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# Let x, y, z be three positive real numbers in a geometric progression

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Manager
Joined: 01 Nov 2017
Posts: 67
Location: India
Let x, y, z be three positive real numbers in a geometric progression  [#permalink]

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Updated on: 28 Feb 2019, 23:43
1
8
00:00

Difficulty:

55% (hard)

Question Stats:

57% (02:16) correct 43% (02:04) wrong based on 30 sessions

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Let x, y, z be three positive real numbers in a geometric progression such that x < y < z. If 5x, 16y, and 12z are in an arithmetic progression then the common ratio of the geometric progression is

(A) 3/6
(B) 3/2
(C) 5/2
(D) 1/6
(E) 3/8

Originally posted by raghavrf on 28 Feb 2019, 23:27.
Last edited by raghavrf on 28 Feb 2019, 23:43, edited 1 time in total.
Intern
Joined: 23 Oct 2013
Posts: 44
Re: Let x, y, z be three positive real numbers in a geometric progression  [#permalink]

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28 Feb 2019, 23:42
I am getting 2/5 and 1/6 as ans

Posted from my mobile device
Manager
Joined: 01 Nov 2017
Posts: 67
Location: India
Re: Let x, y, z be three positive real numbers in a geometric progression  [#permalink]

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28 Feb 2019, 23:54
5x, 16y, and 12z are in an arithmetic progression
this implies
2*16y = 12z + 5x
32y = 12z + 5x....(a)
x, y, z are three positive real numbers in a geometric progression
this implies
y = k*x.....(1)
z = k^2*x.....(2)
Therfore put (1) & (2) in (a)
32*kx = 12(k^2)x + 5x
32k = 12k^2 +5
or, 12k^2 -32k + 5 = 0
which gives k = 5/2 & k = 1/6
For k = 1/6, x < y < z is not satisfied.
Hence, k = 5/2
option(C)
Intern
Joined: 01 Jul 2018
Posts: 20
Re: Let x, y, z be three positive real numbers in a geometric progression  [#permalink]

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28 Apr 2019, 01:35
raghavrf wrote:
5x, 16y, and 12z are in an arithmetic progression
this implies
2*16y = 12z + 5x
32y = 12z + 5x....(a)
x, y, z are three positive real numbers in a geometric progression
this implies
y = k*x.....(1)
z = k^2*x.....(2)
Therfore put (1) & (2) in (a)
32*kx = 12(k^2)x + 5x
32k = 12k^2 +5
or, 12k^2 -32k + 5 = 0
which gives k = 5/2 & k = 1/6
For k = 1/6, x < y < z is not satisfied.
Hence, k = 5/2
option(C)

Could you please elabote more on this part?

this implies
2*16y = 12z + 5x
32y = 12z + 5x....(a)

thanks,
Manager
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Posts: 95
Location: India
Concentration: Operations, General Management
GMAT Date: 04-25-2015
WE: Engineering (Manufacturing)
Re: Let x, y, z be three positive real numbers in a geometric progression  [#permalink]

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01 May 2019, 01:17
raghavrf wrote:
5x, 16y, and 12z are in an arithmetic progression
this implies
2*16y = 12z + 5x
32y = 12z + 5x....(a)
x, y, z are three positive real numbers in a geometric progression
this implies
y = k*x.....(1)
z = k^2*x.....(2)
Therfore put (1) & (2) in (a)
32*kx = 12(k^2)x + 5x
32k = 12k^2 +5
or, 12k^2 -32k + 5 = 0
which gives k = 5/2 & k = 1/6
For k = 1/6, x < y < z is not satisfied.
Hence, k = 5/2

ption(C)

Hi,

32y = 12z + 5x ----- divide by y ....32= 12 (z/y) + 5 ( x/y)----------> 32= 12 (y/x) + 5 ( x/y)
You know y>x.....then I think Plugging in values will solve quickly..
_________________
- Sachin

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Manager
Joined: 29 Dec 2018
Posts: 57
Location: India
Schools: HBS '22, Wharton '22
Re: Let x, y, z be three positive real numbers in a geometric progression  [#permalink]

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17 May 2019, 19:21
If x,y,z are in Geometric progression then
X = a
Y = ar
Z = ar^2

Now,
5x, 16y and 12z are in Arithmetic progression.
So,
16y - 5x = 12z - 16y
16ar - 5a = 12ar^2 - 16ar

Cancelling out “a” from both the sides and simplifying, we get
12r^2 - 32r + 5 = 0
12r^2 - 30r - 2r + 5 = 0
6r(2r - 5) - 1(2r - 5) =0
(2r-5) (6r-1) = 0

So, r = 5/2
Or r = 1/6

But, as it’s said that z>y>x, it’s an increasing GP. So, r must be greater than 1.

sameeruce08
alpacino
sach24x7

Kudos if you support my answer

Posted from my mobile device
Re: Let x, y, z be three positive real numbers in a geometric progression   [#permalink] 17 May 2019, 19:21
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