Question

The sum of two numbers is 9. The sum of squares of the numbers is 41. Find the smaller of the two numbers.

• A. 4
• B. 3
• C. 2
• D. 1
• E. None of these

Answer 1

The correct option is A

4

Let the numbers be a and b.

a + b = 9 a${}^2$ + b${}^2$ = 41

From the first equation, b = (9-a)

Now substitute this in the second equation.

a${}^2$ + (9-a)${}^2$ = 41

a${}^2$ + 81 - 18a + a${}^2$ = 41

2a${}^2$ - 16a + 81 = 41

2a${}^2$ - 16a + 40 = 0

a${}^2$ - 8a + 20 = 0

(a - 5)(a -4) =0

(a - 5) = 0 or (a -4) =0

a = 5 or a = 4

Substitute in the first equation, b = 5 or 4 .

The numbers are 5 and 4.