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

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

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26 Nov 2015, 05:33
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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
[Reveal] Spoiler: OA

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

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26 Nov 2015, 06:16
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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26 Nov 2015, 06:26
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

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

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26 Nov 2015, 11:45
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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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26 Nov 2015, 11:50
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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 the values = x and y

The sum of two positive integers is 24
So, x + y = 24

The difference of their squares is 48
So, x² - y² = 48
Factor to get: (x + y)(x - y) = 48
Replace to get: (24)(x - y) = 48
So, (x - y) = 2

We now have:
x + y = 24
x - y = 2
Add these equations to get: 2x = 26, which means x = 13
If x = 13, then y = 11

So, xy = (13)(11) = 143 = E

Cheers,
Brent
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Re: If the sum of two positive integers is 24 and the difference of their [#permalink]

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26 Nov 2015, 15:33
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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.

[Reveal] Spoiler:
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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Re: If the sum of two positive integers is 24 and the difference of their   [#permalink] 22 Oct 2017, 00:39
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