Question 1

Find the HCF using prime powers Compute the highest common factor (HCF) of 2^3, 3^2, and 15 by interpreting the compressed notation correctly.

Accepted Answer

Introduction / Context:The raw stem “2 3, 3 2 and 15” is a compressed way to denote powers: 2^3 and 3^2, along with 15. We must carefully recover the intended values to apply HCF rules correctly. Given Data / Assumptions: Numbers: 2^3 = 8, 3^2 = 9, and 15. All are positive integers. Concept / Approach:HCF requires the common prime factors with the smallest exponents across all numbers. If there is no prime common to all, the HCF is 1. Step-by-Step Solution:Prime factorisations: 8 = 2^3; 9 = 3^2; 15 = 3 * 5.Compare across all three: there is no prime that appears in all numbers (2 is missing from 9 and 15; 5 is missing from 8 and 9; 3 is missing from 8).Hence HCF = 1. Verification / Alternative check:Compute gcd sequentially: gcd(8, 9) = 1; then gcd(1, 15) = 1. Why Other Options Are Wrong:23 and 32 misread the notation as concatenation rather than powers; 360 is a product not a common factor; 3 is not a factor of 8. Common Pitfalls:Misinterpreting the compressed stem; assuming some non-existent shared factor; forgetting HCF must divide every number. Final Answer:1

Question 2

Find the HCF of 42, 63, and 140 accurately Use either prime factorisation or the Euclidean algorithm to determine the highest common factor.

Accepted Answer

Introduction / Context:Finding the HCF (gcd) among three integers can be done sequentially: gcd(a, b, c) = gcd(gcd(a, b), c). This keeps the computation simple and organised. Given Data / Assumptions: a = 42, b = 63, c = 140. We want gcd(a, b, c). Concept / Approach:Either factor each number or apply the Euclidean algorithm pairwise. We will use a quick gcd chain and confirm by factorisation. Step-by-Step Solution:gcd(42, 63) = 21 (since 42 = 2 * 3 * 7 and 63 = 3^2 * 7).Now gcd(21, 140) = 7 (140 = 2^2 * 5 * 7).Therefore HCF = 7. Verification / Alternative check:Prime factorisations: 42 = 2 * 3 * 7; 63 = 3^2 * 7; 140 = 2^2 * 5 * 7. The only prime common to all three is 7 with exponent 1. Why Other Options Are Wrong:21 divides 42 and 63 but not 140; 14 divides 42 and 140 but not 63; 9 does not divide 42 or 140; 28 fails on 63. Common Pitfalls:Stopping after gcd(42, 63) = 21 without checking against 140; overlooking that a common factor must divide every number. Final Answer:7

Question 3

HCF of several fractions Find the highest common factor (HCF) of the fractions 1/2, 2/3, 3/4, and 4/5 using the rule HCF = HCF(numerators) / LCM(denominators).

Accepted Answer

Introduction / Context:For fractions, the HCF is not found by simple integer gcd on numerators and denominators together. Instead, apply the special rule: HCF of fractions equals HCF of numerators divided by LCM of denominators. Given Data / Assumptions: Fractions: 1/2, 2/3, 3/4, 4/5. All positive. Concept / Approach:Compute HCF(1, 2, 3, 4) and LCM(2, 3, 4, 5), then divide. Step-by-Step Solution:HCF of numerators: gcd(1, 2, 3, 4) = 1.LCM of denominators: LCM(2, 3, 4, 5) = 60.Therefore HCF of fractions = 1 / 60. Verification / Alternative check:Each given fraction divided by 1/60 yields an integer: (1/2) ÷ (1/60) = 30; (2/3) ÷ (1/60) = 40; (3/4) ÷ (1/60) = 45; (4/5) ÷ (1/60) = 48. Why Other Options Are Wrong:1 and 12 are integers, not common fractional divisors for all; 4/5 is larger than most inputs; 1/30 is a common divisor but not the greatest by the rule. Common Pitfalls:Mixing up rules for HCF vs LCM on fractions; canceling across numbers incorrectly. Final Answer:1/60

Question 4

Compute the HCF using prime factorizations quickly Find the highest common factor (HCF) of 1485 and 4356.

Accepted Answer

Introduction / Context:HCF (gcd) between two integers is easiest via either the Euclidean algorithm or prime factorisation. Recognising square structures can speed factorisation substantially. Given Data / Assumptions: Numbers: 1485 and 4356. We aim to find gcd(1485, 4356). Concept / Approach:Factor both: 1485 ends with 5, so divisible by 5; 4356 is 4 * 1089 and 1089 is 33^2, which is 3^2 * 11^2. Take minimum exponents for common primes. Step-by-Step Solution:1485 = 5 * 297 = 5 * (27 * 11) = 3^3 * 5 * 11.4356 = 4 * 1089 = 2^2 * (33^2) = 2^2 * 3^2 * 11^2.Common primes: 3 and 11. Minimum exponents: 3^2 and 11^1.Therefore HCF = 3^2 * 11 = 9 * 11 = 99. Verification / Alternative check:Divide: 1485 ÷ 99 = 15; 4356 ÷ 99 = 44. Both are integers, confirming gcd = 99. Why Other Options Are Wrong:189 = 9 * 21 includes factor 7 (not common); 89 and 83 are primes that do not divide 1485; 27 divides 1485 but not 4356 exactly. Common Pitfalls:Stopping after finding a small common factor like 3 or 11; forgetting to include both common primes at their minimum exponents. Final Answer:99

Question 5

Find the greatest number that divides each of the four integers 210, 315, 147, and 168 exactly (that is, with zero remainder).

Accepted Answer

Introduction / Context:This question asks for the largest common divisor (also called the highest common factor, HCF or greatest common divisor, GCD) of four given integers: 210, 315, 147, and 168. Finding the GCD ensures the number divides each integer without leaving a remainder. Given Data / Assumptions: Numbers: 210, 315, 147, 168. We need the largest positive integer that divides all four exactly. Standard arithmetic properties of divisibility and GCD apply. Concept / Approach:The GCD of multiple integers can be computed stepwise: gcd(a, b, c, d) = gcd(gcd(gcd(a, b), c), d). Prime factorization or the Euclidean algorithm can be used at each step. Euclidean steps are typically quicker for larger numbers. Step-by-Step Solution: Step 1: gcd(210, 315) = 105 (since 315 = 210 + 105 and 210 = 2*105).Step 2: gcd(105, 147): 147 mod 105 = 42; 105 mod 42 = 21; 42 mod 21 = 0 ⇒ gcd = 21.Step 3: gcd(21, 168) = 21 (since 168 = 8*21 exactly).Therefore, gcd(210, 315, 147, 168) = 21. Verification / Alternative check:Prime factors: 210 = 2*3*5*7; 315 = 3^2*5*7; 147 = 3*7^2; 168 = 2^3*3*7. Common primes are 3 and 7, but 3 appears in all, 7 appears in all; the minimum powers across all are 3^1 and 7^1, hence 3*7 = 21. Matches the Euclidean result. Why Other Options Are Wrong: 4410: That is a product-like distractor; it does not divide any of the numbers. 14 or 7 or 3: These divide the set but are not the greatest possible; 21 is larger and still divides all. Common Pitfalls: Stopping at a common divisor that is not the greatest (e.g., 7 or 3). Mixing up LCM and GCD. Final Answer:21

Question 6

Two integers have H.C.F. (greatest common divisor) equal to 12, and the difference between the two numbers is also 12. Which pair of numbers satisfies both conditions?

Accepted Answer

Introduction / Context:We are given two constraints for a pair of integers: their highest common factor (HCF) is 12, and their difference is 12. We must select the pair that meets both conditions at once. Given Data / Assumptions: HCF = 12. Absolute difference between the numbers = 12. All choices are positive integer pairs. Concept / Approach:If HCF is 12, then each number is a multiple of 12. Let the numbers be 12m and 12n, with gcd(m, n) = 1 (since all common factors are captured by 12). The difference is |12m − 12n| = 12|m − n| and is given as 12 ⇒ |m − n| = 1, so m and n must be consecutive coprime integers. Step-by-Step Solution: Check each option quickly for both properties.66, 78 ⇒ difference 12 but gcd(66, 78) = 6, not 12.70, 82 ⇒ difference 12 but gcd = 2, not 12.94, 106 ⇒ difference 12 but gcd = 2, not 12.84, 96 ⇒ difference 12 and gcd(84, 96) = 12 (since 84 = 12*7, 96 = 12*8 and gcd(7,8)=1). Verification / Alternative check:Factor 84 = 2^2*3*7 and 96 = 2^5*3; common prime factors have minimum exponents 2^2*3 = 12. Difference is 12. Both constraints satisfied. Why Other Options Are Wrong: 66, 78; 70, 82; 94, 106: Correct difference but HCF is not 12. 72, 84 (added distractor): HCF is 12, but difference is also 12; however gcd(72,84)=12 is true; if present as an option it would also satisfy. Among the provided original options, 84, 96 is the correct listed pair. Common Pitfalls: Forgetting that both numbers must be multiples of 12. Not checking gcd precisely, especially when both are even. Final Answer:84, 96

Question 7

The H.C.F. (greatest common divisor) of three numbers is 12, and the numbers are in the ratio 1 : 2 : 3. Find the three numbers.

Accepted Answer

Introduction / Context:Numbers in a given ratio with a specified HCF can be constructed by multiplying the ratio terms by a common positive integer equal to the HCF divided by the HCF of the ratio terms. Here the ratio is simple and the HCF is provided directly. Given Data / Assumptions: HCF of the three numbers = 12. Numbers are proportional to 1 : 2 : 3. Concept / Approach:If numbers are 12*k1, 12*k2, 12*k3 with gcd(k1, k2, k3) = 1, and in the ratio 1:2:3, then (k1, k2, k3) = (1, 2, 3). Therefore the actual numbers are 12*1, 12*2, 12*3. This ensures the overall HCF stays at 12 because the inner triple has no common factor beyond 1. Step-by-Step Solution: Let numbers be 12*1, 12*2, 12*3.Compute: 12, 24, 36.Check gcd(12, 24, 36) = 12. Verification / Alternative check:Any common factor larger than 12 would imply a common factor among 1, 2, 3, which is impossible. Hence 12, 24, 36 is consistent and minimal with the given ratio. Why Other Options Are Wrong: 10, 20, 30; 5, 10, 15; 4, 8, 12; 6, 12, 18: These do not all have HCF equal to 12, or they do not adhere to the required HCF and ratio simultaneously. Common Pitfalls: Multiplying the ratio terms by an arbitrary number without ensuring the final HCF is exactly 12. Final Answer:12, 24, 36

Question 8

Find the L.C.M. (least common multiple) of the fractions 2/7, 3/14, and 5/3. (Assume positive values.)

Accepted Answer

Introduction / Context:For positive fractions, a standard rule is used to compute the least common multiple (LCM). Unlike integers, the LCM of fractions depends on both the numerators and denominators, combined through a specific relationship involving LCM and HCF (GCD). Given Data / Assumptions: Fractions: 2/7, 3/14, 5/3. Assume all fractions are positive. Concept / Approach:The formula for the LCM of fractions a/b, c/d, e/f is: LCM = LCM(numerators) / HCF(denominators). We therefore find LCM of 2, 3, 5 and HCF (GCD) of 7, 14, 3. Step-by-Step Solution: LCM of numerators 2, 3, 5 is 30 (since 2, 3, 5 are primes).HCF of denominators 7, 14, 3 is gcd(gcd(7, 14), 3) = gcd(7, 3) = 1.Therefore LCM of the fractions = 30 / 1 = 30. Verification / Alternative check:Each fraction should divide 30 (as a rational) exactly: 30 ÷ (2/7) = 105; 30 ÷ (3/14) = 140; 30 ÷ (5/3) = 18. All results are integers, validating 30 as a common multiple. Being constructed via the formula guarantees minimality. Why Other Options Are Wrong: 35, 45, 25, 15: Do not satisfy the fraction LCM formula, and at least one of the fractions will not divide these cleanly in the rational sense. Common Pitfalls: Using LCM(denominators) instead of HCF(denominators) in the fraction-LCM formula. Attempting integer-style LCM directly on fractions without the proper rule. Final Answer:30

Question 9

Among the fractions 1/2, 3/7, 3/5, and 4/9, which is the smallest in value?

Accepted Answer

Introduction / Context:Comparing fractions can be done by converting to decimal, using cross-multiplication, or reducing them to a common denominator. We are to identify the smallest fraction among the given set. Given Data / Assumptions: Fractions: 1/2, 3/7, 3/5, 4/9. All are positive proper fractions. Concept / Approach:Multiple quick methods exist. Decimal approximation: 1/2 = 0.5; 3/5 = 0.6; 4/9 ≈ 0.444…; 3/7 ≈ 0.42857…. The smallest decimal corresponds to the smallest fraction. Alternatively, compare via cross-multiplication pairwise if decimals are not desired. Step-by-Step Solution: Compute approximate values:1/2 = 0.5.3/5 = 0.6.4/9 ≈ 0.444….3/7 ≈ 0.42857….Hence, 3/7 is the smallest. Verification / Alternative check:Use common denominator 630 (LCM of 2, 5, 7, 9): 1/2 = 315/630; 3/5 = 378/630; 4/9 = 280/630; 3/7 = 270/630. The smallest numerator is 270 for 3/7, confirming the result. Why Other Options Are Wrong: 4/9 and 1/2 are larger than 3/7 but close. 3/5 equals 0.6, clearly the largest here among the listed fractions. Common Pitfalls: Mistakenly comparing only denominators for proper fractions without considering numerators. Rounding too coarsely when using decimals. Final Answer:3/7

Question 10

Which sequence of fractions is in descending order (from largest to smallest)?

Accepted Answer

Introduction / Context:Ordering fractions requires reliable comparison. The goal is to list the given three fractions in strictly descending order (largest first, smallest last). Given Data / Assumptions: Fractions to compare: 7/9, 5/7, 3/5. All are positive proper fractions. Concept / Approach:Use decimal approximations or cross-multiplication. For quick intuition, compare values roughly: 7/9 ≈ 0.777…, 5/7 ≈ 0.714…, 3/5 = 0.6. This suggests 7/9 > 5/7 > 3/5. Step-by-Step Solution: Compute decimal values: 7/9 ≈ 0.777…, 5/7 ≈ 0.714…, 3/5 = 0.6.Descending order: 0.777… > 0.714… > 0.6 ⇒ 7/9, 5/7, 3/5. Verification / Alternative check:Cross-multiplication: Compare 7/9 and 5/7: 7*7 = 49 vs 5*9 = 45 ⇒ 7/9 > 5/7. Compare 5/7 and 3/5: 5*5 = 25 vs 3*7 = 21 ⇒ 5/7 > 3/5. Chain gives 7/9 > 5/7 > 3/5. Why Other Options Are Wrong: Any sequence not matching 7/9 > 5/7 > 3/5 either reverses one comparison or is fully incorrect. Common Pitfalls: Assuming a larger denominator always means a smaller value without checking the numerator. Mixing up descending with ascending order. Final Answer:7/9, 5/7, 3/5

Question 11

Find the smallest integer N such that N − 3 is divisible by each of 21, 28, 36, and 45.

Accepted Answer

Introduction / Context:The statement “N − 3 is divisible by 21, 28, 36, and 45” means N − 3 must be a common multiple of all four numbers. The smallest such N is found by adding 3 to the least common multiple (LCM) of the set. Given Data / Assumptions: Divisors: 21, 28, 36, 45. We want the least N such that N − 3 is a multiple of each divisor. Concept / Approach:Compute LCM(21, 28, 36, 45) using prime factorization. Then take N = LCM + 3. Prime factorization captures the highest needed power of each prime across the set. Step-by-Step Solution: 21 = 3*7; 28 = 2^2*7; 36 = 2^2*3^2; 45 = 3^2*5.LCM collects max powers: 2^2, 3^2, 5, 7.Compute: 2^2 * 3^2 * 5 * 7 = 4 * 9 * 5 * 7 = 36 * 35 = 1260.Therefore N − 3 = 1260 ⇒ N = 1263. Verification / Alternative check:Check divisibility: 1260 ÷ 21 = 60; ÷ 28 = 45; ÷ 36 = 35; ÷ 45 = 28. All are integers, confirming 1260 is the LCM. Hence N = 1263 is minimal. Why Other Options Are Wrong: 1260, 1257: These are N values themselves, not N − 3. Using them would not satisfy the “minus 3” condition. 420, 840: Smaller than the LCM; their +3 versions would not be multiples of all four divisors. Common Pitfalls: Forgetting to add 3 after obtaining the LCM. Computing LCM with incorrect prime powers (e.g., missing 3^2 or 2^2). Final Answer:1263

Question 12

Consider the number 2^222 * 3^333 * 5^555. How many prime factors does it have when counted with multiplicity (i.e., total prime-power exponents summed)?

Accepted Answer

Introduction / Context:When a number is written in its prime factorization form, such as p1^a * p2^b * p3^c, the total number of prime factors with multiplicity is a + b + c. This is distinct from the number of distinct prime factors, which would simply be the count of different primes involved (here, three primes). Given Data / Assumptions: Number: 2^222 * 3^333 * 5^555. We are asked for the total count of prime factors with multiplicity. Concept / Approach:For N = 2^222 * 3^333 * 5^555, the prime-power exponents are 222, 333, and 555, respectively. The required count equals the sum of these exponents. Step-by-Step Solution: Sum the exponents: 222 + 333 + 555.222 + 333 = 555.555 + 555 = 1110. Verification / Alternative check:“With multiplicity” means each prime factor is counted as many times as its exponent indicates. Thus, the result is the exponent sum, not the number of distinct primes (which would be 3). Why Other Options Are Wrong: 3: Counts only distinct primes, not with multiplicity. 1107, 1113, 1272: Incorrect sums or distractors; none equal 222 + 333 + 555. Common Pitfalls: Confusing “distinct prime factors” with “prime factors counted with multiplicity.” Arithmetic slips when adding large exponents. Final Answer:1110

Question 13

Two numbers have H.C.F. (greatest common divisor) 44 and L.C.M. (least common multiple) 264. The first number is 88 (since 88 ÷ 2 = 44). Find the other number.

Accepted Answer

Introduction / Context:The product of two positive integers equals the product of their GCD and LCM. Given the GCD (H.C.F.), LCM, and one number, we can find the other number directly from this relationship. Here, the first number is clarified as 88 via the quotient condition “first number divided by 2 equals 44.” Given Data / Assumptions: HCF = 44. LCM = 264. First number a = 88 (since 88 ÷ 2 = 44). Second number b = ? Concept / Approach:Use the identity a*b = HCF * LCM. Rearranging gives b = (HCF * LCM) / a. Substitute the known values and compute b. This method is standard and robust for such problems. Step-by-Step Solution: a*b = 44 * 264.Compute 44 * 264 = 11616.Therefore, b = 11616 / 88 = 132. Verification / Alternative check:Check gcd(88, 132) = 44 and lcm(88, 132) = (88*132)/44 = 264. Both match the given, confirming correctness. Why Other Options Are Wrong: 264, 66, 33, 176: Do not satisfy both gcd and lcm simultaneously with 88 under the product identity. Common Pitfalls: Mistakes in multiplying HCF and LCM or dividing by the first number. Forgetting that a and b must be integers compatible with the given gcd and lcm. Final Answer:132

Question 14

The L.C.M. of two numbers is 48. The numbers are in the ratio 2 : 3. Find the sum of the two numbers.

Accepted Answer

Introduction / Context:When two numbers are in a given ratio, they can be represented as multiples of a common factor k. Their LCM then depends on both the ratio and k. Given the LCM numerically, we can solve for k and then compute the actual numbers and their sum. Given Data / Assumptions: LCM = 48. Numbers are in the ratio 2 : 3. Let the numbers be 2k and 3k, with k a positive integer. Concept / Approach:For 2k and 3k, the LCM is 6k (because 2 and 3 are coprime, the LCM of 2k and 3k is lcm(2,3)*k = 6k). Set 6k = 48 to find k. Then compute the numbers and their sum. Step-by-Step Solution: 6k = 48 ⇒ k = 8.Numbers: 2k = 16 and 3k = 24.Sum = 16 + 24 = 40. Verification / Alternative check:LCM(16, 24) = 48 indeed (since 16 = 2^4, 24 = 2^3*3, LCM has 2^4*3 = 48). The ratio 16:24 simplifies to 2:3. Everything is consistent. Why Other Options Are Wrong: 28, 32, 64, 36: Not equal to the sum of any pair in the ratio 2:3 whose LCM is 48. Common Pitfalls: Computing LCM of 2k and 3k incorrectly as 3k or 2k. Picking numbers with the correct ratio but wrong LCM. Final Answer:40

Question 15

Greatest measuring length (HCF application): Find the greatest possible length (in centimetres) that can exactly measure each of the following three lengths: 4 m 3 cm, 4 m 34 cm, and 4 m 65 cm.

Accepted Answer

Introduction / Context:This is a classic Highest Common Factor (HCF) or Greatest Common Divisor (GCD) problem framed as measuring equal segments. The greatest length that measures several lengths exactly is the HCF of those lengths when all are in the same unit. Given Data / Assumptions: Lengths: 4 m 3 cm, 4 m 34 cm, 4 m 65 cm. Convert to a single unit (centimetres) to compute the HCF. Goal: greatest length that exactly divides all three. Concept / Approach:When a rod or tape is to measure several lengths exactly, the required length is the HCF. Converting metres to centimetres simplifies arithmetic. Also, when numbers form an arithmetic progression, their HCF must divide the common difference, which can simplify checks. Step-by-Step Solution: Convert units: 4 m 3 cm = 403 cm; 4 m 34 cm = 434 cm; 4 m 65 cm = 465 cm.Observe differences: 434 − 403 = 31; 465 − 434 = 31.Since the three numbers are in arithmetic progression with common difference 31, the HCF divides 31.Check divisibility by 31: 403/31 = 13, 434/31 = 14, 465/31 = 15 (all integers).Therefore HCF = 31 cm. Verification / Alternative check: Compute HCF(403, 434) = 31, then HCF(31, 465) = 31, confirming the result. Why Other Options Are Wrong: 29 cm, 28 cm, 32 cm do not divide all three lengths 403 cm, 434 cm, and 465 cm without remainder. Common Pitfalls: Forgetting to convert metres to centimetres consistently. Taking LCM by mistake instead of HCF. Final Answer: 31 cm

Question 16

Least number with a fixed remainder: Find the smallest number that leaves remainder 2 when divided by each of 12, 15, 20, and 24.

Accepted Answer

Introduction / Context:Numbers leaving the same remainder upon division by several divisors are handled using the Least Common Multiple (LCM). The pattern is N ≡ r (mod each divisor), hence N − r is a common multiple of all divisors. Given Data / Assumptions: Divisors: 12, 15, 20, 24 Common remainder: r = 2 We seek the least positive N such that N ≡ 2 for all divisors. Concept / Approach:Compute LCM of the divisors, then set N = LCM + r for the least positive solution. In general, all solutions are N = LCM * k + r, k ≥ 1 integer, but k = 1 gives the smallest positive answer beyond r itself (which will not satisfy divisibility here). Step-by-Step Solution: Prime factors: 12 = 2^2 * 3; 15 = 3 * 5; 20 = 2^2 * 5; 24 = 2^3 * 3.LCM = 2^3 * 3 * 5 = 120.Least N = 120 + 2 = 122. Verification / Alternative check: 122 ÷ 12 leaves 2; 122 ÷ 15 leaves 2; similarly for 20 and 24, remainder is 2 each time. Why Other Options Are Wrong: 120 is divisible by the moduli (remainder 0, not 2). 124, 126 do not leave remainder 2 with all the given divisors simultaneously. Common Pitfalls: Mistaking LCM itself as the answer rather than adding the fixed remainder. Final Answer: 122

Question 17

LCM of algebraic multiples: If the three numbers are 2a, 5a, and 7a (with a a positive integer), find their LCM in terms of a.

Accepted Answer

Introduction / Context:The Least Common Multiple (LCM) of numbers that are all multiples of a common factor can be computed by factoring that common part out and taking the LCM of the remaining coefficients. Here each term has a factor a, and the coefficients are small integers. Given Data / Assumptions: Numbers: 2a, 5a, 7a a is a positive integer so that divisibility is well-defined. Concept / Approach:LCM(k1*a, k2*a, k3*a) = a * LCM(k1, k2, k3) because each number contains a single factor a to the first power and the LCM keeps the highest power of each prime across the list. The coefficients 2, 5, and 7 are pairwise coprime, so their LCM is their product. Step-by-Step Solution: Coefficients: 2, 5, 7 are pairwise coprime ⇒ LCM(2,5,7) = 2 * 5 * 7 = 70.Hence LCM(2a, 5a, 7a) = a * 70 = 70a. Verification / Alternative check: 70a is divisible by each of 2a, 5a, and 7a and is the least such positive multiple, since any smaller candidate would fail divisibility by at least one coefficient. Why Other Options Are Wrong: 65a, 75a: Coefficients 65 or 75 are not multiples of all of 2, 5, and 7. 70a^3: Exaggerates the power of a; the LCM includes a only to the first power here. Common Pitfalls: Multiplying the a factor too many times rather than keeping its highest power across the numbers (which is 1 here). Final Answer: 70a

Question 18

Find the HCF of three composite products: Compute the HCF of (2 × 3 × 7 × 9), (2 × 3 × 9 × 11), and (2 × 3 × 4 × 5).

Accepted Answer

Introduction / Context:The Highest Common Factor of several numbers is obtained by taking the minimum exponent of each prime that appears in all numbers. Breaking each product into primes makes the comparison straightforward. Given Data / Assumptions: A = 2 × 3 × 7 × 9 B = 2 × 3 × 9 × 11 C = 2 × 3 × 4 × 5 Concept / Approach:Prime factorize each: 9 = 3^2 and 4 = 2^2. Then take the intersection of primes with their smallest exponents across A, B, and C. Any prime not present in all three drops out of the HCF. Step-by-Step Solution: A = 2^1 * 3^3 * 7^1.B = 2^1 * 3^3 * 11^1.C = 2^3 * 3^1 * 5^1.Common primes across all: 2 and 3.Minimum exponents: 2^min(1,1,3) = 2^1; 3^min(3,3,1) = 3^1.Therefore HCF = 2 * 3 = 6 (shown as 2 × 3). Verification / Alternative check: Divide each number by 6 to confirm no larger common factor exists: quotients are integers, and any attempt to include another 2 or 3 or include 5/7/11 fails for one of the numbers. Why Other Options Are Wrong: 2 × 3 × 7 or 2 × 3 × 9: Include primes or powers not common to all three numbers. 2 × 7 × 9 × 11: Contains primes not present across all three simultaneously. Common Pitfalls: Using maximum exponents (LCM logic) instead of minimum (HCF logic). Final Answer: 2 × 3

Question 19

LCM of fractions (use LCM of numerators / GCD of denominators): Find the LCM of the fractions 1/3, 2/9, 5/6, and 4/27.

Accepted Answer

Introduction / Context:The LCM of several fractions is computed by taking the LCM of their numerators and dividing by the GCD of their denominators. This rule is the fractional analogue of the integer LCM/GCD definitions and ensures each fraction divides the LCM exactly. Given Data / Assumptions: Fractions: 1/3, 2/9, 5/6, 4/27. All are positive rational numbers. Concept / Approach:Use the formula: LCM of fractions = LCM(numerators) / GCD(denominators). Compute each component carefully by prime factorization or mental arithmetic with small numbers. Step-by-Step Solution: Numerators: 1, 2, 5, 4 ⇒ LCM = 20.Denominators: 3, 9, 6, 27 ⇒ GCD = 3.Therefore LCM = 20 / 3. Verification / Alternative check: Each given fraction divides 20/3: e.g., (20/3) ÷ (1/3) = 20, (20/3) ÷ (2/9) = 30, (20/3) ÷ (5/6) = 8, (20/3) ÷ (4/27) = 45; all integers. Why Other Options Are Wrong: 1/54, 10/27, 3/20 are either too small or inverted and will not be exact multiples of all the given fractions. Common Pitfalls: Confusing GCD and LCM roles for numerators and denominators. Reducing prematurely and losing track of divisibility checks. Final Answer: 20/3

Question 20

LCM of fractions with coprime denominators: Compute the LCM of 2/3, 3/5, 4/7, and 9/13 using the LCM(numerators)/GCD(denominators) rule.

Accepted Answer

Introduction / Context:For fractions, LCM = LCM of numerators divided by GCD of denominators. When denominators are pairwise coprime overall (GCD = 1), the LCM simplifies to just the LCM of numerators as an integer. Given Data / Assumptions: Fractions: 2/3, 3/5, 4/7, 9/13. All denominators 3, 5, 7, 13 are pairwise coprime, so overall GCD is 1. Concept / Approach:Compute LCM of numerators 2, 3, 4, 9. Then divide by GCD of denominators (which is 1), meaning the LCM is simply that numerator LCM. Step-by-Step Solution: LCM(2, 3, 4, 9) = 36 (since 4 contributes 2^2 and 9 contributes 3^2).GCD(3, 5, 7, 13) = 1.LCM of the fractions = 36 / 1 = 36. Verification / Alternative check: Check divisibility: (36) ÷ (2/3) = 54; ÷ (3/5) = 60; ÷ (4/7) = 63; ÷ (9/13) = 52; all integers. Why Other Options Are Wrong: 1/36 and 1/1365 are reciprocals-like values, not suitable here. 12/455 does not produce integer quotients with all given fractions. Common Pitfalls: Treating the LCM of denominators instead of their GCD for the fraction-LCM formula. Final Answer: 36

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