In a positive fraction, the denominator is greater than the numerator by 3 . If 1 is subtracted from both the numerator and the denominator, the fraction is decreased by 1 14 . Find the fraction.

Let no. of fraction = x Then, denominator = x + 3 So, fraction = x x + 3 Now, According to Question, x - 1 x + 3 - 1 = x x + 3 - 1 14 x - 1 x + 2 = x x + 3 - 1 14 ⇒ x - 1 x + 2 = 14 x - x - 3 14 ( x + 3 ) = 13 x - 3 14 ( x + 3 ) ⇒ 14 ( x + 3 ) ( x - 1 ) = ( 13 x - 3 ) ( x + 2 ) ⇒ 14 ( x + 3 ) ( x - 1 ) = 13 x 2 + 26 x - 3 x - 6 ⇒ 14 x 2 - 14 x + 42 x - 42 = 13 x 2 + 26 x - 3 x - 6 ⇒ 14 x 2 - 14 x + 42 x - 42 - 13 x 2 - 26 x + 3 x + 6 = 0 ⇒ x 2 + 5 x - 36 = 0 ⇒ x 2 + 9 x - 4 x - 36 = 0 ⇒ x ( x + 9 ) - 4 ( x + 9 ) = 0 ⇒ ( x + 9 ) ( x - 4 ) = 0 ∴ Either x + 9 = 0 then, x = - 9 Which is not possible as negative, or x - 4 = 0 , then x = 4 So, Fraction = x x + 3 = 4 4 + 3 = 4 7

Find two consecutive odd positive integers, sum of whose squares is 394 .

Let one of the number be x which is an odd positive integer then the other number be x + 2 . Then according to question, we have x 2 + x + 2 2 = 394 ⇒ x 2 + x 2 + 4 x + 4 = 394 ⇒ 2 x 2 + 4 x - 390 = 0 ⇒ x 2 + 2 x - 195 = 0 ⇒ x 2 + 15 x - 13 x - 195 = 0 ⇒ x x + 15 - 13 x + 15 = 0 ⇒ x - 13 x + 15 = 0 ⇒ x - 13 = 0 or x + 15 = 0 ⇒ x = 13 [ x ≠ - 15 , since we have let x to be is an odd positive integer] The other number is x + 2 = 13 + 2 = 15 Hence, the required numbers are 13 , and 15

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

Consider, the greater number be x and the smaller number be y . According to the question: x 2 - y 2 = 45 … … … ( 1 ) y 2 = 4 x … … … ( 2 ) From ( 1 ) and ( 2 ) , we get: x 2 - 4 x = 45 ⇒ x 2 - 4 x - 45 = 0 ⇒ x 2 - 9 x + 5 x - 45 = 0 ⇒ x ( x - 9 ) + 5 ( x - 9 ) = 0 ⇒ ( x - 9 ) ( x + 5 ) = 0 ⇒ x - 9 = 0 or x + 5 = 0 ⇒ x = 9 ( ∵ x is a natural number) Putting the value of x in equation ( 2 ) , we get: y 2 = 36 ⇒ y = 6 Hence, the required numbers are x = 9 and y = 6 .