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The four integers a , b , c , and d are such that a < b [#permalink]
14 Jul 2011, 10:06

00:00

A

B

C

D

E

Difficulty:

5% (low)

Question Stats:

45% (02:23) correct
55% (01:19) wrong based on 22 sessions

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?

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

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?