If x and y are positive integers

John Robertson

Used test cases, SQR(X)+SQR(Y)=?

1) x+y = 15, could be for x & y, 10 and 5, 9 and 4, etc. Can't get a lock on what x and y are, IS

2) SQR(xy)=6. Indicates xy=36, as only solution for that perfect square to =6 is SQR(36).
But, you can get for xy: 6x6, 9x4, 12x3, etc.
IS

1&2) knowing x+y=15, and SQR(xy)=6, 12+3=15, satisfies (1), and 12x3=36, satisfies (2). Without solving for SQR(x)+SQR(y), you know x and y are definitely 12 and 3 in either order, and it could be solved out.
S, (D)

Feb 16, 2017 • Comment

What is the value of X^0.5 + Y^0.5?

Recognise that squaring the original (X^0.5 + Y^0.5?)^2 leads to X + 2(XY)^0.5 + Y. Knowing the value of the latter is one step away from answering the question.

1: X + Y = 15
In X + 2(xy)^0.5 + Y, this statement does not answer the value of 2(XY)^0.5
Insufficient

2: (XY)^0.5 = 6
In X + 2(XY)^0.5 + Y, this statement does not answer the value of X + Y
Insufficient

1 & 2: Using both statements we can solve X + 2(XY)^0.5 + Y
-> 15 + 2*(6) = 27
-> X^0.5 + Y^0.5 = 27^0.5 = (3*9)^0.5 = 3 * (3)^0.5

Sep 23, 2016 • Comment