# 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 1 of 4.
ANSHUL JYOTI said:
5 months ago

@Nirmala.

7/16 + 5/24 = LCM ( 16, 24) = 48.

Therefore (7x3)/(16x3) + (5x2)/(24x2) = 31/48.

7/16 + 5/24 = LCM ( 16, 24) = 48.

Therefore (7x3)/(16x3) + (5x2)/(24x2) = 31/48.

Maha said:
2 years ago

@Balamurugan.

By Cross multiply, you well get this.

By Cross multiply, you well get this.

Balamurugan said:
3 years ago

1/a+1/b=a+b/ab. How is come??

Aakash Choudhary said:
3 years ago

Let the number be 5a and 5b.

Given HCF= 5 LCM =120.

Sum=55.

5a+5b=55.

=> a+b=11 ------> (1)

By property product of numbers= LCM*HCF.

5a*5b= 5*120

=> 5ab= 120 --------> (2).

Now sum of reciprocal of numbers:-

(1/5a) + (1/5b) = a+b/5ab.

Putting values from (1) and (2).

(1/5a) + (1/5b)= 11/120.

Given HCF= 5 LCM =120.

Sum=55.

5a+5b=55.

=> a+b=11 ------> (1)

By property product of numbers= LCM*HCF.

5a*5b= 5*120

=> 5ab= 120 --------> (2).

Now sum of reciprocal of numbers:-

(1/5a) + (1/5b) = a+b/5ab.

Putting values from (1) and (2).

(1/5a) + (1/5b)= 11/120.

(1)

Milan said:
5 years ago

@Bhuttoo & @Lucas.

1st of all you should find the lcm that is 180 second or 3 minutes. It means they flash together at every 3miunites.

They will flash together.

(a) at 10:48 for the second time.

(b) at 11:00 for 5th time.

1st of all you should find the lcm that is 180 second or 3 minutes. It means they flash together at every 3miunites.

They will flash together.

(a) at 10:48 for the second time.

(b) at 11:00 for 5th time.

Bhuttoo said:
6 years ago

Please, anyone help me to solve this question for me.

Two bulbs flash at regular intervals of 30 and 36 seconds respectively. They flash together at 10.45. At what time will they flash together?

(a) a second time.

(b) a 5th time.

Two bulbs flash at regular intervals of 30 and 36 seconds respectively. They flash together at 10.45. At what time will they flash together?

(a) a second time.

(b) a 5th time.

Abhi said:
6 years ago

The product of two no=2448 and the difference of the two no is equal to their HCF and the HCF is a perfect square. Find the no?

Subasini Barik said:
6 years ago

Two numbers are in the ratio of 5: 8. If 12 is added to each, they are in the ratio of 3: 4. Find the sum of two numbers? Please give me the answer and explanation.

Lucas said:
7 years ago

Please solve this question for me.

Two bulbs flash at regular intervals of 30 and 36 seconds respectively. They flash together at 10.45. At what time will they flash together?

(a) a second time.

(b) a 5th time.

Two bulbs flash at regular intervals of 30 and 36 seconds respectively. They flash together at 10.45. At what time will they flash together?

(a) a second time.

(b) a 5th time.

Avhi said:
7 years ago

The product of LCM and HCF of two no is 24 and the difference of two no is 2 then find the sum of two no?

Please give me the solution for this question.

Please give me the solution for this question.

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