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The sum of squares of two numbers is 68 and the squares of their diffe
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20 Dec 2021, 01:33
20 Dec 2021, 01:33
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The sum of squares of two numbers is 68 and the squares of their difference is 36. What is the value of the product of the two numbers ?
(A) 16
(B) 32
(C) 58
(D) 104
(E) 108
(A) 16
(B) 32
(C) 58
(D) 104
(E) 108
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Re: The sum of squares of two numbers is 68 and the squares of their diffe
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20 Dec 2021, 06:45
20 Dec 2021, 06:45
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Carcass wrote:
The sum of squares of two numbers is 68 and the squares of their difference is 36. What is the value of the product of the two numbers ?
(A) 16
(B) 32
(C) 58
(D) 104
(E) 108
(A) 16
(B) 32
(C) 58
(D) 104
(E) 108
Let x and y be the two numbers
The sum of squares of two numbers is 68
We can write: x²+ y² = 68
The square of their difference is 36
We can write: (x - y)² = 36
What is the value of the product of the two numbers ?
We have the following system of equations:
x²+ y² = 68
(x - y)² = 36
Expand and simplify the bottom equation to get:
x²+ y² = 68
x² - 2xy + y² = 36
Subtract the bottom equation from the top equation to get: 2xy = 32
Divide both sides of the equation by 2 to get: xy = 16
Answer: A
Re: The sum of squares of two numbers is 68 and the squares of their diffe
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20 Dec 2021, 10:43
20 Dec 2021, 10:43
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Alternatively, X2-2XY+Y2=36 can be rewritten as X2+Y2-2XY=36. Since X2+Y2=68 the equation becomes 68-2XY=36 from there XY=16.
