Find the least number which being divided by 9, 12, 16 and 30 leaves in each case remainder 3.
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Solution
Required number = LCM of (9, 12, 16 and 30) + 3 = 720 + 3 = 723
The LCM of x2 – 4 and x2 + x – 6 will be.
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Solution
Breaking the polynomials
x2 – 4 = (x + 2) (x – 2)
x2 + x – 6 = (x + 3) (x – 2)
So LCM = (x – 2) (x + 2) (x + 3)
The LCM of 5, 8, 12, 20 will not be a multiple of.
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Solution
To be a multiple of 9, A number has to be multiple of 3 by twice.
Which is the greatest number of four digits that when divided by 6, 9, 12 and 17 leaves a remainder of 1 in each case ?
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Solution
Greatest number of four digits = 9999
LCM of 6, 9, 12 and 17 = 612
Dividing 9999 by 612 remainder = 207
So required number = 9792 + 1 = 9793
The HCF of polynomials will be.
x2 – 5x + 6, x2 – 7x + 10
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Solution
Breaking the polynomials x2 – 5x + 6 and x2 – 7x + 10
x2 – 5x + 6 = (x – 2) (x – 3)
x2 – 7x + 10 = (x – 2) (x – 5)
HCF = x – 2
Find the LCM of .
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Solution
LCM of
= 440
Two numbers are in the ratio of 5 : 11. If their HCF is ‘8’ then LCM will be.
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Solution
Let numbers are 5x and 11x
HCF = 8
So numbers are 40 and 88
LCM = 11 × 5 × 8 = 440
3 bells commence separately after intervals of 2, 3 and 5 seconds respectively. If they commence at 12 O’ noon together then the no. of times they will commence together till one o’clock ?
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Solution
Intervals of bells = 2, 3, 5 seconds
LCM = (2, 3, 5) = 30
So they together will ring at a cyclic period of 30 seconds.
In hour =
Total = 120 + 1 = 121
The HCF of 0.9 kg and 1005 gram will be.
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Solution
0.9 kg = 900 gram and 1005 gram
HCF of (900 and 1005) = 15 gram
The HCF of terms will be.
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Solution
HCF of