Aptitude - Problems on H.C.F and L.C.M - Discussion
Discussion Forum : Problems on H.C.F and L.C.M - General Questions (Q.No. 6)
6.
The product of two numbers is 4107. If the H.C.F. of these numbers is 37, then the greater number is:
Answer: Option
Explanation:
Let the numbers be 37a and 37b.
Then, 37a x 37b = 4107
ab = 3.
Now, co-primes with product 3 are (1, 3).
So, the required numbers are (37 x 1, 37 x 3) i.e., (37, 111).
Greater number = 111.
Discussion:
76 comments Page 5 of 8.
Prakruthi said:
1 decade ago
a*b = HCF*LCM;
4107 = 37*LCM;
LCM = 4107/37;
LCM = 111;
4107 = 37*LCM;
LCM = 4107/37;
LCM = 111;
Sandip said:
1 decade ago
We know that HCF*LCM = Product of two number.
So the other no is = 4107/37 = 111.
So the other no is = 4107/37 = 111.
Ramesh said:
1 decade ago
Given HCF = 37.
Product of two numbers is = 4107.
As HCF is greatest divisor so the number will be 4107/37 = 111.
The greatest number is 111.
Product of two numbers is = 4107.
As HCF is greatest divisor so the number will be 4107/37 = 111.
The greatest number is 111.
DHEERAJ SINGH said:
1 decade ago
HCF = 37.
Product of two numbers are = 4107.
HCFxLCM = Product of two nos.
37xLCM = 4107.
LCM = 111.
Since LCM is the least no which will be divided by the number.
So the number will be a multiple of 111, so from above options 111 is the only multiple of 111. So answer is 111.
Product of two numbers are = 4107.
HCFxLCM = Product of two nos.
37xLCM = 4107.
LCM = 111.
Since LCM is the least no which will be divided by the number.
So the number will be a multiple of 111, so from above options 111 is the only multiple of 111. So answer is 111.
Durga said:
1 decade ago
But how we will know 37 is the H.C.F first only.
Prasad said:
1 decade ago
Generally ((hcf*lcm)/(n1*n2)) = 1.
So here the hcf is=37;
Let lcm is =x;
The product of two numbers = 4107.
So,
Now ((hcf*lcm)/(n1*n2)) = 1.
(37*x)/(4107) = 1.
By solving above expression we get x value as 111.
So here the hcf is=37;
Let lcm is =x;
The product of two numbers = 4107.
So,
Now ((hcf*lcm)/(n1*n2)) = 1.
(37*x)/(4107) = 1.
By solving above expression we get x value as 111.
Ramkumar LS said:
1 decade ago
Simplest method in few secs.
Formula:
Product of two numbers = HCF X LCM.
Given:
Product of two numbers = 4107, HCF = 37 and LCM = unknown.
Now, simply substitute the given values.
4107 = (37 X unknown).
Lets keep unknown on one side and bring 37 below.
Unknown = 4107/37.
Therefore, Unknown= 111 //simply divide 4107 by 37.
Note: When you do this on paper this will only take 2 secs to find.
Formula:
Product of two numbers = HCF X LCM.
Given:
Product of two numbers = 4107, HCF = 37 and LCM = unknown.
Now, simply substitute the given values.
4107 = (37 X unknown).
Lets keep unknown on one side and bring 37 below.
Unknown = 4107/37.
Therefore, Unknown= 111 //simply divide 4107 by 37.
Note: When you do this on paper this will only take 2 secs to find.
Prasann said:
1 decade ago
Direct formula we had please check.
Product of two numbers (a x b) = L.C.M x H.C.F.
Given H.C.F = 37 and product = 4107.
Therefore, L.C.M = Product/H.C.F = 4107/37 = 111.
Product of two numbers (a x b) = L.C.M x H.C.F.
Given H.C.F = 37 and product = 4107.
Therefore, L.C.M = Product/H.C.F = 4107/37 = 111.
Rups said:
1 decade ago
H.C.F*L.C.M = a*b.
37*L.C.M = 4107.
L.C.M = 4107/37.
L.C.M = 111.
37*L.C.M = 4107.
L.C.M = 4107/37.
L.C.M = 111.
Mukesh said:
1 decade ago
HCF means greater number divisible. 37 is dividable by 1 and 37. Apart from this 37 is highest. So divide 4107/39=111.
Post your comments here:
Quick links
Quantitative Aptitude
Verbal (English)
Reasoning
Programming
Interview
Placement Papers