Question

2. The sum of two numbers is 50. Find the numbers
if the sum of their reciprocals is
8 who will answer this question with in 2minutes I will mark him brainlist​

Answer 1

Step-by-step explanation:

let the two numbers be x and y

according to question-

x+y=50(equation 1)

and

1/x +1/y =8(equation 2)

from equation 1

x=50-y

putting value of x in equation 2

1/50-y +1/y =8

(y+50-y)/y(50-y)=8

= -8y2 + 400y-50=0

after solving you will get the answer

Answer 2

$\dag \: \underline{\sf Correct \: Question:}$

• The sum of two numbers is 50. if the sum of their reciprocals is 1/8. Find the numbers.

$\dag \: \underline{\sf Answer:}$

• In the question we are provided that,The sum of two numbers is 50. So, first assume the two numbers as M and N. Mathematically ;

$:\implies\sf M + N = 50 \qquad\Bigg\lgroup\textbf{Equation (I)}\Bigg\rgroup$

From equation (I) transpose M from LHS to RHS and we get :

$:\implies\sf N = 50 - M \qquad\Bigg\lgroup\textbf{Equation (II)}\Bigg\rgroup$

• We are also given that,the sum of their reciprocals is 1/8. mathematically;

$:\implies\sf \dfrac{1}{M}+ \dfrac{1}{ N} = \dfrac{1}{8} \qquad\Bigg\lgroup\textbf{Equation (III)}\Bigg\rgroup$

$:\implies\sf \dfrac{M + N}{MN} = \dfrac{1}{8}$

• Now, cross multiply both the sides and we get :

$:\implies\sf 8(M + N)= 1(MN)$

• Now, by replacing the value of N from equation (II) we get :

$:\implies\sf 8(M + 50 -M )= M(50 -M )$

$:\implies\sf 8 \times 50= 50M -M^{2}$

$:\implies\sf 400= 50M -M^{2}$

$:\implies\sf M^{2} - 50 M + 400 = 0$

$:\implies\sf M^{2} - 10 M - 40 M + 400 = 0$

$:\implies\sf M( M - 10) - 40 (M + 10) = 0$

$:\implies\sf ( M - 10) (M- 40 )= 0$

$:\implies \underline{ \boxed{\sf M = 10 \: or \: 40}}$

$\therefore\:\underline{\textsf{The required two numbers are \textbf{10 and 40}}}.$