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Re: Four dollar amounts w,x ,y, and z, were invested in a [#permalink]
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Now we see that
statement 1: y < z < x what about w? so this statement alone is not sufficient.
statement 2: x = 0.25(w+x+y+z). Even this statement lone is not sufficient.
Combining both,
x=0.25w + 0.25x + 0.25y + 0.25z
0.75x = 0.25(w+y+z)

now by equation 1, y<x and z<x
Therefore, y+z<2x
So 0.25(w+y+z) < 0.25(w+2x)
So 0.75x < 0.25(w+2x)
0.75x < 0.25w + 0.5x
0.25x<0.25w implies x<w.....so y < z < x < w
Answer is C

Please correct me if I am wrong.
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Four dollar amounts w,x ,y, and z, were invested in a [#permalink]
Thanks for the explanation karthik. I keep forgetting you can add inequalities together.
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Re: Four dollar amounts w,x ,y, and z, were invested in a [#permalink]
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ada453 wrote:
Thanks for the explanation karthik. I keep forgetting you can add inequalities together.


Check this: How to manipulate inequalities (adding, subtracting, squaring etc.).
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Re: Four dollar amounts w,x ,y, and z, were invested in a [#permalink]
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The trick here is to realise what 25% of the total of four investments actually means.

x = 25% (x+y+z+w) = (x+y+z+w)/4 = average of the four investments. So ... by combining the statements we know what: x=average >z>y, thus ... w must be greater than the average for otherwise we could not have an average. It sounds complex but try to plug in values. Let's say ... x=average=100 and y=1 and z=2 ... or that y=98 and z=99 .... you will realise that in order to have an average of 100 .... we must add a value that is greater than the average ... hence w>average>100 = the greatest amount invested.

Cheers!
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Re: Four dollar amounts w,x ,y, and z, were invested in a [#permalink]
Hi,

another way of considering both statements combined.

1) not sufficient


2) not sufficient

1&2)

x = 1/4 * ( x + y + z + w)
3/4 * x = 1/4 (y + z + w)
3*x= y + z + w
x=(y+z+w)/3

Since y < z < x, this means that x is the average of y,z and w, thus, w is greater than x. Sufficient
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Re: Four dollar amounts w,x ,y, and z, were invested in a [#permalink]
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ksharma12 wrote:
Four dollar amounts w,x ,y, and z, were invested in a business. Which amount was greatest?

(1) y < z < x
(2) x was 25 percent of the total of the four investments.


Question Stem Analysis:

We need to determine the greatest amount among w, x, y, and z.

Statement One Alone:

\(\Rightarrow\) y < z < x

Using this statement, we can determine that neither y nor z is the greatest amount, but the greatest amount can still be x or w. Statement one alone is not sufficient.

Eliminate answer choices A and D.

Statement Two Alone:

\(\Rightarrow\) x was 25 percent of the total of the four investments.

Without knowing anything about y, z, or w, we cannot say anything about the greatest amount. Statement two alone is not sufficient.

Eliminate answer choice B.

Statements One and Two Together:

Let T = total of the four investments. Then, we can write x = 0.25T, y < 0.25T, and z < 0.25T. Adding the two inequalities together, we obtain y + z < 0.5T. Adding x to one side and 0.25T to the other side, we obtain x + y + z < 0.75T. Then:

x + y + z < 0.75T

-(x + y + z) > -0.75T

-(x + y + z) + T > -0.75T + T

T - (x + y + z) > 0.25T.

Since T = x + y + z + w, it follows that the left hand side of the last inequality is equal to w. Thus, w > 0.25T. Since x = 0.25T, y < 0.25T, and z < 0.25T, w is the greatest among the four numbers.

Answer: C
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Re: Four dollar amounts w,x ,y, and z, were invested in a [#permalink]
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Re: Four dollar amounts w,x ,y, and z, were invested in a [#permalink]
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