odd. Common Pitfalls: A frequent mistake is to try to guess the multiplier instead of factorising the number. Some learners also think in terms of approximate square roots without checking the prime factorisation, which can be misleading. Working directly with prime exponents is the most reliable approach. Final Answer: The least integer by which 1200 should be multiplied is 3.

Question 20

Which of the following digits cannot appear in the units place of a perfect square?

Accepted Answer

Introduction / Context: This is a conceptual question about the possible last digits of perfect squares. It asks which digit cannot appear in the units place of a perfect square. Knowing the pattern of units digits for squares of integers from 0 to 9 allows us to answer quickly. Given Data / Assumptions: - We are dealing with whole numbers whose squares are perfect squares. - We only care about the units digit of the square. - Options are the digits 4, 6, 8, and 9. Concept / Approach: Any integer can be written in terms of its last digit from 0 to 9. We can square each of these digits and observe the units digit of the result. This reveals the complete set of possible units digits of any perfect square. Any digit not in this set cannot be the units digit of a perfect square. Step-by-Step Solution: Step 1: List digits 0 to 9 and compute their squares. 0^2 = 0 → units digit 0. 1^2 = 1 → units digit 1. 2^2 = 4 → units digit 4. 3^2 = 9 → units digit 9. 4^2 = 16 → units digit 6. 5^2 = 25 → units digit 5. 6^2 = 36 → units digit 6. 7^2 = 49 → units digit 9. 8^2 = 64 → units digit 4. 9^2 = 81 → units digit 1. Step 2: Collect possible units digits of any perfect square. These are 0, 1, 4, 5, 6, 9. Step 3: Compare with the options. Among 4, 6, 8, 9, we see that 8 does not appear in the list above. Verification / Alternative check: Any perfect square is the square of some integer ending with a digit 0 to 9. Since none of these give a units digit of 8 when squared, no perfect square can end with 8. This confirms that 8 is impossible as the units digit of a perfect square. Why Other Options Are Wrong: - 4: Appears from 2^2 and 8^2, so it is a valid units digit of a perfect square. - 6: Appears from 4^2 and 6^2, so it is also valid. - 9: Appears from 3^2 and 7^2, so it is valid as well. Common Pitfalls: Some learners try to reason with a few examples and may miss some digits. Others wrongly assume that all digits can occur as units digits of squares. It is safer to systematically compute squares of 0 to 9, which fully covers all possibilities due to how place value works. Final Answer: The digit that cannot appear in the units place of a perfect square is 8.

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