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[DS][numbers] HSPA posts

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[DS][numbers] HSPA posts

by HSPA » Sat Apr 16, 2011 6:06 am
If r, s, and w are positive numbers such that w = 60r + 80s and r + s = 1, is w < 70?
(1) r > 0.5
(2) r > s
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Source: — Data Sufficiency |
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by force5 » Sat Apr 16, 2011 6:19 am
i liked doing this question without algebra

lets consider r= s then each would have been 1/2

which makes the value of w 70.

but thats when the greater quantity is half ( 80)

if s is less than half then--- 80s can never be more than 40.

which means for any combination of s and r ( as long as s+r=1)
60r+80s will always be less than 70

this can also be understood by the concept of wighted mean.

now both the statements are same. r>0.5 and r>s
hence D
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by Anurag@Gurome » Sat Apr 16, 2011 6:20 am
HSPA wrote:If r, s, and w are positive numbers such that w = 60r + 80s and r + s = 1, is w < 70?
(1) r > 0.5
(2) r > s
(1) r > 0.5
Let us take r = 0.6. Then s = 0.4 (as r + s = 1). So, w = 60r + 80s = 60(0.6) + 80(0.4) = 68, which is less than 70.
If we take r = 0.5. Then s = 0.5 (as r + s = 1). So, w = 60r + 80s = 60(0.5) + 80(0.5) = 70.
So, the answer to the main question is "yes".
Hence, (1) is SUFFICIENT.

(2) r > s implies that r should always be greater than 0.5 (as r + s = 1), which is the same as (1).
Hence, (2) is SUFFICIENT.

The correct answer is D.
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by pemdas » Sat Apr 16, 2011 6:29 am
given: r,s,w>0; w=60r+80s, r+s=1, is w<70
80*(r+s)=80*1 Less 60r+80s=w is equal to 20r=80-w OR w=80-20r. If w<70 then 80-20r<70 and r>1/2. If w>=70 then r=<1/2

st(1) r>1/2,Sufficient, as r>1/2 and then w<70
st(2) r>s, :( Sufficient, as r,s >0 and r+s=1 gives condition for r

IOM d
HSPA wrote:If r, s, and w are positive numbers such that w = 60r + 80s and r + s = 1, is w < 70?
(1) r > 0.5
(2) r > s
Last edited by pemdas on Sat Apr 16, 2011 6:33 am, edited 1 time in total.
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by GMATGuruNY » Sat Apr 16, 2011 6:31 am
HSPA wrote:If r, s, and w are positive numbers such that w = 60r + 80s and r + s = 1, is w < 70?
(1) r > 0.5
(2) r > s
This is a weighted average question.
If r=s, then w=70:
w = (.5)(60) + (.5)(80) = 30+40 = 70.
If more weight is given to 60 -- in other words, if r>s -- then w<70:
w = (.75)60 + (.25)80 = 65.

Question rephrased: Is r>s?

Statement 1: r > .5.
Since r+s = 1, if r > .5, then r>s.
Sufficient.

Statement 2: r > s.
Sufficient.

The correct answer is D.
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by force5 » Sat Apr 16, 2011 6:34 am
great Mitch. I'm happy to see that we are learning from you. Thanks and keep up the good work.
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by Ian Stewart » Sat Apr 16, 2011 12:26 pm
If r, s, and w are positive numbers such that w = 60r + 80s and r + s = 1, is w < 70?
(1) r > 0.5
(2) r > s

One can also approach this algebraically. If r + s = 1, then s = 1 - r, so by substitution, w = 60r + 80(1 - r) = 60r + 80 - 80r = 80 - 20r.

Now if w = 80 - 20r, then if r > 0.5, then 20r > 10, so w will certainly be less than 70. Similarly, if r > s and r+s=1, then r > 0.5, so Statement 2 gives the same information as Statement 1, and each Statement is sufficient alone.

Still, I also find it fastest to look at this as a weighted average question, as a couple of people have done above. There's a similar (official) question at the link below - one which on the surface looks like an abstract algebra problem, but which is fundamentally based on weighted averages:

https://www.beatthegmat.com/tough-algebr ... 13568.html
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