LCM and HCF problem

Question : There are three numbers such that the product of the first two numbers is 91
                 and that of the last two numbers is 221. Find all three numbes and their sum

Answer :
Suppose numbers are A , B AND C.

Also given   A x B  = 91 and B x C = 221

Since the numbers are prime number, they contain only 1 as the common factor.

Also, above two products have B in common.

So, B = H.C.F. of ( 91 and 221 ) = 13;

A  = 91/13 = 7;  C  = 221/13 = 17.

Required sum = (7 + 13 + 21 ) = 41.

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Question : The ratio of two numbers is 3 : 4. The L.C.M. of two numbers is 48.
                  Find sum of the number

Solution : Let the numbers be 3x and 4x.

Then, their L.C.M. = 12x.

So, 12x = 48 or x = 4

The numbers are 12 and 16.

Hence, required sum = (12 + 16) = 28.

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Important points regarding LCM and HCF must remember for quantity exams like Bank PO

1) Least number which when divided by numbers gives same remainder in each case will be LCM of given number plus remainder 

      Example :  Find least number which when divided by 5 , 6 , 7 and 8 leaves remainder 2 ?

                        LCM ( 5 , 6 , 7 , 8 ) is 840

                         So  our answer is 840 + 2 = 842

2)  Largest number which divides to given numbers leaves same reminder , arrange the number in ascending order like a , b ,c then find out HCF of ( b-a , c-b . c-a )

 Example : Find the greatest number that will divide 42, 90 and 178 so as to leave the same remainder in each case. 

Answer is HCF of ( 178-90 , 90-42 , 178-42 ) = HCF of ( 88 , 48 , 136 ) =   8 and remainder 2

3) Largest number which divides X , Y and Z to leave remainder A ,B and C to find this

 calculate HCF of ( X - A  ,  Y - B , Z-c )

LCM and HCF defination  : http://maths4ur-kid.blogspot.in/2017/01/lcm-and-hcf-problem.html

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