Find two natural numbers, the sum of whose squares is 25 times their sum and also equal to 50 times their difference.

Question:

Find two natural numbers, the sum of whose squares is 25 times their sum and also equal to 50 times their difference.

Solution:

Let the two natural numbers be $x$ and $y$.

According to the question:

$x^{2}+y^{2}=25(x+y) \ldots$ (i)

$x^{2}+y^{2}=50(x-y) \ldots$ (ii)

From (i) and (ii), we get:

$25(x+y)=50(x-y)$

$\Rightarrow x+y=2(x-y)$

$\Rightarrow x+y=2 x-2 y$

$\Rightarrow y+2 y=2 x-x$

$\Rightarrow 3 y=x \quad \ldots$ (iii)

From (ii) and (iii), we get:

$(3 y)^{2}+y^{2}=50(3 y-y)$

$\Rightarrow 9 y^{2}+y^{2}=100 y$

$\Rightarrow 10 y^{2}=100 y$

$\Rightarrow y=10$

From (iii), we have:

$3 \times 10=x$

$\Rightarrow 30=x$

Hence, the two natural numbers are 30 and 10 .

Leave a comment

Close
faculty

Click here to get exam-ready with eSaral

For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.

Download Now