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PRODUCT

by grandh01 » Wed Aug 29, 2012 2:42 pm
If the sum of two positive integers is
24 and the difference of their squares
is 48, what is the product of the two
integers?
(A) 108
(B) 119
(C) 128
(D) 135
(E) 143

OA is E
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by Anurag@Gurome » Wed Aug 29, 2012 5:28 pm
grandh01 wrote:If the sum of two positive integers is
24 and the difference of their squares
is 48, what is the product of the two
integers?
(A) 108
(B) 119
(C) 128
(D) 135
(E) 143

OA is E
Let us assume that the two integers are x and y.
x + y = 24 ... Equation 1
x² - y² = 48 ... Equation 2

Substitute the value of y from equation 1 in equation 2, we get,
x² - (24 - x)² = 48
x² - (576 + x² - 48x) = 48
48x = 576 + 48
48x = 624
x = 13
y = 24 - 13 = 11

Therefore, x * y = 13 * 11 = 143

The correct answer is E.
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by coolhabhi » Wed Aug 29, 2012 9:17 pm
grandh01 wrote:If the sum of two positive integers is
24 and the difference of their squares
is 48, what is the product of the two
integers?
(A) 108
(B) 119
(C) 128
(D) 135
(E) 143

OA is E
a² - b² = 48
a + b = 24
We know that a² - b² = (a + b)(a - b)
So 48 = 24(a - b)
=>(a - b) = 2
Solving (a + b) = 24 and (a - b) = 2 We will get a = 13 and b = 11
So the product of the two integers is 143
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