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. 30)
30.
If the sum of two numbers is 55 and the H.C.F. and L.C.M. of these numbers are 5 and 120 respectively, then the sum of the reciprocals of the numbers is equal to:
Answer: Option
Explanation:
Let the numbers be a and b.
Then, a + b = 55 and ab = 5 x 120 = 600.
 The required sum = |
1 | + | 1 | = | a + b | = | 55 | = | 11 |
| a | b | ab | 600 | 120 |
Discussion:
37 comments Page 3 of 4.
Abhishek said:
10 years ago
Thanks it helped me a lot.
Vighneshvinu said:
10 years ago
Dear @Kishan you will require options to find the answer for if HCF of two numbers is 98 and LCM of numbers is 2352 then what is the sum of these numbers?
From the options find a number which is divisible by 98. Which will be your required answer? Example 1372, 1398, 1426, 1484.
Out of these options a number which is exactly divisible by 98 is 1372.
So this will be your required answer (HCF is a common factor for both the numbers. So if x & y are the nos then the sum will be x+y = 98*(a+b). Here x = 98*a & y = 98*b).
Hope this was useful to you.
From the options find a number which is divisible by 98. Which will be your required answer? Example 1372, 1398, 1426, 1484.
Out of these options a number which is exactly divisible by 98 is 1372.
So this will be your required answer (HCF is a common factor for both the numbers. So if x & y are the nos then the sum will be x+y = 98*(a+b). Here x = 98*a & y = 98*b).
Hope this was useful to you.
Narendra Pandey said:
1 decade ago
@Vishwajit.
As HCF and LCM is given.
Standard Formula:
Product of two number = HCF*LCM.
Suppose two number be x and y.
xy = 5*12.
x+y = 55.
Now x = 55-y put the value of x in xy = 600.
Now (55-y)y = 600.
y^2-55y+600 = 0.
By factorization we get.
y^2-40y-15y+600 = 0.
y(y-40)-15(y-40) = 0.
Now y=40 and 14.
As in question reciprocal of number 1/x+1/y.
1/40+1/15 = 55/600 = 11/120.
That's it.
As HCF and LCM is given.
Standard Formula:
Product of two number = HCF*LCM.
Suppose two number be x and y.
xy = 5*12.
x+y = 55.
Now x = 55-y put the value of x in xy = 600.
Now (55-y)y = 600.
y^2-55y+600 = 0.
By factorization we get.
y^2-40y-15y+600 = 0.
y(y-40)-15(y-40) = 0.
Now y=40 and 14.
As in question reciprocal of number 1/x+1/y.
1/40+1/15 = 55/600 = 11/120.
That's it.
Jedi Knight Ani said:
1 decade ago
@Amol.
Reciprocal is calculated by dividing 1 by that number. Sum of the reciprocals = 1/a + 1/b.
Reciprocal is calculated by dividing 1 by that number. Sum of the reciprocals = 1/a + 1/b.
Viswajith said:
1 decade ago
The HCF of two numbers is 15. Find its LCM? Anyone please help with steps.
Math_lover said:
1 decade ago
@Tom.
There are 2 answers for this:
12 and 108.
36 and 84.
There are 2 answers for this:
12 and 108.
36 and 84.
Tom said:
1 decade ago
The sum of two numbers is 120 and their hcf is 12. Find the numbers?
Babita loky said:
1 decade ago
The hcf and lcm of two numbers are 4 and 48 respectively. One of the number is 12. Find the other number?
Kishan said:
1 decade ago
If HCF of two numbers is 98 and LCM of numbers is 2352 then what is the sum of these numbers?
Raj said:
1 decade ago
How to find those two numbers?
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