The sum of two numbers is 18 and the sum of their reciprocals

Q) The sum of two numbers is 18 and the sum of their reciprocals is 9/40. Find the numbers.

Ans:

Let the numbers be X and Y

Step 1: By first condition:

X + Y = 18 ………… (i)

Step 2: By second condition:

$\frac{1}{\times} + \frac{1}{Y} = \frac{9}{40}$

∴ $\frac{ Y + \times}{\times Y} = \frac{9}{40}$

∴ 40 (Y + X) = 9 X Y

e put value of (X + Y) from equation (i), we get:

∴ 40 (18) = 9 X Y

∴ X Y = 40 x $\frac{18}{9}$

∴ X Y = 80

∴ X = $\frac{80}{Y}$ ……… (ii)

Step 3: Next, we substitute value of X in equation (i):

X + Y = 18

∴ $\frac{80}{Y} + Y = 18$

∴ $\frac{80 + Y^2}{Y} = 18$

∴ 80 + Y2 = 18 Y

∴ Y² – 18 Y + 80 = 0

∴ Y² – 10 Y – 8 Y + 80 = 0

^{∴ Y (Y – 10) – 8 (Y – 10) = 0}

∴ (Y – 10) ( Y – 8) = 0

∴ Y = 10 or Y = 8

Step 4: We put values of Y in equation (i):

X + Y = 18

∴ X + (10) = 18

∴ X = 18 – 10

∴ X = 8

Similarly, for Y = 8, we get X = 10

Therefore, the two numbers are 8 and 10.

Check: if numbers are 8 and 10, then 8 + 10 = 18 …. 1st condition satisfied

$\frac{1}{8} + \frac{1}{10} = \frac{9}{40}$ …. 2nd condition also satisfied.

Since both the given condition are satisfied, hence our answer is correct.

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