If the sum of two positive integers is 24 and the difference of their

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



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

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

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 \(b\) \(=\)\(11\)




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

13 x 11 =>143

hence answer is (E)

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

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

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,
Rich

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.

in which book i can find this question?

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