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# What is the ratio of the average (arithmetic mean) height

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What is the ratio of the average (arithmetic mean) height [#permalink]

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12 Aug 2007, 20:22
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What is the ratio of the average (arithmetic mean) height of students in class X to the average height of students in class Y?

(1) The average height of the students in class X is 120 centimeters.

(2) The average height of the students in class X and class Y combined is 126 centimeters.

A. Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.

B. Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient.

C. BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.

D. EACH statement ALONE is sufficient.

E. Statements (1) and (2) TOGETHER are NOT sufficient.

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Intern
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12 Aug 2007, 20:59
I would be tempted to pick E. The key is that you don't know how many students are in each class. It would seem that they want you to pick C though. I think they're trying to trick you into thinking that you can just algebraically determine the avg of the height in class Y by saying (120 + y) /2 = 126 which you can't do without knowing the number of students in each class (it's not an average of the two averages). What's the OA?

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Director
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12 Aug 2007, 21:06
entranced wrote:
I would be tempted to pick E. The key is that you don't know how many students are in each class. It would seem that they want you to pick C though. I think they're trying to trick you into thinking that you can just algebraically determine the avg of the height in class Y by saying (120 + y) /2 = 126 which you can't do without knowing the number of students in each class (it's not an average of the two averages). What's the OA?

I agree with answer E. They are setting up a trap. We do not know the weighted average!

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Senior Manager
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12 Aug 2007, 22:10
E.

Unless we know the number of students in each class. Its impossible to find the avg ht for class B.

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Director
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12 Aug 2007, 23:40

Let ∑Hx = Sum of heights of all students in class x
Nx = number of students in class x
The average height of students in class x = ∑Hx / Nx

Let ∑Hy = Sum of heights of all students in class y
Ny = number of students in class y
The average height of students in class y = ∑Hy / Ny

As per question, we need to find: ∑Hx / Nx / ∑Hy / Ny = (∑Hx / Nx) * (Ny * ∑Hy)

(1) The average height of the students in class X is 120 centimeters.
∑Hx / Nx = 120
This is not sufficient to solve the problem as we have no information on class y.

(2) The average height of the students in class X and class Y combined is 126 centimeters.
∑Hx / Nx + ∑Hy / Ny = 126

This is also not sufficient as it is not providing any information on Class x or Class y students information.

Considering both 1 and 2:

∑Hx / Nx + ∑Hy / Ny = 126 divide both sides by ∑Hx / Nx

1 + (∑Hy / Ny) / ∑Hx / Nx = (126) / (∑Hx / Nx)
1 + (∑Hy / Ny) / ∑Hx / Nx = 126 / 120
1 + (∑Hy / Ny) / ∑Hx / Nx = 1.05

(∑Hy / Ny) / ∑Hx / Nx = 0.05
Inverse both sides:
∑Hx / Nx * Ny / ∑Hy = 20

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Director
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12 Aug 2007, 23:54
hanumayamma wrote:

∑Hx / Nx + ∑Hy / Ny = 126 divide both sides by ∑Hx / Nx

I guess bold portion should be (∑Hx + ∑Hy)/(Nx+Ny) = 126 as we are talkng abt combined average of whole class.

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Intern
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12 Aug 2007, 23:54
I think one of your statements may be inaccurate.

Statement 2 should be

(∑Hx + ∑Hy) / (Nx + Ny) = 126 rather than

∑Hx / Nx + ∑Hy / Ny = 126 which would give you a sum of the two averages, rather than a combination. Unfortunately, the rest of your calculations seem to hinge on this definition so I'm afraid it disconnects at this piont. Also, I think the OP mentioned OA was E.

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Manager
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16 Dec 2009, 03:10
Let Average height of students in class X = X/N
Let Avg height of students in class Y = Y/M

Total weighted average height of both classes = X/N + Y/M / 2

Statement 1 --------Insufficient

We only know the avg height in class X i.e. X/N

Statement 2 --------Insufficient

We are told that X/N + Y/M / 2 = 126

Combining both,

120 + Y/M / 2 = 126

We need M, or Y to calculate it

Hence E

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Re: GMATSets   [#permalink] 16 Dec 2009, 03:10
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