The sum of two natural numbers is 15 and the sum of their reciprocals is 3/10. Find the numbers.

Let the required natural numbers x and (15 – x)

According to given condition,



taking LCM



cross multiplying





Using the splitting middle term – the middle term of the general equation is divided in two such values that:


Product = a.c


For the given equation a = 1 b = – 15 c = 50


= 1.50 = 50


And either of their sum or difference = b


= – 15


Thus the two terms are – 10 and – 5


Sum = – 10 – 5 = – 15


Product = – 10. – 5 = 50



x(x – 10) – 5(x – 10) = 0


(x – 5) (x – 10) = 0


(x – 5) = 0 or (x – 10) = 0


x = 5 or x = 10


Case I: when x = 5


15 – x = 15 – 5 = 10


Case II: when x = 10


15 – x = 15 – 10 = 5


Hence required numbers are 5 and 10.


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