If the LCM of two numbers is 138. But their GCD is 23. The numbers are in a ratio 1:6. Which is the largest number amongst the two?
Answer: B
Let the common factor be K
Therefore, the numbers are K and 6K
Therefore, K X 6K = 23 X 138
Therefore, K = 23
Therefore, larger number = 6K = 6 X 23 = 138
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Rajesh had to arrange his books in uniform groups. He makes groups of 4 books each. But 3 books are left. He tries it with groups of 5 books each. But still 3 books are left. 3 books are still left when he tried with groups of 9 or 10 books each. How many books does he have?
Answer: D
No answer description available for this question.
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If the product of two numbers is 84942 and their H.C.F. is 33, find their L.C.M.
Answer: A
\(H.C.F. \times L.C.M. = 84942\), because we know
Product of two numbers = Product of H.C.F. and L.C.M.
\(L.C.M. = 84942 \div 33\)
\(= 2574\)
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There are three numbers, these are co-prime to each other are such that the product of the first two is 551 and that of the last two is 1073. What will be the sum of three numbers :
Answer: C
As given the questions these numbers are co primes, so there is only 1 as their common factor.
It is also given that two products have the middle number in common.
So, middle number = H.C.F. of \(551\) and \(1073 = 29\)
So first number is : \(\frac{551}{29} = 19\)
Third number = \(\frac{1073}{29} = 37\)
So sum of these numbers is,
\(= \left ( 19 + 29 + 37 \right ) = 85\)
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A, B and C start at the same time in the same direction to run around a circular stadium. A completes a round in 252 seconds, B in 308 seconds and c in 198 seconds, all starting at the same point. After what time will they again at the starting point ?
Answer: D
L.C.M. of \(252, 308\) and \(198 = 2772\)
So, \(A, B\) and \(C\) will again meet at the starting point in \(2772\) seconds.
i.e. \(46\) minutes \(12\) seconds.
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