In this HCF and LCM worksheets 1,we give the questions on HCF and LCM whose qualities are in high standard and practicing these questions will definitely make the students to score marks in HCF and LCM.
Question No.1 Question No.2 Question No.3 Question No.4 Question No.5 Question No.6 Question No.7 Question No.8 Question No.9 Question No.10 |
2^{3}x3^{2}x5x7^{4}, 2^{2}x3^{5}x5^{2}x7^{6}, 2^{3}x5^{3}x7^{2}
In the above given numbers, we find 2, 5 and 7 in common. Take 2, 5 and 7 with minimum power. That is 2^{2}, 5, 7^{2} Now, H.C.F = 2^{2}x5x7^{2} = 980
2^{2}x3^{3}x5
, 2^{3}x3^{2}x5^{2}, 2x3x5^{2}
In the above given numbers, take all prime factors with maximum power. They are, 2^{3}, 3^{3}, 5^{2} Now, L.C.M = 2^{3}x3^{3}x5^{2} = 5400
We we look in to the given numbers 0.63, 1.05, and 2.1, maximum number of digits after the decimal is 2.
So, let us multiply each number by 100 to avoid decimal. When the given numbers are multiplied by100, we get 63, 105, 210 63 = 3^{2}x7 105 = 5x3x7 210 = 2x5x3x7 In the prime factors (63, 105, 210), we find 3 and 7 in common Take 3 and 7 with minimum power. They are 3^{1} and 7^{1} Now, H.C.F of (63, 105, 210) = 3^{1}x7^{1} = 21 To get H.C.F of (0.63, 1.05, 2.1), divide 21 by 100. 21/100 = 0.21 Therefore, H.C.F of (0.63, 1.05, 2.1) = 0.21
From the given ratio, two numbers are 15x and 11x.
H.C.F of 15x and 11x = x ---(1) Given : H.C.F of two numbers = 13 ---(2) From(1)& (2), we get x = 13 15x = 15(13) = 195 11x = 11(13) = 143 Hence, the two numbers are 195 and 143
Given: One of the numbers = 77, H.C.F = 11 and L.C.M = 693
Let "x" be the other number. Product of two numbers = Product of their H.C.F and L.C.M 77x = 11X693 x = 99 Hence, the other number is 99
Given numbers are 16, 24, 36, 54
16 = 2^{4} 24 = 2^{3}x3 36 = 2^{2}x3^{2} 54 = 2x3^{3} In the above prime factors of (16, 24, 36, 54), take all prime factors with maximum power. They are 2^{4} and 3^{3} Now, L.C.M = 2^{4}x3^{3} = 432
Given numbers are 513, 1134, 1215
513 = 3^{3}x19 1134 = 2x3^{4}x7 1215 = 3^{5}x5 In the above prime factors of (513, 1134, 1215), we find 3 in common.So, take 3 with minimum power. That is 3^{3} Now, H.C.F = 3^{3} = 27
Required greatest possible length = H.C.F of (495, 900, 1665)
495 = 3^{2}x5x11 900 = 2^{2}x3^{2}x5^{2} 1665 = 3^{2}x5x37 In the prime factors of (495, 900, 1665), we find 3 and 5 in common. Take 3 and 5 with minimum power. They are 3^{2} and 5 Now, H.C.F = 3^{2}x5 = 45 Hence, the required greatest possible length is 45 cm
Required greatest number = H.C.F. of (1657 - 6) and (2037 - 5)
= H.C.F. of (1651, 2032) = 127.
Required Number = H.C.F of [(132-62), (237-132), and (237-62)]
= H.C.F of [70, 105, 175] = 35 |