Q21 of 293 Page 491

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

Let the two natural numbers be x and y.

According to the question

x² + y² = 25(x + y) – – – – – (1)

x² + y² = 50(x – y) – – – – (2)

From (1) and (2) we get

25(x + y) = 50(x – y)

x + y = 2(x – y)

x + y = 2x – 2y

y + 2y = 2x – x

3y = x – – – – – (3)

From (2) and (3) we get

(3y)² + y² = 50(3y – y)

9y² + y² = 50(3y – y)

10 y² = 100y

y = 10

From (3) we have,

x = 3y = 3.10 = 30

Hence the two natural numbers are 30 and 10.

Divide 27 into two parts such that the sum of their reciprocals is 3/20.

Divide 16 into two parts such that twice the square of the larger part exceeds the square of the smaller part by 164.

The difference of the squares of two natural numbers is 45. The square of the smaller number is four times the larger number. Find the numbers.

Three consecutive positive integers are such that the sum of the square of the first and the product of the other two is 46. Find the integers.