Question

Two positive integers differ by 4 and sum of their reciprocals is 10/21. Then, one of the numbers is ?

Answer 1

Answer:

Let two positive integers be x and y respectively.

So by statement ,

x-y = 4.........................eq (1)

1/x +1/y = 10/21

(x + y)/xy = 10/21

(x + y)21 = 10xy

21x +21y = 10xy...........eq (2)

So , from eq (1)

x = y+4

Put above equation in eq (2)

21(y+4) +21y = 10(y+4)y

21y +84 +21y = 40y + 10y^2

-10y^2+2y+84= 0

10y^2 -2y -84 =0

Or 5y^2 -y -42 =0

By solving the quadratic equation we will get the value of y as

y = 3 , -2.8

Therefore , x=7 , 1.2

Bt since it is given in the question that integers should be positive

Therefore y = 3 and x = 7 ( neglect y = -2.8 as it is negative )

Answer 2

Answer

Let two positive integers be x and y respectively.

So by statement ,

x-y = 4.........................eq (1)

1/x +1/y = 10/21

(x + y)/xy = 10/21

(x + y)21 = 10xy

21x +21y = 10xy...........eq (2)

So , from eq (1)

x = y+4

Put above equation in eq (2)

21(y+4) +21y = 10(y+4)y

21y +84 +21y = 40y + 10y^2

-10y^2+2y+84= 0

10y^2 -2y -84 =0

Or 5y^2 -y -42 =0

By solving the quadratic equation we will get the value of y as

y = 3 , -2.8

Therefore , x=7 , 1.2

Bt since it is given in the question that integers should be positive

Therefore y = 3 and x = 7 ( neglect y = -2.8 as it is negative )