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Four dollar amounts w,x ,y, and z, were invested in a
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15 Apr 2010, 12:19
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67% (01:24) correct 33% (01:20) wrong based on 230 sessions
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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.
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Re: Four dollar amounts w,x ,y, and z, were invested. DS problem
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16 Apr 2010, 01:51
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. IMO CStatement 1: is not sufficient as we dont know anything of w. Statement 2 : alone not sufficient as we just know x = 1/4 of total. but if we combine the both, since x = 1/4 of total that means we can distribute 3/4 among w, y and z. since y and z are both less than 1/4 that means y+z < 2/4 so x+y+z < 3/4 => w>1/4 of total, thus w is the greatest. Thus C, both taken together are sufficient.
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Re: Four dollar amounts w,x ,y, and z, were invested in a
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07 Aug 2014, 20:07
Quote: 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. Statement 1: not sufficent since we don't know about w Statement 2: not sufficent since we only know about x When combine the both, X was 25% of total, that meant "X was average of 4 numbers, in arithmetic mean". Therefore, if X was greater than Y and Z, so defenitely X would be less than W, or W was the greatest of four. Answer C



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Re: Four dollar amounts w,x ,y, and z, were invested in a
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27 Aug 2015, 08:24
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
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01 Oct 2015, 19:14
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
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01 Oct 2015, 22:44
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
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09 Mar 2018, 04:38
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.
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Re: Four dollar amounts w,x ,y, and z, were invested in a
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20 Mar 2019, 05:08
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