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The four integers a , b , c , and d are such that a < b

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Manager
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Joined: 26 Mar 2007
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The four integers a , b , c , and d are such that a < b [#permalink]

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New post 14 Jul 2011, 11:06
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A
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C
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E

Difficulty:

  75% (hard)

Question Stats:

55% (02:02) correct 45% (01:19) wrong based on 31 sessions

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The four integers a , b , c , and d are such that a < b < c < d , d = 7 a , and b-a = c-b = d-c. What is the average (arithmetic mean) of the four integers?

(1) c = 25

(2) \(\frac{c}{a}\) = 5
[Reveal] Spoiler: OA
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Re: Average [#permalink]

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New post 14 Jul 2011, 11:44
kannn wrote:
The four integers a , b , c , and d are such that a < b < c < d , d = 7 a , and b-a = c-b = d-c. What is the average (arithmetic mean) of the four integers?

(1) c = 25

(2) \(\frac{c}{a}\) = 5


A it is

we know from 1. that c = 25
c-b = d-c
2c = d+b
d+b = 50..............................1

b-a = c-b
b-a = 25-b
2b-a = 25............................2
we know d = 7a, substitute in 1

7a+b= 50
- a +2b = 25
solve for a; a = 5
we know b = 15
5<15<25<35......................sufficient

St2: insufficient
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Re: Average [#permalink]

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New post 03 Nov 2011, 13:23
b-a = c-b = d-c: the numbers are equidistant from each other....this implies avg{a,b,c,d}=(a+d)/2=(a+7a)/2=4a
this means we only need value of a.
stmnt1: c=25. if we hv a relationship b/w a and c , then our work is done. Imagine the 4 numbers on a number line.* distance btween a and d is 6a.... therefore distance between each pair of numbers is 6a/3=2a......distance between a and c is 4a=>c-a=4a....c=25...a=5...avg=20. SUFFECIENT
stmnt2: we can deduce this from qs itself. its not a new info.....unless we know the value of c we cant find a and hence cant find avg.INSUFECIENT

Ans:A

*i wish i could draw a number line here...but I cant seem to be able to insert an image in my post.....and god knows I am no computer whiz....can some1 help me with this?
Re: Average   [#permalink] 03 Nov 2011, 13:23
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The four integers a , b , c , and d are such that a < b

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