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Re: If the sum of two positive integers is 24 and the difference of their [#permalink]
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Bunuel wrote:
If the sum of two positive integers is 24


\(a\) \(+\) \(b\)\(=\) \(24\)-------->(I)

Bunuel wrote:
difference of their squares is 48


\(a^{2} - b^{2}\) \(=\) \(48\)

\(a^{2} - b^{2}\) = \(( a + b )(a - b)\)

Or, 48 = 24 \((a - b)\)

Or, \(a - b\) = 2 -------->(II)

Adding (I) & (II)

\(2a\) \(=\) \(26\)
or, \(a\) \(=\) \(13\)

Substituting \(a\) \(=\) \(13\) in (II) we get \(b\) \(=\)\(11\)


Bunuel wrote:
If what is the product of the two integers?


Product of the integers is \(a\) x \(b\)

13 x 11 =>143

hence answer is (E)
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Re: If the sum of two positive integers is 24 and the difference of their [#permalink]
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Let the two numbers be X and Y.
As given in the problem.
X+Y=24---------(1)
Also, (X+Y) (X-Y)=48----------(2)
Solving (1) in (2), we get
X-Y=2-----(3)

Solving (1) and (3), we get X=13 and Y=11.

Hence XY=143.....Option E
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Re: If the sum of two positive integers is 24 and the difference of their [#permalink]
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Let the 2 positive numbers x and y

x+ y = 24 -- 1
x^2 - y^2 = 48
=> (x+y)(x-y)=48 -- 2
Using equation 1 in 2 , we get
=> x-y = 2 -- 3

Solving equation 1 and 3 , we get
x= 13
y= 11
Product = 13*11 = 143

Answer E
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Re: If the sum of two positive integers is 24 and the difference of their [#permalink]
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Hi All,

This question can be solved with 'brute force' and a bit of logic.

We're given a few facts about 2 numbers:
1) They're both POSITIVE INTEGERS.
2) Their sum is 24.
3) The difference in their squares is 48.

We're asked for the product of the two integers.

Let's see what happens when we TEST a couple of options...

IF the numbers are 1 and 23
The difference in their squares is 529-1 = 528, which is TOO BIG.

IF the numbers are 4 and 20
The difference in their squares is 400 - 16 = 384, which is still TOO BIG.

This proves that the two numbers are likely fairly close to one another...

IF... the numbers are 11 and 13
The difference in their squares is 169 - 121 = 48, which is a MATCH for what we were told.

The PRODUCT of 11 and 13 is 143.

Final Answer:

GMAT assassins aren't born, they're made,
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Re: If the sum of two positive integers is 24 and the difference of their [#permalink]
Bunuel 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


let x and y >0 where x+y= 24, substituting x in second equation:

x^2 - y^2 = 48
(24-y)^2 = 48 + y^2
so, y=11 and x is 13.
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Re: If the sum of two positive integers is 24 and the difference of their [#permalink]
Bunuel 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


in which book i can find this question?
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Re: If the sum of two positive integers is 24 and the difference of their [#permalink]
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Re: If the sum of two positive integers is 24 and the difference of their [#permalink]
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