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 6 of 8.
ZUBAYRUL HASAN said:
8 years ago
As 37 is the HCF of these numbers. So, we can take the smaller number is 37 and greater is 37x.
A.C, 37*37x=4107,
or, x= 4107÷(37*37),
or, x =3,
Now, the greatest number is 37*3=111 (Ans).
A.C, 37*37x=4107,
or, x= 4107÷(37*37),
or, x =3,
Now, the greatest number is 37*3=111 (Ans).
Arun said:
8 years ago
Thank you all.
Aparna said:
8 years ago
No need of these complications.
As per formula PRODUCT OF TWO NOS. = LCM * HCF.
hence, substitute from the question wherein,
4107=lcm * 37.
4107/37 = 111 =lcm.
As per formula PRODUCT OF TWO NOS. = LCM * HCF.
hence, substitute from the question wherein,
4107=lcm * 37.
4107/37 = 111 =lcm.
Vinay gudipati said:
7 years ago
Consider the equation be like x*y=4107.
Here x and why are the taken when product them we get 4107.
And also given that HCF of the two numbers is 37 so.
37*x*37*y=4107.
x*y= (4107/ (37*37) ) =3.
xy=3.
x*y=3*1.
x=3.
y=1.
Therefore 37*3=111.
37*1=37 so the answer is 111.
Here x and why are the taken when product them we get 4107.
And also given that HCF of the two numbers is 37 so.
37*x*37*y=4107.
x*y= (4107/ (37*37) ) =3.
xy=3.
x*y=3*1.
x=3.
y=1.
Therefore 37*3=111.
37*1=37 so the answer is 111.
(1)
Abitha said:
7 years ago
Here for ab=3, its possibilities are (1, 2) & (2, 1) Only right?
Please, anyone, help me.
Please, anyone, help me.
Manikanta said:
7 years ago
For a*b =3. So.
1*3 =3.
And.
3*1 = 3.
(May be a=1 then b=2).
(May be b=1 then a=2).
Then how you are deviating to 1, 2 and 2, 1? Please tell me.
1*3 =3.
And.
3*1 = 3.
(May be a=1 then b=2).
(May be b=1 then a=2).
Then how you are deviating to 1, 2 and 2, 1? Please tell me.
Manikanta said:
7 years ago
For a*b =3. So.
1*3 =3.
And.
3*1 = 3.
(May be a=1 then b=2).
(May be b=1 then a=2).
Then how you are deviating to 1, 2 and 2, 1? Please tell me.
1*3 =3.
And.
3*1 = 3.
(May be a=1 then b=2).
(May be b=1 then a=2).
Then how you are deviating to 1, 2 and 2, 1? Please tell me.
Nidhu said:
6 years ago
We know that;
H.C.F * L.C.M = product of two numbers.
37* x = 4107,
x=4107/37,
x = 111.
H.C.F * L.C.M = product of two numbers.
37* x = 4107,
x=4107/37,
x = 111.
Harwani Payal said:
6 years ago
Assume Two numbers p and q.
Multiply them with 37.
37p*37p.
Now, 37p*37p = 4107.
1369pq = 4107.
pq = 4107/1369.
pq = 3.
Multiply them with 37.
37p*37p.
Now, 37p*37p = 4107.
1369pq = 4107.
pq = 4107/1369.
pq = 3.
Sai vamsi said:
6 years ago
x*y = 4107.
Hcf of two nums is 37. So these two nums are divisible by 37.
37x*37y = 4107.
37(x*y) = 4107.
x*y = 4107/37 = 111.
Hcf of two nums is 37. So these two nums are divisible by 37.
37x*37y = 4107.
37(x*y) = 4107.
x*y = 4107/37 = 111.
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