📘 Quiz

Solution: Step 1: Let the two whole numbers be A and B. Given that their ratio is x:y, we can represent them as A = zx and B = zy, where z is their HCF and x, y are co-prime integers from the simplified ratio. Step 2: Recall the fundamental relationship between two numbers, their HCF, and their LCM: Product of the two numbers = HCF × LCM Step 3: Substitute the expressions for A, B, and HCF into the formula: (zx) × (zy) = z × LCM Step 4: Simplify the equation: z^2xy = z × LCM Step 5: Solve for LCM: LCM = z^2xy / z LCM = xyz Step 6: The LCM of the two numbers is xyz.

Solution: Step 1: Convert all decimal numbers into fractions with the same denominator. The highest number of decimal places is 3 (in 0.175), so convert to fractions with a denominator of 1000. 2.5 = 2.500 = 2500/1000 0.5 = 0.500 = 500/1000 0.175 = 175/1000 Step 2: To find the LCM of fractions, use the formula: LCM(a/b, c/d, e/f) = LCM(a, c, e) / HCF(b, d, f). To find the HCF of fractions, use the formula: HCF(a/b, c/d, e/f) = HCF(a, c, e) / LCM(b, d, f). Step 3: Identify the numerators and denominators: Numerators: 2500, 500, 175 Denominators: 1000, 1000, 1000 Step 4: Calculate the LCM of the numerators (2500, 500, 175): Prime factorization: 2500 = 2² × 5⁴ 500 = 2² × 5³ 175 = 5² × 7 LCM(2500, 500, 175) = 2² × 5⁴ × 7¹ = 4 × 625 × 7 = 2500 × 7 = 17500. Step 5: Calculate the HCF of the denominators (1000, 1000, 1000): HCF(1000, 1000, 1000) = 1000. Step 6: Calculate the LCM of the decimal numbers: LCM = LCM(Numerators) / HCF(Denominators) = 17500 / 1000 = 17.5. Step 7: Calculate the HCF of the numerators (2500, 500, 175): HCF(2500, 500, 175) = 5² = 25. Step 8: Calculate the LCM of the denominators (1000, 1000, 1000): LCM(1000, 1000, 1000) = 1000. Step 9: Calculate the HCF of the decimal numbers: HCF = HCF(Numerators) / LCM(Denominators) = 25 / 1000 = 0.025. Step 10: The question asks for LCM and HCF, and the correct option provided (17.5) is the LCM. So, the LCM is 17.5 and the HCF is 0.025.

Solution: Step 1: Find the prime factorization of each number. 132 = 2 * 66 = 2 * 2 * 33 = 2^2 * 3 * 11 204 = 2 * 102 = 2 * 2 * 51 = 2^2 * 3 * 17 228 = 2 * 114 = 2 * 2 * 57 = 2^2 * 3 * 19 Step 2: Identify the common prime factors and their lowest powers present in all factorizations. The common prime factors are 2 and 3. The lowest power of 2 common to all is 2^2. The lowest power of 3 common to all is 3^1. Step 3: Multiply these common prime factors with their lowest powers to find the HCF. HCF = 2^2 * 3 = 4 * 3 = 12. Step 4: The HCF of 132, 204, and 228 is 12.

Solution: Step 1: If a number divides 989 and leaves a remainder of 5, it means that the number perfectly divides (989 - 5). 989 - 5 = 984. Step 2: If the same number divides 1327 and leaves a remainder of 7, it means that the number perfectly divides (1327 - 7). 1327 - 7 = 1320. Step 3: The greatest number that divides both 984 and 1320 perfectly is their Highest Common Factor (HCF). Step 4: Find the HCF of 984 and 1320. Step 5: Using the Euclidean algorithm: 1320 = 1 * 984 + 336 984 = 2 * 336 + 312 336 = 1 * 312 + 24 312 = 13 * 24 + 0 Step 6: The last non-zero remainder is the HCF. Thus, HCF(984, 1320) = 24.

Solution: Step 1: Let the two numbers be A and B. Given their ratio A:B = 1:4. We can represent them as A = k and B = 4k for some constant k. Step 2: We are given HCF = 21 and LCM = 84. Step 3: Use the fundamental property that the product of two numbers is equal to the product of their HCF and LCM. A * B = HCF * LCM k * 4k = 21 * 84 4k^2 = 1764. Step 4: Solve for k^2: k^2 = 1764 / 4 = 441. Step 5: Solve for k: k = sqrt(441) = 21 (since we are looking for positive numbers). Step 6: Find the two numbers: A = k = 21 B = 4k = 4 * 21 = 84. Step 7: The larger of the two numbers is 84.

Solution: Step 1: Find the prime factorization of each number: 36 = 2 × 2 × 3 × 3 = 2^2 × 3^2 84 = 2 × 2 × 3 × 7 = 2^2 × 3^1 × 7^1 Step 2: To find the H.C.F., take the lowest power of each common prime factor. Common prime factors are 2 and 3. Lowest power of 2 is 2^2. Lowest power of 3 is 3^1. Step 3: Multiply these lowest powers together. H.C.F. = 2^2 × 3^1 = 4 × 3 = 12.

Solution: Step 1: Let the numbers be 2x, 3x, and 4x Step 2: LCM of 2x, 3x, and 4x is 12x Step 3: Given LCM = 360, so 12x = 360 Step 4: Solve for x: x = 360 / 12 = 30 Step 5: Largest number = 4x = 4 * 30 = 120

Solution: Step 1: The least common multiple (LCM) of two prime numbers is equal to their product. Step 2: Given LCM(A, B) = 209. Therefore, A × B = 209. Step 3: Find the prime factors of 209. 209 is not divisible by 2, 3, 5, or 7. Divide 209 by 11: 209 / 11 = 19. So, the prime factors are 11 and 19. Step 4: Since A and B are prime numbers and A > B, we can identify A = 19 and B = 11. Step 5: Calculate the value of the expression A^2 - B: A^2 - B = (19)^2 - 11 = 361 - 11 = 350.

Solution: Step 1: Let the numbers be a and b, with a > b Step 2: Division steps: a = 4b + r1, b = 2c + r2, c = 8d + r3, d = 72 Step 3: From the last step: c = 72 * 8 = 576 Step 4: b = 2 * 576 + r2, since r2 < 72, let r2 = 0 for HCF Step 5: b = 2 * 576 = 1152 Step 6: a = 4 * 1152 = 4608 Step 7: Difference = a - b = 4608 - 1584 = 3024

Solution: Step 1: Use the fundamental relationship between HCF, LCM, and two numbers: The product of two numbers is equal to the product of their HCF and LCM. Number1 × Number2 = HCF × LCM Step 2: Let the given numbers be Number1 = 72 and Number2 = N. We are given: HCF = 4 LCM = 936 Step 3: Substitute these values into the formula: 72 × N = 4 × 936 Step 4: Solve for N: N = (4 × 936) / 72 Step 5: Simplify the expression: N = (4 × (72 × 13)) / 72 (since 936 divided by 72 is 13) N = 4 × 13 N = 52. Step 6: The other number is 52.

Solution: Step 1: Recall the fundamental property for two numbers, A and B: Product of A and B = HCF(A, B) × LCM(A, B). Step 2: Identify the given values: Product of two numbers = 396 × 576. LCM = 6336. Let HCF be 'H'. Step 3: Substitute the values into the formula: 396 × 576 = H × 6336. Step 4: Solve for H. H = (396 × 576) ÷ 6336. Step 5: Perform the calculation. H = 228096 ÷ 6336 H = 36.

Solution: Step 1: Observe the differences between each divisor and its respective remainder: 4 - 1 = 3 5 - 2 = 3 6 - 3 = 3 7 - 4 = 3 8 - 5 = 3 Step 2: Since the difference is constant (3), the required number 'N' can be expressed in the form (LCM of divisors × k - common difference). Step 3: Calculate the LCM of 4, 5, 6, 7, and 8. Prime factorization: 4 = 2^2 5 = 5 6 = 2 × 3 7 = 7 8 = 2^3 Step 4: Identify the highest power of each prime factor present: 2^3 (from 8) 3^1 (from 6) 5^1 (from 5) 7^1 (from 7) Step 5: Calculate the LCM = 2^3 × 3 × 5 × 7 = 8 × 3 × 5 × 7 = 840. Step 6: So, the required number 'N' is of the form 840k - 3. Step 7: We need to find the greatest four-digit number. The largest four-digit number is 9999. Step 8: Find the largest multiple of 840 that is less than or equal to 9999. Divide 9999 by 840: 9999 ÷ 840 = 11 with a remainder. The largest multiple is 840 × 11 = 9240. (The next multiple, 840 × 12 = 10080, is a five-digit number). Step 9: Substitute this multiple into the formula from Step 6: Required number = 9240 - 3 = 9237. Step 10: The greatest four-digit number satisfying the conditions is 9237.

Solution: Step 1: Let the two numbers be represented as 3x and 4x, where x is a common factor. Step 2: The H.C.F. (Highest Common Factor) of 3x and 4x is x, because 3 and 4 are co-prime. We are given that the H.C.F. is 4. Therefore, x = 4. Step 3: Calculate the actual values of the two numbers: First number = 3 × x = 3 × 4 = 12 Second number = 4 × x = 4 × 4 = 16 Step 4: Find the L.C.M. (Least Common Multiple) of 12 and 16. Prime factorization: 12 = 2^2 × 3 16 = 2^4 L.C.M. = 2^4 × 3 = 16 × 3 = 48.

Solution: Step 1: Let the two numbers be N1 and N2. Step 2: Given that their H.C.F. is 13, we can represent the numbers as N1 = 13a and N2 = 13b, where 'a' and 'b' are coprime integers. Step 3: The product of the two numbers is 2028: N1 * N2 = 2028 (13a) * (13b) = 2028 169 * a * b = 2028 Step 4: Solve for the product 'a * b': a * b = 2028 / 169 a * b = 12 Step 5: Now, we need to find pairs of coprime integers (a, b) such that their product is 12. Step 6: List all factor pairs of 12: (1, 12) (2, 6) (3, 4) Step 7: Check which of these pairs consist of coprime integers (i.e., their H.C.F. is 1): - For (1, 12): H.C.F.(1, 12) = 1. This pair is coprime. - For (2, 6): H.C.F.(2, 6) = 2. This pair is NOT coprime. - For (3, 4): H.C.F.(3, 4) = 1. This pair is coprime. Step 8: The coprime pairs (a, b) are (1, 12) and (3, 4). Step 9: Each coprime pair corresponds to a unique pair of numbers (N1, N2): - For (a, b) = (1, 12), the numbers are (13*1, 13*12) = (13, 156). - For (a, b) = (3, 4), the numbers are (13*3, 13*4) = (39, 52). Step 10: Therefore, there are 2 such distinct pairs of numbers.

Solution: Step 1: Recall the rule for finding the HCF of fractions: HCF of fractions = (HCF of Numerators) ÷ (LCM of Denominators). Step 2: Identify the numerators and denominators. Numerators: 3, 5, 6. Denominators: 4, 6, 7. Step 3: Calculate the HCF of the numerators (3, 5, 6). Prime factorization: 3 = 3 5 = 5 6 = 2 × 3 The only common factor is 1. Therefore, HCF(3, 5, 6) = 1. Step 4: Calculate the LCM of the denominators (4, 6, 7). Prime factorization: 4 = 2² 6 = 2 × 3 7 = 7 LCM(4, 6, 7) = 2² × 3 × 7 = 4 × 3 × 7 = 84. Step 5: Apply the formula for HCF of fractions. HCF(3/4, 5/6, 6/7) = HCF(3, 5, 6) ÷ LCM(4, 6, 7) = 1 ÷ 84.

Solution: Step 1: Let the number to be subtracted be X. The problem states that (5834 - X) must be exactly divisible by 20, 28, 32, and 35. This means (5834 - X) must be a multiple of the Least Common Multiple (LCM) of these numbers. Step 2: Find the prime factorization of each divisor: 20 = 2² × 5 28 = 2² × 7 32 = 2⁵ 35 = 5 × 7 Step 3: Calculate the LCM of 20, 28, 32, and 35 by taking the highest power of each prime factor: LCM(20, 28, 32, 35) = 2⁵ × 5¹ × 7¹ = 32 × 5 × 7 = 1120. Step 4: So, (5834 - X) must be a multiple of 1120. To find the *greatest* number X to be subtracted, the resulting number (5834 - X) must be the *smallest positive* multiple of 1120. Step 5: The smallest positive multiple of 1120 is 1120 itself. Step 6: Set up the equation: 5834 - X = 1120 Step 7: Solve for X: X = 5834 - 1120 = 4714. Step 8: The greatest number that must be subtracted is 4714.

Solution: Step 1: Find the prime factorization of each divisor: 8 = 2³ 12 = 2² × 3 16 = 2⁴ 20 = 2² × 5 Step 2: Calculate the Least Common Multiple (LCM) of these divisors by taking the highest power of each prime factor: LCM(8, 12, 16, 20) = 2⁴ × 3¹ × 5¹ = 16 × 3 × 5 = 240 Step 3: The number that leaves a remainder of 5 when divided by 8, 12, 16, and 20 will be 5 more than their LCM. Step 4: Add the remainder to the LCM: Required number = 240 + 5 = 245.

Solution: Step 1: Let the two numbers be represented as 2x and 3x, given their ratio is 2:3. Step 2: The L.C.M. (Least Common Multiple) of two numbers of the form (ax) and (bx), where a and b are co-prime, is a × b × x. For numbers 2x and 3x (where 2 and 3 are co-prime), their L.C.M. = 2 × 3 × x = 6x. Step 3: We are given that the L.C.M. of the two numbers is 48. So, 6x = 48. Step 4: Solve for x: x = 48 ÷ 6 = 8. Step 5: Determine the actual values of the two numbers: First number = 2x = 2 × 8 = 16. Second number = 3x = 3 × 8 = 24. Step 6: Calculate the sum of the numbers: Sum = 16 + 24 = 40.

Solution: Step 1: Let L be the LCM and H be the HCF of the two numbers. Step 2: Formulate equations based on the given information: L + H = 512 (Equation 1) L - H = 496 (Equation 2) Step 3: Add Equation 1 and Equation 2 to solve for L: (L + H) + (L - H) = 512 + 496 2L = 1008 L = 504 Step 4: Substitute the value of L back into Equation 1 to solve for H: 504 + H = 512 H = 512 - 504 H = 8 Step 5: Use the property that the product of two numbers (N1 and N2) is equal to the product of their LCM and HCF: N1 × N2 = L × H Step 6: Given N1 = 72, substitute the values: 72 × N2 = 504 × 8 Step 7: Solve for N2: N2 = (504 × 8) / 72 N2 = 504 / 9 N2 = 56