Question 1

The highest common factor (HCF) of 3240, 3600, and a third number is 36, and their least common multiple (LCM) is 2^4 × 3^5 × 5^2 × 7^2. What is the value of this third number?

Accepted Answer

Introduction: This is a challenging number theory problem involving both the highest common factor (HCF) and least common multiple (LCM) of three numbers. You must use prime factorisation and the relationships between HCF, LCM, and the given numbers to identify the unknown third number. Given Data / Assumptions: First number = 3240 Second number = 3600 Third number = unknown HCF of all three numbers = 36 LCM of all three numbers = 2^4 × 3^5 × 5^2 × 7^2 Concept / Approach: For any set of numbers, the HCF is obtained by taking common primes with minimum exponents, while the LCM is obtained by taking all primes that appear with maximum exponents. We factorise the known numbers, compare them with the given LCM and HCF, and deduce the prime factorisation of the unknown third number so that all conditions are satisfied. Step-by-Step Solution: First factorise the known numbers: 3240 = 2^3 × 3^4 × 5 3600 = 2^4 × 3^2 × 5^2 Given LCM of the three numbers = 2^4 × 3^5 × 5^2 × 7^2 Given HCF of the three numbers = 36 = 2^2 × 3^2 Notice that 7 and 7^2 appear only in the LCM, so they must come entirely from the third number. Also, the LCM has 3^5, while current maximum from 3240 and 3600 is 3^4, so the third number must contribute at least 3^5. Because the HCF is 2^2 × 3^2, the third number must contain 2^2 and at least 3^2, but it must not add any factor of 5 beyond the given minimum pattern. From the HCF condition: gcd(360, third number) = 36. This forces the exponent of 2 in the third number to be exactly 2 and the exponent of 5 to be 0. Therefore the third number must be: 2^2 × 3^5 × 7^2 Compute: 2^2 = 4, 3^5 = 243, 7^2 = 49 Third number = 4 × 243 × 49 = 47628 Verification / Alternative check: If we compute gcd(3240, 3600, 47628), we get 36. If we compute lcm(3240, 3600, 47628), we obtain 2^4 × 3^5 × 5^2 × 7^2, exactly as given. This double check confirms that 47628 satisfies both HCF and LCM conditions. Why Other Options Are Wrong: The other options do not produce the required prime factorisation conditions. For example, they either include extra prime factors of 5, miss the factor 7^2, or do not respect the HCF constraint of 36. None of them give the correct combination of HCF and LCM simultaneously. Common Pitfalls: Students often try to solve such questions using only approximate divisibility without doing complete prime factorisation. Another mistake is to ignore the HCF condition after matching the LCM, or vice versa. Both constraints must be checked carefully because they strongly restrict the possible third number. Final Answer: The required third number that satisfies the given HCF and LCM conditions is 47628.

Question 2

Three tanks have capacities 98 litres, 182 litres, and 266 litres. What is the capacity, in litres, of the largest measuring cylinder that can exactly fill each of the three tanks an integer number of times?

Accepted Answer

Introduction: This question applies the concept of the highest common factor (HCF) to a real life situation involving measuring liquids. We want the largest possible measuring cylinder that can measure all three given quantities exactly, without leaving any remainder in any tank. Given Data / Assumptions: Tank capacities: 98 litres, 182 litres, and 266 litres We need the largest possible cylinder capacity that will divide each capacity exactly. Units of all capacities are in litres. Concept / Approach: To find the largest measure that can exactly measure all given quantities, we need the HCF of the three numbers. The HCF is the greatest positive integer that divides each number without leaving a remainder. Prime factorisation or the Euclidean algorithm can be used to find this HCF. Step-by-Step Solution: Step 1: Find HCF of 98 and 182: 98 = 2 × 7^2 182 = 2 × 7 × 13 Common primes: 2 × 7 = 14, so HCF(98, 182) = 14 Step 2: Find HCF of the result with 266: 266 = 2 × 7 × 19 HCF(14, 266): 14 = 2 × 7, and 266 has 2 and 7 as factors, so HCF = 14 Therefore, HCF(98, 182, 266) = 14 So the largest measuring cylinder that can be used is 14 litres Verification / Alternative check: We can verify by division: 98 ÷ 14 = 7, 182 ÷ 14 = 13, and 266 ÷ 14 = 19. All results are integers, confirming that 14 litres exactly measures each capacity. No larger common divisor exists, because any larger number would not divide at least one of the capacities without remainder. Why Other Options Are Wrong: 7 litres divides all three capacities, but it is not the largest possible measure. 42 litres does not divide all the capacities exactly (for example, 182 ÷ 42 is not an integer). 98 litres also fails because it does not divide 182 or 266 evenly. Hence only 14 litres is correct. Common Pitfalls: Learners may confuse HCF with LCM and attempt to find a common multiple instead of a common divisor. Another mistake is to choose a measure that works for two of the tanks but not for the third. Always confirm that the chosen measure divides every given capacity exactly. Final Answer: The capacity of the largest measuring cylinder that can exactly measure all three tanks is 14 litres.

Question 3

If N is the greatest number that divides 1305, 4665, and 6905 leaving the same remainder in each case, what is the sum of the digits of N?

Accepted Answer

Introduction: This question tests your understanding of how to use the highest common factor (HCF) when a divisor leaves the same remainder in several divisions. This is a standard trick used in many aptitude questions involving remainders and common divisors. Given Data / Assumptions: N divides 1305, 4665, and 6905 leaving the same remainder each time. We must find the greatest possible N. We finally need the sum of the digits of N. Concept / Approach: If a number N divides several numbers leaving the same remainder in each case, then N will exactly divide the difference between any pair of those numbers. Therefore, we first compute the pairwise differences and then find the HCF of those differences. That HCF is the greatest possible N. Step-by-Step Solution: Compute differences: 4665 − 1305 = 3360 6905 − 4665 = 2240 6905 − 1305 = 5600 Now N must divide 3360, 2240, and 5600 exactly. Find HCF of 3360 and 2240: HCF(3360, 2240) = 1120 Now find HCF of 1120 and 5600: HCF(1120, 5600) = 1120 Therefore N = 1120 Sum of digits of N = 1 + 1 + 2 + 0 = 4 Verification / Alternative check: To verify, we can calculate the remainders when 1305, 4665, and 6905 are divided by 1120. In each case, the remainder is the same. Also, any larger number would fail to divide at least one of the differences 3360, 2240, or 5600 exactly, so 1120 is indeed the greatest such number. Why Other Options Are Wrong: Options 6, 8, and 9 are just different possible digit sums but do not match the actual digit sum of 1120. They would correspond to different hypothetical divisors, which do not satisfy the requirement of being the greatest number that leaves the same remainder in all three divisions. Common Pitfalls: A common error is to try to find a divisor of the given numbers directly instead of looking at their differences. Another mistake is to compute only one difference and take its factors without checking the other differences. Always use all given numbers to ensure the divisor works for every pair. Final Answer: The greatest number N has digit sum equal to 4.

Question 4

Find the greatest common divisor (GCD or HCF) of the decimal numbers 1.08, 0.36, and 0.90.

Accepted Answer

Introduction: This question checks your ability to handle greatest common divisor (GCD or HCF) calculations when numbers are given in decimal form. The key idea is to convert the decimals into whole numbers, find the HCF, and then convert back if necessary. Given Data / Assumptions: Numbers: 1.08, 0.36, and 0.90 We must find their greatest common divisor. All values are given with two decimal places. Concept / Approach: Dealing directly with decimals in divisibility problems is inconvenient. Instead, multiply all numbers by 100 to remove the decimal points. This does not change their ratio. Then find the HCF of the resulting integers. Finally, divide that HCF by 100 to restore the original scale and get the GCD of the decimal numbers. Step-by-Step Solution: Multiply each number by 100 to avoid decimals: 1.08 × 100 = 108 0.36 × 100 = 36 0.90 × 100 = 90 Now find HCF of 108, 36, and 90. First find HCF(108, 36): 108 = 2^2 × 3^3 36 = 2^2 × 3^2 So HCF(108, 36) = 2^2 × 3^2 = 4 × 9 = 36 Now find HCF(36, 90): 90 = 2 × 3^2 × 5 Common primes: 2^1 × 3^2 = 2 × 9 = 18 So HCF(108, 36, 90) = 18 We had multiplied by 100, so divide by 100: GCD of original decimals = 18 ÷ 100 = 0.18 Verification / Alternative check: You can check that 0.18 divides each of the original numbers exactly: 1.08 ÷ 0.18 = 6, 0.36 ÷ 0.18 = 2, and 0.90 ÷ 0.18 = 5. No larger decimal will divide all three numbers without leaving a remainder, so 0.18 is indeed the greatest common divisor. Why Other Options Are Wrong: 0.03 and 0.108 are smaller common divisors but not the greatest. 0.90 is larger than some of the numbers and clearly cannot divide all of them. Therefore only 0.18 satisfies the definition of GCD for these decimals. Common Pitfalls: Students sometimes forget to scale all numbers by the same power of 10, or they attempt to find the HCF directly on decimals, which can be confusing. Remember that multiplying by a common factor and later dividing by the same factor preserves divisibility relationships. Final Answer: The greatest common divisor of 1.08, 0.36, and 0.90 is 0.18.

Question 5

Find the greatest number that will divide 964, 1238, and 1400, leaving remainders 41, 31, and 51 respectively. What is this greatest number?

Accepted Answer

Introduction: This question is another classic remainder and HCF problem. A single number divides several given numbers but leaves different remainders. We must use the differences between the numbers and their remainders to determine the greatest possible divisor. Given Data / Assumptions: On division by the required number N: 964 leaves remainder 41 1238 leaves remainder 31 1400 leaves remainder 51 We must find the greatest such N. Concept / Approach: If N divides 964 and leaves remainder 41, then 964 − 41 is exactly divisible by N. Similarly for the other numbers. Therefore, N divides the adjusted values 964 − 41, 1238 − 31, and 1400 − 51. These adjusted values are all multiples of N, so N must be a common divisor of them. The greatest such N is the HCF of those adjusted values. Step-by-Step Solution: Compute adjusted values: 964 − 41 = 923 1238 − 31 = 1207 1400 − 51 = 1349 Now we need HCF(923, 1207, 1349). Observe divisibility: 923 = 13 × 71 1207 = 17 × 71 1349 = 19 × 71 All three numbers have the common factor 71 and no higher common factor. Therefore HCF(923, 1207, 1349) = 71 Hence, the greatest number that satisfies the given conditions is 71 Verification / Alternative check: Check by division: 964 ÷ 71 gives quotient 13 and remainder 41. 1238 ÷ 71 gives quotient 17 and remainder 31. 1400 ÷ 71 gives quotient 19 and remainder 51. All remainders match the problem statement, confirming that 71 works. No larger common divisor exists because 923, 1207, and 1349 share only 71 as a common prime factor. Why Other Options Are Wrong: 64, 69, and 58 do not divide the adjusted values 923, 1207, and 1349 exactly. At least one of the divisions would leave a non zero remainder, which violates the requirement that N must divide each adjusted value perfectly. Common Pitfalls: One common mistake is to try to directly find a number that fits the three original remainders without subtracting them. Another error is to compute the HCF of the original numbers 964, 1238, and 1400, which does not incorporate the remainder information and leads to the wrong answer. Final Answer: The greatest number that satisfies all the given conditions is 71.

Question 6

Find the least common multiple (LCM) of the two numbers 15 and 12.

Accepted Answer

Introduction: This is a straightforward question on least common multiple (LCM) of two numbers. Questions like this commonly appear in the HCF and LCM section of aptitude examinations and are meant to test basic factorisation skills. Given Data / Assumptions: Two numbers: 15 and 12 We want the least positive integer that is exactly divisible by both 15 and 12. Concept / Approach: To find the LCM using prime factorisation, we express each number as a product of prime factors. The LCM is obtained by taking each distinct prime with the highest exponent that occurs in any of the numbers. Multiplying these together gives the LCM. Step-by-Step Solution: Prime factorise each number: 15 = 3 × 5 12 = 2^2 × 3 Collect all primes involved: 2, 3, and 5 Take the highest power of each prime: For 2: highest power is 2^2 For 3: highest power is 3^1 For 5: highest power is 5^1 LCM = 2^2 × 3 × 5 Calculate step by step: 2^2 = 4 4 × 3 = 12 12 × 5 = 60 So, LCM(15, 12) = 60 Verification / Alternative check: Check divisibility: 60 ÷ 15 = 4 and 60 ÷ 12 = 5. Both quotients are integers, so 60 is a common multiple. Also, no smaller number than 60 can be simultaneously divisible by both 15 and 12, because any smaller candidate would lack sufficient powers of 2 or 5. Why Other Options Are Wrong: 3 is a common factor, not a common multiple. 12 is divisible by 3 but not by 15, so it cannot be the LCM. 30 is a common multiple of 15 but not of 12, because 30 ÷ 12 is not an integer. Hence, only 60 satisfies all the conditions for being the least common multiple. Common Pitfalls: Students sometimes confuse HCF and LCM and mistakenly select a common factor like 3 or 12 instead of a common multiple. Another typical error is to add or subtract the two numbers rather than use factorisation. Always remember that LCM refers to multiples, not factors. Final Answer: The least common multiple of 15 and 12 is 60.

Question 7

Three numbers are in the ratio 3 : 4 : 5 and their least common multiple (LCM) is 3600. What is the highest common factor (HCF) of these three numbers?

Accepted Answer

Introduction: This problem combines the concepts of ratios, least common multiple (LCM), and highest common factor (HCF). You are given that three numbers are in a simple ratio and their LCM is known. From this, you must determine the HCF of the numbers. Given Data / Assumptions: The three numbers are in the ratio 3 : 4 : 5. The LCM of the three numbers is 3600. We need to find the HCF of the three numbers. Concept / Approach: If three numbers are 3k, 4k, and 5k, where k is their common factor, then their HCF is k (because 3, 4, and 5 themselves are pairwise co prime). Their LCM will then equal 3 × 4 × 5 × k, because the LCM of 3, 4, and 5 is 60. Given the LCM value, we can solve for k and thus find the HCF. Step-by-Step Solution: Let the numbers be 3k, 4k, and 5k. Since 3, 4, and 5 are co prime, their LCM is 3 × 4 × 5 = 60. LCM of the actual numbers 3k, 4k, and 5k is therefore 60 × k. We are given that this LCM is 3600. So we have: 60 × k = 3600 Solve for k: k = 3600 / 60 = 60 Hence, the HCF of the three numbers is k = 60 Verification / Alternative check: With k = 60, the numbers are 3 × 60 = 180, 4 × 60 = 240, and 5 × 60 = 300. The HCF of these numbers is clearly 60. The LCM of 180, 240, and 300 is 3600, which matches the given value. This confirms that the reasoning and computation are correct. Why Other Options Are Wrong: 40, 100, and 120 do not satisfy the equation 60 × k = 3600 if treated as k. If they were the HCF, we would obtain inconsistent numbers or an incorrect LCM. Only k = 60 makes the ratio and the given LCM work together correctly. Common Pitfalls: Some learners mistakenly treat 3, 4, and 5 as the actual numbers rather than scaled versions. Others confuse the relationship between LCM and HCF for ratio based problems. Remember that when numbers are expressed as multiples of k, the common factor k is typically the HCF, and the LCM scales accordingly. Final Answer: The highest common factor of the three numbers is 60.

Question 8

A palace has 18,000 gold coins, 9,600 silver coins, and 3,600 bronze coins. All rooms must contain the same number of coins, and each room can only contain coins of a single type. What is the minimum number of rooms required?

Accepted Answer

Introduction: This problem applies the concept of highest common factor (HCF) in a practical distribution scenario. The palace has three types of coins, and each room should hold an equal number of coins, but only one type per room. We must minimise the number of rooms while respecting these rules. Given Data / Assumptions: Gold coins = 18,000 Silver coins = 9,600 Bronze coins = 3,600 Each room contains coins of only one type. Every room must contain the same number of coins, say K. We need the minimum number of rooms. Concept / Approach: If each room contains K coins and K is the same for all rooms, then K must divide 18,000, 9,600, and 3,600 exactly. To minimise the number of rooms, we should maximise K, which means K must be the HCF of the three counts. Once K is known, we can find the number of rooms by dividing each total by K and adding them up. Step-by-Step Solution: Find HCF of 18,000, 9,600, and 3,600. Factor 18,000 = 180 × 100 = 18 × 10 × 100 = 2^4 × 3^2 × 5^3 Factor 9,600 = 96 × 100 = 2^7 × 3 × 5^2 Factor 3,600 = 36 × 100 = 2^4 × 3^2 × 5^2 HCF is product of common primes with smallest exponents: For 2: min power = 2^4 For 3: min power = 3^1 For 5: min power = 5^2 So HCF = 2^4 × 3 × 5^2 = 16 × 3 × 25 = 1200 Thus each room should contain K = 1200 coins. Number of rooms for gold = 18,000 ÷ 1200 = 15 Number of rooms for silver = 9,600 ÷ 1200 = 8 Number of rooms for bronze = 3,600 ÷ 1200 = 3 Total rooms = 15 + 8 + 3 = 26 Verification / Alternative check: We can check that no larger value than 1200 can divide all three numbers. Increasing K would make at least one division non integer. Since 1200 is the HCF, it is the largest possible K, which ensures the minimum possible number of rooms. The computed 26 rooms therefore satisfies all conditions. Why Other Options Are Wrong: Choosing a smaller K, such as 600, would increase the number of rooms. That would contradict the requirement of minimum rooms. The options 12, 18, and 24 correspond to such smaller K values and therefore are not optimal. Only 26 matches the HCF based calculation. Common Pitfalls: A common error is to find the least common multiple (LCM) instead of the HCF, which has no meaning here. Another mistake is to divide by an arbitrary common factor instead of the highest one, leading to a higher number of rooms than necessary. Final Answer: The minimum number of rooms required is 26.

Question 9

Three containers hold mixtures of milk and water in quantities 1365 litres, 1560 litres, and 1755 litres respectively. What is the largest capacity measure (in litres) that can exactly measure all three quantities?

Accepted Answer

Introduction: This question is a standard application of the highest common factor (HCF) concept to liquid measuring problems. We want a single largest measuring vessel that can be used to measure each of the three quantities exactly, without any fraction left over in any container. Given Data / Assumptions: Container 1: 1365 litres Container 2: 1560 litres Container 3: 1755 litres We need the largest capacity measure that divides all three exactly. Concept / Approach: The largest measure that can be used to measure each quantity exactly is the HCF of the three numbers. We can find this using prime factorisation or the Euclidean algorithm. Once we compute the HCF, that value directly gives the required maximum capacity in litres. Step-by-Step Solution: We will use successive HCF calculations. First find HCF of 1365 and 1560. Compute difference: 1560 − 1365 = 195 Now HCF(1365, 1560) = HCF(1365, 195) 1365 ÷ 195 = 7 exactly, so HCF(1365, 1560) = 195 Next, find HCF of 195 and 1755. 1755 ÷ 195 = 9 exactly, so HCF(195, 1755) = 195 Therefore HCF(1365, 1560, 1755) = 195 So the largest measure that can measure all three quantities exactly is 195 litres Verification / Alternative check: Check divisibility directly: 1365 ÷ 195 = 7, 1560 ÷ 195 = 8, and 1755 ÷ 195 = 9. All results are integers, confirming that 195 litres exactly measures each quantity. No larger number can divide all three values without remainder because 195 is already the greatest common divisor. Why Other Options Are Wrong: 37, 97, and 129 litres are all factors of some numbers but not common to all three. They either fail to divide at least one quantity exactly or are smaller than the true HCF. Therefore, they do not represent the largest possible measure. Common Pitfalls: Many test takers mistakenly try to use the least common multiple (LCM), which would give a huge value and has no relevance to the measuring problem. Others try trial and error with small divisors without checking carefully. Using the Euclidean algorithm for HCF is both systematic and reliable. Final Answer: The largest capacity measure that can exactly measure all three quantities is 195 litres.

Question 10

Find the highest common factor (HCF) of the two numbers 865 and 2595.

Accepted Answer

Introduction: This is a basic question on the highest common factor (HCF) of two numbers. It is designed to test whether you can recognise when one number is an exact multiple of another and hence immediately see the HCF. Given Data / Assumptions: Two numbers: 865 and 2595 We are asked to find their HCF. Concept / Approach: The HCF of two numbers is the largest positive integer that divides both numbers without leaving any remainder. If one number divides the other exactly, then the smaller number itself is the HCF. We can check divisibility either by simple multiplication or by the Euclidean algorithm. Step-by-Step Solution: Check whether 2595 is a multiple of 865. Compute 865 × 3 = 2595 Since 2595 equals 865 × 3 exactly, 865 divides 2595 without remainder. Therefore, any common factor of 865 and 2595 must also be a factor of 865. The largest such factor is clearly 865 itself. So HCF(865, 2595) = 865 Verification / Alternative check: Using the Euclidean algorithm: divide 2595 by 865. The quotient is 3 and the remainder is 0. When the remainder becomes 0, the divisor at that step is the HCF. Here that divisor is 865, confirming the answer. Why Other Options Are Wrong: 5 and 25 are factors of both numbers but they are not the highest common factor. 2595 cannot be the HCF because it is greater than 865 and obviously does not divide 865. Therefore only 865 satisfies the requirement of being the greatest common divisor. Common Pitfalls: Students sometimes choose a small common factor like 5 or 25 because they do not check whether one number is an exact multiple of the other. Always look for the possibility that the smaller number might divide the larger number completely, which immediately gives you the HCF. Final Answer: The highest common factor of 865 and 2595 is 865.

Question 11

Find the least common multiple (LCM) of the two numbers 72 and 84.

Accepted Answer

Introduction: This question tests your ability to compute the least common multiple (LCM) of two numbers using prime factorisation. The numbers 72 and 84 are typical values used in aptitude questions to practice LCM calculations. Given Data / Assumptions: Two numbers: 72 and 84 We must find their LCM. Concept / Approach: The LCM of two numbers is the smallest positive integer that is a multiple of both numbers. Using prime factorisation, we express each number as a product of primes and then take each prime with the highest exponent found in either number. Multiplying those prime powers gives the LCM. Step-by-Step Solution: Prime factorise the numbers: 72 = 2^3 × 3^2 84 = 2^2 × 3 × 7 List all primes involved: 2, 3, and 7. Select the highest powers: For 2: highest power is 2^3 For 3: highest power is 3^2 For 7: highest power is 7^1 So LCM = 2^3 × 3^2 × 7 Compute step by step: 2^3 = 8 3^2 = 9 8 × 9 = 72 72 × 7 = 504 Therefore LCM(72, 84) = 504 Verification / Alternative check: Check divisibility: 504 ÷ 72 = 7 and 504 ÷ 84 = 6. Both results are integers, so 504 is indeed a common multiple of 72 and 84. Any smaller candidate would not contain enough factors of 2, 3, or 7 to be divisible by both numbers. Why Other Options Are Wrong: 252 is divisible by 84 but not by 72. 360 and 420 are also not divisible by both 72 and 84 simultaneously. Hence none of these can be the LCM, even though they are smaller than 504. Only 504 satisfies the definition of the least common multiple. Common Pitfalls: Some students may mistakenly multiply the two numbers directly, leading to 6048, which is a common multiple but not the least. Others may forget to take the highest powers of primes and instead use lower powers, which leads to a number that is not divisible by one of the originals. Final Answer: The least common multiple of 72 and 84 is 504.

Question 12

A natural number gives a remainder of 3 when divided by 9, 10, 12, or 15. What is the smallest possible value of this natural number?

Accepted Answer

Introduction: This question examines your understanding of modular arithmetic and the use of least common multiple (LCM) to handle conditions involving the same remainder for several divisors. It is a typical number system problem in aptitude tests. Given Data / Assumptions: A natural number N. N leaves remainder 3 when divided by 9, 10, 12, and 15. We must find the smallest such N. Concept / Approach: If a number N leaves the same remainder r when divided by several divisors, then N − r is exactly divisible by each of those divisors. Therefore, N − 3 must be a common multiple of 9, 10, 12, and 15. The smallest possible N occurs when N − 3 is the LCM of these divisors. Step-by-Step Solution: Let N be the required number. Given: N ≡ 3 (mod 9), N ≡ 3 (mod 10), N ≡ 3 (mod 12), N ≡ 3 (mod 15) So N − 3 is divisible by 9, 10, 12, and 15. Compute LCM of 9, 10, 12, and 15. 9 = 3^2 10 = 2 × 5 12 = 2^2 × 3 15 = 3 × 5 LCM needs highest powers: 2^2, 3^2, 5 LCM = 2^2 × 3^2 × 5 = 4 × 9 × 5 = 180 Thus, N − 3 = 180 for the smallest N. So N = 180 + 3 = 183 Verification / Alternative check: Check divisibility: 183 ÷ 9 leaves remainder 3. 183 ÷ 10 leaves remainder 3. 183 ÷ 12 leaves remainder 3. 183 ÷ 15 leaves remainder 3. These verifications confirm that 183 satisfies all conditions. Any smaller number would correspond to a smaller common multiple than 180, which is impossible because 180 is the LCM. Why Other Options Are Wrong: 63, 123, and 153 do not leave remainder 3 for all four divisors simultaneously. At least one of the divisors gives a different remainder for each of these numbers, so they cannot satisfy the given conditions. Only 183 works for every divisor. Common Pitfalls: Students sometimes attempt random trial and error or only check some of the divisors. Another common mistake is to confuse HCF with LCM and use the HCF instead of the LCM, which leads to much smaller and incorrect values of N. Final Answer: The smallest natural number satisfying all the conditions is 183.

Question 13

The LCM and HCF of two numbers are 4284 and 32 respectively. If one of the numbers is 672, what is the other number?

Accepted Answer

Introduction: This question uses a key relationship between two numbers, their highest common factor (HCF), and their least common multiple (LCM). Given the LCM, HCF, and one number, you are asked to find the second number. Given Data / Assumptions: LCM of two numbers = 4284 HCF of the two numbers = 32 One number = 672 We must find the other number. Concept / Approach: For two positive integers a and b, there is a direct relation: a × b = HCF(a, b) × LCM(a, b). This formula allows us to compute the missing number when the other three values are known. We simply rearrange it to solve for the unknown number. Step-by-Step Solution: Let the unknown number be x. Given: HCF = 32, LCM = 4284, known number = 672. Use the relation: a × b = HCF × LCM So 672 × x = 32 × 4284 Compute the right side: 32 × 4284 = 137088 So 672 × x = 137088 Now solve for x: x = 137088 ÷ 672 137088 ÷ 672 = 204 Therefore, the other number is 204 Verification / Alternative check: We can check that HCF(672, 204) is 32 and LCM(672, 204) is 4284. Using prime factorisation: 672 = 2^5 × 3 × 7, 204 = 2^2 × 3 × 17. The HCF is 2^2 × 3 = 32, and the LCM is the product of HCF and co-prime parts, which indeed yields 4284. This confirms that 204 is correct. Why Other Options Are Wrong: The options 64, 92, and 102, when multiplied with 672, do not satisfy the product equation HCF × LCM. They either give the wrong HCF, wrong LCM, or both. Therefore they cannot be the other number in this pair. Common Pitfalls: Students sometimes forget the formula relating HCF, LCM, and the product of two numbers, and instead try to guess factors. Others may make arithmetic mistakes while multiplying or dividing large numbers. Writing the product relation clearly and simplifying step by step reduces these errors. Final Answer: The second number whose HCF and LCM with 672 are 32 and 4284 is 204.

Question 14

If a and b are two co-prime natural numbers (with a greater than b), what is the least common multiple (LCM) of a and b?

Accepted Answer

Introduction: This conceptual question tests your understanding of co-prime numbers and their least common multiple (LCM). Co-prime numbers are pairs of integers whose highest common factor (HCF) is 1, even though each number may be greater than 1. Given Data / Assumptions: a and b are natural numbers. a and b are co-prime, so HCF(a, b) = 1. a is greater than b. We need the LCM of a and b. Concept / Approach: For any two positive integers a and b, there is a fundamental relationship: a × b = HCF(a, b) × LCM(a, b). If a and b are co-prime, their HCF is 1. Substituting this into the formula directly gives the LCM of two co-prime numbers. Step-by-Step Solution: Start with the general formula: a × b = HCF(a, b) × LCM(a, b) Given that a and b are co-prime, HCF(a, b) = 1. So the formula becomes: a × b = 1 × LCM(a, b) Therefore, LCM(a, b) = a × b Thus, for co-prime numbers, the LCM is simply the product ab Verification / Alternative check: Take a simple example: a = 3 and b = 4. These numbers are co-prime. The multiples of 3 are 3, 6, 9, 12, 15, and so on. The multiples of 4 are 4, 8, 12, 16, and so on. The smallest common multiple is 12, which is 3 × 4. This matches the rule LCM = ab for co-prime numbers. Why Other Options Are Wrong: a and b by themselves cannot be the LCM because the LCM must be a number that is a multiple of both a and b. a + b is not guaranteed to be divisible by either a or b, so it is not a general formula for the LCM of co-prime numbers. Only ab always works. Common Pitfalls: Learners sometimes think that the LCM of co-prime numbers is their sum or the larger of the two numbers. This is incorrect. The product ab is the smallest number that has all the prime factors of both a and b with correct multiplicities when they share no common prime factor. Final Answer: For two co-prime natural numbers a and b, the least common multiple is ab.

Question 15

The product of two two-digit numbers is 2160 and their highest common factor (HCF) is 12. What are these two numbers?

Accepted Answer

Introduction: This question combines the concepts of HCF and factor pairs of a product. You know the product of two numbers and their HCF and must determine the two original two digit numbers. This type of reasoning is common in aptitude exams. Given Data / Assumptions: The product of the two numbers is 2160. The HCF of the two numbers is 12. Both numbers are two digit numbers. Concept / Approach: Let the two numbers be 12x and 12y, because 12 is their HCF. Then their HCF is 12 × HCF(x, y). For the HCF to be exactly 12, x and y must be co-prime. Also, the product 12x × 12y must equal 2160. This gives an equation in terms of x and y that we can solve to find co-prime factors. Step-by-Step Solution: Let the numbers be 12x and 12y. Their product is 12x × 12y = 144xy. We are given that 144xy = 2160. So xy = 2160 ÷ 144 Compute: 2160 ÷ 144 = 15 Thus x and y are co-prime factors of 15. The positive factor pairs of 15 are (1, 15) and (3, 5). If x = 1, y = 15, the numbers are 12 and 180 (but 180 is not a two digit number). If x = 3, y = 5, the numbers are 36 and 60, both two digit numbers. Therefore, the required numbers are 36 and 60 Verification / Alternative check: Check product: 36 × 60 = 2160, which matches the given product. Check HCF(36, 60): 36 = 2^2 × 3^2, 60 = 2^2 × 3 × 5, so common primes with smallest powers give HCF = 2^2 × 3 = 12. Both conditions are satisfied. Why Other Options Are Wrong: 60 and 72 have HCF 12 but their product is 4320, not 2160. 12 and 60 have product 720 and also 12 is not a two digit number. 72 and 30 have product 2160 but HCF(72, 30) is 6, not 12. Only 36 and 60 satisfy both the HCF and product conditions. Common Pitfalls: Students may only match the product 2160 or only match the HCF 12, forgetting that both conditions must simultaneously hold true. Others may ignore the two digit requirement, accidentally choosing numbers like 12 and 180, which would be invalid here. Final Answer: The two required two digit numbers are 36 and 60.

Question 16

A student can arrange her books into groups of 5, groups of 9, or groups of 13, with no books left over in each case. What is the smallest possible total number of books she can have?

Accepted Answer

Introduction: This problem tests your understanding of the least common multiple (LCM) in a practical context. The student can divide her books into equal groups of various sizes without any remainder, which implies that the total number of books is a common multiple of these group sizes. Given Data / Assumptions: The student can arrange the books in groups of 5 with none left over. She can also arrange them in groups of 9 with none left over. She can also arrange them in groups of 13 with none left over. We need the smallest possible total number of books. Concept / Approach: If a number can be divided exactly by 5, 9, and 13, it must be a common multiple of 5, 9, and 13. The smallest such positive integer is the least common multiple (LCM). By calculating the LCM of 5, 9, and 13, we obtain the smallest possible total number of books. Step-by-Step Solution: Prime factorise the numbers: 5 is prime. 9 = 3^2 13 is prime. The LCM is the product of all primes with their highest powers present: LCM = 5 × 3^2 × 13 Compute step by step: 3^2 = 9 5 × 9 = 45 45 × 13 = 585 Therefore, the smallest possible number of books is 585 Verification / Alternative check: Check divisibility: 585 ÷ 5 = 117, 585 ÷ 9 = 65, and 585 ÷ 13 = 45. All results are integers, so the student can indeed arrange her books into groups of 5, 9, or 13 without leftovers. No smaller number can satisfy all three divisibility conditions simultaneously because 585 is the LCM. Why Other Options Are Wrong: 487, 635, and 705 are not divisible by all three numbers 5, 9, and 13. For instance, 487 is prime and not divisible by 5. 635 is divisible by 5 but not by 9. 705 is divisible by 5 and 9 but not by 13. Hence they cannot be the smallest such total. Common Pitfalls: Learners sometimes only check one or two of the divisibility conditions or rely on random trials. Another mistake is to compute the HCF instead of the LCM. Always remember that when a number must be divisible by several given numbers, the LCM is the fundamental tool to use. Final Answer: The smallest possible number of books the student can have is 585.

Question 17

Find the highest common factor (HCF) of the two numbers 3341 and 3328.

Accepted Answer

Introduction: This problem tests your ability to use the Euclidean algorithm to find the highest common factor (HCF) of two numbers that are close in value. The Euclidean method is often faster and more reliable than full prime factorisation for large numbers. Given Data / Assumptions: Two numbers: 3341 and 3328 We must find their HCF. Concept / Approach: The Euclidean algorithm states that HCF(a, b) = HCF(b, a − b) when a is greater than b. Repeatedly applying this rule with smaller and smaller pairs eventually leads to a remainder of 0, and the last non zero remainder is the HCF. This is especially convenient when the two numbers are relatively close together. Step-by-Step Solution: Let a = 3341 and b = 3328, with a greater than b. First compute the difference: a − b = 3341 − 3328 = 13 So HCF(3341, 3328) = HCF(3328, 13) Now divide 3328 by 13: 3328 ÷ 13 = 256 exactly (13 × 256 = 3328) Since the remainder is 0, the divisor 13 is the HCF. Therefore, HCF(3341, 3328) = 13 Verification / Alternative check: We can also check that 3341 is divisible by 13: 13 × 257 = 3341. Both numbers 3341 and 3328 are thus multiples of 13, and no larger number divides both because their difference is 13, and any common divisor must divide this difference as well. Hence 13 is indeed the greatest common divisor. Why Other Options Are Wrong: 31, 257, and 337 do not divide both 3341 and 3328 exactly. In particular, 257 divides 3341 but not 3328, and 337 is larger than the difference between the numbers, so it cannot be a common divisor. Hence they cannot be the HCF. Common Pitfalls: A common mistake is to attempt factorisation of each number separately, which is time consuming and prone to errors. Another mistake is to overlook the simple use of the difference a − b when the numbers are close. Using the Euclidean algorithm streamlines the process significantly. Final Answer: The highest common factor of 3341 and 3328 is 13.

Question 18

In number theory, the product of two positive integers is 3026 and their least common multiple (LCM) is 89. What is the highest common factor (HCF) of these two numbers?

Accepted Answer

Introduction:In this aptitude question, you are asked to relate the product of two numbers with their least common multiple (LCM) in order to find their highest common factor (HCF). This is a standard number system concept that frequently appears in competitive exams and placement tests. Given Data / Assumptions: Product of the two integers = 3026. Least common multiple (LCM) of the two numbers = 89. Both numbers are positive integers. We need to determine their highest common factor (HCF). Concept / Approach:The key concept used here is the fundamental relationship between HCF, LCM and the product of two numbers. For any two positive integers a and b, we use:a * b = HCF(a, b) * LCM(a, b)This allows us to compute HCF if we know the product and the LCM. Step-by-Step Solution:Step 1: Write the relationship.a * b = HCF * LCMStep 2: Substitute the known values.3026 = HCF * 89Step 3: Solve for HCF by dividing the product by the LCM.HCF = 3026 / 89Step 4: Perform the division.89 * 34 = 3026, so HCF = 34 Verification / Alternative check:We can also think of the two numbers as 89 * k and 34 * m with suitable integers k and m such that their product equals 3026. Since 3026 / 89 is exactly 34, it is reasonable that 34 is the common factor that fits the relation. The fact that the division is exact confirms the correctness. Why Other Options Are Wrong:33: If HCF were 33, then HCF * LCM = 33 * 89 = 2937, which is not equal to 3026.35: 35 * 89 = 3115, again not matching the given product 3026.29: 29 * 89 = 2581, also not equal to 3026.38: 38 * 89 = 3382, which is greater than 3026 and incorrect. Common Pitfalls:Many students confuse the relationship between HCF, LCM and the product, or try to factorize the numbers directly, which is unnecessary here. Another common mistake is to think that HCF is simply a divisor of the LCM, which may not always help without using the proper formula. Final Answer:The highest common factor (HCF) of the two numbers is 34.

Question 19

For a given set of positive integers, how does the least common multiple (LCM) always compare to the product of all those numbers?

Accepted Answer

Introduction:This conceptual number theory question tests your understanding of how the least common multiple (LCM) of several positive integers is related to the product of those integers. Understanding this relationship is crucial for quickly eliminating wrong options in LCM and HCF problems. Given Data / Assumptions: We have a set of positive integers. We consider their least common multiple (LCM). We also consider the product of all these numbers. We need a statement that is always true when comparing LCM and the product. Concept / Approach:For any two positive integers a and b, we have the relationship:a * b = HCF(a, b) * LCM(a, b)Since HCF(a, b) is at least 1, and usually greater than 1, this already suggests that LCM is less than or equal to the product. For more than two numbers, similar reasoning applies: common factors are shared between the numbers and counted fewer times in the LCM than in the full product. Step-by-Step Solution:Step 1: Consider two numbers a and b.We know: a * b = HCF * LCM.Step 2: Note that HCF(a, b) ≥ 1.So, LCM = (a * b) / HCF ≤ a * b.Step 3: Extend the idea to more numbers.When you have several integers, their LCM is formed by taking prime factors at the highest powers appearing in any one number, not multiplying all prime powers together as in the full product.Step 4: Conclusion.Therefore, the LCM of any collection of positive integers is always less than or equal to the product of all those integers. Verification / Alternative check:Example: numbers 4 and 6. Product = 24, LCM = 12, which is less than the product. For coprime numbers, such as 4 and 9, product = 36 and LCM = 36, which equals the product. In no case will the LCM exceed the full product. Why Other Options Are Wrong:Always greater than the product: impossible because LCM divides the product when considering all factors.Always less than the product: false for coprime numbers where LCM equals the product.Always equal to the product: false whenever numbers share common factors (like 4 and 6).Cannot be determined in general: incorrect because we have a clear rule, less than or equal to the product. Common Pitfalls:Students often think LCM is always the product or confuse it with HCF. Another mistake is forgetting that for coprime numbers, LCM equals the product, not less. Final Answer:The correct comparison is that the LCM is always less than or equal to the product of the given numbers.

Question 20

The highest common factor (HCF) of two numbers is 11 and their least common multiple (LCM) is 330. If one of the numbers is 55, what is the other number?

Accepted Answer

Introduction:This problem checks your understanding of the relationship between the highest common factor (HCF), the least common multiple (LCM), and the product of two numbers. It is a very common type of question in aptitude and bank exams. Given Data / Assumptions: HCF of the two numbers = 11. LCM of the two numbers = 330. One of the numbers = 55. We must find the other number. Concept / Approach:For any two positive integers a and b, the fundamental relationship is:a * b = HCF(a, b) * LCM(a, b)This allows us to find the missing number when we know the HCF, LCM, and one of the numbers. Step-by-Step Solution:Step 1: Use the formula.a * b = HCF * LCMStep 2: Substitute the known values.55 * (other number) = 11 * 330Step 3: Compute the right-hand side first.11 * 330 = 3630Step 4: Solve for the other number.Other number = 3630 / 55Step 5: Perform the division.55 * 66 = 3630, so the other number is 66 Verification / Alternative check:Let the two numbers be 55 and 66. Their HCF is 11 (since 55 = 5 * 11 and 66 = 6 * 11), and their LCM is (55 * 66) / 11 = 3630 / 11 = 330. This matches the given HCF and LCM, confirming our answer. Why Other Options Are Wrong:33: The product 55 * 33 = 1815, which does not equal 11 * 330 = 3630.44: 55 * 44 = 2420, again not equal to 3630.77: 55 * 77 = 4235, does not satisfy the relation.99: 55 * 99 = 5445, also incorrect. Common Pitfalls:Some learners try to guess the other number or confuse HCF and LCM. Others forget the crucial formula connecting product, HCF and LCM. Always start with the formula to avoid mistakes. Final Answer:The other number is 66.

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