Question

sum of two numbers is 11 and sum of their reciprocals is 11/28 Find the number?

Answer 1

let the two numbers are x and y.
therefore according to question their sum is 11
x+y=11---------i
now.
sum of their reciprocal is 11/28
1/x+1/y=11/28
x+y/by=11/28
28(x+y)=11xy-------ii
xy=28.
now
(x-y)^2=x^2+y^2-2xy
(x-y) ^2=(x+y)^2-4xy
(x-y) ^2=121-112
x-y=3--------iii
on adding I and IIi, we get
x+y+x-y=3+11
2x=14
x=7
on putting x=7 in first equation we get
y=4
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Answer 2

Answer:

$\large{\underline{\boxed{\sf 7\: and \: 4}}}$

Step-by-step explanation:

Given that -

The sum of two numbers is 11 and the sum of their reciprocal is 11/28.

Let the numbers be x and y respectively.

Sum of numbers is 11.

⇒ x + y = 11.... (i)

⇒ y = 11 - x .... (ii)

Sum of reciprocals is 11/28.

⇒ $\dfrac{1}{\text{x}} + \dfrac{1}{\text{y}}= \dfrac{11}{28}$.... (iii)

Now, on solving (iii),

$\dfrac{1}{\text{x}} + \dfrac{1}{\text{y}}= \dfrac{11}{28} \\ \\ \frac{ \text {x + y }}{ \text{xy}} = \frac{11}{28} \\ \\ \bf on \: cross \: multiplying : \\ \\28( \text{x + y) = 11xy}$

Putting the value of (i) and (ii) here, we get -

$28 * 11 = 11 \text{x(11 - x)} \\ \\ \implies 28 = \text{x(11 - x)} \\ \\ \implies 28 = 11 \text{x - x}{}^{2} \\ \\ \implies \text{x} {}^{2} - 11\text x + 28 \\ \\ \implies \text{x} {}^{2} - 7\text{x - 4x} + 28 = 0 \\ \\ \implies \text{x(x - 7) - 4(x - 7) = 0 } \\ \\ \implies \text{(x - 7)(x - 4) = 0} \\ \\ \boxed{\therefore \bf x = 7 \: or \: x = 4}$

Hence, the required numbers are 7 and 4.