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. 22)
22.
Three numbers which 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. The sum of the three numbers is:
Answer: Option
Explanation:
Since the numbers are co-prime, they contain only 1 as the common factor.
Also, the given two products have the middle number in common.
So, middle number = H.C.F. of 551 and 1073 = 29;
| First number = |  |
551 |  |
= 19; Third number = | |
1073 | |
= 37. |
| 29 | 29 |
Required sum = (19 + 29 + 37) = 85.
Discussion:
53 comments Page 2 of 6.
Animo said:
6 years ago
Thanks @Safeer.
Kalpit said:
7 years ago
Good explanatiion, Thanks @Uttam.
Shaurya said:
7 years ago
Let the three no be a,b,c.
a*b=551
a*b=19*29
Now
b*c=1073
b*c=29*37
So we get a=19,b=29 and c=37.
On adding a+b+c=19+29+37=85.
a*b=551
a*b=19*29
Now
b*c=1073
b*c=29*37
So we get a=19,b=29 and c=37.
On adding a+b+c=19+29+37=85.
(2)
Tanu said:
7 years ago
Why we take HCF why not LCM? Please explain me.
Abhoi said:
7 years ago
Thanks @Asutosh.
Dinu said:
7 years ago
@Jyoti.
30-1 = 29.
So,
80+5=85.
30-1 = 29.
So,
80+5=85.
Anamika said:
7 years ago
Thanks all for the answer.
(1)
Muhammad safeer ahmed said:
7 years ago
a*b=551
b*c=1073
So, b is =1073-551=522,
522-551=29,
29 is middle and common term exist in both and b =29.
Put the value
a*b=551
a*29=551
a=551/29=19
a=19.
Put the value
b*c=1073
29*c=1073
C=1073/29=37.
So
a=19
b=29
c=37
19+29+37=85.
b*c=1073
So, b is =1073-551=522,
522-551=29,
29 is middle and common term exist in both and b =29.
Put the value
a*b=551
a*29=551
a=551/29=19
a=19.
Put the value
b*c=1073
29*c=1073
C=1073/29=37.
So
a=19
b=29
c=37
19+29+37=85.
(6)
Asaduzzaman said:
8 years ago
Thanks.
Surya Deep said:
8 years ago
Consider numbers as X, Y, Z respectively.
Now,
Consider product XY = 551 ------eq1
Consider product YZ = 1073 -----eq2
So, clearly common highest factor btw XY and YZ should be Y
i.e HCF (551,1073) = 29.
From eq 1, X=19.
from eq 2, Z=37.
Now, Add this numbers (X+Y+Z)
(19+29+37)=85 And.
Now,
Consider product XY = 551 ------eq1
Consider product YZ = 1073 -----eq2
So, clearly common highest factor btw XY and YZ should be Y
i.e HCF (551,1073) = 29.
From eq 1, X=19.
from eq 2, Z=37.
Now, Add this numbers (X+Y+Z)
(19+29+37)=85 And.
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