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Question 1 / 20
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1
Three traffic lights at different signal points change after every 45 seconds, 75 seconds, and 90 seconds, respectively. If all of them change simultaneously at 7:20:15 hours, at what time will they change simultaneously again?
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Solution: Step 1: To find when the lights will change simultaneously again, calculate the least common multiple (LCM) of their individual change intervals: 45, 75, and 90 seconds. Prime factorization of each interval: 45 = 3^2 × 5 75 = 3 × 5^2 90 = 2 × 3^2 × 5 LCM = 2 × 3^2 × 5^2 = 2 × 9 × 25 = 18 × 25 = 450 seconds. Step 2: Convert the LCM from seconds to minutes and seconds. 450 seconds = 450 / 60 minutes = 7 minutes and 30 seconds (since 450 = 7 × 60 + 30). Step 3: Add this duration to the initial time they changed simultaneously, which was 7:20:15 hours. Starting time: 7 hours 20 minutes 15 seconds Add duration: 0 hours 07 minutes 30 seconds ------------------------------------------------ Next simultaneous change: 7 hours 27 minutes 45 seconds.
2
There are 21 mango trees, 42 apple trees, and 56 orange trees to be planted. They must be arranged in rows such that each row contains an equal number of trees of only one variety. What is the minimum total number of rows required?
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Solution: Step 1: To minimize the number of rows, the number of trees in each row must be maximized. This maximum number of trees per row will be the HCF of the total number of trees of each variety. Step 2: Find the HCF of 21, 42, and 56. Prime factorization: 21 = 3 × 7 42 = 2 × 3 × 7 56 = 2³ × 7 HCF(21, 42, 56) = 7. So, each row will contain 7 trees. Step 3: Calculate the number of rows for each type of tree. Rows for mango trees = 21 ÷ 7 = 3 rows. Rows for apple trees = 42 ÷ 7 = 6 rows. Rows for orange trees = 56 ÷ 7 = 8 rows. Step 4: Calculate the total minimum number of rows. Total rows = 3 + 6 + 8 = 17 rows.
3
Two numbers have a ratio of 3:4. If their Highest Common Factor (HCF) is 4, what is their Least Common Multiple (LCM)?
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Solution: Step 1: If two numbers are in the ratio A:B and their HCF is H, then the numbers can be represented as H*A and H*B (assuming A and B are co-prime). Step 2: Given ratio is 3:4 and HCF = 4. Step 3: The two numbers are: First number = 4 * 3 = 12. Second number = 4 * 4 = 16. Step 4: For two numbers N1 and N2, LCM(N1, N2) = (N1 * N2) / HCF(N1, N2). Step 5: Alternatively, if numbers are H*A and H*B (with A, B co-prime), then LCM = H * A * B. Step 6: Using this property, LCM = 4 * 3 * 4 = 48. Step 7: The LCM of the two numbers is 48.
4
If two numbers have a ratio of 3:4 and the product of their HCF (Highest Common Factor) and LCM (Least Common Multiple) is 2028, what is the sum of these two numbers?
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Solution: Step 1: Let the two numbers be 3x and 4x, where x is their HCF. Step 2: The LCM of 3x and 4x is 12x (since 3 and 4 are coprime, LCM = x × 3 × 4 = 12x). Step 3: Apply the property: Product of two numbers = HCF × LCM. (3x) × (4x) = x × (12x) => 12x² = 12x² (This confirms the relationship). Step 4: We are given that the product of their HCF and LCM is 2028. So, 12x² = 2028. Step 5: Solve for x. x² = 2028 ÷ 12 x² = 169 x = √169 = 13 (Since x represents HCF, it must be positive). Step 6: Find the two numbers using x = 13. First number = 3x = 3 × 13 = 39. Second number = 4x = 4 × 13 = 52. Step 7: Calculate the sum of the numbers. Sum = 39 + 52 = 91.
5
Determine the smallest multiple of 7 that yields a remainder of 4 when divided by 6, 9, 15, and 18.
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Solution: Step 1: Find the L.C.M. of 6, 9, 15, and 18. Prime factorization: 6 = 2 × 3 9 = 3^2 15 = 3 × 5 18 = 2 × 3^2 L.C.M. = 2 × 3^2 × 5 = 2 × 9 × 5 = 90. Step 2: The required number must be of the form (L.C.M. × k) + remainder. So, the number is 90k + 4. Step 3: This number must also be a multiple of 7. Test values for k to find the least k for which (90k + 4) is divisible by 7. If k = 1, number = 90(1) + 4 = 94 (94 ÷ 7 = 13 R 3) If k = 2, number = 90(2) + 4 = 184 (184 ÷ 7 = 26 R 2) If k = 3, number = 90(3) + 4 = 274 (274 ÷ 7 = 39 R 1) If k = 4, number = 90(4) + 4 = 360 + 4 = 364. Step 4: Check if 364 is divisible by 7. 364 ÷ 7 = 52. It is exactly divisible by 7. Step 5: The least number satisfying all conditions is 364.
6
Calculate the highest common factor (HCF) of the numbers 9, 12, 18, and 21.
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Solution: Step 1: Find the prime factorization of each number: * 9 = 3^2 * 12 = 2^2 × 3 * 18 = 2 × 3^2 * 21 = 3 × 7 Step 2: Identify the common prime factors among all the numbers. The only common prime factor is 3. Step 3: Take the lowest power of the common prime factor. The lowest power of 3 present in the factorizations is 3^1. Step 4: Therefore, the HCF = 3. Step 5: The highest common factor of 9, 12, 18, and 21 is 3.
7
What is the smallest number that, when divided by 4, 6, 8, 12, and 16, always results in a remainder of 2?
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Solution: Step 1: The least number that leaves a common remainder 'R' when divided by a set of numbers (N1, N2, ...) is given by the formula (L.C.M. of N1, N2, ...) + R. Step 2: Find the L.C.M. of 4, 6, 8, 12, and 16. Prime factorization: 4 = 2^2 6 = 2 × 3 8 = 2^3 12 = 2^2 × 3 16 = 2^4 L.C.M. = 2^4 × 3 = 16 × 3 = 48. Step 3: The given remainder is 2. Step 4: Apply the formula to find the required number. Required number = L.C.M. + remainder = 48 + 2 = 50.
8
What is the smallest number that needs to be subtracted from 1294 so that the resulting number yields a remainder of 6 when divided by 9, 11, and 13 individually?
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Solution: Step 1: Let the remaining number after subtraction be N. The problem states that N leaves a remainder of 6 when divided by 9, 11, and 13. Step 2: This implies that (N - 6) must be perfectly divisible by 9, 11, and 13. Step 3: Therefore, (N - 6) must be a common multiple of 9, 11, and 13. To find the smallest such N, (N - 6) must be the Least Common Multiple (LCM) of 9, 11, and 13. Step 4: Calculate the LCM of 9, 11, and 13. Since 9 (3^2), 11 (prime), and 13 (prime) are pairwise co-prime, their LCM is simply their product. LCM(9, 11, 13) = 9 × 11 × 13 = 99 × 13 = 1287. Step 5: So, N - 6 = 1287. Step 6: Solve for N: N = 1287 + 6 = 1293. Step 7: The question asks for the least number that must be subtracted from 1294 to get N. Let this number be 'X'. Step 8: X = 1294 - N = 1294 - 1293 = 1. Step 9: Therefore, the least number that must be subtracted is 1.
9
Determine the smallest number such that if 5 is added to it, the resulting sum is perfectly divisible by 24, 32, 36, and 54.
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Solution: Step 1: Let the required least number be N. Step 2: The problem states that (N + 5) is divisible by 24, 32, 36, and 54. This means (N + 5) is the Least Common Multiple (LCM) of these numbers. Step 3: Calculate the LCM of 24, 32, 36, and 54. Prime factorization: 24 = 2^3 × 3 32 = 2^5 36 = 2^2 × 3^2 54 = 2 × 3^3 Step 4: Identify the highest power of each prime factor present: 2^5 (from 32) 3^3 (from 54) Step 5: Calculate the LCM = 2^5 × 3^3 = 32 × 27 = 864. Step 6: So, N + 5 = 864. Step 7: Solve for N: N = 864 - 5 = 859. Step 8: The least number is 859.
10
The ratio of two numbers is 3:4. Given that their Least Common Multiple (LCM) is 240, what is the value of the smaller number?
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Solution: Step 1: Let the two numbers be in the ratio 3:4. We can represent them as 3x and 4x, where 'x' is their Highest Common Factor (HCF). Step 2: The Least Common Multiple (LCM) of two numbers can be found using the formula: LCM(a, b) = (a * b) / HCF(a, b). For numbers 3x and 4x, the HCF is x, and their co-prime parts are 3 and 4. Step 3: Thus, LCM(3x, 4x) = x * (3 * 4) = 12x. Step 4: We are given that the LCM is 240: 12x = 240 Step 5: Solve for x: x = 240 / 12 = 20. Step 6: Now find the actual numbers: First number = 3x = 3 × 20 = 60 Second number = 4x = 4 × 20 = 80 Step 7: The smaller of the two numbers is 60.
11
Five bells start tolling simultaneously. They then toll at individual intervals of 6, 5, 7, 10, and 12 seconds. Excluding the initial simultaneous toll, how many times will all five bells toll together within a one-hour period?
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Solution: Step 1: To find when the bells will toll together again, calculate the Least Common Multiple (LCM) of their individual tolling intervals: 6, 5, 7, 10, and 12 seconds. Step 2: Find the prime factorization of each number: 6 = 2 * 3 5 = 5 7 = 7 10 = 2 * 5 12 = 2^2 * 3 Step 3: The LCM is found by taking the highest power of all prime factors present: LCM = 2^2 * 3 * 5 * 7 = 4 * 3 * 5 * 7 = 420. Step 4: The bells will toll together every 420 seconds. Step 5: Convert 420 seconds to minutes: 420 seconds / 60 seconds/minute = 7 minutes. Step 6: The total duration for which we need to count is one hour. Convert one hour to minutes: 1 hour = 60 minutes. Step 7: To find how many times they toll together in one hour (excluding the start), divide the total time by the interval time: 60 minutes / 7 minutes/toll. Step 8: 60 / 7 = 8 with a remainder. This means they will toll together 8 additional times within the 60-minute period after the initial simultaneous toll.
12
Calculate the minimum number of square tiles needed to cover a room floor that is 15 meters 17 cm long and 9 meters 2 cm broad.
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Solution: Step 1: Convert room dimensions to centimeters: Length (L) = 15 m 17 cm = (15 * 100) + 17 = 1517 cm. Breadth (B) = 9 m 2 cm = (9 * 100) + 2 = 902 cm. Step 2: For the least number of square tiles, the side length of each tile must be the largest possible, which is the HCF (Highest Common Factor) of the room's length and breadth. Step 3: Find HCF(1517, 902) using the Euclidean Algorithm: 1517 = 1 * 902 + 615 902 = 1 * 615 + 287 615 = 2 * 287 + 41 287 = 7 * 41 + 0 The HCF is 41 cm. Step 4: The side length of each square tile is 41 cm. Step 5: Calculate the area of one tile = 41 cm * 41 cm = 1681 sq. cm. Step 6: Calculate the area of the room floor = 1517 cm * 902 cm = 1368334 sq. cm. Step 7: Number of tiles required = (Area of room floor) / (Area of one tile). Step 8: Number of tiles = (1517 * 902) / (41 * 41) = (1517/41) * (902/41) = 37 * 22 = 814 tiles.
13
Determine the Highest Common Factor (HCF) of 132, 204, and 228.
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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.
14
Count how many numbers below 1000 are divisible by both 10 and 13.
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Solution: Step 1: A number divisible by both 10 and 13 must be a multiple of their Least Common Multiple (LCM). Step 2: Since 10 and 13 are coprime (they have no common factors other than 1), their LCM is their product. Step 3: Calculate LCM(10, 13) = 10 × 13 = 130. Step 4: We need to find how many multiples of 130 are less than 1000. Step 5: Divide the upper limit (1000) by the LCM (130): 1000 ÷ 130 ≈ 7.69. Step 6: The number of multiples less than 1000 is the floor of this result, which is 7. Step 7: The multiples are 130, 260, 390, 520, 650, 780, and 910. Step 8: There are 7 such numbers.
15
Four individuals begin running simultaneously around a circular track from the same starting point on the circumference at 9:00 am. They complete one round in 40, 50, 60, and 30 minutes, respectively. At what time will they all meet together at the starting point again?
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Solution: Step 1: To find when they will all meet together again at the starting point, calculate the Least Common Multiple (LCM) of their individual times to complete one round. Step 2: The given times are 40 minutes, 50 minutes, 60 minutes, and 30 minutes. Step 3: Calculate LCM(40, 50, 60, 30): - Prime factorization: 40 = 2^3 * 5, 50 = 2 * 5^2, 60 = 2^2 * 3 * 5, 30 = 2 * 3 * 5. - LCM = 2^3 * 3^1 * 5^2 = 8 * 3 * 25 = 600. Step 4: They will meet again after 600 minutes. Step 5: Convert 600 minutes to hours: 600 minutes / 60 minutes/hour = 10 hours. Step 6: Since they started at 9:00 am, they will meet again at 9:00 am + 10 hours = 7:00 pm.
16
Calculate the highest common factor for the integers 36 and 84.
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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.
17
Three individuals, A, B, and C, begin running simultaneously from the same starting point and in the same direction around a circular stadium. A finishes a lap in 252 seconds, B in 308 seconds, and C in 198 seconds. How long will it take for them to all meet again at their starting point?
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Solution: Step 1: To find when A, B, and C will all meet again at the starting point, calculate the Least Common Multiple (L.C.M.) of their individual lap times. The lap times are 252 seconds, 308 seconds, and 198 seconds. Step 2: Find the prime factorization of each number: 252 = 2^2 × 3^2 × 7 308 = 2^2 × 7 × 11 198 = 2 × 3^2 × 11 Step 3: Calculate the L.C.M. by taking the highest power of all prime factors present: L.C.M. = 2^2 × 3^2 × 7 × 11 = 4 × 9 × 7 × 11 = 36 × 77 = 2772. Step 4: They will meet again after 2772 seconds. Convert this time into minutes and seconds. Number of minutes = 2772 ÷ 60 = 46 with a remainder of 12. So, 2772 seconds = 46 minutes and 12 seconds.
18
Calculate the Least Common Multiple of 24, 36, and 40.
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Solution: Step 1: Find the prime factorization of each number: - 24 = 2^3 * 3^1 - 36 = 2^2 * 3^2 - 40 = 2^3 * 5^1 Step 2: To find the L.C.M., take the highest power of all unique prime factors that appear in any of the numbers. - The highest power of 2 is 2^3. - The highest power of 3 is 3^2. - The highest power of 5 is 5^1. Step 3: Multiply these highest powers together to get the L.C.M.: - L.C.M. = 2^3 * 3^2 * 5^1 = 8 * 9 * 5. Step 4: Perform the multiplication: L.C.M. = 72 * 5 = 360.
19
Given that the H.C.F. of two numbers is 11 and their L.C.M. is 7700, and one of the numbers is 275, determine the other number.
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Solution: Step 1: Recall the fundamental relationship between the H.C.F. and L.C.M. of two numbers: Product of two numbers = H.C.F. × L.C.M. Step 2: Let the two numbers be A and B. We are given: H.C.F. = 11 L.C.M. = 7700 One number (A) = 275 We need to find the other number (B). Step 3: Substitute the given values into the formula: 275 × B = 11 × 7700 Step 4: Solve for B: B = (11 × 7700) / 275 B = (11 × 7700) / (11 × 25) B = 7700 / 25 B = 308. Step 5: The other number is 308.
20
Given that the HCF and LCM of two numbers are 8 and 48, respectively, and one of the numbers is 24, determine the value of the other number.
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Solution: Step 1: Apply the fundamental property that for any two positive integers, the product of the numbers is equal to the product of their H.C.F. and L.C.M. Product of numbers = H.C.F. × L.C.M. Step 2: Let the two numbers be 'A' and 'B'. We are given: H.C.F. = 8 L.C.M. = 48 One number (A) = 24 We need to find the other number (B). Step 3: Substitute the known values into the formula: 24 × B = 8 × 48 Step 4: Solve for B: B = (8 × 48) ÷ 24 B = 8 × (48 ÷ 24) B = 8 × 2 B = 16. Step 5: The other number is 16.
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