Question

The sum of two numbers is 55 and the H.C.Fand L.C.M of these numbers 5 and 120 respectively.then,find out the sum of the reciprocal of these two numbers?​

Answer 1

Answer:-

Let the numbers be "a" and "b".

given,

Sum of the numbers = 55

a + b = 55 -- equation (1)

HCF of both the numbers = 5

LCM of both the numbers = 120.

We know that, Product of two numbers = HCF * LCM.

Hence, a*b = 5*120

ab = 600 -- equation (2)

We know that, (a - b)² = (a + b)² - 4ab

So, (a - b)² = (55)² - 4(600)

(a - b)² = 3025 - 2400

(a - b)² = 625

a - b = √625

a - b = 25 -- equation (3)

Adding equations (1) and (3),

a + b + a - b = 55 + 25

2a = 80

a = 80/2

a = 40

Substitute "a" value in equation (1)

a + b = 55

40 + b = 55

b = 55 - 40

b = 15

Reciprocal of "a" = 1/a = 1/40

Reciprocal of "b" = 1/b = 1/15

$sum \: of \: the \: reciprocals = \frac{1}{40} + \frac{1}{15} \\ \\ \frac{1}{a} + \frac{1}{b} = \frac{3 + 8}{120} \\ \\ \frac{1}{a} + \frac{1}{b} = \frac{11}{120}$

Answer 2

Let two numbers be M and N.

Given that, the sum of two numbers is 55.

$\implies\:\sf{M+N\:=\:55}$

$\implies\:\sf{M\:=\:55-N}$

Also given that, H.C.F. and L.C.M. of these numbers are 5 and 120.

→ H.C.F. = 5 and L.C.M. = 120

We know that,

Product of two number = L.C.M. × H.C.F.

$\implies\:\sf{MN\:=\:120 \times 5}$

$\implies\:\sf{MN\:=\:600}$

We have to find the sum of the reciprocal of these two numbers.

Sum of reciprocal of numbers = $\sf{\dfrac{1}{M}+\dfrac{1}{N}}$

$\implies\:\sf{\dfrac{N+M}{MN}}$

Now,

From above we have M + N = 55 and MN = 600

Substitute these values to find the sum of the numbers.

$\implies\:\sf{\dfrac{55}{600}}$

On simplifying we get,

$\implies\:\sf{\dfrac{11}{120}}$

Therefore, sum of the reciprocal of these numbers (M and N) is 11/120.