Question

The sum of two numbers is 55 and the HCF and LCM of these numbers are 5 and 120 respectively then find sum of the reciprocals of the numbers .

Answer 1

Step-by-step explanation:

Given:-

The sum of two numbers is 55 and the HCF and LCM of these numbers are 5 and 120 respectively .

To find:-

Find the sum of the reciprocals of the numbers ?

Solution:-

Let the required numbers be a and b

Given that

The sum of two numbers = 55

a + b = 55 ----------(1)

The HCF of the two numbers = 5

The LCM of the two numbers = 120

The product of their LCM and HCF

= 5×120

= 600

We know that

The product of the LCM and the HCF of two numbers = The product of the numbers

a×b = 600 ----------(2)

We know that

(a-b)^2 = (a+b)^2-4ab

=> (a-b)^2 = 55^2-4(600)

=> (a-b)^2 = 3025-2400

=> (a-b)^2 = 625

=> a-b = ±√625

=> a-b=±25

Since it can be negative

a-b = 25------------(3)

On adding (1)&(3)

a + b = 55

a - b = 25

(+)

_________

2a + 0 = 80

__________

=> 2a = 80

=>a = 80/2

=>a = 40

In Substituting the value of a in (3)

40-b = 25

=> b = 40-25

=> b = 15

Therefore, a = 40 and b = 15

The numbers are 40 and 15

Reciprocal of a = 1/40

Reciprocal of b = 1/15

Sum of their reciprocals

= (1/40)+(1/15)

LCM of 40 and 15 = 120

= (3+8)/120

= 11/120

Answer:-

The sum of the reciprocals of the numbers = 11/120

Check:-

The numbers = 40 and 15

Their sum = 40+15=55

40 = 2×2×2×5

15 = 3×5

Their LCM = 2×2×2×3×5 = 120

Their HCF = 5

Verified the given relations in the problem.