PSA_PRACTICE_QNR38. Solve the below questions:
1.) The sum of two numbers is 15. The sum of their squares is 9 more than 13 times the larger number. Find the natural numbers.
2.) The product of two natural numbers is 10 less than 16 times the smaller number. If twice the smaller number is 5 more than the larger number, find the two numbers.
1) Let the numbers be x and y, the latter being the larger of the two.
x^2 + y^2 = 13y —-(B)
Squaring equation (A), we get
x^2 + y^2 + 2xy = 225
=> 13y + 2xy = 225
=> y= 225/(13+2x)
13+2x must be a multiple of 5.
If x=1, y=15 . [Rejected as sum is 15]
If x=6, y= 225/25 = 9. [Accepted].
[Assumption, both x and y are natural numbers here].
From equation (B) x^2 + y^2 = 13y
=> x^2 =13y-y^2
=> x^2 =y(13-y)
Now, minimum value of y can be 8 (as 7+8=15 and y has to be larger than x always).
By trial and error,
if y=8; y(13-y)= 8*5. Not a square. [Rejected as x^2 is in LHS]
if y=9; y(13-y)= 9*4. A square. [Accepted; and value of x=6].
2) Let the numbers be x and y, the latter being the larger of the two.
Also, 2x= y+5 —-(B)
Multiplying (A) by 2,
=>y(y+5) = 16(y+5) -20
=>y^2 + 5y = 16y + 80 -20
=>y^2 – 11y -60 = 0
=> y^2 – 15y +4y -60 =0
=> y=15, -4
(y,x)= (15,10) is the correct set.