Question 11

The sum of the number and its positive square root is 7/36. Find the number.

Solution:

Let "x" be the required number

The sum of the number and its positive square root is 7/36.

x + √x = 7/36

√x = (7/36) - x

x = [(7/36) - x]²

x = (7/36)² - 2 (7/36) x + x²

x = (49/1296) - (7x/18) + x²

1296 x = 49 - 504 x + 1296 x²

49 - 504 x + 1296 x² = 1296 x

1296 x² - 504 x - 1296 x + 49 = 0

1296 x² - 1800 x + 49 = 0

1296 x² - 1764 x - 36 x + 49 = 0

36 x (36 x - 49) - 1 (36 x - 49) = 0

(36 x - 1) (36 x - 49) = 0

36 x - 1 = 0               36 x - 49 = 0

36 x = 1                      36 x = 49

x = 1/36                     x = 49/36

Therefore the required number is 1/36

Verification:

The sum of the number and its positive square root is 7/36.

(1/36) + √(1/36) = 7/36

(1/36) + (1/6) = 7/36

taking L.C.M we get

(1 + 6)/36 = 7/36

7/36 = 7/36

Question 12

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

Solution:

Let "x" and "y" are the required two numbers

The sum of two numbers is 15

x + y = 15

y = 15 - x -----(1)

the sum of its reciprocals is 3/10

(1/x) + (1/y) = 3/10

(y + x)/x y = 3/10

10 (x + y) = 3 x y

10 x + 10 y = 3 x y  ---- (2)

Now we are going to apply the value of y in the second equation

10 x + 10 (15 - x) = 3 x (15 - x)

10 x + 150 - 10 x = 45 x - 3 x²

3 x² - 45 x + 150 = 0

3 x² - 15 x - 30 x + 150 = 0

3 x (x - 5) - 30 (x - 5) = 0

(3 x - 30) (x - 5) = 0

3 x - 30 = 0              x - 5 = 0

3 x = 30                     x = 5

x = 30/3

x = 10

Therefore the two numbers are 10 and 5.

Verification:

The sum of two numbers is 15

10 + 5 = 15

15 = 15

the sum of its reciprocals is 3/10

(1/10) + (1/5) = 3/10

Taking l.C.M we get,

(1 + 2)/10 = 3/10

3/10  = 3/10