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If a + b = x, and a – b = y, then ab =

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Senior Manager
Joined: 02 Jan 2017
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If a + b = x, and a – b = y, then ab =  [#permalink]

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25 Feb 2017, 22:58
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Question Stats:

76% (01:27) correct 24% (01:45) wrong based on 115 sessions

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If a + b = x, and a – b = y, then ab =

(A) (x^2 – y^2)/2
(B) (x + y)(x – y)/2
(C) (x^2 – y^2)/4
(D) xy/2
(E) (x^2 + y^2)/4
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Joined: 19 Apr 2016
Posts: 271
Location: India
GMAT 1: 570 Q48 V22
GMAT 2: 640 Q49 V28
GPA: 3.5
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Re: If a + b = x, and a – b = y, then ab =  [#permalink]

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25 Feb 2017, 23:56
1
2
vikasp99 wrote:
If a + b = x, and a – b = y, then ab =

(A) x^2 – y^2/2
(B) (x + y)(x – y)/2
(C) x^2 – y^2/4
(D) xy/2
(E) x^2 + y^2/4

a + b = x
$$(a+b)^2 = x^2$$
$$a^2 + b^2 + 2ab = x^2$$ ---------------I

a - b = y
$$(a-b)^2 = y^2$$
$$a^2 + b^2 - 2ab = y^2$$ ---------------II

I - II
$$4ab = x^2 - y^2$$
$$ab = (x^2 - y^2)/4$$

vikasp99 : Could you please add brackets to the numerators of the options as it will make the options more clear ?

Hence option C is correct
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Posts: 56304
Re: If a + b = x, and a – b = y, then ab =  [#permalink]

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26 Feb 2017, 03:33
vikasp99 wrote:
If a + b = x, and a – b = y, then ab =

(A) x^2 – y^2/2
(B) (x + y)(x – y)/2
(C) x^2 – y^2/4
(D) xy/2
(E) x^2 + y^2/4

Please pay attention to formatting when posting a question. Thank you.
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Re: If a + b = x, and a – b = y, then ab =  [#permalink]

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26 Feb 2017, 03:34
vikasp99 wrote:
If a + b = x, and a – b = y, then ab =

(A) (x^2 – y^2)/2
(B) (x + y)(x – y)/2
(C) (x^2 – y^2)/4
(D) xy/2
(E) (x^2 + y^2)/4

Similar question to practice: https://gmatclub.com/forum/if-a-b-x-and ... 54248.html
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Re: If a + b = x, and a – b = y, then ab =  [#permalink]

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26 Feb 2017, 07:46
Point to Note:

a*b = [(a+b)^2 - (a-b)^2 ] /4
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Re: If a + b = x, and a – b = y, then ab =  [#permalink]

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26 Feb 2017, 08:18
vikasp99 wrote:
If a + b = x, and a – b = y, then ab =

(A) (x^2 – y^2)/2
(B) (x + y)(x – y)/2
(C) (x^2 – y^2)/4
(D) xy/2
(E) (x^2 + y^2)/4

Hi,

Squaring on both side

$$(a + b)^2 = x^2$$, and $$(a – b)^2 = y^2$$

On simplification

$$x^2-y^2=4ab$$

$$ab=(x^2-y^2)/4$$

Hence C
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Posts: 1209
If a + b = x, and a – b = y, then ab =  [#permalink]

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26 Feb 2017, 10:52
vikasp99 wrote:
If a + b = x, and a – b = y, then ab =

(A) (x^2 – y^2)/2
(B) (x + y)(x – y)/2
(C) (x^2 – y^2)/4
(D) xy/2
(E) (x^2 + y^2)/4

subtracting, 2b=x-y
multiplying, 4ab=(x+y)(x-y)
ab=(x^2-y^2)/4
C
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Joined: 06 Jun 2017
Posts: 10
Re: If a + b = x, and a – b = y, then ab =  [#permalink]

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12 Mar 2018, 08:08
Does anybody mind explaining to me why I should automatically think to square "a+b=x" & "a-b=y" upon seeing these clues? What about the given information should indicate to a test taker "Ya know what... lets square those clues in order to solve this problem"?
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Re: If a + b = x, and a – b = y, then ab =  [#permalink]

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12 Mar 2018, 12:41
1
aaronhew wrote:
Does anybody mind explaining to me why I should automatically think to square "a+b=x" & "a-b=y" upon seeing these clues? What about the given information should indicate to a test taker "Ya know what... lets square those clues in order to solve this problem"?

[(a+b)^2 - (a-b)^2 ] = 4*a*b

[(a+b)^2 + (a-b)^2 ] = a^2 + b^2

These are two of the relevant and standard formula to be at your fingertips. As far as this problem is concerned, asking for the value of a*b is the obvious clue and there is nothing more to research on this problem.
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Re: If a + b = x, and a – b = y, then ab =  [#permalink]

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12 Mar 2018, 12:46
$$a+b = x$$
$$a-b = y$$

when we square both .. we will have the squared terms in positive and the product of a& b term with opposite signs in the two equations.

We can subtract to get rid of the quadratic terms

so 1 - 2 leads to

$$4ab = x^2 - y^2$$

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If a + b = x, and a – b = y, then ab =  [#permalink]

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12 Mar 2018, 14:10
1
vikasp99 wrote:
If a + b = x, and a – b = y, then ab =

(A) (x^2 – y^2)/2
(B) (x + y)(x – y)/2
(C) (x^2 – y^2)/4
(D) xy/2
(E) (x^2 + y^2)/4

aaronhew wrote:
Does anybody mind explaining to me why I should automatically think to square "a+b=x" & "a-b=y" upon seeing these clues? What about the given information should indicate to a test taker "Ya know what... lets square those clues in order to solve this problem"?

aaronhew Hilarious. Levity is good.

You may know the algebraic identities but may not have seen the clue.
Bunuel lists them here, Algebraic Identities (scroll down)

If you draw a blank, another way to answer is to assign values.

Let a = 3, b = 2
x = (a + b) = (3 + 2) = 5
y = (a - b) = (3 - 2) = 1

(ab) = (3*2) = 6

Using x = 5 and y = 1, find the answer that yields 6

(A) (x^2 – y^2)/2: $$\frac{(5^2-1^2)}{2}=\frac{24}{2}=12$$. NO

(B) (x + y)(x – y)/2: $$\frac{(5+1)(5-1)}{2}=\frac{24}{2}=12$$. NO

(C) (x^2 – y^2)/4: $$\frac{(5^2-1^2)}{4}=\frac{24}{4}=6$$. MATCH

(D) xy/2: $$\frac{(5*1)}{2}=\frac{5}{2}$$. NO

(E) (x^2 + y^2)/4: $$\frac{(5^2+1^2)}{4}=\frac{26}{4}=\frac{13}{2}$$. NO

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If a + b = x, and a – b = y, then ab =   [#permalink] 12 Mar 2018, 14:10
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