If three positive real numbers x, y, z satisfy y – x = z – y and x y z

Question

If three positive real numbers x, y, z satisfy y – x = z – y and x y z = 4, then what is the minimum possible value of y?

(a) 2^1/3 
(b) 2^2/3 
(c) 2^1/4 
(d) 2^3/4
(e) 2^1/2

TeamGMATIFY · Accepted Answer

Given  : y – x = z – y and x y z = 4
Or 2y = x + z (i) and xyz = 4 (ii)

Required  : Minimum value of y

From (i), y is the Arithmetic Mean of x and z

There is a property about Arithmetic mean and Geometric mean that says:

AM  \geq  GM

Therefore,
(x+ y + z)/3  \geq  (xyz)^1/3
Substituting the values from (i) and (ii)

(y + 2y)/3  \geq   4^{1/3} 
Or y  \geq   2^{2/3}

Therefore the minimum value would be  2^{2/3} 
Option B

However, there is very little chance that you would see such question on the GMAT.

chetan2u · Answer

Hi,
the Q involves the formula between Arithmetic mean and Geometric mean..
ie AM>=GM..

But I am yet to see this Q in GMAT or anything close to this..
so the Q is more based on CAT in India ...
so spending time on such Qs may not help you, if you are targeting GMAT..

davedekoos · Answer

First of all, I completely agree that this problem would not be seen on the GMAT, except for some high level experimental question.

BUT, it is totally solvable, even if you don't know the property that  AM \geq GM

If we put the three number x, y, z on a number line, we would see that y is in the middle and is evenly spaced between x and z.

x-----y-----z
or
z-----y-----x

So the distance between x and y is equal to the distance between y and z (that is what the first equation in the question tells us) and for simplicity we can call that distance  a .

|--a--|--a--|
x-----y-----z

So then we can say that  x=y-a  and  z=y+a 
Using that in the equation  xyz=4  we get  (y-a)y(y+a)=4

=y(y^2-a^2)=4

Now, we want to minimize  y . We can do so by maximizing the term  (y^2-a^2) . And to do that, we need to minimize  a . The smallest  a  can be is 0 (since it is just a distance between numbers), therefore, to minimize y, our equation becomes

y(y^2-0)=4

y^3=4

y=4^{1/3}

y=2^{2/3}

Answer: B

For the more visually inclined, you could also imagine the number line, with  y  on a slider. You can slide  y  anywhere you want to within the positive numbers. Now, wherever you decide to put  y , you need to adjust  x  and  z  so that  xyz=4 . Keep in mind that  x  and  z  must also be positive and equidistant from  y .

What happens if  y  gets really big? Can you still choose values for  x  and  z  such that the equation  xyz=4  remains true? The answer is yes. Assuming that  x<y , then you can always set  x  close enough to 0 to compensate for the size of  y  and  z . No matter how big  y  gets, you can always find a value of  x  and  z  so that the equation  xyz=4 .

For example, if we take  y=1000 , then  x  would have to be very close to 0, and then  z  would be almost 2000. In this case,  x  would be just slightly larger than  \frac{4}{(1000*2000)}=0.000002 . You can see that no matter how big  y  gets,  x  can always get closer to 0 to keep the equation true. We can also see that, in terms of  a , as  y  increases,  a  increases. Conversely, as  y  decreases,  a  decreases. Therefore, when  y  is at its minimum,  a  will be at its minimum, and the minimum value of  a  is 0. When  a=0 ,  xyz=y^3=4  and  y=2^{2/3}

rohit8865 · Answer

Chetan 
thanks....

rohitm1984 · Answer

i solved by plugging in the options for the equation:
2y=x + z
and xyz= 4

so if y is 2^1/3, then x and z are the same. when we multiply all 3, we don't get 4. so eliminated.

But, B satisfies. So answer is B

davedekoos · Answer

Rohitm,

Even though you chose the correct answer, you did so by pure chance. The question is asking for the  minimum possible  value of y. Just because one value of y in the answer choices satisfies the equations doesn't mean it's the smallest possible value. If you look at the decimal values of the answer choices you get:
A) 1.260
B) 1.587
C) 1.189
D) 1.682
E) 1.414

If you had tried option D first, you would have seen that, after some complicated and non-GMAT tested exponent algebra, it would also satisfy the equations. Try solving  y(y^2-a^2)=2^{3/4}((2^{3/4})^2-a^2)=4  in under 4 minutes without a calculator... Of course, if you ever run into something like that, you can be almost certain that you don't need to solve it to answer the question, there must be another way.

(BTW, in this case x = 1.011, y = 1.682 and z = 2.353. These satisfy the equations, but that is not the minimum value of y.)

In your solution you plugged in a value for y, and then set x and z equal to the same value. (I think that's what you were doing), but how did you know that x and z should equal y? And if you did already know that, then the answer becomes simple, because then 
 xyz=y^3=4 
 y=2^{2/3}

The point is, when you're asked to find the  minimum possible  value, plugging in the answer choices can be dangerous unless you know the relative size of each of the answer choices. Then, plug in answers starting with the smallest until one works. In this case it's not super obvious which are the largest and smallest values, so just because one value satisfies the equation doesn't mean it's the smallest possible. There might be other answer choices that also satisfy the equations but are even smaller.

Cheers

Arjunarocks · Answer

This can be easily solved by AM and GM rule of AM>=GM

So, (x+y+z)/3 >= (xyz)^1/3

and it is given that x+z=2y
and also, that, xyz=4

So, 3y/3>=4^1/3
hence the least value of y is 2^(2/3) ie option B.

nipunjn · Answer

is there any other way to solve it

QuantMadeEasy · Answer

y-x=z-y
y=(x+z)/2

xyz=4 (Given)
xz(x+z)/2 = 4

xz(x+z) = 8
to minimise y we need to maximise x and z , which can be done when x=z
From above relation it can be said 
2x^3 =8
x^3 = 4
x=2^2/3 = z

Substituting x and z in the given relation
2^2/3 x y x 2^2/3 = 4
2^4/3 x y = 2^2
y = 2^2/3

B is correct.

This approach can be used

ZIX · Answer

Here is how I used simple substitution to solve this:

First we translate the options into something simple:
 2/3; 4/3; 1/2; 2; 1

Now,
 y-x = z-y ==> 2y = z+x  --- (1)
 xyz = 4 ==> y = \frac{4}{xz}  --- (2)

Plug equation 2 into 1:
 8 = xz^2 + x^2z 
 8 = xz(z+x)  --- (3)

Now let's quickly scan through easiest choices:

Part 1:
As we know that xyz = 4. We will test integer values that multiply to 4.

As given in answer choices, we will test  y = 2 and 1 . 
(1,2,2)(2,1,2)

Using Equation 1:
2(2) ≠ 2 +1
2(1) ≠ 2+2

so we eliminate options D and E.

Part 2:
Now let's use equations 2 and 3 and try substituting remaining 3 values:

y = 1/2  
1/2 = 4/xz (eq 2)
xz = 8

However, we know that 8 = xz(z+x), so it can't be 8 = 8(z+x). Eliminate option C.

y = 2/3  
2/3 = 4/xz (eq 2)
xz = 6

8 = 6(z+x) (eq 3)
4 = 3z + 3x

However, we know that 3z and 3x must at least be 6. Eliminate option A.

y = 4/3  
4/3 = 4/xz (eq 2)
xz = 3

8 = 3(z+x) (eq 3)
8 = 3z + 3x

Hence, option B is correct.

Although this is a simpler method but as you see it is too cumbersome and somewhat lengthy. But the calculations are easy so you might be able to pull it off.

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If three positive real numbers x, y, z satisfy y – x = z – y and x y z

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