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If 1 < x < y < z, which of the following has the greatest value?

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If 1 < x < y < z, which of the following has the greatest value?  [#permalink]

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New post 16 Jun 2017, 03:58
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If 1 < x < y < z, which of the following has the greatest value?

A. z(x + 1)
B. z(y + 1)
C. x(y + z)
D. y(x + z)
E. z(x + y)

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If 1 < x < y < z, which of the following has the greatest value?  [#permalink]

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New post 16 Jun 2017, 05:14
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Plug values x=2,y=3,z=4 , which satisfies the condition 1<x<y<z

If we use this values in the answer options
A. z(x + 1) = 4(3) = 12
B. z(y + 1) = 4(4) = 16
C. x(y + z) = 2(7) = 14
D. y(x + z) = 3(6) = 18
E. z(x + y) = 4(5) = 20

Therefore, z(x+y) - Option E has the greatest value.
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Re: If 1 < x < y < z, which of the following has the greatest value?  [#permalink]

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New post 15 Nov 2017, 15:43
AbdurRakib wrote:
If 1 < x < y < z, which of the following has the greatest value?

A. z(x + 1)
B. z(y + 1)
C. x(y + z)
D. y(x + z)
E. z(x + y)


We can let x = 2, y = 3, and z = 4, and we have:

A) z(x+1)

4(3) = 12

B) z(y+1)

4(4) = 16

C) x(y+z)

2(7) = 14

D) y(x+z)

3(6) = 18

E) z(x+y)

4(5) = 20

Answer: E
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Re: If 1 < x < y < z, which of the following has the greatest value?  [#permalink]

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New post 07 Apr 2018, 19:15
whenever we see that there are variables in all answer choices and variables in the question (without clear values) then we can pick easy number(s), and apply the numbers across all the options and see which option gives the required result. In this case, as the above posters have pointed out, we can pick numbers as x=2, y=3, and z=4 and apply then to all the options, finding that option (E) gives the largest result. So E is the answer
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Re: If 1 < x < y < z, which of the following has the greatest value?  [#permalink]

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New post 15 Aug 2018, 05:57
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I'm curious. Do we have other ways to solve this problem WITHOUT plug in the number?
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Re: If 1 < x < y < z, which of the following has the greatest value?  [#permalink]

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New post 28 Aug 2018, 06:14
Since 1<x<y<z , so x,y,z are all positive real numbers
Now assume x= 1+m
since y>x , so y can be assumed as y= x+ n
and z = y + o.
Putting in the corresponding values of x & y in z, we have
Now z= 1+m+n+o

Since all are positive numbers
so the greatest value can only be achieved by multiplying the greatest number by the sum of other two( This approach is subjective to the Options given in answers because answer is expressed as only the product of one number with the sum of the other two). So mathematically it can be concluded that z has to be individual number and the other should be x+y.

Note: We can also conclude the answer by substituting the value of z , y, x in answer choices.

Hope I am right
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Re: If 1 < x < y < z, which of the following has the greatest value?  [#permalink]

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New post 29 Aug 2018, 07:07
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septwibowo wrote:
I'm curious. Do we have other ways to solve this problem WITHOUT plug in the number?


Yeah really good question. If you distribute the multiplication you can get really direct comparisons between the quantities:

(A) zx + z

(B) zy + z

(C) xy + xz

(D) yx + yz

(E) zx + zy

(A) and (B) should be easy to eliminate since in each case you can see that choice (E) includes one of the two terms (zx or zy) but then adds something bigger than z to it.

Then for (C) vs. (D), note that each includes xy, but then (D) adds a bigger second term (since y > x, then yz > xz). So (D) is bigger than (C). And when you compare (E) to (D), you can use the same logic: each includes a yz term, but then (E) adds something bigger to it (zx > yx since z > y). So now you have algebraic proof that (E) is greater than all the other four.
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Re: If 1 < x < y < z, which of the following has the greatest value? &nbs [#permalink] 29 Aug 2018, 07:07
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