Latest IBM Aptitude Question SOLUTION: If 19 and 1140 are the respective HCF and LCM of two numbers, which are greater than 19 then what will be the possible number of such pair?

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If 19 and 1140 are the respective HCF and LCM of two numbers, which are greater than 19 then what will be the possible number of such pair?

Asked In IBM AISHWARYA (10 years ago)

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PriyaAmmu  (10 years ago)

Product of HCF and LCM = product of the numbers
Then, product of the numbers = 19 x 1140
Let 19a and 19b be the numbers.
19a x 19b = 19 x 1140
ab = 19 x 1140 / 19 x 19 = 60

If ab = 60 then (a,b) = (1,60), (2,30), (3,20), (4,15), (5,12) and (6,10).
Since a and b are co-primes then (a,b) = (1,60), (4,15) and (5,12)
Hence the number of such pairs = 3a

Is this solution Helpfull? Yes (43) | No (21)  

Kunal Patel  (9 years ago)

3 pairs are possible.
How:
product of two numbers= 19 x 1140
both the numbers should be greater than 19. So take two numbers as (19xa=19a) and (19xb=19b)
19a x 19b = 19x1140 ......(1)
ab=60.
If ab=60 then we can have pairs like (1,60) ,(2,30),(3,20),(4,15),(5,12),(6,10).
Now both the numbers will be prime numbers because if they are not then their HCF will not be 19.
So possible pair of (a,b) are (1,60),(3,20),(4,15),(5,12).
Not (1,60) is not allowed because one of the two number will then become 19 and we required both number must be > 19.
Use (a,b)=(3,20) in Eq. (1)...We got two numbers 57 and 380
Use (a,b)=(4,15) in Eq. (1)... We got two numbers 76 and 285
Use (a,b)=(5,12) in Eq. (1)... We got two numbers 95 and 228
So,at the end we have 3 possible pairs of two numbers which are >19 are (57,380),(76,285),(95,228)

Is this solution Helpfull? Yes (20) | No (1)  

Raunak MAnchanda  (10 years ago)

Let nos be 19a,19b. Then since Hcf=19 and Lcm=1140.
We get a*b=60. Also given in question,the numbers are greater than 19,(1,60) will not be a pair.Possible pairs for which a*b=60 is (2,30),(3,20),(4,15),(5,12),(6,10) =5

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sairam  (10 years ago)

19a*19b=19*1140
a*b=19*1140/361
3 such pairs are possible

Is this solution Helpfull? Yes (11) | No (9)  

Monalisa Sarkar  (10 years ago)

The numbers should be greater than 19 right?? so how can any pair be the possible answer???

Is this solution Helpfull? Yes (9) | No (3)  

alekya  (10 years ago)

product of two numbers=hcf*lcm..let two numbers be a & b then a*b=19*1140
since hcf is 19 number may 19a & 19b where a and b are coprimes.
so, 19a*19b=19*1140.
a*b=60.
so,the possible pairs are (1,60)(2,30)(3,20),(4,15),(5,12),(6,10)..total 6 pairs

Is this solution Helpfull? Yes (6) | No (5)  

VIVEK KUMAR  (9 years ago)

since hcf=19
hence no. are 19a and 19 b.
product of two numbers= HCF * LCM
19a *19b = 19 * 1140
hence ab= 60
hence pair are (1,60),(2,30), (3,20),(4,15)(5,12),(6,10)
but the numbers should be higher than 19.
hence no such pairs exists..

Is this solution Helpfull? Yes (5) | No (5)  

Sudipta Ghosh  (10 years ago)

there are three possible co-prime pairs,but (1,60) is not possible because numbers are greater than 19.so ans will be (4,15) and (5,12)

Is this solution Helpfull? Yes (3) | No (3)  

vishal  (10 years ago)

SAIRAM U ARE CRRECT RAUNAK U ARE WRONG;;;;;THE WRITE ANSWER IS 3

Is this solution Helpfull? Yes (2) | No (1)  

riya  (6 years ago)

5 pairs like if u agree..

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