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 1 of 6.
Ram said:
9 years ago
29 is the HCF of 551 and 1073.
To find HCF of 551 and 1073 follow these steps.
551)1071(1
551
----------------
522)551(1
522
------------
29)522(18
29
---------
232
232
------------
XXX
So,HCF of 551 and 1073 is 29.
To find HCF of 551 and 1073 follow these steps.
551)1071(1
551
----------------
522)551(1
522
------------
29)522(18
29
---------
232
232
------------
XXX
So,HCF of 551 and 1073 is 29.
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.
Uttam said:
1 decade 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, Y=37.
Pls comment if my assumption is wrong.
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, Y=37.
Pls comment if my assumption is wrong.
Aparna said:
1 decade ago
Nothing to worry,
see in HCF we get a number which is common to both numbers.
So find the hcf of 551 and 1073, we get 29
which is one number among three nos.
Let other two numbers be a and b
Given
a*29=551
a=551/29
a=19
Now,
b*29=1073
b=37
Hence three numbers. are 29,19,37
see in HCF we get a number which is common to both numbers.
So find the hcf of 551 and 1073, we get 29
which is one number among three nos.
Let other two numbers be a and b
Given
a*29=551
a=551/29
a=19
Now,
b*29=1073
b=37
Hence three numbers. are 29,19,37
(16)
Sathiya priya said:
1 decade ago
Let assume three numbers are a, b and c.
Given:
The product of a and b is 551.
The product of b and c is 1073.
b is common in both the product
b = H.C.F. of 551 and 1073 = 29;
a = (551/29) = 19.
c = (1073/289) = 37.
The sum of three number is 85.
Given:
The product of a and b is 551.
The product of b and c is 1073.
b is common in both the product
b = H.C.F. of 551 and 1073 = 29;
a = (551/29) = 19.
c = (1073/289) = 37.
The sum of three number is 85.
Nivi said:
5 years ago
Let me explain.
Co-prime : factors are 1 and itself.
Given: XY=551.
YZ=1073.
Since why is a common factor in both the equations and the as the no's are caprice then the other factor becomes the highest common factor.
So, Y is the HCF of XY and YZ.
Co-prime : factors are 1 and itself.
Given: XY=551.
YZ=1073.
Since why is a common factor in both the equations and the as the no's are caprice then the other factor becomes the highest common factor.
So, Y is the HCF of XY and YZ.
(7)
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)
Neharika said:
9 years ago
Four prime numbers are arranged in ascending order according to their magnitude. The product of first Three is 20,677 and the product of last three 33,263. Find the numbers.
Can anyone solve and explain this question ?
Can anyone solve and explain this question ?
Divyanshu Priyadarshi said:
8 years ago
Let the three numbers be x, y and z.
Hence, xy=551 and yz=1073.
Middle number =HCF of 551 and 1073 = 29.
First number =551/29= 19.
Third number =1073/29= 37.
Required numbers are 19,29 and 37.
Hence, xy=551 and yz=1073.
Middle number =HCF of 551 and 1073 = 29.
First number =551/29= 19.
Third number =1073/29= 37.
Required numbers are 19,29 and 37.
Ranjith said:
9 years ago
xy = 551 & yz = 1073; clearly y is a HCF.
Multiply xy * yz = 1073 * 551 we get xzy^2 = 29sqroot703;
So 29 is y; then it is easy to proceed .
Multiply xy * yz = 1073 * 551 we get xzy^2 = 29sqroot703;
So 29 is y; then it is easy to proceed .
(1)
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