Question 3

The product of two successive multiples of 3 is 810.Find the multiples

Solution:

Let x and (x + 3) are successive multiples of 3

The product of two successive multiples = 810

x (x + 3) = 810

x² + 3 x = 810

x² + 3 x - 810 = 0

x² + 30 x - 27 x  - 810 = 0

x (x + 30) - 27 (x + 30) = 0

(x + 30) (x - 27) = 0

x  + 30 = 0         x - 27 = 0

x = -30            x = 27

Therefore the required multiples are 27 and 30.

Verification:

The product of two successive multiples of 3 is 810

=  27 (27 + 3)

=  27 (30)

=  810

Question 4

The sum of the squares of two natural numbers is 34 and sum of 5 times the smaller and 3 times the larger is 30. Find the numbers.

Solution:

Let x be the smaller natural number

and y be the larger natural number

The sum of their squares = 34

x² + y² = 34

sum of 5 times the smaller and 3 times the larger is 30

5 x + 3 y = 30

5 x = 30 - 3y

x = (30 - 3 y)/5

[(30 - 3 y)/5]² + y² = 34

[(30)² - 2 (30) (3y) + (3y)²] /25 + y² = 34

[900 - 180 y + 9 y²] /25 + y² = 34

([900 - 180 y + 9 y²] + 25 y²)/25 = 34

([900 - 180 y + 9 y²] + 25 y²) = 34 (25)

34 y² - 180 y + 900 = 850

34 y² - 180 y + 900 - 850 = 0

34 y² - 180 y + 50 = 0

34 y² - 170 y - 10 y + 50 = 0

34 y (y - 5) - 10 ( y - 5) = 0

(34 y -10) (y - 5) = 0

34 y - 10 = 0             y - 5 = 0

34 y = 10                  y = 5

y =10/34

if y = 5 then x = [30 - 3(5)]/5

x = (30 - 15)/5

x = 15/5

x = 3

Verification:

sum of the squares of two natural numbers is 34

5² + 3² = 34

25 + 9 = 34

34 = 34

sum of 5 times the smaller and 3 times the larger is 30

5 (3) + 3 (5) = 30

15 + 15 = 30

30 = 30  