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# If c + d = x and c – d = y, then cd

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Joined: 30 Aug 2017
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If c + d = x and c – d = y, then cd [#permalink]

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27 Sep 2017, 23:40
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If c + d = x and c – d = y, then cd

A. $$\frac{X^2}{Y^2}$$

B. $$\frac{X-Y}{2}$$

C. $$\frac{(X+Y)^2}{4}$$

D. $$\frac{X^2 - Y^2}{4}$$

E. $$\frac{X^2 - Y^2}{2}$$

Can someone show how this works? I used different numbers to the explanation given by Kaplan and they dont work. It seems that you have to use numbers divisble by 4 for y where c - d = y, but how would I know this with out plugging a series of numbers.
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Re: If c + d = x and c – d = y, then cd [#permalink]

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28 Sep 2017, 00:07
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nmargot wrote:
If c + d = x and c – d = y, then cd

A. $$\frac{X^2}{Y^2}$$

B. $$\frac{X-Y}{2}$$

C. $$\frac{(X+Y)^2}{4}$$

D. $$\frac{X^2 - Y^2}{4}$$

E. $$\frac{X^2 - Y^2}{2}$$

Can someone show how this works? I used different numbers to the explanation given by Kaplan and they dont work. It seems that you have to use numbers divisble by 4 for y where c - d = y, but how would I know this with out plugging a series of numbers.

c + d = x and c – d = y

Sum the equations: 2c = x + y;
Subtract the equations: 2d = x - y.

Multiply above two: 4cd = x^2 - y^2;

cd = (x^2 - y^2)/4.

Hope it's clear.
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Re: If c + d = x and c – d = y, then cd [#permalink]

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28 Sep 2017, 00:28
nmargot wrote:

Can someone show how this works? I used different numbers to the explanation given by Kaplan and they dont work. It seems that you have to use numbers divisble by 4 for y where c - d = y, but how would I know this with out plugging a series of numbers.

nmargot : Here we have 4 variables. So it will be easy to use variable form instead of using numbers to find the correct answer.
For any question, try to check if variable option makes sense, if its tough,try by putting different values.
We have 2 methods to solve a question : variable or by using numbers. Always try to find which one will work better.

Here
If c + d = x and c – d = y, then cd

To find cd we have to write given equations in form of c and d
c+d =x
c-d =y

Adding both : 2c = x+y => c = (x+y)/2
Subtracting: 2d = x-y => d = (x-y)/2

Now multiple 2 equations to get value of cd => cd = (x+y)/2 * (x-y)/ 2 = (x^2-y^2)/4

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Re: If c + d = x and c – d = y, then cd [#permalink]

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28 Sep 2017, 01:29
The best way is,
let c= 4, d=3
then x = 7, and Y=1
CD = 12
Put X and Y in the answer and check

only satisfying solution is D
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Re: If c + d = x and c – d = y, then cd [#permalink]

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03 Oct 2017, 08:40
NickM83 wrote:
If c + d = x and c – d = y, then cd

A. $$\frac{X^2}{Y^2}$$

B. $$\frac{X-Y}{2}$$

C. $$\frac{(X+Y)^2}{4}$$

D. $$\frac{X^2 - Y^2}{4}$$

E. $$\frac{X^2 - Y^2}{2}$$

If we add the two equations, we have:

2c = x + y

c = (x + y)/2

If we subtract the two equations, we have:

2d = x - y

d = (x - y)/2

Therefore, cd = (x + y)/2 * (x - y)/2 = (x^2 - y^2)/4.

Alternate Solution:

Let’s square the equations:

c^2 + 2cd + d^2 = x^2

c^2 - 2cd + d^2 = y^2

Let’s subtract the second equation from the first:

4cd = x^2 - y^2

cd = (x^2 - y^2)/4

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Re: If c + d = x and c – d = y, then cd   [#permalink] 03 Oct 2017, 08:40
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