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Let <x, y, z> denote 2x + yz. Moreover, let x + y = 2p, y + z = 2q, z

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Math Revolution GMAT Instructor
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Let <x, y, z> denote 2x + yz. Moreover, let x + y = 2p, y + z = 2q, z  [#permalink]

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New post 22 Nov 2019, 02:02
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[GMAT math practice question]

Let \(<x, y, z>\) denote \(2x + yz\). Moreover, let \(x + y = 2p, y + z = 2q, z + x = 2r, xy = a, yz = b\) and \(zx = c. \)

How can \(<x, 3y, z> + <y, 3z, x> + <z. 3x, y>\) be written in terms of \(a, b, c, p, q\) and \(r\)?

A. \(p+q+r+a+b+c\)

B. \(2(p+q+r+a+b+c)\)

C. \(2(p+q+r+a+b+c)\)

D. \(2(p+q+r)+3(a+b+c)\)

E. \(3(p+q+r)+2(a+b+c)\)

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Re: Let <x, y, z> denote 2x + yz. Moreover, let x + y = 2p, y + z = 2q, z  [#permalink]

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New post 22 Nov 2019, 03:22
From the question stem, we can estimate that this is a cyclic expression. A cyclic expression is where the order of the terms follow a cyclic pattern. This can be understood better when we break down the question stem.

It’s given that <x,y,z> = 2x + yz. In layman terms, this means, whatever the first number is has to be multiplied by 2 and added to the product of the other two numbers.

Using this, we get <x,3y,z> = 2x + 3yz = 2x + 3b {because yz = b}. Let this be equation 1.

<y,3z,x> = 2y + 3zx = 2y + 3c { notice how the positions of the variables have shifted by one place to the left and that is why I preferred to call them cyclic}. Let this be equation 2.

<z,3x,y> = 2z + 3xy = 2z + 3a. {notice how the last terms are 3b, 3c and 3a, which are cyclic as well}. Let this be equation 3

Adding the three equations, we have,
<x,3y,z> + <y,3z,x> + <z,3x,y> = 2x + 2y + 2z + 3(a+b+c). Let’s call this equation 4.

We also know that x + y = 2p, y + z = 2q, z + x = 2r; adding the equations, we get,
2x + 2y + 2z = 2(p+q+r). Substituting this value in equation 4, we obtain our required answer as 2(p+q+r) + 3(a+b+c).

The correct answer option is D.
Hope that helps!
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Re: Let <x, y, z> denote 2x + yz. Moreover, let x + y = 2p, y + z = 2q, z  [#permalink]

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New post 24 Nov 2019, 19:22
=>

When we add \(x + y = 2p, y + z = 2q,\) and \(z + x = 2r\), we have \(x + y + y + z + z + x = 2p + 2q + 2r, 2(x + y + z) = 2(p + q + r)\) or \(x + y + z = p + q + r.\)

When we add \(xy = a, yz = b\) and \(zx = c,\) we have \(xy + yz + zx = a + b + c.\)

\(<x, 3y, z> = 2x + 3yz, <y, 3z, x> = 2y + 3zx, <z, 3x, y> = 2z + 3xy.\)

Then \(<x, 3y, z> + <y, 3z, x> + <z, 3x, y> = 2x + 3yz + 2y + 3zx + 2z + 3xy = 2(x +y + z) + 3(xy + yz + zx) = 2(p + q + r) + 3(a + b + c).\)

Therefore, the answer is D.
Answer: D
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Re: Let <x, y, z> denote 2x + yz. Moreover, let x + y = 2p, y + z = 2q, z   [#permalink] 24 Nov 2019, 19:22
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Let <x, y, z> denote 2x + yz. Moreover, let x + y = 2p, y + z = 2q, z

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